"Master Organic Chemistry"
Summary Sheet - Hybridization And Bonding
Hybridization
What is bonding and why does it occur?
In a chemical bond, electrons are shared between atoms.
This is energetically favorable (more stable) because of Coulomb's law
(opposite charges attract; negatively charged electrons held between two
postitively charged nuclei)
Constructive end-on-end overlap between atomic orbitals results in a
σ ("sigma") molecular orbital
1s
σ
1s
+
constructive overlap
electron density shared between both nuclei (stabilizing)
What is antibonding?
Destructive overlap between atomic orbitals gives an antibonding molecular orbital. This
means that two positively charged nuclei are held closely together without any electrons
between them to "glue" them together; repulsion between nuclei makes this a very unstable
situation
+
1s
Sigma bonds are formed through end-on overlap of two atomic orbitals, as distinguished
from pi bonds (see later section).
Molecular Orbital Diagram for H 2
σ∗ (molecular orbital)
antibonding
(high E)
This is the lowest unoccupied
molecular orbital (LUMO)
Type
s
sp3
1
3
0
tetrahedral
sp2
1
2
1
trigonal planar
sp
1
1
2
linear
σ (molecular orbital)
Note that this is the
highest occupied molecular
orbital (HOMO)
What is Pi (π) bonding?
Pi bonding occurs when orbitals overlap in a "side-on" fashion.
This is possible when neighboring atoms have electrons in unhybridized p orbitals
(like alkenes and alkynes)
Pi bonding occurs when orbitals overlap in a "side-on" fashion.
Electron density is shared between neighboring p orbitals,
H resulting in a π bond.
H
C C
H
H Atoms involved in pi bonding cannot be rotated, since this
would destroy overlap.
H
There are also π* orbitals (not shown)
It is possible for two pi bonds to be present; this involves the side-on overlap
of two p orbitals, each at 90° to each other
H
C
C
H
H
H
H
C
C
H
H
H
H
H
H
H
H
H
104.5°
"Bent" molecular
geometry
H
H
tetrahedral
trigonal pyramidal
H 2O
sp3
tetrahedral
bent
NH 2
sp3
tetrahedral
bent
CH3
sp3
tetrahedral
trigonal pyramidal
BH 3
sp2
trigonal planar
trigonal planar
CH3
sp2
trigonal planar
trigonal planar
BeCl 2
sp
linear
linear
sp2
trigonal planar
C O
sp2
trigonal planar
C N
sp2
trigonal planar
sp2
trigonal planar
H
C
H
H
H
H
H
H
H
H
s orbitals are closer to the nucleus and more tightly held
The more "s-character" a bond has, the stronger (and shorter) it will be
H
H 3C–CH3
H
H
H
H
C
H
C
The Shortcut For Finding The Hybridization Of Any Atom
Count the number of surrounding atoms (A) + lone pairs (LP)
sp3
If A+LP = 4 hybridization is sp3
sp2
If A+LP = 3 hybridization is sp2
N
Try it with Viagra!
(don't forget the "hidden" lone pairs)
Cl
Be
Cl
O
sp2
O
S
sp3
O
N
HN
N
N
sp2
O
Sildenifil (Viagra)
Exception - not always covered - if atom with lone pair next to pi bond, rehybridization
occurs to put that lone pair in a p orbital. So it's actually sp2 not sp3
O
If one s orbital is hybridized with one p orbital we obtain two hybrid orbitals
180° apart. We have two "leftover" unhybridized p orbitals. This is the geometry
of the central atom in beryllium compounds (e.g. BeCl 2) as well as that of atoms
with triple bonds
H
sp–sp3
stronger
(more s-character)
sp2 –sp3
H
sp Hybridization
C
CH3
sp3–sp3
Trigonal planar
C
H C C CH3
N
H
H
H
C C
If A+LP = 2 hybridization is sp
If one s orbital is hybridized with two p orbitals we obtain three
hybrid orbitals that will be 120° apart. We will have one "leftover"
unhybridized p orbital. This is the geometry of the central atom in
trivalent boron compounds, carbocations, and atoms with a single pi bond
B
sp3
H
Hybridization
H
NH 3
weaker
(less s-character)
O
O
"Trigonal pyramidal" Tetrahedral
molecular geometry orbital geometry
H
tetrahedral
How does hybridization affect bond strength?
107.5°
sp2
tetrahedral
H
H
H
Tetrahedral
molecular geometry
N
H
sp3
Molecular geometry
H
Molecules such as NH 3 and H 2O also have four electron
pair domains that are arranged tetrahedrally around the
central atom; some of these are "lone pairs" that are not
bonded to an atom. The lone pairs take up slightly more
space than bonding pairs, which compresses the bond angles
Pi bonding not possible here because these p orbitals are
at 90° to each other
C
H
C
N
CH4
C C
In CH4 the hydrogens are arranged in a tetrahedral
fashion around the central carbon atom. We say CH4 has
"tetrahedral" molecular geometry
H
H
109.5°
H
Orbital Geometry
H
sp3 sp3 sp3 sp3
(Each 109.5° apart - "tetrahedral" orbital geometry)
H
Hybridization
H
H
Tetrahedral
orbital geometry
•In the diagram above for H 2, if we added a third electron, it would
have to go to the antibonding orbital because there's nowhere else to put it.
•Antibonding orbitals can be filled when electrons absorb energy of a certain
frequency (promoting a bonding electron to antibonding) or if an extra electron
is added to a molecule where all orbitals are already full.
C
pz
py
4 hybrid orbitals
Why care about antibonding if it's usually empty?
H
4 orbitals
mix
1s
(atomic
orbital)
bonding
(low E)
Example
molecules adopt a geometry that
maximizes the distance between
electron pairs
C
1s
(atomic
orbital)
H
px
s
overlap of two atomic orbitals
gives two new molecular orbitals
ΔE
Some examples
orbital
geometry
The bond angles in CH4 are 109.5° .
The orbitals containing each pair of electrons
are neither "pure p" or "pure s". They are a hybrid
between p and s. Since we are hybridizing one s orbital
and three p orbitals, we call this "sp3" hybridization
σ∗
destructive overlap
no electron density between nuclei; destabilizing
Can't we just have bonding without antibonding?
No. When atomic orbitals come together, the number of molecular orbitals has to equal
the number of atomic orbitals. Bringing together two atomic orbitals will always result in two
new molecular orbitals, one bonding ("constructive overlap") and one antibonding
("destructive overlap")
1s
masterorganicchemistry.com
leftover
hybridized (unhybridized)
p
p
sp3 Hybridization
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
N
C
H
H
atom with lone pair
adjacent to sp2
hybridized atom; lone
pair will be in p orbital,
not sp3 orbital.
Nitrogen is sp2 not sp3
N
C
O
rehybridization to sp2
allows lone pair to be in
conjugation with p orbitals
of pi bond (allows for resonance)
Omissions, Mistakes, Suggestions?
james@masterorganicchemistry.com
This sheet copyright 2022,, James A.
Ashenhurst masterorganicchemistry.com
"Master Organic Chemistry"
Formal Charge, Dipoles, Boiling Points, and Curved Arrows
masterorganicchemistry.com
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
How To Calculate Formal Charge Of An Atom
Dipole Moments
Curved Arrows
1. Start with the number of valence electrons, "V"
The dipoles of bonds in a molecule add up (as vectors) to
give an overall dipole moment. Example: NH 3
Curved Arrows: A way to show the "movement" of electrons.
"V"
B
C
N
O
F
3
4
5
6
7
N
2. Count the number of non-bonded electrons, "N"
"N"
H
N H
H
H
O H
H
H
C H
H
Cl
F
F B F
F
2
2
2
8
0
H
H
C H
H
H
C
1
2
H
0
+1
–1
–1
–1
H
0
0
Alternative: instead of multiplying the # of bonding electrons "BE"
by 1/2, count the number of bonds, "B" around the atom instead
Formal charge = V – [ N + B ]
Dipoles
2.2
H
H
add up to overall
dipole moment D (shown)
individual dipoles
shown
2.2
2.6 2.6
2.7 3.0
3.0
3.2
3.4
4.0
A consequence of electronegativity is that electrons in bonds between
atoms with different electronegativities are not equally shared. The
more electronegative atom will have a greater "share" of the bonding
electrons, giving it a partial negative charge, and the less electronegative
atom will have a lesser "share" of the bonding electrons, giving it a
partial positive charge.
This is what is known as a dipole
example:
the δ – means that fluorine bears a partial
δ+
δ–
negative charge
H
F
hydrogen
2.2
4.0
bears a
partial
electronegativities
positive
charge
Representative examples (with largest dipoles shown)
+
–
Hδ
Oδ
δ–
–
δ+ δ
δ+
δ+ O δ+ δ+ N δ+
C
H 3C Cl
H
H
H δ– H
Understanding dipoles is KEY for comprehending boiling points
(this sheet) but also many aspects of reactivity!
One important thing to keep in mind: dipoles always give an accurate
portrait of electron density, whereas formal charge may not
δ+H
+
δ
even though oxygen bears a
H O
δ+H O– H δ+
positive
formal charge, it
δ
is still more electron-rich than
hydrogen is
hydrogen is
hydrogen due to electronegativity
electron-poor,
electron-poor,
oxygen is
oxygen is (still)
The hydrogens are still electron-poor
electron-rich
electron-rich
("electrophilic")
head
Take a pair of
electrons from here
(bond / lone pair)
and move them here
(bond / lone pair)
The curved arrow shows "movement" of a pair of electrons... it's an extremely
useful accounting system that lets us keep track of changes in bonding and charge
The tail must be at a source of electrons, either a lone pair or a bond.
The head must be able to accept a pair of electrons without breaking the octet rule
δ–
δ–
Cl
Cl
δ+
δ–
C
Cl
Clδ –
Cl
C
Cl
Cl
Dipole moment:
0D
Cl (i.e. no dipole moment)
Polar molecules: tend to have larger dipole moments, higher
water solubility, lower solubility in non-polar solvents, and relatively
higher boiling and melting points than non-polar molecules.
Factors That Affect Boiling Points
Boiling occurs when enough energy is added to a liquid to overcome the
attractive forces between molecules.
The greater the attractive forces, the higher the boiling point.
Atomic nuclei differ in their ability to pull electrons toward themselves.
This property is called "electronegativity", and is measured on a scale
from roughly 1.0 (low) to 4.0 (high). The higher the electronegativity,
the "greedier" that atom is for electrons.
B
H
P
S
C
I
Br N
Cl
O
F
2.0
H
tail
As bonds become more polarized, the charges on the atoms become greater,
which leads to stronger intermolecular attractions, which leads to higher boiling
points
There are four types of attractive forces between molecules. Here they are in order
from largest to smallest:
Ionic > Hydrogen Bonding > VDW Dipole-Dipole > VDW Dispersion
(VDW = Van der Waals)
There are only three "moves" possible for a curved arrow
Bond → Lone pair
Lone pair→ Bond
Bond → Bond
Lone pair → Lone pair
not possible
Simple example:
Shows bond forming
between O and H →
→
Charge
H
A good rule of thumb: the larger the dipole moment, the
more polar the molecule
H
C
H
C H
H
H
N
Some molecules with dipoles along individual bonds will have
no overall dipole moment since the vectors cancel
3. Count the number of bonding electrons, "BE" [2 electrons per bond]
H
H
H
H
H
F
C
N H
O H
C H
C H
Cl
F B F
H
H
H
H
H
F
"BE" 6
6
6
0
8
6
4
Formal charge = V – [ N + 1 BE ]
2
H
H
H
F
O H
C H
N H
F B F
Cl
H
H
H
F
Dipole moment:
1.42 D
δ–
Electrons move from the tail to the head
[Lone pair→ Bond]
HO
H
Oxygen goes from "owning"
a lone pair to "sharing", so its
formal charge becomes less
negative by 1
Negative charges can stand in
for lone pairs. It's OK to draw
the tail from a negative charge,
so long as the atom has a lone
pair of electrons
Shows bond breaking
between H and Cl
[Bond → Lone pair]
HO H
Cl
The pair
of electrons
in the bond
will become
a new lone
pair on Cl
Note change
in formal
charge at
oxygen:
–1 → 0
Cl
Note change
in formal
charge at
chlorine:
0 → –1
Cl goes from "sharing"
this pair to "owning" it
so the charge becomes
more negative by 1
Only two charges are changed: the initial tail becomes more positive
and the final head becomes more negative
Example with three arrows:
H
H O
"final head"
Cl
+ Cl
HO
1. Ionic forces
Attraction between point charges. Ionic compounds have the highest melting
H↑
and boiling points.
initial tail (oxygen) more positive by 1 0 → +1
"initial tail"
O
final head (chlorine) more negative by 1 0 → –1
CH3CH2O Na
NH 4 Cl
O
Resonance example:
NH 3
2. Hydrogen bonding
Formal charge
Formal charge
The next-strongest intermolecular force. Look for O-H, N-H, or F-H bonds.
= 6 – [6 + 1/2(2)]
=
6
–
[4
+
1/2(4)]
These are highly polarized and lead to relatively large boiling points.
Bonds formed Bonds broken
= –1
=0
O
O
O
+ + –
O
B
B O–C (π)
δ– +
δ+
A
A O–C (π)
δ+
– δ
–
δ
δ
δ
δ
H
δ H
O
H
O
O
H
H F
N
N
O
Charges
δ–
+
H
Hδ
+
δ
Formal charge
Formal charge
O –1 → 0
3. Van der Waals Dipole interactions
=
6
–
[6
+
1/2(2)]
= 6 – [4 + 1/2(4)]
Look for bonds between C and highly electronegative elements like O, N, or
C 0→0
= –1
=0
halogens. Smaller differences in electronegativity than with hydrogen bonding,
O 0 → –1
but still significant
–
–
δ–
δ
Examples of how the 4 forces lead to boiling point differences:
O
O
+
+ δ δ+
δ+ δ –
δ
+
+ δ
O
δ
δ
H 3C F
ONa >
OH >
CH
H 3C
CH3
>
O
Nδ – 3
δ+ CH3
b.p. >200 °C
b.p. 97 °C
b.p. 7°C
b.p. –1°C
4. Van der Waals Dispersion interactions
Temporary (instantaneous) dipoles; weakest attractive interactions
[Ionic]
[H-bonding] [Dipole-dipole] [Dispersion]
Present in all molecules (even hydrocarbons and noble gases)
Omissions, Mistakes, Suggestions?
Boiling
point
also
increases
with
chain
length
Generally will increase with surface area
james@masterorganicchemistry.com
>
>
This is why boiling points increase with increasing chain lengths
This sheet copyright 2022, James A.
heptane
hexane
pentane
Ashenhurst masterorganicchemistry.com
b.p. 36°
b.p. 98°C
b.p. 68°C
↑
e.g.
Introduction to Alkane Nomenclature
A. Determining the Priority of Functional
Groups.
What's in a name?
3-ethyl-5-(1-methylpropyl)-4,4-dimethylnonane
Too big a subject to cover on one sheet! This
paper will focus on alkanes. Determining
functional group priority will be the subject of
a subsequent sheet.
suffix
https://masterorganicchemistry.com
D. Applying the Lowest Locator Rule
F. Dealing With Branched Substituents
(the IUPAC Way)
Number the chain from one end so as to provide
the lowest locator possible for the first substituent.
Treat each branched substituent as its own
naming problem.
Carbon #1 of the branched substituent will be
where it meets the main chain.
2
1
3
4
5
6
6
7
7
5
2
4
1
3
B. Applying the Chain Length Rule
the NUMBERS are called "locants" (sometimes "locators")
items in BLUE are called "substituents".
the name in RED at the end is called the suffix.
9
7
8
6
5
3-methyl heptane
4
3
This also applies for subsequent substituents,
if either direction would give the same number.
2
1
The purpose of this sheet is to demonstrate the rules by which alkanes are named.
4-ethyl-6-methylnonane
ORDER OF BUSINESS
A. Determine the priority of functional groups (not covered here since
we're dealing with alkanes only)
B. Find the longest linear chain of your molecule, or the largest ring (whichever is
greatest). This is the Chain length rule which defines both the "main chain" and also
the suffix.
tiebreaker: where more than one "path" along the molecule leads to the
longest chain, the main chain is the one that contains the most substituents.
C. Identify the substituents along your main chain. Substituents are classified
according to length of carbon chain and the suffix "yl" is attached.
D. Number your chain from one of the ends. The LOWEST LOCATOR RULE determines
which end is chosen as carbon #1: "Number the chain such as to provide the lowest
possible locators for the chain."
tiebreaker for lowest-locator rule: alphabetization
E. Multiple instances of substituents are given the prefixes di, tri, tetra, etc.
note: must have locator for all substituents. Example: 2,2-dimethyl is correct. 2dimethyl is incorrect.
F. Branched substituents are numbered and named seperately from the main chain,
and put in brackets.
G. The FINAL name is assembled such as to arrange the substituents in
alphabetical order.
-"di", "tri", "tetra" are ignored for alphabetization purposes.
-prefixes like "n", "tert", "i" and "sec" are ignored for alphabetization purposes.
THE EXCEPTION is "isopropyl" and "isobutyl". For some reason these count as "i"
- not covered here, but this is also where one puts in descriptors like "cis", "trans", (R), (S)
(E), (Z) and so on.
Names for Hydrocarbon Chains and Rings
1
methane
2
ethane
3
butane
5
pentane
6
hexane
7
heptane
8
octane
9
nonane
10
decane
undecane
12
dodecane
7
6
4
Chains vs. rings
Rings take priority over chains, assuming there
are only alkyl groups in the chain. ***see note below
5
7
5
7 6
4
3
1
2
Where more than one "longest chain" exists,
the more substituted chain is chosen as
the "longest chain"
6
7
4
5
8
R
( )n
6
7
5
4
1
3
2
heptane
methyl
2
3
4
5
6
6
7
7
5
2
4
1
3
3-ethyl-5-methylheptane NOT 3-methyl-5-ethylheptane
1
2
2
1
NOT
1-ethyl-2-methylcyclobutane
1-methyl-2-ethylcyclobutane
Multiples of the same substituent are given the
prefixes "di", "tri", "tetra", etc. The lowest locator
rule still applies.
2
1
3
4
5
6
7
sec-butyl
isobutyl
( )n
1
3-ethyl-5-methyloctane
propyl
tert-butyl
( )n
1
2
3
4
6
5
e.g. 4-isopropyl-3-methyl-5-tert-pentylnonane.
The official IUPAC nomenclature system will never let
you down. It would probably be best to just go with that.
6
5
4 3
2
1
1. Put your substituents together in alphabetical order.
For alphabetization purposes:
IGNORE •"di", "tri", "tet", etc.
• sec-, tert-, n-,
•cycloDO NOT IGNORE
• "iso". For some reason "isopropyl" is alphabetized
under "i" and not "p".
2. Affix the locators.
4. Attach the suffix at the end.
1
2
4
3
5
6
7
8
9
7
2,3,5-trimethylheptane
Also applies to substituents on the same carbon:
2
3
4
5
6
7
NOT 4-dimethylheptane
NOT 4,4-methylheptane
*** IUPAC 61.2 says, "Choice between these methods [either choosing
rings or chains as the root] is made according to the more appropriate of
the following principles: (a) the maximum number of substitutions into a
single unit of structure; (b) treatment of a smaller of structure as a
substituent into a larger."
This sheet copyright
James A. Ashenhurst
https://masterorganicchemistry.com
Errors/omissions/suggestions?
james@masterorganicchemistry.com
4,4-dimethylheptane
R
tert-pentyl
In certain instances, you may see the trivial names isopropyl,
isobutyl, tert-butyl, tert-pentyl used.
3-ethyl-4,4-dimethyl-5-(1-methylpropyl)-nonane
3,5-dimethylheptane
4-propylheptane
R
9
8
3. Make sure any branched substituents are in parentheses
1
n-pentane
R
7
6
3-methyl 4-(1-methylethyl) 5-(1,1-dimethylpropyl) nonane
3
Substituents are carbon fragments branching off the
main chain. They are named according to the number
of carbons like the main chain would be, except the
"ane" is dropped and replaced with "yl"
Trivial names for substituents
4
5
ethyl
butane
3
methylcyclohexane the same as 1-methylcyclohexane
E. Multiples of the Same Substituent
7 and higher follow
the same pattern.
We put the name in parentheses to avoid confusing the
numbers of the branched substituent with the numbers
of the main chain.
G. Putting the Name Together.
3-ethyl-2-methyloctane
(NOT - 3-isopropyloctane)
cyclohexane
3
•Longest chain is ethyl
•Methyl substituent is on carbon #1
•Name of substituent is (1-methylethyl)
1
2
Tiebreaker: Alphabetization
7
isopropyl
6
5
NOT 2,5,6-trimethylheptane
For rings with one substituent, the locator "1"
can be dropped. Why? Imagine a street that had
only one house on it. Would the house really need
a number?
1
cyclopentane
n-hexane
( )n
4
2,3,6-trimethylheptane
propylcyclobutane cyclopropylbutane
C. Identifying Substituents
eicosane
R
3
If the same locators are obtained from either
direction of the chain, the chain is numbered
according to alphabetical order of the substituents.
Sometimes you will see "n" in front to
indicate that it is a straight-chain alkane
( )n
2
Tiebreaker: Alphabetization
8
cyclobutane
1
•Longest chain is propyl
•Methyl substituents are
on carbon #1
•Name of substituent is
(1,1-dimethylpropyl)
1 2
2-methyloctane
Longest chain is 8 carbons - suffix
will be octane
cyclopropane
6
11
3 2
5 1
4
2
propane
4
20
3
Longest chain is 9 carbons - suffix
will be nonane
Watch out! Longest chain might not
be drawn as a "straight chain"
8
1
NOT 5-methyl heptane
2
Extremely common
mistakes!
For more comoplete resources on nomenclature consult:
1) "Organic Chemistry Online" by William Reusch:
http://www2.chemistry.msu.edu:80/faculty/reusch/VirtTxtJml/intro
1.htm
2) IUPAC "Blue Book"
http://www.acdlabs.com/iupac/nomenclature/
Stereoisomerism
Nomenclature 2
Stereoisomerism is the first part of every name. It identifies the configuration of any double bonds or stereocenters. If there is no double
bonds or stereocenters in the molecule, then we do not need to include this part of the name. Indentifying the configuration of a stereocenter
requires a separate sheet to itself so we will only focus on double bonds here.
Every name can be broken down into five parts:
Stereoisomerism
Functional group
Unsaturation
Parent
Substituents
• Stereroisomerism indicates whether double bonds are cis or trans (E or Z), and whether stereocenters are R or S.
• Substituents are groups connected to the parent chain.
• Parent is the main chain.
• Unsaturation indicates whether ther are any double or triple bonds.
• Functional group is the group after which the compound is named.
The best way to name a compound systematcially is to start at the end (functional groups) and work backwards to the first part
(stereoisomerism) because the position of the functional group affects which parent chain you choose.
There are two ways to name them:
1) if the compound has identical groups on either side of the double bond, we can name them cis/trans according to the following way. If the
two identical groups are on the same side of the double bond, we call them cis. If they are on the opposite sides, we call them trans.
2) If the compound has no identical groups (all four groups are different), then we cannot use the first way but instead we name them E/Z
according to some rules. First we compare the priorities of groups on the left side of the alkene based on atomic number of the first element
connected to the alkene. Then we compare the prioritis of groups on the right side of the alkene based on the same thing. If both the high
priority groups are on the same side, we call them Z. If they are on the opposite side, we call them E.
H
Functional Group
Cl
When a compound has one of the functional groups listed below, we include it in the name by adding a suffix to the end of the name. The
table below is listed according to the priority of those functional groups with the highest on the left and the lowest on the right. When there is
more than one functional group present in the compound, the one with the highest priority will go in the suffix of the name and the others will
be prefixes in the substituent part of the name.
Name
R
O
O
O
Structure
H
O
R
Prefix
carboxy-oic acid
R
O
N(H, R’)2
R
Amide
Aldehyde
alkoxycarbonyl-
carbamoyl-amide
-oate
R
R O H
R’
R N(H, R’)2
di = 2
3 = prop
oxo-
hydroxy-
amino-
Bring Them All Together
-al
-one
-ol
-amine
To correctly name a compound, we need to bring those pieces together. The numbering tells us where those substituents or unssturations are
located on the parent chain. To number the parent chain properly, we use the same hierarchy that we use to choose the parent in the first place:
(1) the highest priority functional group, (2) double bond, and (3) triple bond.
tri = 3
tetra = 4
penta = 5
hexa = 6
1) If there is a functional group, then number the parent chain so that the functional group gets the lower numer.
2) If there is no functional group, then number the parent chain so that the double bond gets the lower number.
3) If there is no functional group or double bond, then number the parent chain so that the triple bond gets the lower number.
OH
1
6
3
5 6
1
3
5 6
2 4
4
2 1
2
4
5
3
6 = hex
7 = hept
8 = oct
9 = non
Cl
6
prop
pent
hex
hept
Substituents
Everything else connected to the parent chain is called a substituent. To name an alkyl substituent, we follow the same terminology that we
used for naming parent chains, but we add “yl” to the end to indicate that it is a substituent. Here is how we name an alkyl substituents based
on the number of carbons:
1 = methyl
2 = ethyl
3 = propyl
4 = butyl
5 = pentyl
6 = hexyl
7 = heptyl
8 = octyl
9 = nonyl
10 = decyl
If there is a branched susbtituent (substituent that is not connected in one stright line to the parent chain), some common names are used.
the triple bond gets 1 instead of 5
4) If there is no functional group, double bond, or triple bond, then number the parent chain so that the substituent gets the lower number.
5) If there is more than one susbtituent on the parent chain, then number the parent chain so the substituents get the lowest number possible.
10 = dec
If there is a ring, we add the term “cyclo”, so a ring of six carbons is called “cyclohex-” and a ring of five carbons is called “cyclopent-.” In order
to correctly identify which carbon chain is the parent, we have to make sure to include (1) the highest priority functional group, (2) double bond,
and (3) triple bond. In the case where there is no functional group, use the longest chain that includes the double bond. If there is no double
bond, use the longest chain that includes the triple bond. If there are no functional groups, no double bonds, no triple bonds, then we simply
use the longest chain possible. Here are some examples:
O
OH
O
OH
OH
HO
OH
OH
hex
the double bond gets 1 instead of 5
OH gets 2 instead of 5
5 = pent
F
formyl-
pentyne
4 = but
two priority groups (OCH3 and F) are E
two priority groups (ethyl and F) are Z
The parent of the compound is the carbon chain that is going to be the root of the name. Everything else is connected to this chain at a
specific location, designated by numbers. Here is how we name the parent based on the number of carbons:
2 = eth
F
*If the groups are different but the atoms are the same, then we need to follow the same rules that we use for assigning priorities with R and S.
Amine
Parent
1 = meth
two hydrogens (not shown) are trans
H
H3C
Alcohol
A compound with a double or triple bond is said to be “unsaturated”. The presence of unsaturation is indicated in the following way, where “en”
represents a double bond, and “yn” represents a triple bond. The absence of unsaturation is indicated by “an.” If there are more than one
double/triple bonds in a compound, a prefix will be added such as “-dien-“ represents 2 double bonds and “-triyn-“ represents 3 triple bonds.
Here are some examples:
pentene
two chloros are cis
H3CO
Ketone
Alkenes and Alkynes (Unsaturation)
pentane
Br
H
1. If the atoms are the smae on one side, then just move further out and analyze again.
2. One oxygen beats three carbons (look for the first point of difference).
3. A double bond counts as two individual bonds.
O
H
Ester
Carboxylic acid
Suffix
R’
O
two methyls are trans
Cl
6
5 4 3 2 1
5
Cl Cl
4
1
3
2
Cl
3,3,4-trichloro instead of 3,4,4-trichloro
Cl gets 3 instead of 4
• For functional group, the number is generally palced directly before the suffix (eg. hexan-2-ol). If the functional group gets number 1, then the
number does not need to be included in the name (eg. hexanol). It is assumed that the functional group is at number 1 position.
• For unsturation (double and triple bonds), the number indicates the lower number of two carbons and that number is placed directly before
the suffix (eg. hex-2-ene). If there are two double bonds or triple bonds, both of them should be numbered and placed directly before the suffix
(eg. hexa-2,4-diene).
• For substituents, the numbers that the substituents get are placed directly before the substituent.
We arrange all those pieces in an alphabetical way (di, tri, tetra…are not counted as a part of alphabetization). Here are some examples to
name a compound with 2 or 3 functional groups.
O
2
O
O
Cl
9
1 6
5
4
8
4 5
HO 1
3
5
7
2
4
6 7
3
6
2
1
OH
3
OH
Cl
Cl
(trans)-4,5-dichloro-6,6-dimethylhept-4-en-2-one
4-hydoxynon-7-ynoic acid
(E)-4-chloro-5-hydroxycyclohex-3-enone
A Special Functional Group: Ethers
Parent
isopropyl
(1-methylethyl)
Parent
Parent
Parent
sec-butyl
(1-meythylpropyl)
isobutyl
(2-methylpropyl)
tert-butyl
(1,1-dimethylethyl)
Parent
isopentyl or isoamyl
(3-methylbutyl)
Parent
neopentyl
(2,2-dimethylpropyl)
Lower priority fucntional groups not included in the parent chain are also substituents. We name them by adding the prefix as shown in the
table above to the name. Some other important sustituents are halogens, —N3, and —NO2. They are always prefixes and named in the
following way:
—Br = bromo
—Cl = chloro
—F = fluoro
—I = iodo
—N3 = azido
—NO2 = nitro
Where there are multiple substituents of the same kind, we use the same prefix (di, tri, tetra…)that we used for classifying the number of
double and triple bond.
IUPAC allows two different ways to name ethers:
1) A common name is constructed by identifying the R group on each side, arranging them in alphabatical order, and then add “ether” in the
end. If two R groups are identical, then it is named as dialkyl ether.
2) A systematical name is constructed by choosing the larger group to be the
parent chain and naming the smaller group as an alkoxy sustituent. Complex ethers
Practice Questions on This Material
must be named using the systematic way.
4
O
ethyl propyl ether
(ethoxypropane)
O
diethyl ether
(ethoxyethane)
5
3
1
2
Cl Cl
O
1,1-dichloro-3-ethoxycyclopentane
Practice questions:
https://bit.ly/3Ce2UdX
Reach out with feedback:
james@masterorganicchemistry.com
Introduction to Conformations
"Master Organic Chemistry"
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
masterorganicchemistry.com
What are conformations?
•Sigma bonds (single bonds) are free to rotate.
•Through the rotation of bonds, molecules can
adopt different 3-dimensional shapes. These
are called conformations. Two identical
molecules with different 3-D shapes are called
conformational isomers.
•Not all conformations are equal in energy
Because electron clouds repel, conformations
which maximize the distance between atoms
are generally favored (i.e. lower in energy).
H
H
H
H
C
H6
H5
C
H6
H5
(
H2
H1
C
H
H
C
H
C
H
(
C
H1
Take this molecule,
and look at it from
the side...
...and you get this → H1
H5
CH 3
H 3C
H
H
H3
Note how the hydrogens
from the front carbon are directly
in front of the hydrogens from
the back carbon, hence "eclipsed"
"eclipsed"
H6
H2
Take this molecule, H 2
and look at it from
the side...
...and you get this→ H 5
H4
H3
H3
"Time"
The dihedral angle is the
angle between groups on the
front and the back carbons.
CH 3
H
H
CH 3
H
CH 3
H
H
CH 3
H 3C
H
"staggered"
Note how the hydrogens are
"spaced" out from each other,
hence "staggered"
CH 3
H 3C
0° : "syn"
H
H
2
6
Each 2 hour increment represents
60 degrees.
CH 3
H
H
H
C(CH3)3 > CH(CH 3)2 > CH 3 > Cl, OH > H
H
H
H
CH 3
H
H
H
8:00
10:00
0°
60°
120°
180°
240°
300°
2.9
–
–
–
–
–
–
–
2 x 1.4
–
2 x 1.4
–
–
1.0
–
1.0
–
–
0.9
–
–
–
0.9
4.9
0.9
3.8
0.0
3.8
0.9
2 x 1.0
Gauche CH3 CH 3
Total Cost
(kcal/mol)
Least
"expensive"
(lowest-energy conformation)
Step 3: Make a graph
The Graph: Energy versus angle (in 60 degree increments)
4.9
0.9
3.8
0.0
3.8
0.9
240°
300°
5.0
4.0
CH 3
smallest
largest
H 3C
H
6:00
Eclipsed H – H
H
These repulsions between groups are called "steric
interactions". The larger the group involved, the
greater the steric interactions, or "cost".
H
H
4:00
Most
"expensive"
(highest-energy conformation)
60° : "gauche"
French for "awkward"
H
H
CH 3
2:00
Dihedral angle
HH
CH 3
4
8
H
H
12:00
Eclipsed CH3 H
H6
10
60°
H
H
12
H6
CH 3
H
CH 3
HH
Eclipsed CH3 CH 3
Analogy: think of the hands of a clock
H1
Example: Adding up the "costs" of each conformation for butane
difference: 2.4 kcal/mol (10 kJ/mol)
Key Concept: The "Dihedral Angle"
H5
Step 2: Calculate the "costs" for each of the six conformations
more stable
less stable
H1
H2
Step 1: like a clock, pick one group as the "hour hand" and one as the "minute hand". Then rotate that "hand" through 360°, in 60° (2h) increments.
In this example, it's best to pick the two methyl groups as your "hands" since there is only one on each carbon.
bonds are offset
from each other:
"staggered"
bonds are lined
up next to each other:
"eclipsed"
H4
H3 2
H
H4
H
Example: how to figure out the rotational energies for butane (CH 3CH2CH2CH 3) ?
H4
H3
C
H
H
Key Tool: The Newman projection
H4
C
rotate 60°
H
H
180° : "anti"
H
CH 3
Energy
kcal/mol
Most
"expensive"
3.0
2.0
Least
"expensive"
1.0
interestingly, CH 3 is larger than Cl or OH, since it "sweeps out" a larger area
Repulsions between groups cause strain, which can be relieved through bond rotation; this is often
called "torsional strain".
Example of a Price List: 4 interactions with different "costs"
CH 3
H 3C
H
H
H 3C
H
H
H
"Syn" (eclipsed) CH3 CH 3
Cost: 2.9 kcal/mol
(12.1 kJ/mol)
HH
CH 3
H 3C
H
H
H
CH 3
= 2.8 kcal/mol (total)
60°
CH 3
H 3C
120°
CH 3
H
CH 3
H
CH 3
H
H
180°
CH 3
H
CH 3
H
H
CH 3
H 3C
H
"Syn" (eclipsed) H H
Cost: 1.0 kcal/mol (each)
H
H
H
H
HH
H
CH 3
H
(4.18 kJ/mol each)
= 2.0 kcal/mol (total)
H
H
12:00
2:00
4:00
H
H
H 3C
H
H
H
CH 3
6:00
8:00
H
H
H
10:00
CH 3
"Syn" (eclipsed) CH3 H
Cost: 1.4 kcal/mol (each)
(5.86 kJ/mol (each)
0°
CH 3 Gauche ("awkward") CH 3 CH 3
H
H
H
H
Cost: 0.9 kcal/mol
(3.8 kJ/mol)
Omissions, Mistakes, Suggestions?
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This sheet copyright 2022, James A.
Ashenhurst masterorganicchemistry.com
"Master Organic Chemistry"
Introduction to Cycloalkanes
Cyclic molecules
Molecules in organic chemistry don't just exist as chains - they can also exist as rings. Compare
propane and cyclopropane:
H2
H2
C
Cyclopropane
Propane
C
CH3
H 3C
H 2C
CH2
C3H 8
C3H 6
old "footrest"
is now "headrest"
Energy of activation
H
H
Note how all axial groups became equatorial and all
H
move "headrest" H
for chair flip:
2H
H
equatorial groups became axial....
4
H
down
~
10.7
kcal/mol
H 3
3
H
2
H
1
H
HH
5 H
BUT all "up" groups are still "up", and all "down" groups are
6
HH
1 H
5
H
still "down" !!!
4
6
H
H
H H
move "footrest" up
old "headrest"
H
H
is now "footrest"
Stereochemistry
When groups are on a ring, they are constrained - they can't rotate in the same way that linear molecules can.
This can give rise to "cis" and "trans" isomers - stereoisomers.
H 3C
X
CH3
H 3C
cis-1,2-dimethylcyclopropane
For substituted cyclohexanes, these two forms are not generally equal in energy
axial steric interactions between
CH
H
3 and H 5 / H 3
H
CH3
H
2H
H
4
H
H 3
3
6
5
H
2
H
1
1
H 3C
5 H
HH
CH
6
H
3
1 H
H
4
5
H
4
6
2
H
H
3
H H
H
H
"dash"
group points
behind page
These molecules can't interconvert
through bond rotation!
CH3
trans-1,2-dimethylcyclopropane
stereoisomers: same molecular formula, different boiling points/melting points
Other examples:
H 3C
CH3
CH3 H 3C
CH3
CH3
CH3
cis-1,3-dimethylcyclopentane
trans-1,3-dimethylcyclopentane
CH3 axial (less stable by
1.8 kcal/mol)
CH3
Cl
In contrast to chains, where sp3 hybridized carbons can adopt their ideal geometry of 109.5°,
the geometry of carbons in some cycloalkanes can be far from ideal, leading to ring strain
H 3CO
2 3
1
4
6
Cyclobutane
90°
27.6 kcal/mol 26.4 kcal/mol
Ring strain (kcal/mol)
(kJ/mol)
115.5 kJ/mol
110.4 kJ/mol
Cyclopentane
108°
0
27.2 kJ/mol
0
The Chair Conformation of Cyclohexane
The cyclohexane ring is not a flat hexagon. Why not? Two reasons: 1) the bond angles would be
120° (creating angle strain) and 2) the neighboring hydrogen atoms would be eclipsed with each other.
Instead cyclohexane adopts a chair conformation where all bond angles are close to 109 degrees and
all neighboring groups are staggered.
Groups that point "straight" up and down
are said to be "axial" - in red
H
H
axial
H
H
CH2
"headrest"
Newman
H
H
projection
H H
H
of chair
H
H equatorial
CH2
H
H
"footrest"
H H
H
H
H
Groups that point "somewhat" up or down
are said to be "equatorial" - in blue
In the chair form, all substitutents are staggered
along every C–C bond
H
H
H
H
H
H
H
H
H
H
H
H H
H
H
H
H
H
H
H
H
H
H
H
H
H H
Planar
"Half-Chair"
angle strain (120°)
eclipsing strain
eclipsing strain
steric crowding
(two hydrogens omitted 10.7 kcal/mol less
for clarity)
stable than chair
H
H H
H
H
"Boat"
eclipsing strain
steric crowding
6.9 kcal/mol less
stable than chair
(Technically these are "gauche" interactions)
Each interaction between axial CH 3 and axial H "costs" ~0.9
kcal/mol, for a total of about 1.75 kcal/mol.
Using this value we can calcuate the % of 1-methylcyclohexane
that is present in the axial form at equilibrium:
ΔG = –RT ln K
The size of the A value is related to the magnitude of the steric
interactions between axial substitutents
CH3 equatorial
OCH 3 equatorial
Cl axial
Group
A Value
Group
A Value
CH 3
1.74
Cl
0.43
CH2CH3
1.75
OH
0.87
i-Pr
2.15
OCH 3
0.6
t-Bu
4.7
NH 2
1.6
This table gives an idea of the preference of the substituent
for the equatorial position. For example t-Bu is the bulkiest
of these groups, whereas Cl is the least bulky.
Using the Gibbs equation it's possible to calculate the
position of the equilibrium for each substituent.
For example
Equilibrium constant H
CH3
2H
H
4
K ~ 10 5
3
H 3C
H
t-Bu
5 H
H CH3
6
1 H
H
H
H
t-Bu
H
H H 3
3
4
2
HH
2
1
6
1
6
5
H
5
CH3 H
4
H
t-Bu axial (strongly disfavored)
CH3 axial
CH3 equatorial
t-Bu equatorial
CH 3 equatorial
CH 3 axial
Application in substitution and elimination reactions (SN2 and E2)
H
HH
H
H
3
4
2
H
H
H
H H
H H
"Twist boat"
eclipsing strain
steric crowding
5.3 kcal/mol less
stable than chair
1
6
1
6
t-Bu
5
H
H
1
5
3
4
2
Cl
2H
Cl
3
6
H
H
H
4
5
H
H
H
t-Bu
H
H H 3
Cl
HH
2
1
6
H
HH
5
H
4
H
Cl
K = e –ΔG/RT
This value: 1.75 kcal/mol is sometimes referred to as
the "A" value of CH 3
Understanding the proportions of different chair forms is important because SN 2 and E2 reactions will
only occur when the leaving group is in the axial position.
H
H
all red groups remain up
all blue groups remain down
Cl
H
2H
H
4
3
H
CH3
H 3CO
5 H
6
1 H
H
H
H
The t-butyl group is so bulky that the axial position is extremely disfavored. Equilibrium
favors the equatorial position to such an extent that one can consider the chair to be
"locked" in the conformation with t-Bu equatorial
H
Other conformations of cyclohexane (higher in energy than the chair)
H
5
all red groups are up
all blue groups are down
OCH 3
H
H 3
Cl
2
1
HH
HH
5
4
6
H H
H
CH3
CH3 axial
OCH 3 axial
Cl equatorial
The axial position is less favored for substitutents since
it can undergo unfavorable steric interactions with other axial
substitutents. These are avoided in the equatorial position.
At 1.75 kcal/mol this gives us K = 0.052, or approximately 95:5
ratio in favor of the equatorial CH 3
Important: the t-butyl group is conformationally "locked" in the equatorial position
Ring strain is a combination of angle strain (where geometry of the carbon deviates from the
ideal) and torsional strain (arising from eclipsing interactions)
H
CH3
Cyclohexane
* see below
6.5 kcal/mol
CH3 equatorial
Another example: note the positions of axial and equatorial groups
cis-1,2-dimethylcyclohexane
trans-1,2-dimethylcyclohexane
Ring Strain
Name:
Cyclopropane
60°
Interior angle:
masterorganicchemistry.com
Cyclohexane rings can be drawn as two different "chair" forms
note how there are two fewer hydrogens in the molecular
formula of cyclopropane than for propane
"wedge"
group points
out of page
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
only this conformation will undergo SN 2
or E 2 reactions (leaving group axial)
t-Bu is equatorial (favored)
only this conformation will undergo SN 2
or E 2 reactions (leaving group axial)
t-Bu is axial (disfavored)
Substitution and elimination reactions
will be extremely slow for this molecule
A value
% equatorial (at 25°C)
0
50%
0.5
70%
1
84%
1.5
93%
2
97%
2.5
98.5%
3
99.4%
3.5
99.7%
4.0
99.9%
4.5
99.95%
5
99.98%
Omissions, Mistakes, Suggestions?
james@masterorganicchemistry.com
This sheet copyright 2022 James A. Ashenhurst
masterorganicchemistry.com
Introduction to Resonance
"Master Organic Chemistry"
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
masterorganicchemistry.com
1. What is Resonance?
3. When Resonance Forms are Not Equal: How to tell
which are most significant?
The true distribution of electrons in a molecule cannot
always be accurately depicted through a single Lewis structure.
Resonance is a tool to show this.
Rule #4 Stabilizing positive charge
If the resonance form has a positive charge, try and find a
resonance form in which all octets are full (Rule #2)
Rule #1 - The Fewer Charges, the Better
Resonance forms decrease in importance as their charges increase
If all octets are not full, place positive charge on the atom best
able to support a carbocation (i.e. the most substituted carbon)
So what about applying this concept to these molecules?
O
C
The Lewis structure of this molecule appears to tell us that one oxygen
is more negative than the other (bears a greater electron density). Our model
based on pure electronegativity would suggest that it should be more reactive
than the neutral oxygen.
O
2
Similarly, the Lewis structure of this molecule appears to tell us that one carbon
bears the positive charge and the others are neutral
3
1
A
X
O
2
O
3
1
H
H
H
H
H
2 charges
less significant
O
O
0 charges
more significant
Best
2 charges
less significant
O
C
O
C
H
H
O
O
O
full octets
More significant
The "double-headed" arrow
denotes that two molecules
are resonance isomers
(NOT in equilibrium)
C
O
Resonance hybrid
Acetate ion resonance forms
O
H
O
carbon lacks a full octet
Less significant
2
3
1
2
O
N
H
3
1
Allyl carbocation resonance forms
1
3
Resonance hybrid
The true bond order is an average
of the bond order in each resonance
hybrid, weighted by the importance
of the hybrid to the overall structure
O
BAD!
O
1) π−donation
Note: never have less than
a full octet on oxygen or
nitrogen.
Best
There are only three legal "moves" we can do with curved arrows to show resonance.
Alkenes attached to an atom with a lone pair such as O, N, S, etc.
(often called "heteroatoms") have an important resonance form
with a negative charge adjacent to the carbon-atom bond.
H
O
Example:
sp2
1
2
1
1
2
2
3
3
3
Tail is at π bond between C1 and C2
Head is at C2
Arrow shows movement of electrons
from the C1-C2 π bond
to become a lone pair on carbon 2.
1
2
1
2
3
2
3
*only possible if an
or sp hybridized carbon is
next door
(i.e. don't break the octet rule)
1
3
CH2
CH2
Best
Second-best
negative charge
is on least basic
atom (O)
*only possible if an sp2
or sp hybridized carbon is
next door
(i.e. don't break the octet rule)
Tail is at π bond between C1 and C2
Head is at π bond between C2 and C3
Arrow shows movement of electrons from the C1-C2 π bond
(which breaks) to the C2-C3 π bond (which forms).
H
O
O
This effect makes these alkenes more nucleophilic (electron-rich).
–
–
–
δ
–
δ
δ
δ S
O
N
N
–
δ
–
–
–
δ
δ
δ
H
negative charge
is on most basic
atom (C)
2) π−accepting
When double bonds are connected to a polarized π bond, the
molecule will have a significant resonance form where there is
positive charge on the adjacent carbon.
O
Use acidity trends to determine the "best" resonance form
in these cases
Polarized
π-bond
(C=O)
• Electronegativity
Tail is on lone pair of C2
Head is between C1 and C2.
Arrow shows movement of electrons
from the lone pair on C2
to become a π bond between C1 and C2
π Bond → π Bond
negative charge
is on most basic
atom (N)
O
If given a choice, the best place to put a negative
charge on a resonance form is on the least basic atom
Lone pair→ π Bond
π Bond → Lone pair
Second-best
negative charge
is on least basic
atom (O)
2. How to Use Curved Arrows
O
NH
NH
Here it would be 1.5
N
O
H
aromatic
4. Applications of Resonance
"Enolates"
O
O
aromatic
Rule #3 - Place a negative charge on the least basic atom
"Amide anions"
2nd-best
A resonance form that is aromatic will have a disproportionate
contribution to the resonance hybrid
H
2
Best
Rule #5 - Aromaticity (Org 2 Topic)
this contributes
more to the
resonance hybrid
Instead, the "true" structure of the molecule is a hybrid of these two forms.
O
2nd-best
A resonance form in which all atoms have a full octet is more significant than one
where at least one atom lacks a full octet.
numbering is for
placeholding, not
IUPAC
Acetate ion
H
Rule #2 - Full Octets Are Preferred
2
X
3
1
H
this contributes
more to the
resonance hybrid
Important: this is not an equilibrium!
O
C
H
0 charges
more significant
In reality, the charge in both of these molecules is distributed equally between the different ends
O
C
B
O
O
• Polarizability
This effect makes these alkenes more electrophilic
(electron-poor)
–
–
–
–
δ
δ
O
+ δ
Oδ
N
δ N
+
+
+
δ
Sδ
C
δ
• Induction
Polarizabiity
O
Electron withdrawing groups
O
S
Best
negative charge
is on least basic
atom (S)
δ
F
S
Second-best
negative charge
is on most basic
atom (O)
F
Best
negative charge is
on least basic atom
+
δ
+
δ
+
δ
+
F
F
Second-best
negative charge is
on most basic atom
Omissions, Mistakes, Suggestions?
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"Master Organic Chemistry"
Summary Sheet - Free Radicals
Free radicals contain an unpaired electron in their valence shell
Properties:
Chlorine radical
masterorganicchemistry.com
Electron movement is shown with "single-barbed" arrows
Chlorination Reactions Are Less Selective Than Bromination Reactions
Example: Chlorination can lead to substitution at primary, secondary, and tertiary positions
As free radicals contain less than a full octet of Example #1 - homolytic cleavage
electrons, they are electron-deficient and
therefore are highly reactive
light (hν)
Cl Cl
Cl
Cl
H
H
H
Methyl radical
Cl2
Cl
Bromination tends to be very selective for substitution at tertiary C–H
1. Substitution
Stability increases with the number of attached carbons. Neighboring alkyl
groups can donate electron density to the electron-poor radical
H
<
H
CH3
C
CH3
H
<
Primary radical
Methyl radical
Secondary radical
<
C
H
H
H 2C
H
C
CH2
Less stable
More stable
Iodination doesn't work well (thermodynamically unfavorable) and
fluorination is difficult to control
The Three Key Mechanistic Steps In Free-Radical Halogenation
Are Initiation, Propagation, and Termination
<
H 2C
F
<
H 2C
OH
<
H 2C
NH 2
most stable
least stable
Stability increases with increasing donor ability of adjacent
atom with lone pairs
Other Factors Affecting Free Radical Stability
<
Cl
<
Br
<
most stable
least stable
Hybridization: Stability increases with decreasing s-character,
which moves the free radical farther from the nucleus
H
<
< H 3C–CH2
C C
R C C
H
H
most stable
least stable
sp3
sp
sp2
Electron-Withdrawing Groups: Groups which withdraw electron
density from the free radical (without being able to donate a lone pair)
destabiize radicals
F 3C–CH2
<
less stable
H 3C–CH2
most stable
Bond strengths are a useful guide to free radical stability
Since bond dissociation energies measure homolytic cleavage
It's possible to discern trends in free radical stability by examining
CH 3
bond dissociation energies.
CH 3
H–C CH3
H–C
H–CH3
H–CH2CH3
CH3
CH3
H
Strongest bond
Weakest bond
104 kcal/mol
98 kcal/mol
96 kcal/mol
93 kcal/mol
(methyl)
(primary)
Cl
H
19,400
Cl •
0.004
1
4.3
6.0
Allylic And Benzylic Bromination
CH2 –H
Free-Radical Addition of H–Br To Alkenes
Cl
Note that each side of the equation
has one free radical (no change in #)
CH3
Note that the product side of the
equation has no free radicals
Cl
Cl
CH3
H–Br
R
RO–OR
heat or light
H Br
Free Radical Initiators
Free radical "initators" generate free radicals when heated. They generally have a weak
bond (such as O–O) that readily undergoes homolytic cleavage.
N
Examples:
O
O
C
N
N
di-t-butyl peroxide
C
benzoyl peroxide
N
AIBN
Often just abbreviated as "RO–OR" or "peroxides"
R
R
R
O O
Note that Br adds to least substituted
end of double bond. Why?
Br
CH3
Avoid the common mistake of writing this step as Propagation Step 2!
Note how the number of free radicals decreases from 2 to to zero; therefore
it is a termination step
Br
R
After initiation forms peroxide radical, and peroxide
radical removes H from H–Br to give Br• , this key step occurs:
other termination reactions also possible
O–O
Br
NBS is N-bromo succinimide,
a source of reactive bromine
allylic C–H
This chlorine radical can then participate
in Propagation Step 1 with a new molecule
of alkane; therefore, this is a chain reaction!
Cl
Allylic and benzylic C–H bonds
are relatively weak, since removal
will result in a resonance-stabilized
free radical
hγ or Δ
Propagation (Step 2)
CH3
CH2 –Br
NBS
hγ or Δ
benzylic C–H
Cl
CH3
CH3
(tertiary)
220
CH2 –H
Cl–H
Cl–Cl
(secondary)
(CH3)3C–H
1
Note that this is an equilibrium!
Not all the Cl 2 is consumed!
Propagation (Step 1)
Cl
(CH3)2CH–H
NBS
Termination
I
CH 3CH2 –H
0.007
Δ (heat)
or h γ (light)
Polarizability: Stability increases with increasing distance of the
half-filled orbital from the nucleus; in other words, with increasing size
of the atom (stability increases going down the periodic table)
F
CH3–H
Br •
• Initiation steps have a net increase in the number of free radicals
• Propagation steps have no net change in the number of free radicals
ª Termination steps have a net decrease in the number of free radicals
More stable
3. (Advanced) Atoms With Adjacent Lone Pairs
Adjacent lone pairs stabilize free radicals through "hyperconjugation"
H 3C
+ H–Br
R–Br
Cl–Cl
Less stable
Selectivity of Radicals Towards Alkyl Hydrogens
Radical
Br 2
Initiation
H
Major product
How much more selective is Br• than Cl• ?
+ H–Cl
hγ or Δ
2. Free Radicals Are Stabilized By Resonance
Radicals are stabilized if they can be delocalized over several atoms (as in resonance)
H 3C
R–Cl
hγ or Δ
Tertiary radical
CH2
<
Cl2
R–H
Br
hν or Δ
Key Reaction: Halogenation of Alkanes With Cl2 or Br 2
R–H
Most stable
H
C
resonance forms
CH3
<
C
H 3C
CH3
Least stable
H H
C
Br 2
H
The same factors which stabilize carbocations. Keep these two general principles in
mind: 1) electron-poor species (like radicals) are stabilized by electron-rich neighbors,
or 2) by being able to spread electron density out over a greater volume.
Cl
Cl
4 possible products formed here; chlorination occurs at primary,
secondary and tertiary positions
What Factors Stabilize Free Radicals?
CH3
C
H
H
Cl
hν or Δ
Example #2 - showing interconversion of resonance forms
geometry of methyl radical is a "shallow" trigonal pyramid
H
C
Cl
Each barbed arrow shows the movement of one electron
Neutral (formal charge of zero)
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
Br
Br• adds to the end of
the alkene such that the
more stable secondary
radical is formed
Br
In the second propagation
step, the alkyl radical
removes a hydrogen from
H–Br, regenerating Br•
H
Br
R
+ Br
How To Calculate Selectivity? An Example
Cl
H
H 3C C CH3
H 3C C CH2Cl
CH 3
CH3
43%
57%
Here there are two types of C-H bonds; a tertiary C-H and a primary C-H
How selective is the reaction for tertiary C-H ?
We can't compare by the yields directly [43% and 57%] because there is one tertiary C-H
and 9 primary C-H . To adjust for this statistical factor we divide 57 by 9.
H
H 3C C CH3
CH 3
Cl2
This gives us 6.3 .
Now we divide 43 by 6.3, which gives
us 6.8 .
Therefore the reaction is 6.8 times
more selective for tertiary C-H than
primary C-H
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Introduction to Acids and Bases
"Master Organic Chemistry"
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
masterorganicchemistry.com
1. Introduction
Factor #1: The less charge, the better
2
Two definitions of acids:
<
O
<
O H
least stable
(i.e. most basic)
Brønsted definition: An acid donates a proton (H+)
Lewis definition: An acid accepts a lone pair of electrons
<
H O
Two definitions of bases:
Brønsted definition: A base accepts a proton (H+)
Lewis definition: A base donates a lone pair of electrons
H O H
most stable
(least basic)
<
H OH
Factor #6: Negative charge is stabilized as the hybridization includes
greater s character
Factor #2: Going across the periodic table, greater
electronegativity stabilizes negative charge.
Simple example of an acid-base reaction
H2
H 3C C H
H
<
H–OH
+ Na OH
+
Na Cl
↑
Acid
Base
Conjugate
acid
Form
Break
H OH
H Cl
Conjugate
base
The four actors:
Every acid base
reaction will involve
the formation and
breakage of a bond
to hydrogen
Acid: donates a proton (or - accepts a lone pair)
Base: accepts a proton (or - donates a lone pair)
"Conjugate base": is what remains of the acid after it donates H+
"Conjugate acid" is the species formed when the base accepts H+
H–I
acid
H–OH
acid
H–CH 3
acid
H
H
H
+
I
conjugate
base
K a = 1 × 1010
+
K a ~ 1 × 10 -15
OH
conjugate
base
+
CH3
conjugate
base
K a = 1 × 10 -50
–10
Strong acid!
The stronger the acid, the weaker the conjugate base
The weaker the acid, the stronger the conjugate base.
The best way to understand acidity/basicity is to understand the
factors that stabilize negative charge (i.e. high electron density)
There are seven important factors (placed here in roughly decreasing
order of importance).
H–CH 3
<
F
H–OH <
F
50
Extremely weak
acid!
Br
<
least polarizable
(least stable)
(most basic)
H–F
<
H
<
H–Br <
Weakest acid
Stronger acid
4. What makes an acid-base reaction favorable (or not?)
Key principle: In favorable acid-base reactions, a stronger acid adds to a
stronger base to give a weaker acid and a weaker base
Key questions to ask: Is the "acid" stronger than the "conjugate acid" ?
Is the "base" stronger than the "conjugate base" ?
If so, the acid-base reaction will be favorable.
H–I
strongest acid
How do you tell which acid is stronger or weaker?
A pKa table collates experimental measurements of acidity, and thus incorporates
all the factors mentioned above.
Factor #4: Negative charge is stabilized by resonance,
which delocalizes electron density
O
H
<
Weaker acid
I
most polarizable
(most stable)
(least basic)
H–Cl
Strongest acid
aromatic! 10 33 more stable than the molecule on the left!
(more stable)
(less basic)
not aromatic
(less stable)
(more basic)
H–F
<
H C C H
<<
strongest acid
Cl
<
<
H
Factor #7: A Special case - negative charge is especially stable if the
electrons are part of an aromatic π-system
most electronegative
(most stable)
(least basic)
H–NH 2 <
O
Strong acids have low pKa values. Weak acids have high pKa values.
<
A very abridged pKa table:
resonance stabilization
(more stable)
(less basic)
no resonance
(less stable)
(more basic)
15.7 Weak acid
Generally, the strength of an acid is related to the stability of the conjugate base.
The greater the stability of the conjugate base, the stronger the acid.
The lower the stability of the conjugate base, the weaker the acid.
Stable conjugate base = a "weak" base
Unstable conjugate base = a "strong" base
<
least electronegative
(least stable)
(most basic)
pKa
(–log K a)
3. What factors make an acid strong or weak?
A different way to put it: stability is inversely related to basicity.
OH
Factor #3: Going down the periodic table, greater
polarizability stabilizes negative charge.
Acidity is defined as the position of the equilibrium between an acid and its conjugate base.
We obtain this number through experimental measurement.
solvent is technically
For a generic acid-base reaction:
The acidity constant, K a is:
a part of the acidbase equation, but we
+ A
[A ] [ H ] [solv]
H–A
H
can ignore this
Ka =
because
all K a's are
acid
conjugate
[H–A] [solv]
base
expressed in the
same solvent
These values can differ by an extremely large amount (10 60) so it is convenient to work
with the logarithm of this number, defined as the pKa:
pK a = – log K a
3 concrete examples:
<
NH 2
Weakest acid
2. How Do We Define and Measure Acidity?
experimental
measurement
<
CH3
H
H
H C C
sp hybridized
(most stable)
(least basic)
C C
Weakest acid
H–Cl
<
C C
H
H
2
sp
sp3 hybridized
(least stable)
(most basic)
H OH 2
strongest acid
weakest acid
H
<
H 3C CH 2
OH
strongest
acid
OH
<
Factor #5: Negative charge is stabilized by adjacent electron
withdrawing groups (inductive effects)
H
H C
H
<
Cl
H
C
H
<
Cl
Cl
C
H
<
No electron
withdrawing groups
(least stable)
(most basic)
H
H C H
H
Weakest acid
<
Cl
Cl
C
Cl
3 electron
withdrawing
groups (Cl)
(most stable)
(least basic)
Cl
H
C H
H
<
pKa
H–Cl
–8
Cl
H–OH2
O
H O CCH3
–1.7
10
H 2O
O
O CCH3
NH 3
15.7
OH
H–OCH2CH3
16
OCH 2CH3
H C C H
25
C C H
H–NH2
38
NH 2
H–CH3
~50
CH 3
H–NH3
H–OH
Stronger acid
Weaker acid
Acid
Cl
Cl
C H
H
<
Cl
Cl
C H
Cl
Strongest acid
weakest
acid
Conjugate Base
4.5
weakest
base
strongest
base
Example of a favorable acid-base reaction:
H–Cl
pKa –8
+ Na OH
H–OH
+
Cl
pKa ~15
Here, we're going from a stronger
acid (HCl, pKa –8) to a weaker acid
(H 2O, pKa 15) so this acid-base
reaction is favorable.
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"Master Organic Chemistry"
Summary of Functional Groups (1)
What Are Functional Groups?
Functional groups are collections of atoms that have a characteristic pattern of chemical
reactivity
Alkane
a hydrocarbon with no multiple bonds is an alkane
C
H 3C
H2
C
H 3C
OH
Methanol
Butane
C C
Alkene
Decane
a hydrocarbon with at least one C–C double
bond ( π bond) is an alkene
Examples:
H
OH
A carbonyl (C=O) attached to a hydrogen
and another carbon is an aldehyde
Examples:
Ethanol
2-propanol ("Isopropanol")
a primary alcohol
a secondary alcohol
2-methyl-2-propanol
(t-butanol)
a tertiary alcohol
δ – δ+
R O R
Ether
H 3C
O
C
O
O
H
O
H
H
H
H
H
H
Propene
trans-2-butene
[or (E)-2-butene]
cis-2-butene
(E)-3,5-dimethylhex-2-ene
[or (Z)-2-butene]
O
Characteristics: nonpolar. Molecule cannot rotate along double bond.
Geometry: trigonal planar (sp 2 hybridized)
Reactivity: undergo addition reactions, as well as oxidative cleavage
Stability increases with increasing # of carbons attached
Alkyne
C C
a hydrocarbon with at least one C–C
triple bond ( π bond) is an alkyne
Suffix: "-yne". As a substituent: "alkynyl"
Examples:
O
Dimethyl ether
O
Ethyl methyl ether
(or "methoxyethane")
Methyl phenyl ether
(or "methoxybenzene",
or "anisole")
Alkyl halide
–
δ+ δ
R Cl
–
δ+ δ
R Br
–
δ+ δ
R I
Tetrahydrofuran (THF)
a cyclic ether
2-butyne
3-heptyne
internal alkynes
Alkynes with a C–H bond are called "terminal" alkynes
Geometry: linear (sp hybridized)
Characteristics: non polar
Reactivity: addition reactions
oxidative cleavage reactions
acid-base reactions (terminal alkynes are unusually acidic)
A six-membered ring containing
3 alternating double bonds is a
benzene ring
Benzene ring
Suffix: "benzene". As a substituent: "phenyl"
Examples:
An alkyl group attached to
a halogen is an alkyl halide
H 3C
Fluoroethane
2-chloropropane
2-bromo-2-methylpropane
(ethyl fluoride)
(isopropyl chloride)
(t-butyl bromide)
a primary alkyl
a secondary alkyl halide
a tertiary alkyl halide
halide
Characteristics: generally considered non polar (but more polar than alkanes)
Reactivity: substitution reactions (Cl, Br, I can be good leaving groups)
elimination reactions (Cl, Br, I can be good leaving groups)
Amine
substitution reactions (e.g. electrophilic aromatic substitution
or nucleophilic aromatic substitution)
Less reactive than normal alkenes due to aromatic stability
δ – δ+
R N H
H
Suffix: "-amine". As a substituent: "amino"
Examples:
NH 2
NH 2
Ethylamine
a primary amine
O
C
O δ−
+ δ+
Cδ H
O −
R
δ
A carbonyl (C=O) adjacent to a hydroxyl
(OH) and an R group is a carboxylic acid
O
O
O
OH
OH
OH
Methanoic acid
("formic acid")
Ethanoic acid
("acetic acid")
OH
Butanoic acid
O δ−
+
Cδ
R
O R
A carbonyl (C=O) adjacent to an alkoxy
(OR) and an R group is an ester
Suffix: "-oate"
Examples:
H
N
Dimethylamine
a secondary amine
N
Triethylamine
a tertiary amine
Characteristics: polar (N-H group participates in hydrogen bonding, although not as
much as a hydroxy group
Reactivity: acid-base reactions (tend to act as bases)
substitution reactions (can act as nucleophiles)
benzoic acid
Reactivity: acid-base reactions (the O–H is acidic)
acyl substitution reactions (can replace OH with other groups under
acidic conditions)
Ester
A nitrogen attached to simple carbon or
hydrogen atoms is an amine
phenyl methyl ketone
("acetophenone")
addition reactions (the carbonyl carbon reacts easily with nucleophiles)
acid-base reactions (carbons adjacent to the ketone can be deprotonated
to give enolates)
Carboxylic acid
H
Iodomethane
(methyl iodide)
3-hexanone
Characteristics: the C=O bond is somewhat polar (less so than O-H however)
Br
F
O
2-propanone
2-butanone
("acetone")
("methyl ethyl ketone")
Examples:
Methylamine
1,4-dimethylbenzene
(para-dimethylbenzene)
O
Examples:
I
Benzaldehyde
A carbonyl (C=O) flanked by two carbons
is a ketone
O
Suffix: "-oic acid"
H 3C
Methylbenzene
(toluene)
O δ−
+
Cδ
R
R
Suffix: "-ane". As a substituent: "haloalkyl"
H 3C C C H
Propyne
a terminal alkyne
Ketone
Reactivity:
Cl
Butanal
Characteristics: the C=O bond is somewhat polar
Reactivity: addition reactions (the carbonyl carbon reacts easily with nucleophiles)
Characteristics: borderline between nonpolar and polar (due to dipole-dipole)
Reactivity: acid-base reactions (oxygen can act as a very weak base
–
δ+ δ
R F
Propanal
Examples:
O
C
H 3C
CH3
As a substituent: "alkoxy"
Examples:
O
H 3C
CH3
Ethanal
Suffix: "-one". As a substituent: "oxo"
An oxygen flanked by two carbons is an ether
H
C C
Reactivity:
O δ−
+
Cδ
R
H
Suffix: "-al" (if attached to ring: carbaldehyde) As a substituent: "oxo"
Characteristics: polar (O-H group participates in hydrogen bonding)
Reactivity: acid-base reactions (can act as acids or bases)
substitution reactions (can act as nucleophiles)
oxidation reactions (primary and secondary alcohols (and methanol)
can be oxidized to aldehydes, ketones, or carboxylic acids, depending
on structure and reagent used)
2-methylpentane
Suffix: "-ene". As a substituent: "alkenyl", or "vinyl" in the case of C2H 3
Benzene
OH
Aldehyde
8
Characteristics: nonpolar
Geometry: tetrahedral (sp3 hybridized)
Reactivity: free radical reactions (e.g. free radical chlorination or bromination)
H
OH
( )
CH3
Propane
H 3C
"OH" attached to an alkyl group is referred to as an
alcohol. OH attached to a benzene ring is a "phenol" (not
shown).
Suffix: "-ol". As a substituent: "hydroxy"
Examples:
Suffix: "-ane". As a substituent: "alkyl", or "methyl" for CH 3, "methylene" for CH2, and
Examples:
"methine" for R 3C–H
masterorganicchemistry.com
δ – δ+
R O H
Alcohol
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
H
O
C
O
O
O
O
OCH3
O
Methyl methanoate
Reactivity:
Methyl ethanoate
O
Ethyl butanoate
Methyl benzoate
acyl substitution reactions (can replace OR with other functional
groups under acidic conditions)
addition reactions (the carbonyl carbon reacts easily with nucleophiles)
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Ashenhurst http://masterorganicchemistry.com
"Master Organic Chemistry"
Summary of Functional Groups (2)
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
masterorganicchemistry.com
What Are Functional Groups?
Functional groups are collections of atoms that have a characteristic pattern of chemical
reactivity
Imine
δ – δ+
R S H
Thiol
δ –R
N
+
Cδ
R
H
Thiols are the sulfur-containing
analogs of alcohols
Imines are the nitrogen-containing
analogues of ketones and aldehydes
Nitro
Examples:
N
C
Examples:
SH
H 3C
SH
N
C
SH
SH
Methanethiol
Ethanethiol
Thiophenol
Isoprenyl mercaptan
"eau de skunk"
acid-base reactions (thiols can act as weak acids)
substitution reactions (thiols are good nucleophiles)
Reactivity:
Sulfide
δ – δ+
(Thioether) R S R
A sulfur flanked by two carbon
atoms is a sulfide
N
C
H
an aldimine
Reactivity:
Suffix: "-oyl halide"
Examples:
Examples:
H 3C
CH3
S
H 3C
tetrahydrothiophene
Phenyl methyl
sulfide
reduction reactions (will reduce the intermediate of ozonolysis
reactions); sulfide is oxidized to "sulfoxide"
Dimethylsulfide
Reactivity:
–
O δ
C C
+
δ+ δ
Epoxide
Epoxides are cyclic ethers
consisting of a 3-membered ring
containing an oxygen and two
carbons.
Suffix: "-oxide"
O
H 2C CH2
Reactivity:
Nitrile
Nitromethane
Reactivity:
Enol
Acyl halides are carboxylic acid derivatives where -OH has been
replaced by a halogen
OH
Cl
OH
Cl
Propanoyl bromide
Reactivity:
Benzoyl chloride
enols are good nucleophiles. They will perform addition
reactions to aldehydes and ketones, among other reactions.
Enols will tautomerize to aldehydes/ketones
acyl substitution reactions (halides are good leaving groups and can be replaced
by good nucleophiles)
δ–
O
Enolate
δ–
O
+
Cδ
R
NHR
δ–
O
O
Examples:
O
Enolates are the conjugate bases of enols
δ–
R
Amides are the nitrogen containing analogues of esters
Examples:
O
Ph
Propylene oxide
NH 2
Styrene oxide
epoxides have considerable ring strain, and will undergo
ring opening when treated with various types of nucleophiles
(e.g. Grignard reagents, organolithium reagents)
δ+ δ –
R C N
A molecule containing the CN group is a nitrile
acetamide
a primary amide
O
O
O
Examples:
CN
CN
Propionitrile
O
N
H
N-methyl acetamide
a secondary amide
Reactivity:
N-ethyl N-methyl acetamide
a tertiary amide
Enamine
O
O
Benzonitrile
reduction reactions (nitriles can be reduced to amines with strong
reductants such as LiAlH 4 )
hydrolysis reactions (nitriles can be converted to carboxylic acids
with aqueous acid)
O
Acetic anhydride
(Ethanoic anhydride)
Reactivity:
O
Propionic anhydride
O
O
O
O
Succinic anhydride
Ph
O
O
enolates are excellent nucleophiles (better than enols) They will
perform addition reactions to aldehydes and ketones, as well as
substitution reactions with alkyl halides (among others)
R
R
δ–
R
N
Enamines are alkenes attached to
an amino substitutent.
δ–
Examples:
N
Reactivity:
O
O
H
reduction reactions (amides can be reduced to amines with strong reductants
such as LiAlH 4
hydrolysis reactions (amides can be hydrolyzed to carboxylic acids with aqueous acid)
δ–
δ–
O
O
+
+
Anhydride
δ C
Anhydrides are oxygen atoms flanked by two acyl groups
Cδ
R
O
R
Examples:
O
N
Reactivity:
Suffix: "-anhydride"
Reactivity:
OH
H
Br
Suffix: "-nitrile"
Acetonitrile
Enols are alkenes attached to a
hydroxyl substitutent. They are
obtained from tautomerization of
aldehydes/ketones
Examples:
O
Amide
Nitrobenzene
reduction reactions (nitro groups can be reduced to amines)
acid-base reactions (C-H bonds adjacent to the NO 2 group are acidic)
δ – H δ+
O
δ–
R
N
H 3C CN
Nitroethane
O
Ethanoyl chloride
("Acetyl chloride")
Reactivity:
H 3C
Suffix: "-amide"
Examples:
Ethylene oxide
O
C
NO 2
NO 2
NO 2
reduction reactions (imines can be reduced to amines)
hydrolysis reactions (imines can be hydrolyzed to give back aldehydes or ketones)
Suffix: "-sulfide".
S
•Imines formed from aldehydes are
sometimes called "aldimines"
•Imines formed from ketones are
sometimes called "ketimines"
Imines are also sometimes called "Schiff bases"
a ketimine
δ–
O
+
Acyl halide
Cδ δ –
Cl
("Acid halide") R
S
A molecule containing the NO 2 group is a
nitro compound
Examples:
Suffix: "-imine".
Suffix: "-thiol". As a substituent: "mercapto"
δ–
O
+
Nδ
R
O –
δ
N
N
enamines are excellent nucleophiles. They will perform addition
reactions to aldehydes and ketones, as well as substitution
reactions with alkyl halides (among others)
Ph
Benzoic anhydride
acyl substitution reactions (anhydrides can be cleaved with good nucleophiles)
reduction reactions (anhydrides can be reduced to aldehydes with various reducing agents)
Omissions, Mistakes, Suggestions?
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Ashenhurst masterorganicchemistry.com
"Master Organic Chemistry"
Introduction to Stereochemistry
Symbols
Example
"wedge"
Cl
denotes a group pointing "out of the page"
Types of isomers
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
masterorganicchemistry.com
Isomers: same empirical formula, different structural formula.
Type
Connectivity? Configuration? 3-D Shape?
Conformational isomers
Same
Same
Different
Cl
Physical properties
Example
Identical, as long as they can interconvert
through bond rotation
staggered
eclipsed
denotes a group pointing "into the page"
"dash"
Cl
Different
Constitutional isomers
Different boiling points, melting points, and
other physical properties
"squiggle" undefined (mixture of stereoisomers) Rarely used.
Same
Stereoisomers
OH
1-butanol
OH
2-butanol
OH
OH
Different
Terms
anti
CH 3
H
H
syn
H 3C H
H
CH 3
H
H
Br
Diastereomers
CH 3
Stereoisomers that are NOT nonsuperimposable mirror images
Different boiling points, melting points, etc.
(R )-2-butanol
trans-2-butene
The R,S convention
two groups on adjacent carbons oriented at 60° to each other
CH 3
H
Generally, gauche interactions between
bulky groups are intermediate in torsional
strain between the anti and syn relations
H
when all 3 substituents overlap completely with all 3 substituents on a
neighboring carbon.
Generally, eclipsed conformations
H 3C H
H 3C CH 3
have the highest torsional strain
H
H
H
H
H
H
CH 3
H
CH 3
H
H
H
H
H
H
CH 3
H
H
CH 3
the 3-D arrangement of bonds around a carbon.
racemic mixture
Meso compound:
A molecule with chiral centers, but a plane of symmetry that makes
the molecule achiral
plane of
symmetry
plane of
symmetry
on the same side of a double bond or ring
Trans
on opposite sides of a double bond or ring.
Chiral molecule
a molecule with an enantiomer; cannot possess a plane of symmetry
Chiral Center
Has 4 different constituents
Strain that arises from the proximity of bonds (and the electrons in them)
- generally eclipsing
Torsional Strain
3
2
1
1,2,3 goes CW: R
1,2,3 goes CCW: S
1
2
H
swap H and OH
(S)-2-pentanol
H
OH
(R)-2-pentanol
4
1,2, and 3 go counterclockwise.
4 ranked substituent is in front
Recognizing Enantiomers/Diastereomers Using Only Their Names
Example:(2S, 3R)-2-bromo-3-chlorobutane and (2S, 3S)-2-bromo-3-chlorobutane
(2S, 3S, 4S, 5R)-2-bromo-3-chloro-4-methyl-5-propyldecane
and (2S, 3R, 4R, 5S)-2-bromo-3-chloro-4-methyl-5-propyldecane
also, diastereomers will have the same name, but differ in E/Z (or cis/trans).
a 50:50 mixture of two enantiomers
Cis
4
HO
3
Diastereomers: same name, but R/S designations are not exactly opposite
two molecules with the same molecular formula, but different in their structures
configuration
2
Single swap rule: switching any two groups on a
chiral center will flip (R) to (S) and vice versa.
works for any other two groups as well
(e.g. OH and CH 3, H and CH 3, OH and CH 2CH 2CH 3, etc.)
Therefore, flip: this is R !
staggered refers to the orientation when all 3 substituents on a carbon are arranged at
a 60 deg angle to all 3 substituents on another carbon.
H 3C
4
What if #4 is in the front? One approach is to
trace 1,2 and 3 as you normally would. Then flip!
3
H
H
OH
The single swap rule:
Rank according to CIP protocol
Put #4 ranked substituent in back.
1
cis-2-butene
OH
OH
(trans)-cyclohexane 1,2-diol
(cis)-1,2-cyclohexanediol
H 3C H H H
H
H 3C
(S)-2-butanol
OH
Note: another name for stereoisomers is "configurational isomers":
they have the same connectivity, but differ in the configurations of
the carbons.
CH 3
H
isomers
Identical. Differ in optical rotation
two groups on adjacent carbons oriented at 0° to each other
H
eclipsed
Stereoisomers that are nonsuperimposable mirror images
CH 3
HH
H
H
H 3C CH 3
gauche
Enantiomers
two groups on adjacent carbons oriented at 180° to each other
e.g. (Z)-2-butene and (E)-2-butene.
cis-1,2-dimethylcyclohexane and trans-1,2-dimethylcyclohexane
Enantiomers: same name, but have all chiral centers have opposite R/S designations
Example: (R)-2-butanol, (S)-2-butanol
(2S, 3R)-2-bromo-3-chlorobutane and (2R, 3S)-2-bromo-3-chlorobutane
(2S, 3S, 4S, 5R)-2-bromo-3-chloro-4-methyl-5-propyldecane
and (2R, 3R, 4R, 5S)-2-bromo-3-chloro-4-methyl-5-propyldecane
Important exception: If the molecule has a mirror plane, then it is meso, and the two "enantiomers"
are in fact the same molecule
Cyclohexane chair conformations
in the cyclohexane chair conformation: all C–C bonds staggered
CH 3
H 3C
CH 3 axial
(disfavored)
CH 3 equatorial
(favored)
Chair flips: all axial groups become equatorial, and all equatorial
groups become axial.
BUT all groups that are "up" stay "up" and all groups that
are "down" stay "down.
Bulky groups prefer the equatorial position.
The bulkier the group, the greater the energy difference will be.
Bulkiness:
tertiary C (3°) > secondary C (2°) > primary C (1°), methyl (CH 3) >> H
The energy difference between two different chair forms
(in ΔG) is related to the equilibrium constant by
K = e –(ΔG/RT)
Example: (2R,3S)-2,3-dichlorobutane and (2S,3R)-2,3-dichlorobutane are the same
Cl
Cl
mirror plane
Cl
Cl
mirror plane
Omissions, Mistakes, Suggestions?
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Ashenhurst masterorganicchemistry.com
Copyright 2022, Master Organic Chemistry
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Substitution and Elimination - Page 1 of 2
Key points:
•Nucleophilic substitution is swapping of one functional group for another
•Structure of alkyl halide is key: primary > secondary >> tertiary
•Mechanism proceeds through backside attack
•Stereochemistry proceeds with inversion
•Rate depends on concentration of both nucleophile and alkyl halide
•Reaction occurs faster in polar aprotic solvents
Example:
Na
OH
CH 3CH2OH + Na
+ CH 3CH2Br
One main thing: Good Leaving Groups are Weak Bases
I
pKa of
conjugate
acid
Br
–9
–8
pKa of
conjugate
acid
,
The SN2 Reaction is Sensitive to Steric Effects.
In the SN 2 the electrophile is the antibonding orbital.
as you populate the antibonding orbital with electrons R groups shield
backside from nucleophilic
you weaken the bond to the leaving group
attack
HH
HR
Nu:
X
X
R
Primary alkyl halide
Intermediate
Nu
Me
Et H
Br
Me
Nu
+ Br
Me
100% ee
one enantiomer
inversion of configuration
100% ee
(one enantiomer)
The Rate Limiting Step of the SN2 Is Bimolecular
[R–X]
x
[Nu]
1
1
2x
2
1
2x
1
2
4x
2
2
Doubling the
concentration of either
component
will double the rate.
R
R
O
O
H
H
F
H
H
H
O
O
R
R
H
O
O
R
R
Note that in polar aprotic solvent
reactivity of halides reverses
because there is no hydrogen
bonding to solvent
F > Cl > Br > I
H 2N
16-18
~38
Examples of polar protic solvents:
H 2O, CH 3OH, CH 3CH2OH,
CH 3COOH, i-PrOH
CH 3CH2CH2NH 2
Polar aprotic solvents:
O
S
DMSO
H
O
Me C N
42
Me 2N
Fluorine forms very strong bonds and is thus
a very bad leaving group
LG
secondary alkyl (2°)
O
P
NMe 2
NMe 2
H H
α
H
β
β H
Br
+ HOC2H 5
NaOC 2H 5
C2H 5OH,
95° C
SN2 (slower than primary)
E 2 (provided a proton
is on the β carbon)
SN1/E1 : can occur when LG
SN1/E1: not observed is OH and a strong acid is added
watch out for rearrangements!
Generally poor when LG is a
halide.
Note that hybridization in all cases here is sp3
Trisubstituted alkene
"Zaitsev product"
(more C-C bonds
on alkene)
> H 2S
HS
>
SN1/E1 : strongly favored
if LG = halide, good reaction
if LG = OH, requires acid
Examples of good bases:
:NH 3
In order for the E2 reaction to occur one must use a good base,
HO
CH 3O
RO
H 2N
R 2N
R 3N:
I > Br > Cl > F
H 2N
N3
N C
H C C
HS
CH 3S
I
Br
Cl
F
O
F > Cl > Br > I
H
H
H
H
LG
B: = base
LG = leaving group
H
H
H
Why? The lone pair of the C-H bond
must be lined up with the σ*
(antibonding orbital) of the C-LG bond
H
LG
Good Substrates
Br
Bad Substrates
Br
Br
Br
All of these substrates are effective for the E2
No H on
β-carbon
H 3C
CH 3
Br
No H on β−carbon
that is anti to LG
The Rate Limiting Step of the E2 Is Bimolecular
Rate = k [Substrate ] [Base]
"Weak Nucleophiles" "Non-Nucleophiles"
CH 3OH
HSO 4
H
O
O
(CH 3)3C–O
The E2 Can Occur With Primary, Secondary, and Tertiary Substrates
4. Steric bulk: If a sterically bulky group surrounds the nucleophilic atom, this will decrease its
nucleophilicity.
H 3C
Example: CH 3O is more nucleophilic than
for this reason
H 3C C O
H 3C
H 2O
B:
B:
Note that in polar aprotic solvent reactivity of halides reverses
because there is no hydrogen bonding to solvent
CH 3O
(LDA)
The H And Leaving Group MUST be Anti-Periplanar
3. Solvent: Polar protic solvents will hydrogen bond with nucleophiles (see column on left)
which decreases their nucleophilicity.
•More polarizable nucleophiles will be less affected by polar protic solvents, since hydrogen
bonding ability is greatest for most electronegative nucleophiles.
•Polar aprotic solvents will not hydrogen bond: therefore they are good solvents for the SN 2
"OK Nucleophiles"
Li
N
H
H 3C
R
Basicity increases going to the left along the periodic table: C: > N: > O: > F:
Another term for how "tightly held" electrons are is polarizability.
Nucleophilicity increases with polarizability.
HO
HMPA
M = Na, K
Exception #2 is when the leaving group is bulky (e.g. NR 3 )
A Strong Base Is Required
CN > HCN
:PH 3 >
HO
Monosubstituted alkene
"Hoffmann product"
O M
Basicity increases going up the periodic table:
HS
MINOR
Exception: Bulky bases will give less of the
Zaitsev product because deprotonation on
the less substituted carbon becomes favored.
SN 2: Not observed
E 2 (provided a proton
is on the β carbon)
These reactions do not occur for alkenyl and alkynyl halides under normal conditions.
> H 2O
+ NaBr
MAJOR
The Major Product Is the Most Substituted Alkene
Why? Because alkenes become more thermodynamically stable as C-H
bonds are replaced with C-C bonds
CH3
CH3
tertiary alkyl (3°)
(most sterically hindered)
(most stable carbocations)
SN2 (unhindered)
E 2 (provided a proton
is on the β carbon)
"Good Nucleophiles"
H
NMe 2
DMF
>45
H 3C
Acetone Acetonitrile
O
alkyl, alkenyl
H
CH3
LG
CH3
primary alkyl (1°)
(least sterically hindered)
(least stable carbocations)
e.g.
Polar protic solvents make a "jacket" around nucleophiles through
Why? hydrogen bonding, decreasing the reactivity of the nucleophile.
H
16
RO
Nucleophilicity increases as we go to the LEFT along the periodic table: C: > N: > O: > F:
Nucleophilicity increases as we go DOWN the periodic table:
Two types of polar solvents: polar protic (have O-H or N-H bonds)
Polar aprotic (has dipoles but cannot hydrogen bond).
H
Conjugate base of the strong acid
tosic acid, similar in strength to
H 2SO 4 (sulfuric acid)
2. Electronegativity: The less tightly held a pair of electrons are, the more readily they can
be donated.
Because of the charges involved in the SN 2 mechanism polar solvents are better
than nonpolar solvents.
R
O
LG
e.g. HO
Polar aprotic solvents are the best for the SN2
R
O
4
What Factors Influence Nucleophilicity?
1. Charge: The conjugate base is always a better nucleophile than the conjugate acid
Rate = k [Substrate ] [Nucleophile]
Rate
–1.7
Note how HO is not a good leaving group but H 2O is: therefore, HO can be made into
a good leaving group by adding a strong acid
Tertiary alkyl halide.
So slow it doesn't happen.
H Et
Br
–3
H 3C
The nucleophile (electron pair) attacks the empty antibonding orbital
which is on the backside of the carbon-leaving group bond
H Et
–7
Key points:
• Elimination is the loss of a leaving group from the α-carbon and a hydrogen from the
β-carbon resulting in the formation of a double bond
•The major product is the more substituted alkene (Zaitsev product) and the bulkiest
groups will be trans
•Use of a bulky base will result in formation of more of the less substituted alkene
•Requires strong base
•The leaving group and the hydrogen must be oriented anti-periplanar
•Can occur with primary, secondary, tertiary substrates provided there is a hydrogen on
the β-carbon that can orient itself anti to the leaving group
•Rate determining step is bimolecular
•Solvent is generally polar protic
•Reaction is favored with heat
Example: Elimination of HBr
How The Substrate Influences Reactivity
The SN2 Mechanism: The Backside Attack Leads to Inversion
Nu:
AcO
R
Secondary alkyl halide
Fast
H 2O
With the exception of F , these are all strong bases
X
R
TsO
OH
1
RR
Nu:
Cl
Bad Leaving Groups
Br
substitution of Br for HO
O
H 3C
S O
O
TsO = p-methyltoluenesulfonate
Good Leaving Groups
F
Nu:
Elimination, Bimolecular (E2)
What Makes A Good Leaving Group?
Nucleophilic Substitution Bimolecular (SN2)
OH
These are examples only, not comprehensive
(as NaH)
Rate
[R–LG]
[B]
x
1
1
2x
2
1
2x
1
2
4x
2
2
Doubling the
concentration of either
component
will double the rate.
Solvent is Generally Polar Protic - not in all instances, but polar protic will favor E2
in cases where it competes with the SN2.(see left)
The E2 Is Favored By Heat
•Keep this in mind when considering substrates that could go E2 or SN2
•Heat favors elimination due to higher entropy
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Substitution and Elimination - Page 2 of 2
All About Carbocations
• Carbocations are flat (trigonal planar) - the central carbon is sp2 hybridized
•They are stabilized by 1) increasing substitution at carbon
and 2) resonance
•Tertiary carbocations > secondary carbocations >> primary carbocations
•They can be formed in several ways: 1) dissolving a tertary
alkyl halide in a polar protic solvent 2) adding acid to a
secondary or tertiary alcohol
•They are more stable in polar solvents
•Secondary carbocations can rearrange to tertiary
carbocations through hydride or alkyl shifts
What influences carbocation stability?
1) Carbon substituents stabilize carbocations
Tertiary > secondary >> primary > methyl
R
R
Tertiary carbocation:
Cl
Fast
most stable
R
R Cl
R
R
Substitution Nucleophilic Unimolecular (SN1) and Elimination, Unimolecular (E1)
• These reactions have a lot in common and often occur together.
They represent different fates of the same carbocation
CH 3
Nu
H 3C
CH3
CH 3
Nu:
H 3C
SN1
deprotonation
CH3
H 3C
E1
CH3
The E1 is deprotonation of the carbon
adjacent to the carbocation
The SN1 is nucleophilic attack at the
carbocation
What the SN1 and E1 Have In Common
• The rate-limiting step is formation of a carbocation
•The rate-limiting step is unimolecular and only depends on the concentration of
the substrate (e.g. alkyl halide)
• The rate is also proportional to the stability of the carbocation (see column on right)
• Tertiary carbocations > secondary carbocations >> primary carbocations > methyl carbocations
• Carbocations are ions (polar) so according to "like dissolves like", polar solvents
are favored.In particular, polar protic solvents help
SN1 Example:
Br
H 3C
CH 3
H
Cl
R
E1 Example:
CH 3
H 2O
H 3C
CH 3
H 2O
CH 3
SN1
HO
H 3C
CH 3
CH 3
H 3O Br
HO
H 3C
CH 3
CH 3
H 2SO 4
H 3C
CH 3 -[H2O]
H 2O
CH 3
E1
H 3C
CH 3
+ H 2O
The E1 also involves two steps that occur in sequence:
1) departure of the leaving group 2) deprotonation of the carbon adjacent to the carbocation
The SN1 involves two steps that occur in sequence:
1) departure of the leaving group 2) attack of the nucleophile at the carbocation
•The reaction proceeds with weak/moderate nucleophiles (e.g. H 2O, ROH)
•If the carbon is chiral, the chiral carbon becomes racemized
•The SN1 is always accompanied by some E1 (~95:5)
Why does the SN1 lead to a mixture of stereoisomers?
Because the carbocation is flat!
H
OCH 3
OCH3
retention
Br
CH 3OH
Pr
Me
Et
Pr
Pr
Me
(+ HBr)
Me
Et
+
Et
Pr
100% ee
Me
Et
inversion
OCH
3
one enantiomer
OCH 3
H
Nucleophilic attack occurs
equally from either face
•The proton is removed by a weak base
•The more substituted double bond is favored (Zaitsev product)
•The major product will have the bulkiest groups trans to each other across the double bond
•The reaction is favored by heat
•The base can be the solvent, the leaving group, or the counter-ion of the acid
(whichever is strongest)
Rate = k [Substrate ]
Rate
[R–X]
[Nu]
x
1
1
2x
2
1
x
1
2
2x
2
2
Doubling the concentration of
the substrate doubles the rate.
Doubling the concentration of
the nucleophile/base has no
effect
Br
:Nu
H
H
Br
Equatorial
No hydrogens
anti-periplanar
to the Br (only
carbons are)
base
(E2)
H
Cl
R
Br
H
Why?
When the leaving group is equatorial, the antibonding
orbital points into the center of the ring, making
it sterically inaccessible
When the leaving group is axial, the antibonding
orbital points above the ring, where it can be
attacked by a nucleophile
Relative Rate of SN2 Reactions where X is a leaving group
Br
base
CH 3X
CH3
CH 3CH2 X
30
1
H
CH3CH2CH2 X
Br
only product
One hydrogen
are anti-periplanar
to the Br
Therefore only one
double bond can form
Primary carbocation:
least stable
(methyl carbocations are
even more unstable)
H Cl
R
2) Resonance stabilizes carbocations
PhCH 2
benzyl carbocation
allyl carbocation
>>
primary alkyl carbocation
3) Vinyl and aryl carbocations are very unstable
Br
Br
Br
Br
How are Carbocations formed?
Three ways.
1) dissolve a tertiary alkyl halide in a polar protic solvent.
Cl
H 3C
HO
H 3C
CH 3
H 3C
H 2O
H 3C
CH 3
Cl
CH 3
H 3C
CH 3
HCl
H 3C
CH 3
Cl
CH 3
Can also be done for secondary alcohols - watch out for rearrangements!
allowed!
not allowed!
If there is no hydrogen anti to the leaving
group, then the double bond cannot form
CH3 H
Secondary carbocation
Cl
H
Cl
H 3C
CH 3
note that even though X is on
a primary carbon, it is adjacent
to a quaternary carbon, which
results in a very slow reaction
0.4
X
0.025
X
1 x 10 -5
H 3C
AgNO3
CH 3
H 3C
H 2O
[–H 2O]
CH 3
hydride
shift
Example of alkyl shift:
rate
decreases
with
increasing
steric bulk
NO 3
+ AgCl
silver halides form
insoluble precipitates,
driving reaction forward
Secondary Carbocations Can Rearrange
Driving force is formation of
Example of hydride shift:
more stabilized carbocation
H
H
H
HCl
HO H
CH3
R
3) (Less common) - addition of silver salts to an alkyl halide
:Nu
Br
Br
Axial
Two hydrogens
are anti-periplanar
to the Br. The double
bond can form through
removal of either hydrogen
R
2) Add an acid to a secondary or tertiary alcohol:
The SN2 can only occur on a cyclohexane when the leaving group is axial
The E2 can only occur when the leaving group
is axial:
H
R
The Rate Limiting Step of the SN1 and E1 Is Unimolecular
A racemic mixture is obtained
Further Notes on the SN2 / E 2 (more advanced topics)
H
Slow
HCl
CH3
[–H 2O]
HO H
This can also lead to ring expansions
HCl
OH
[–H 2O]
CH3
alkyl
shift
H
"Master Organic Chemistry"
Note: applying this framework mindlessly without understanding
the concepts and variables behind each of the four major reactions
is unlikely to lead to success. Usage of this framework assumes you
are familiar with the following concepts:
•What makes a good nucleophile
•What makes a good leaving group
•Strength of acids and bases
•Factors that affect carbocation stability
SN1/SN2/E1/E2 Flowchart
Do you see Br, Cl, I, OTs, OMs,
or other similar good leaving
group present?
It also goes without saying that it is absolutely
necessary to be able to draw the products of
each reaction, including stereochemistry
masterorganicchemistry.com
Is an alcohol (OH) present
NO
and attached to
YES
CH3
YES
1°
Identify Carbon Attached To
Leaving Group
Primary
YES
hybridized carbon
Here, HCl, HBr, and HI have nucleophilic
counter-ions, whereas H 2SO4, TsOH,
and H 3PO 4 do not
NO
Is it attached to an sp3 hybridized
carbon (i.e. a carbon with no
multiple bonds)
SSNN22
sp3
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
NO
Is acid being added?
If so, identify
NO
NO REACTION
H 2SO4, TsOH, H 3PO 4
Alkenyl and alkynyl halides
do not participate in these reactions
3°
HCl, HBr, HI
Tertiary
Tertiary
alcohol?
YES
E1
NO
2°
YES
SN1
E2 (but with
formation of least
substituted alkene)
Strong base present? YES
Secondary
Is the nucleophile a strong,
bulky base (e.g. tBuO )
YES
NO
Slow SN2
Heat generally favors
elimination reactions
YES
Heat will also favor
E2 over SN 2
YES
Is the solvent
polar aprotic?
YES
NO
SN1
NO
Carbocation formation:
Is rearrangement possible
to give a more stable
carbocation?
NO
E1
YES
NO
Carbocation formation:
Is rearrangement possible
to give a more stable
carbocation?
Rearrangement (hydride or
alkyl shift)
YES
* this can be very instructor-dependent!!
best to check this with your instructor.
Rearrangement (hydride or
alkyl shift)
Base Strength
Excellent
Strong Bases
CN HS
RS
RO
HO
Good
Cl
Br
I
R
H C C
HO
R–SH
O
O
RO
NH 2 H
Leaving Group Ability
Excellent
R
NR 2 (CH3)3CO H C C
R 3N R 2NH RNH 2
O
R–OH
R
HSO 4
TsO
I
Br
Good
R 3N
OH
(CH3)3CO due to steric hindrance
For more background see the
summary sheet on SN1/SN2/E1/E2
Solvents
Polar Protic
Cl
TsO
MsO H 2O
CH3COO
The weaker the
base,the better
F
HO RO
the leaving group
Non leaving groups
NH 2
H
R
Poor
Essentially Non-nucleophilic
H
Secondary
alcohol
Rearrangement (hydride or
alkyl shift)
Nucleophilicity
Poor
H 2O
concerted
H shift w/
subsequent
E1 (rare)
YES
Some variability here; E 2 will
strongly compete
F
NO
E1
SN1
Carbocation formation:
Is rearrangement possible
to give a more stable
carbocation?
SSNN22
N3
Secondary
alcohol
YES
YES
YES
NO
Polar Protic solvent?
Rule of thumb: pKa of
conjugate acid is above 12
Primary
alcohol?
SN1
NO
Is the nucleophile
strongly basic?
YES
YES
Is a good nucleophile or
strong base present?
SSNN22
NO
NO
Primary
alcohol?
NO
SSNN22
Heat applied?
Good nucleophile present?
E2
E2
NO
NO
Tertiary
alcohol?
Polar Aprotic
H 2O
DMSO
CH3OH (MeOH)
Acetonitrile
CH3CN
O
CH3CCH3 Acetone
CH3CH2OH (EtOH)
(CH3)3COH (t-BuOH)
(Dimethyl Sulfoxide)
DMF
N,N-Dimethyl
formamide
HMPA
CH3COOH (AcOH)
Omissions, Mistakes, Suggestions?
In general, solvents
with O-H or N-H
bonds
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Additions to Alkenes
Reaction
Hydroboration
Oxymercuration
Acid-catalyzed addition of H 2O
(hydration)
Addition of HX
Addition of HX
"Regiochemistry"
R
R
R
H
1) BH 3
R
H OH
R
2) NaOH,
H 2O 2
R
H
1) Hg(OAc) 2
H 2O
HO
2) NaBH 4
R
R
R
R
H
R
R
R
H
H 2SO4
R
R
R
H
HCl
R
R
R
H
HBr
R
R
R
H
Halohydrin Formation
Dihydroxylation
Dihydroxylation
R H
R H
R
R
R
H
OsO 4
R
R
R
H
KMnO 4
cold, dilute
syn + anti
Markovnikoff
syn + anti
R
Epoxidation
R
H
Hydrogenation
R
R
R
H
R
H
H
H
Radical addition of HBr
R
syn + anti
N/A
anti addition
H
Br or H 2O/ROH depending on solvent
anti addition
As with bromination, above. Although not depicted, use of water or alcohol as solvent
will also lead to formation of the halohydrin product (also anti).
HO OH
R
R
R
H
N/A
syn addition
HO OH
R
R
R
H
N/A
syn addition
N/A
syn addition
N/A
syn addition
R
Br
H
R
Cl
R
R
R
H
CH3
S
O
H
R
H H
H
Br
Anti-Markovnikoff
H H
O3
H
O
R
R
H
Cyclopropanation
syn + anti
+
O
R
H
CH2I 2
Zn/Cu
R
R
O
+
O
R
H
N/A
Keywords are "cold, dilute". NOTE: If "heat" or "acid" is mentioned in the conditions, the diol will
be cleaved to provide carbonyl compounds (same reaction as ozonolysis with oxidative workup,
below).
NOTE: "anti" hydroxylation can be achieved through epoxidation followed by treatment with NaOH (basic)
or aqueous acid ( H 3O+ )
RCO 3H is a peroxyacid. A common peroxy acid for this reaction is m-CPBA
(m-chloroperoxybenzoic acid). If H 3O+, heat is written afterwards, this
is opening of the epoxide to give the diol (anti-selective)
O
Cl
OH
O
m-CPBA
The catalyst can vary - you might see Pt or Ni as well. All provide the same product with the same
stereochemistry.
Peroxides generate the Br• radical, which adds to the double bond in the way that will generate the
most stable radical (i.e. the radical will go on to the most substituted carbon). This explains the
selectivity for the anti-Markovnikoff product. It gives a mixture of syn and anti because it goes through
a free radical process.
Oxidative workup: Hydrogen peroxide is used to obtain the carboxylic acid instead of the aldehyde.
Can also use KMnO 4 and acid
Omissions, Mistakes, Suggestions?
R
OH
H
C
Osmium is a transition metal. The tools won't be given in this course to fully understand how this reaction
works. Occasionally a second reagent like NaHSO 3, H 2S, or Na 2S2O3 is also given as a reactant in this
reaction - minor detail, it's used to remove the osmium from the hydroxyl groups.
Reductive workup: Zinc (Zn), or dimethyl sulfide (DMS, Me 2S) is a reducing agent. It reduces excess ozone, allowing for isolation
of the aldehyde.
CH3 (or Zn/H+)
H 2O 2
R
H
R
R
R
Ozonolysis (Oxidative Workup)
R
R
R
H
H
O3
Ozonolysis (Reductive workup)
Bromonium ion mechanism
N/A
HBr
peroxides
(RO-OR)
R
H
anti addition
R
R
Pd/C
Br
Markovnikoff
R
R
H2
R
R
The key detail in these reactions is solvent: water and alcohol solvents will form the
halohydrin products (the ones containing the OH and Br). All other solvents (you might
see CCl 4, CHCl 3, hexane, etc. ) provide the dibromide.
OH
O
Strong acid protonates the alkene, generating free carbocation. Watch out for
possibility of rearrangements when a tertiary carbocation could be generated through a 1,2 shift.
HSO 4– anion is not strongly nucleophilic, hence it does not add. Gives a mixture of syn and anti products
due to the free carbocation.
HCl and HBr (as well as HI, not pictured) protonate the alkene to give a free
carbocation which can then be trapped by the halide anion. Gives a mixture of syn and anti
O
R
R
This reaction goes through 3-membered "mercurinium" ion. The NaBH 4 step removes the
mercury. While the addition is anti, the overall reaction is stereorandom because this step involves a carbon
based free radical (usually not discussed). Alternatively, an alcohol used in place of water will produce an ether.
R
Br
Cl
R
R
Cl2
Markovnikoff
R
R
HO
R
R
R
H
R
Sometimes you might see BH 3•THF or B 2H 6 used here: it's the same reagent in a slightly different form.
The base (can be NaOH, KOH, identity unimportant) helps make H 2O2 more reactive. The reaction is anti Markovnikoff because the H–B bond is polarized toward H (electronegativity of H = 2.2, B = 2.0) - the H
adds to the carbon best able to stabilize positive charge (i.e. the most substituted one).
H
Br
R
R
R
R
syn + anti
Markovnikoff
R
Br 2
Markovnikoff
masterorganicchemistry.com
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
H
Br
H 2O
Chlorination
R H
R
syn addition
H
Cl
Br 2
Bromination
R H
R
Anti-Markovnikoff
H
HO
H 2O
"Stereochemistry"
"Master Organic Chemistry"
syn addition
This reaction goes through addition of a carbene (actually, "carbenoid") to
the double bond. The reaction is stereospecific.
Another set of conditions to provide a cyclopropane is CHCl 3 with strong
base (NaOH), which makes the dichlorocyclopropane.
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Ashenhurst masterorganicchemistry.com
Summary of Alkene Reaction Patterns
"Master Organic Chemistry"
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
masterorganicchemistry.com
Pattern 1: The Carbocation Pathway
These reactions proceed through attack of an alkene upon an acid, which results in formation of a carbocation.
The carbocation is then attacked by a nucleophile, which can occur from either face of the carbocation.
Pattern 2: The 3-Membered Ring Pathway
These reactions proceed through formation of a 3-membered ring cation, which is then attacked from the
backside by a nucleophile, on the most substituted carbon.
Example 1: addition of Br 2 to alkenes
Example: addition of H–Br to alkenes
H
C
R
δ+
H
C
H
δ–
Br
H
H
C
R
R
Br
Br
R
Br
CH3
R
R
Example 2: Addition of Br 2/H2O
H
C
CH3
Regiochemistry: Markovnikoff-Selective. In other words, hydrogen ends up bonded to the less substituted
carbon of the alkene, and (in this case) Br ends up attached to the most substituted end of the alkene.
Br
R
Reactions in this category: Addition of H–Cl
Addition of H–Br
Note: since this reaction goes through a carbocation, rearrangements are possible.
Example 1 - hydrogenation
R
R
R
H
H
R
R
H
R
Br
R
R
H
OH 2 R
Br
OH
H
R
Oxymercuration (with an alcohol as solvent) Hg (OAc) 2, ROH, then NaBH 4
Opening of epoxides under acidic conditions
Two reaction patterns of note
Oxidative cleavage: Addition of an oxidant (such as ozone, O3 leads to cleavage of the carbon-carbon double bond and
formation of two new carbon-oxygen double bonds.
Ozonolysis (Reductive workup)
H
H
R
R
R
R
H
O3
CH3
R
R
S
+
O
R
H
O
here, the C=C bond is cleaved to give
two C=O bonds
CH3
[or Zn, with H+]
BH 2
BH 3
R
R
R
R
Reactions in this category: Addition of halogens: Cl–Cl, Br–Br, I–I
Addition of halogens in the presence of nucleophilic solvents: Br 2/H2O, Br 2/R-OH
Oxymercuration: Hg(OAc) 2, H 2O, then NaBH 4
Example 2 - hydroboration
R
Br
Pattern 4: Two miscellaneous reactions
Although these reactions do not share a common mechanism, they each form two new bonds on the
same side of the alkene, a consequence of a "concerted" reaction mechanism.
Pd/C
Br
OH 2
Pattern 3: The "Concerted" Pathway
R
Br
H
R
Stereochemistry: Anti. The two groups add to opposite faces of the alkene.
Addition of H 3O
(sometimes written
as H 2O / H 2SO 4)
H
R
R
Br
R
R
H
R
Regiochemistry: Markovnikoff-Selective (where appropriate). For example, with Br 2/H2O, water adds to the most
substituted carbon of the alkene.
Addition of H–I
H2
Br
OH 2
R
Stereochemistry: Mixture of syn + anti . In other words the H and the other group (e.g. Br) do not add exclusively
"syn" or "anti", but provide a mixture of the two
H
Br
R
R
R
R
Br
H
R
H
R
R
BH 2
NaOH
H 2O 2
H
R
H
R
R
Ozonolysis (Oxidative Workup)
OH
H
R
R
R
R
H
Regiochemistry: anti-Markovnikoff-Selective (for BH 3). Note how hydrogen adds to the most
substituted end of the alkene, and OH ends up on the least substituted end.
O3
R
R
O
+
O
H 2O 2
alternatively, KMnO 4 can be used
to give the same products
R
OH
here, the C=C bond is cleaved to give
two C=O bonds
and the C-H bond is replaced with a
C-OH bond
C–C single bonds are not affected.
Radical addition of HBr
Stereochemistry: Syn. The two bonds are formed on the same side of the alkene
Reactions in this category: Hydroboration: BH 3, then H 2O2/NaOH
Hydrogenation: Pd/C, H 2
Epoxidation: RCO 3H (e.g. mCPBA)
Dihydroxylation: OsO4
Cyclopropanation: CH2I 2, Zn-Cu
Here, addition of HBr to alkenes in the presence of a radical initiator (peroxides) leads to anti-Markovnikoff
addition of H-Br to the alkene; the Br adds to the least substituted carbon of the alkene, and the H adds to the most
substituted.
R
R
R
H
H
HBr
peroxides
(RO-OR)
R
Br
Anti-Markovnikoff
R H
stereorandom
R
Regiochemistry: anti-Markovnikoff-Selective. Note that the Br adds to the less substituted end of the alkene.
Dichlorocyclopropanation CHCl 3, NaOH
Stereochemistry: Mixture of syn + anti.
H and Br do not add exclusively "syn" or
"anti" across the alkene.
Omissions, Mistakes, Suggestions?
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Ashenhurst masterorganicchemistry.com
"Master Organic Chemistry"
Introduction to Reactions of Alkynes
Synthesis of Alkynes
Elimination of Hydrogen halides
Br
Br
NaNH 2
NaNH 2 is a strong base and will lead to elimination of HBr. According
to Zaitsev's rule, the most substituted alkyne should form.
C C
geminal dihalide
Br
NaNH 2
masterorganicchemistry.com
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
C C
Br
vicinal dihalide
Deprotonation
R C C H
Reaction with Electrophiles
R C C
NaNH 2
Alkynes are relatively acidic [pKa = 25] A strong base like NaNH 2 will remove the C-H to provide the acetylide anion. The resulting anion is an excellent nucleophile.
Na
R C C
Can also use strong bases like n-BuLi, or Grignard reagents (RMgBr, etc.)
Na
R
Br
Deprotonated alkynes are excellent, useful nucleophiles and will react with alkyl halides in SN 2 reactions
R C C
R
Addition Reactions to Alkynes
Hydrogenation
H2
H H
R C C
H H
"Lindlar catalyst"
H
H2
(or Ni 2B, see right)
R
Pd/C
R C C
R
Partial Hydrogenation
R C C
R
Reduction (Dissolving metal reduction)
Addition of Hydrogen bromide to
form gem-dibromides
Addition of Hydrogen chloride to
form gem-dichlorides
Acid-catalyzed Hydration to
provide ketones
Hydroboration to form aldehydes
Na/NH 3
R C C
R
R
R
R
H C C
R
H
R
Sodium (Na) is a very strong reducing agent, and will reduce alkynes (but not
alkenes). Selective for the trans-alkene. The NH 3 provides the hydrogens in the final
product.
C C
H
H
C
H
H
C
H
H
H 2O
(same as "H 3O+")
BH 3
Lindlar catalyst is a "poisoned" palladium catalyst for the partial hydrogenation of alkynes. It contains
palladium (Pd), lead (Pb), and quinoline, and is selective for the cis-alkene. Sometimes also see "Pd,
CaCO 3". Nickel boride (Ni 2B) is sometimes used as an alternative.
R
(1 equiv)
H 2SO4
H C C
H
C C
(1 equiv)
HCl
H C C
R
R
HBr
H C C
Can also use Pt, Ni, etc. as hydrogenation catalysts. These will also reduce alkynes to alkanes.
C
H
Br
C
Br
HBr
R
H
H
(1 equiv)
Cl
C
H
H
(1 equiv)
OH
C
HgSO 4
H C C
R
R
R
H
H
Br 2
H C C
R
H 2O 2
R
HO
NaOH
H
R
H
H
HgOH
Chlorination
H C C
R
Ozonolysis
R'
H 2O
R
R
Cl2
Cl
R
Anti-Markovnikoff
H
Note: Hydroboration on internal alkynes will give mixture of ketones.
Aldehyde
In truth, BH 3 is actually not the best reagent to use, as it will often lead to multiple
hydroborations, but textbooks differ on the best alternative (R 2BH, 9-BBN, Sia2BH are
examples).
R
Markovnikoff
H
Br Br
R
first addition is trans
Cl Cl
H Cl
O
OH
H
The first intermediate in each of these three reactions is a cyclic ion, which
undergoes attack by a nucleophile at the more substituted carbon (Markovnikoff )
+
HO
R'
R
X
H
R
H Br Br
O
O3
R C C
Br 2
Br
H
O
Ketone
R
Cl
tautomerization (keto-enol tautomerization in this case) is the
interconversion of enols and aldehydes/ketones through the
movement of a proton and the shifting of bonding electrons. It is
an equilibrium which generally favors the keto form.
O
tautomerization
R
Br
H
H
"enol"
OH
H
tautomerization
X
This carbocation will then be attacked by the nucleophile in question, leading
to the formation of substituted products.
Markovnikoff
H
H
H
H
R
Vinyl carbocation
R
Ketone
H 2B
H
Cl2
R
H X
H
O
H
H 2SO4, H 2O
Markovnikoff
H geminal dichloride
"enol"
Bromination
Cl
"enol"
(or R 2BH, 9BBN,
Sia 2BH - see right)
The key intermediate in all three of these reactions is an intermediate vinyl carbocation
which is formed upon protonation of the alkyne by strong acid. The carbocation forms on
the more substituted side of the carbon, according to Markovnikoff's rule.
Markovnikoff
geminal dibromide
H
tautomerization
syn addition
Oxymercuration to provide ketones
R
Cl
HCl
R
Br
H
R
X = Hg, Br, or Cl
first addition is trans
Cl
If R = H , then one product will be CO2
O C O
Omissions, Mistakes, Suggestions?
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Ashenhurst masterorganicchemistry.com
"Master Organic Chemistry"
Common Reaction Patterns (Org 1)
Pattern #1 - Acid Base Reactions
+
Generic example:
B
+
B H
Base
Conjugate
acid
Bonds
Formed
X
Conjugate
base
Bonds
Broken
+
B
H–X
B–H
Base
Specific example:
H Br +
Acid
masterorganicchemistry.com
Pattern #4 - Elimination Reactions
Generic example:
H X
Acid
H 3C–OH +
H 3C–O
Br
H H
H C C H
H
LG
Substrate
Base
CH3CH2O
Generic example:
Nucleophile
Substrate
Product
LG
Bonds
Formed
Bonds
Broken
C–Nu
C–LG
Leaving
group
Nucleophile
+ CH3CH2 –Br
CH3CH2OH
Substrate
Product
+
Product
H
Leaving
group
Generic example:
Electrophile
Bonds
Formed
A B
R C C H
H
H
Product
C C
H
H
Substrate
Bonds
Broken
Electrophile
H 3C
C–C (π)
A–B
C–A
C–B
C C
H 3C
CH3
H
+
CH3CH2O–H
+
Br
Product
Bonds
Formed
H 3C C C
H
H 3C
Carbocation (secondary)
C–H
Bonds
Broken
C–H
Carbocation (tertiary - more stable)
Alkyl shift example:
Bonds
Formed
CH3
C C CH3
H 3C
H
H 3C
C–C
Bonds
Broken
C–C
Carbocation (tertiary - more stable)
Pattern #6 - Free Radical Substitution
Generic example:
Br
H
R C C H
H
H
H
C C
H
C–H
C–LG
Notes: Rearrangements are favorable when the migration of hydride or alkyl leads to a more stable carbocation. Recall that
carbocation stability proceeds in the order tertiary > secondary >> primary. Resonance can also stabilize carbocations. Hydrogen tends
to migrate preferentially over carbon. Relief of ring strain can also be a major contributing factor with rearrangement reactions.
Specific example:
H–Br
H 3C
H 3C
Carbocation (secondary)
A–B
C–C (π)
B–H
Leaving
group
H
C C CH3
H 3C
H
CH3
H 3C
CH3
H 3C C C
H
H 3C
Pattern #3 - Electrophilic Addition
H
Conjugate
acid
LG
Hydride shift example:
Br
Notes: Substitution reactions can proceed through two pathways. The SN 2 pathway is concerted, and C-Nu bond
formation occurs at the same time as C-LG bond breakage. Steric hindrance is paramount. In the SN1 pathway, the
leaving group leaves first to give a carbocation, which is then attacked by a nucleophile. In the SN1 pathway
rearrangements can occur depending on carbocation stability. Good nucleophiles tend to have a loosely held lone pair of
electrons. Good leaving groups are weak bases.
R
+
H
Bonds
Broken
Pattern #5 - Carbocation Rearrangements
Specific example:
HO
+ B H
C C
Bonds
Formed
Conjugate
Leaving
group
acid
Notes: Note that every elimination reaction involves the formation of a new C–C π bond. Since alkene stability increases as the
number of attached carbons increases, eliminations tend to favor formation of the more substituted alkene ("Zaitsev's rule"). An
important exception is when more hindered bases (such as t-BuO(–) are used, which gives the less substituted alkene. Two dominant
mechanisms for elimination are the concerted pathway (E2) which occurs in one step, requiring anti-periplanar arrangement of H and
leaving group. The second pathway is the stepwise E1 pathway, which occurs with loss of leaving group followed by deprotonation.
A third (rare) pathway involves deprotonation, followed by loss of leaving group, called the E1-conjugate base (E1cb) pathway.
All elimination reactions tend to be favored by heat.
Pattern #2 - Nucleophilic Substitution Reactions
Nu–R +
H
H
H CH3
H 3C C C H
H 3C
Br
Substrate
Base
+ R–LG
H
Specific example:
Conjugate Conjugate
acid
base
Notes: Favorable acid base reactions will involve a stronger acid and a stronger base reacting to give a weaker acid
and weaker base. Relative strengths of acids and bases can be obtained through consulting a pKa table
Nu
Note - this sheet is not meant to be comprehensive. Your course
may provide additional material, or may not cover some of the
reactions shown here. Your course instructor is the final authority.
H
Substrate
H H
H C C H
H
H
Substrate
Product
Notes: Note that every addition reaction involves breakage of a C-C π bond and formation of two new single bonds to C. For
stepwise addition reactions, the tendency will be to place positive charge on the carbon best able to accept it (i.e. the carbon
attached to the most carbon atoms). This gives rise to "Markovnikoff's rule", where H always adds to the least substituted carbon.
[Notable exception is hydroboration-oxidation, which is "anti-Markovnikoff".] Stereochemistry of the two new bonds can be syn,
anti, or a mixture of both, depending on mechanistic pathway.
+ Cl
Cl
Halogen
light (hν)
or heat (Δ)
Cl H
H C C H
H
H
Product
+
H Cl
Hydrohalic acid
Bonds
Formed
Bonds
Broken
C–Cl
H–Cl
C–H
Cl–Cl
Notes: Free radical reactions proceed in three steps; initiation, propagation, and termination. Light (hν) or heat (Δ) is used to initiate the
reaction. Chlorine is quite unselective relative to bromine, which will only react with tertiary and allylic/benzylic C-H bonds (which are
weakest). Iodine does not perform this reaction.
If multiple equivalents of chlorine are present, chlorination can occur multiple times until all C-H bonds are replaced with C-Cl.
Pattern #7 - Oxidative Cleavage
Bonds
Formed
Specific example:
O3
H 3C
C C
CH3
H 3C
H
Electrophile Substrate
reductive
workup
H 3C
C O
+
H 3C
O C
CH3
H
Products
C–O (2)
C–O (π) (2)
Bonds
Broken
C–C (2)
C–C (π) (2)
Notes: Oxidation of C-C multiple bonds (π bonds) can be achieved with ozone (O 3)
or potassium permanganate (KMnO 4). With ozone, reductive workup (using Zn or
dimethyl sulfide) leaves all C-H bonds intact. With KMnO 4or O3 using oxidative
workup, sp2 hybridized C–H bonds are oxidized to C–OH.
A different type of oxidative cleavage can occur with vicinal diols using NaIO 4, HIO 4,
or Pb(OAc) 4. These oxidants will cleave C–C single bonds and form C–O (π) bonds
Omissions, Mistakes, Suggestions?
james@masterorganicchemistry.com
This sheet copyright 2022, James A.
Ashenhurst masterorganicchemistry.com
Introduction to Alcohols and Ethers
https://MasterOrganicChemistry.com
Alcohols
Conversion of Alcohols to Good Leaving Groups
Alcohols to tosylates and mesylates ("sulfonate esters")
Alcohol protecting groups
Alkyl (“aliphatic”) alcohols are classified by the number of
carbons directly bonded to the C (“carbinol” carbon):
OH
H H
H OH
H3C OH
H3C OH
C
C
C
H 3C
OH H3C
CH3 H3C
CH3
methanol primary (1°) secondary (2°) tertiary (3°) aryl alcohol
(phenol)
alcohol
alcohol
alcohol
1°, 2°, 3°
0 C–C
2 C–C
1 C–C
3 C–C
3 C–H
does not apply
1 C–H
2 C–H
0 C–H
The hydroxide group (HO–) of alcohols is a strong base and a poor
leaving group. Converting the OH to a halogen or "sulfonate" (e.g. OTs
or OMs) greatly facilitates substitution & elimination reactions.
Alcohols to alkyl chlorides using SOCl2, PCl3, or PCl5
Note that these reactions proceed with inversion of stereochemistry.
OH
Cl
SOCl2 or PCl3 or PCl5
inversion!
Like halides, sulfonates RSO3– are great leaving groups. Note that the
stereochemistry of the C–O doesn’t change (unlike SOCl2 or PBr3)
The weakly acidic OH group can interfere with various reagents
like this attempted SN2 with acetylide that instead deprotonates OH
Ethers
An ether is a functional group containing oxygen bonded to two
carbon atoms. Carbons can be alkyl, alkenyl, aryl, etc.
H 3C
O
O
CH3
Tetrahydrofuran
(THF)
Dimethyl
ether
O
O
Tetrahydropyran Ethylene oxide
(THP)
(epoxide or oxirane)
The Williamson ether synthesis is by far the most important method
for forming ethers. It is an SN2 reaction between a deprotonated
alcohol (“alkoxide”) and an alkyl halide (or sulfonate, e.g. OTs or OMs)
Br
ONa
1
3
4
NaH
OH
1
2
O
4
What doesn’t work?
NaOEt
Cl
Cl
Primary alcohols go through SN2 with HCl, HBr, and H-I
OH
H Cl
OH2
Cl
Cl
NaOEt
OH
O
p-toluenesulfonyl
CH3 (“tosyl”, Ts)
Alcohols to Alkenes with H2SO4 or H3PO4
Heating alcohols with H2SO4 or H3PO4 usually leads to elimination.
The conjugate base of sulfuric acid (HSO4– ) is a very poor nucleophile.
H
B
OH H OSO3H
OH2 OSO3H
Δ
Synthesis of symmetrical ethers from 1° alcohols through dehydration
The OH group is converted to a good leaving group (OH2) with strong acid.
The carbon is then attacked by another alcohol (SN2), forming a new ether.
OH +
H2SO4
OH
+ H 2O
O
heat
This is only useful for formation of symmetrical ethers from 1° alcohols.
Ethers from alkenes
H
O R
HO R
H2SO4
O R
ROH
[–H ]
Acid leads to the formation of a carbocation, from the alkene, which is
then trapped by the alcohol as solvent.
Carbocation rearrangements (hydride and alkyl shifts) can occur in cases
where a more stable carbocation can be formed.
OR
HO R
H
The key difference is that it does not proceed through a
carbocation, so no rearrangements can occur.
no reaction! SN2
doesn’t work on sp2
hybridized carbons
Tertiary alkyl halides
will give elimination (E2)
not substitution
Reactions of Ethers
Ethers are generally unreactive functional groups and don’t undergo
any significant reactions except for cleavage with strong acid.
O
H–I
(excess)
O
CH3
H–I
(excess)
I
I
OH
this C-O bond will
not break (sp2)
H 3C
I
R–OTs
(E2)
NaOEt
heat
alkene
CH3
N3
alkane
[LiCu(CH3)2]
azide
(NaN3)
O
SCH3
thioether
(NaSCH3)
thiol
(NaSH)
OCH3
O
CH3
H3C Si Cl
CH3
TMS-Cl
OH
OTMS
The advantage of using silicon-based protecting groups
is that they can easily be removed with fluoride ion,
which forms very strong bonds with Si (130 kcal/mol)
TMS-Cl
F
OTMS
TBAF = tetrabutyl
ammonium fluoride
OH
(TBAF)
Bu4N F
Another protecting group for alcohols is tetrahydropyranyl (THP).
This can be formed by treating an alcohol with dihydropyran and
acid, which forms an acetal.
dihydropyran
H
O
THP
=
OH
R
O
O
O
R
H 3O
acetal
Removed with aqueous acid
functional
group
Benzyl ethers are installed through a Williamson ether reaction and
can then be removed through hydrogenation (Pd-C, excess H2)
→
NaH, C6H5CH2Cl
ether
ester
(NaOCH3, DMSO) (NaOCOCH3)
R
OH
O
Pd-C, H2 (excess) R
C
H
H benzyl ether
Epoxides from alkenes
Opening Under Basic Conditions Is Similar To SN2
Under basic conditions, nucleophiles will attack epoxides at the least
sterically hindered position (primary [fastest] > secondary > tertiary [slowest])
The reaction is essentially an SN2 reaction!
OH
O
O
Nu
H
Nu
Nu
H 3C
H 3C
Example: reaction of epoxides with Grignard reagents
OH
1) CH3MgBr
CH3
2) acid workup
H+,
H3
O+,
H 2O
m-CPBA
O
m-CPBA (meta-chloroperoxybenzoic acid, a peroxyacid)
converts alkenes to epoxides. Other peroxyacids can be
used (e.g. CH3CO3H)
Epoxides from halohydrins
Halohydrins (formed from alkenes with X2 / H2O) can form
epoxides upon deprotonation of OH by base (e.g. NaH)
through an intramolecular SN2.
note that OH and Br
are anti
OH
Br2, H2O
base
O
Br
This is an internal SN2 reaction which occurs with backside
attack. If rotation cannot occur (such as in cyclic halohydrins)
then no epoxide can form.
H2O)
Epoxide Opening Under Acidic Conditions Occurs At The
Most Substituted Carbon
Under acidic conditions, the epoxide oxygen is protonated, and
weak nucleophiles (like H2O) can open the epoxide.
O
O
This can be prevented if the OH is “protected” with a blocking group
(“protecting group”) like trimethylsilyl (TMS)
R
C
Cl
C
R C C
TMSO
TMSO
Opening of epoxides
Epoxides are highly reactive towards nucleophiles due to ring strain
(about 13 kcal/mol). They will react with nucleophiles under both
acidic and basic conditions. However the patterns are different.
O
Cl
HO
R C C
Epoxides
(e.g.
Ethers from alkenes through oxymercuration
+ HOAc
OR
1) Hg(OAc)2, ROH
+ Hg (s)
+ NaOAc
2) NaBH4
+ BH3
OTs
TsCl
alcohol
SH
With secondary alcohols, rearrangement through hydride or alkyl
shifts is very common. Watch out!
TsCl
O
MsCl
Cl S CH3 methanesulfonyl
R–OH
R–OMs
(“mesyl”, Ms)
O
O
trifluoromethanesulfonyl
OH Tf O
OTf
X S CF3 (“triflyl”, Tf)
2
(Generally only used for
O
phenols)
Generally speaking, converting OH to OTs / OMs is the best way to convert
alcohols into good leaving groups for subsequent SN2 / E2 reactions.
OH
Secondary alcohols - watch out for rearrangements (SN1) either via
hydride or alkyl shifts. H2SO4 will generally give alkenes via E1.
R–OH
O
O S R
+
O
resonancestabilized
leaving group
→
Cl
O
Cl S
O
O H
base
no epoxide formation, since backside
attack is impossible
Br
OH
H2SO4
The nucleophile will attack the carbon best able to stabilize
positive charge - which is the more substituted carbon. This
attack occurs with inversion.
Very similar to bromonium /
more substituted C
preferred site
mercurinium ions from the
more-stabilized charge
of nucleophilic
alkene chapter!
weaker C–O bond
attack: the "most
++ Br
H
substituted" carbon
++
δ
O
δ
of the epoxide
+
+
δ
δ
has the longer
(and easier to
less substituted C
break) C–O bond
less-stabilized charge
stronger C–O bond
→
OH
OH
→
OH
Br
OH2
3
The Williamson is an SN2 reaction, so the alkyl halide should be
primary or secondary, not tertiary or sp2-hybridized
Br
inversion!
ROH
Intramolecluar Williamson ether synthesis is also possible.
2
H Cl
OH
O
+ H2O + NaBr
base (e.g. NaOH)
Another way to do it is by adding a strong base (e.g. NaH) to an alcohol
1) NaH
CH3
OH
O
2) CH3–I
Cl
Br
PBr5
Note that some courses teach that SOCl2 gives retention and SOCl2
plus base (e.g. pyridine) gives inversion. Check with your teacher!
Alcohols to alkyl halides by using acids
Tertiary alcohols go through SN1 with HCl, HBr, and H-I
Ethers have lower boiling points than alcohols. The O–C bond is
not as polarized as an O–H bond, so there is no hydrogen bonding.
Synthesis of ethers
or
Nu
base (e.g.
pyridine)
→
Alcohols can be synthesized from alkenes via hydroboration,
acid-catalyzed hydration, or oxymercuration (HgOAc2 / H2O)
Oxymercuration happens with no rearrangements.
1) BH3
H
H H
H H
H 3O
2) H2O2
H
H
HO
OH
hydroboration
H acid-catalyzed
H H
H hydration
H
They can also be synthesized from alkyl halides via substitution
reactions (SN1 for tertiary, SN2 for primary)
NaOH
Br H2O
OH
Br
OH
SN1
SN2
Alcohols to alkyl bromides
OH
SOBr2 or PBr3
→
O
δ–
Alcohols are weak acids and will react with strong bases to give
their conjugate bases, “alkoxides”
much stronger
NaH
OH
nucleophile (and stronger
O
Na
(strong
base) than R-OH
pKa 16-18
alkoxide
base)
Alcohols are also weak bases and will react with strong acids
to give oxonium ions, which are great leaving groups.
OH2
OH
OH2 ←oxonium
ion
substitution
H
[–H2O]
and/or elimination
reactions
→
Alcohols have relatively high boiling points.
The hydroxy group (OH) is polar and can participate
in hydrogen bonding.
δ+
H
note that stereochemistry doesn’t change!
O
O
Cl S R
OH
O S R
Nu
O
O
Practice Questions on This Material
Practice questions:
https://bit.ly/34BijTR
Reach out with feedback:
james@masterorganicchemistry.com
Reaction Energy Diagrams
https://MasterOrganicChemistry.com
Reaction Intermediates
→
ΔA = +200 ft
(“uphill”)
Point 2
(1200 ft)
Point 1
(1000 ft)
Point 2
(1200 ft)
Point 1
(1000 ft)
Journey progress
Journey progress
Altitude (A) is a state function, meaning that its value is independent of
the path taken to get there.
For our purposes going forward the two most important state functions
are enthalpy (H) and Gibbs free energy (G)
We can monitor reactions by measuring the changes in enthalpy (ΔH)
or Gibbs energy (ΔG) as they progress.
Camp 1
(A = 1000 ft)
Journey progress
For the journey, we calculate the change in altitude (ΔA) as:
Δ A = A Camp2 – ACamp1 Δ A = 1200 – 1000
Δ A = + 200
The net change is that we went 200 feet “uphill” (+ 200).
Enthalpy (H)
What about the reverse journey? We could plot it by flipping
the graph, like this:
peak
altitude (2000 ft)
ΔA = A point 1 - A point 2
Altitude
ΔA = –200
(A)
(ft)
Camp 2
ΔA = –200 ft
(1200 ft)
Camp 1
(“downhill”)
(1000 ft)
Journey progress
Gibbs free Energy (G)
• Heat energy given off / absorbed
by reaction at a particular
temperature / pressure
• Gibbs free energy spontaneous
or released available to do work
• Reaction that gives off heat
(–ΔH ) is exothermic
• Reaction that releases energy
(–ΔG) is exergonic
→
• Reaction that absorbs heat
(+ΔH) is endothermic
ΔHreaction = ΔHproducts – ΔHreactants
This journey results in a net change of -200 feet (downhill).
• Reaction that requires input
of energy (+ΔG) is
endergonic
ΔGreaction = ΔGproducts – ΔGreactants
Note that knowing the ΔH orΔG of a reaction doesn’t tell you anything
about how fast the reaction goes (reaction rates). That’s where the
height of the barrier comes into play.
Rather than flip the graph, though, it’s usually less work to
use the same graph for both cases and just flip the sign
depending on which direction we’re going.
→
→
→ Product
ΔE
Reactants
Br
(slow)
C
Carbocation
Free
radical
Transition state
Energy maximum
(“peak”)
C
OH2
R
Conjugate
acid
vs.
HO
O
R
R
Nu
‡
δ
H
(fast)
Nu
SN 1
Br
δ
Some examples of reaction intermediates
R
Step 2: Attack
of nucleophile
Carbocation
intermediate
Each step has a transition state; an energy maximum with partial bonds
Since the 1st step is rate-determining (slow), it must have the highest activation energy
Products
R
R C
R
Ea
Conjugate
base
Activation
energy
for step 1
Ea
ΔE
Intermediate
Full bonds
Partial charges
Full (formal) charges
Brief lifetime (10–9 s)
Longer lifetime
Nu
Br
Intermediate
(Carbocation)
Energy minimum
(“valley”)
Partial bonds
C
‡
δ
Nu
δ
Nu
Exothermic
overall
Br
Reaction Progress
Transition states and activation energy
Multiple Intermediates
Hammond’s Postulate
How do we tell by looking at a reaction coordinate diagram whether
the reaction is exothermic, endothermic, endergonic, or exergonic?
First, look at the units on the y-axis of the graph!
All reactions pass through a transition state, which is like the
mountain “peak” on our previous diagrams.
It’s also possible for reactions to have 2 (or more!) intermediates
In an exothermic reaction the
transition state will be closest in energy
to the reactants.
The graph for an exothermic reaction will have the products lower in
energy than the reactants (“downhill”, overall). An endothermic reaction
will have products higher in energy than reactants (“uphill”, overall)
Exothermic reaction (–ΔH)
Endothermic reaction (+ΔH)
One example: SN1 reactions with hydride or alkyl shifts
Transition states are high-energy species with partial bonds, partial
charges, and extremely short lifetimes (10–9 s). They cannot be isolated.
step 1
H
step 2
step 3
H
H
Br
→
Exothermic vs Endothermic
Nu
CH3
net “uphill”
Products
Products
net “downhill”
Transition
state (TS)
Reactants
Reaction progress
Reaction progress
Exergonic vs Endergonic
The reaction coordinate of an exerrgonic reaction will have products
lower in energy than the reactants (“downhill”).
The reaction coordinate of an endergonic reaction will have products higher
in energy than the reactants (“uphill”)
Exergonic reaction (–ΔG)
Activation energy
(forward reaction)
ΔG
Endergonic reaction (+ΔG)
net “uphill”
Reactants
ΔG
Products
Reactants
Products
net “downhill”
Reaction Progress
Reaction Progress
For a reaction at equilibrium, the difference in energy between reactants
and products ΔG is related to the equilibrium constant K by the
equation ΔG = -RT ln K
A difference of 1 kcal/mol gives a ratio of 82:18
H H
δ—
Br
C
δ+
H E
TS
‡
ΔE
3 step reaction: 3 transition states
2 intermediates
Activation
energy
(reverse
reaction)
Reaction Progress
Hints on Sketching Reaction Energy Diagrams
ESM
R Br
ΔG
δ—
Nu
Nu Na
Na Br
R Nu
EProducts
We can use the same graph to calculate Ea for the reverse reaction
since all we have to do is go in the backwards direction:
Ea = ETS – EProducts
The relationship between the rate constant and the activation
energy is described by the Arrhenius equation:
k = Ae(–Ea/RT)
The lower the value of Ea, the faster the reaction.
• Draw out the mechanism and count the number of steps.
(This tells you how many peaks your diagram will need).
• The number of intermediates will be one less than the number
of steps.
• What’s the rate-limiting step? That step should have the largest
activation energy (distance from reactant to peak).
• Label your axis accordingly. Using “E” for energy is usually fine unless
it’s specified that enthalpy (ΔH) or Gibbs energy (ΔG) is required.
• Is the reaction exothermic or endothermic overall? Draw the
relative position of product and reactant accordingly (net “uphill”
or “downhill”).
TS
reactants
products
‡
products
ΔE
→
ΔH
Reactants
In an endothermic reaction the
transition state will be closer in
energy to the products.
TS ‡
CH2
C
‡
the “double dagger” symbol
H H
Transition state
δ—
δ—
H H
means
Br
Nu
for an SN2 reaction
C
Reactant
Intermediate
Intermediate
Product
+
we
are
referring
to
a
ΔE
δ
with (unstable!)
H
transition state
five-coordinate carbon
When drawing these, just remember that each step will have
note the partial bonds and partial charges
its own transition state, and each intermediate will be an energy
minimum.
The activation energy is the difference in energy between
the starting materials (reactants) and the transition state (TS)
Ea = ETS – EReactants
ΔH
Each reaction forward
and back has an
activation energy.
Activation energy
for step 2
Since the activation
energy for step 1 is
larger than the activation
energy for step 2, step 1
is the rate determining
step.
Exothermic/Endothermic vs Endergonic/Exergonic
For ΔH (enthalpy) on the y-axis, exothermic / endothermic are relevant.
For ΔG (Gibbs energy) on the y-axis, use exergonic and endergonic.
R Nu
Drawing a reaction energy diagram for a two-step reaction: The SN1 Reaction
Step 1: Loss of
leaving group
Reaction Progress
R
R C
R
→
+ Nu
R
The step with the highest activation energy (energy barrier) is the rate determining
step.
Br
This is the
intermediate!
(an energy
minimum)
→
Intermediate
→
Altitude
(A)
(ft)
Camp 2
(A = 1200 ft)
Reaction Energy Diagram For a Two-Step Reaction
Transition
Transition
state #2
state #1
(an energy
(an energy
maximum)
maximum)
→
Altitude
(A)
(ft)
We still get the same value for ΔA if the height of the mountain is
maximum
1500 feet, 5,000 feet, or 15,000 feet.
altitude
maximum
(5000 ft)
altitude (1500 ft)
→
Starting at Camp 1 (A = 1000 ft) we pass over the peak
(A = 2000 ft) en route to Camp B (A = 1200 ft).
peak
altitude (2000 ft)
• Any reaction that proceeds in two or more steps
has at least one reaction intermediate.
• An intermediate is a species with full (not partial)
bonds and (at least in theory!) can be isolated.
• Think of it as a “valley” in our mountain climbing analogy.
→
What about the height of the peak?
Note that the height of the peak (barrier) doesn’t matter for the
purposes of calculating ΔA. All we care about is the difference
between the altitude of A to B and vice versa.
→
Reaction Progress = Mountain Climbing
A reaction coordinate diagram (or reaction energy diagram) shows the
changes in energy that occur as a reaction goes from starting materials
(“reactants”) to products via a transition state.
A useful analogy comes from graphing the change in altitude (A)
as one travels from one destination to another via climbing a mountain.
(Changes in altitude correlate with changes in potential energy, which is
why it’s a useful analogy for chemistry).
Rate-Determining Steps
A two-step reaction will have two transition states and one intermediate.
Reactant
Intermediate
+ X
R X
R
reactants
Reaction Progress
Reaction Progress
This also means that the geometry
of the transition state for an
exothermic reaction will resemble
that of the reactants:
And the geometry of the transition
state for an endothermic reaction
will resemble that of the products:
Example: free-radical chlorination
of alkanes
Cl
Cl
H
H ‡
H Cl
Example: bromination of alkanes
→
→
short C-H bond, long H-Cl bond
(resembles reactants)
H Br
H
→
Br
‡
→
H
Br
long C-H bond, short H-Br bond
(resembles free-radical product)
Practice Questions on This Material
Practice questions:
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Reach out with feedback:
james@masterorganicchemistry.com
Too small? Try printing at 200% size.
4-page version also available upon request
Synthesis Tips — Org 1
1) Dihalide → alkyne
SN2 and SN1: SN2 offers more predictable products and stereochemistry, while SN1 typically results in racemization and invloves
carbocation rearrangement. In an SN2 reaction, the nucleophile attacks the α position, where the LG is connected and replaces the
leaving group (LG) with inversion of configuration.
NaSH
NaOR
NaOH
Reagents
5) Terminal alkyne → ketone
Reactions of Alkynes
Reactions of Alkyl Halides
NaN3
NaCN
NaSR
X X
X
or
R1
R2
Geminal dihalide
R1
R2
R1
(X=Cl, Br, I)
C=O at the most substituted position; no rearrangement
R2
Most substituted alkyne
X
6) Terminal alkyne → aldehyde
1) Sia2BH
2) H2O2, H2O
R
C=O at the least substituted position
Vicinal dihalide
HC C Na
H
2) Extension of the carbon chain (C—C bond formation)
R
X
R
R
OH
R
OR
R
SH
R
SR
CN
R
HC C
N3
R2
R1
E2
X
E2
R1
R2
R2
R2
R1
SH
R1
SR
1) NaNH2
H
R1
R1
R2
2) R2X (X=Cl, Br, I)
2) R1X (X=Cl, Br, I)
The alkyl bromide must be primary or methyl (otherwise elmination
will occur)!
R
(X=Cl, Br, I)
H
R2
R1
CN
E2
N3
NaOEt
t-BuOK
X
(X=Cl, Br, I)
NaOEt
X
(X=Cl, Br, I)
Zaitsev
H
α
H
β
X
Zaitsev
(X=Cl, Br, I)
H removed is antiperiplanar to X!
Hofmann
Converting OH to a good LG
Br
PBr3
OTs
SOCl2
OH
Pyridine
1° or 2°
1) Hg(OAc)2, H2O
R2
R3
2) NaBH4
D
HO
R1
HO
R1
H
R1
H
R2
H OH
1) BH3 • THF
H
R1
2) NaOH, H2O2,
R3
R2
H 2O
• BD3 variant, anti-Markovnikov, syn addition
D OH
H
R1
1) BD3 • THF
H
R1
2) NaOH, H2O2,
R3
R2
R3
R2
H 2O
2) Alkene → alkyl halide
• Markovnikov, syn + anti addition
H
R3
R1
H
HX
R2
R3
(X=Cl, Br, I)
X
R1
R2
H
Br2
R2
R3
CH3COOH
Br
R1
R2
+ En
HO
OH
1) OsO4, H2O2
HO
OH
2) NaHSO3/H2O
+ En
• Meso product
Br2
R2
R1
Br
Br
Br
Br
CH3COOH
R2
D
Na/ND3
R1
• Racemic product
H2
Br2
Lindlar’s
catalyst
CH3COOH
D
+ En
H +
R3
+ En
+ En
H
H
R1
H2
Pd/C
R3
H
H
5) Alkene → diol
• Syn addition
R1
H
R2
R3
1) OsO4, H2O2
or
2) NaHSO3/H2O
HO
R1
KMnO4, NaOH
Cold, dilute
R3
+ En
R2
R1
H
m-CPBA
R2
R3
CH2Cl2
R1
O
R2
H
H2SO4
R3
+ En
H 2O
HO
R1
OH
H
R2
R3
+ En
R1
R2
H
R1
R2
H
Br2
HO
R1
R2
R3
H 2O
R2
H
R3
+En
R2
H
+En
R3
1) O3, CH2Cl2
2) DMS
or Zn, CH3COOH, H2O
R2
H
R3
1) O3, CH2Cl2
2) H2O2, HCl, H2O
Cl2
Lindlar’s
catalyst
+ En
NaOEt
H 3C
hν, 25 °C
Cl2
H3C CH3
R3
H2C CH2
H
R1
O
+
R1
OH
+
R2
Oxidizing conditions lead to ketones/carboxylic acids.
m
O
-C
PB
A
Br2
+ En
Br
Br
+ En
H
1) NaNH2
2) H3C
Br
Br2
CH3COOH
NaNH2
Br
CH4
Cl2
hν, 25 °C
H
H
H3C Cl
1) NaNH2
2) CH3Cl
3) NaNH2
4) CH3Cl
3
m
R3
O
H
Na/NH
Lindlar’s
catalyst O 2 O
2
2
, H 3/H
4
sO O
O aHS
)
1 )N
2
R3
R2
O
H2C CH2
Cl
H2
O
NaNH2
Br
Cl
NaOEt
H 3C
hν, 25 °C
Br
Br2
CH3COOH
7) Synthesis of meso and racemic epoxide/dibromide/diol from 2-butyne
Reducing condition leads to ketones/aldehydes.
R1
+ En
H
Br
5) Synthesis of 1-butyne from ethane
Br
7) Alkene → ketones/aldehydes/carboxylic acids
R1
Br
H2
NaNH2
2
Cl
6) Synthesis of 2-butyne from ethane and methane
R1
H
R3
+En
4) Converting trans alkene to cis alkene
Br
Br
H3C CH3
Na/NH3
+ En
CH3COOH
• Markovnikov, anti addition
H
Na/NH3
NaNH2
Br
CH3COOH
OH
H
NaNH2
1) NaNH2
2) CH3Cl
Br
Br2
Br
Br + En
CH3COOH
Br + En
3) Converting cis alkene to trans alkene
R3
R2
Br2
NaNH2
CH3COOH
+ En
6) Alkene → alkyl bromide/alcohol
Br
R3
R2
H
8) Synthesis of trans-2-pentene from 1-butene
Br
Br2
• Anti addition (alkene → epoxide, syn addition)
X
H
• Anti-Markovnikov, syn + anti addition
H
Br
H
R1
HBr
H +
R1
ROOR,
light/heat
R3
R2
R3
R2
+ En
3) Alkene → dibromide
• Anti addition
R1
D2
Lindlar’s
catalyst
R2
R2
2) NaHSO3/H2O
Na/NH3
2) Converting alkene to alkyne
4) Alkene → alkane
• Syn addition
En = enantiomer
• Anti-Markovnikov, syn addition
R2
D
R1
HO
Useful Tips & Tricks (Continued)
H
H +
R3
R2
R3
+En
+ En
No rearrangement! If use H2SO4, H2O, rearrangement may occur.
R1
Na/NH3
1°, 2°, or 3°
1) Alkene → alcohol
• Markovnikov, syn + anti addition
H
R2
• Deuterium variant
Reactions of Alkenes
R1
R2
Lindlar’s
catalyst
• Racemic product
R1
R2
OMs
MsCl
Pyridine
R1
Na/NH3
R1
Hofmann
OH
TsCl
Cl
t-BuOK
O
+
OH
1) OsO4, H2O2
H2
Lindlar’s catalyst
R2
• Trans alkene
R1
1) Consequences of stereochemistry
• Meso product
R2
H2
R1
Hofmann
NaOEt
t-BuOK
R2
• Cis alkene
X
(X=Cl, Br, I)
O
1) Sia2BH
2) H2O2, H2O
R1
4) Alkyne → alkene
t-BuOK
trans Zaitsev
R1
Pd/C
R2
R
Useful Tips & Tricks
H2
R1
E2 and E1: E2 is the most reliable way to make alkenes from alkyl halides beacuase it offers more predictable products and
stereochemistry than the E1 does. In an E2 reaction, the Zaitsev (more substututed) product is generally favored over the Hofmann
(less substituted) product unless a sterically hindered base is used, in which case the Hofmann product will be favored. A trans
alkene will also be favored over a cis alkene and the proton removed must be antiperiplanar to the LG.
O
H
7) Alkyne → carboxylic acid
3) Alkyne → alkane
(X=Cl, Br, I)
NaOEt
1) NaNH2
H
R
HgSO4, H2O
R
NaNH2
O
H2SO4
H
HO
O
OH
+ En
Meso
+ En
-C
PB
O2 O
2
2
, H /H
4 O3
O
s
O aHS
)
1 )N
2
A
Br2
Br
Br
+ En
HO
+ En
OH
Racemic
Practice Questions on This Material
Practice questions:
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