Chapter Objectives
Present reactions of alkenes and alkynes
Reactions related to those found in biology
Must
know reactions
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1
Preparation of Alkenes
Precursors
Alcohols (especially in biological chemistry)
Alkyl Halides (industrial chemistry)
O H
s tr o n g a c id
H
d e h y d r a t io n
X
s tro n g b a s e
H
d e h y d r o h a lo g e n a t io n
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2
Biological Dehydration
Rarely done on free alcohol
Generally done on molecules containing carbonyl and
hydroxyl groups
H2O
-
HO
CO2CO2-
O2C
Fall, 2007
aconitase
CO2O2C
CO2-
3
1
Reaction with X2
Halogenation
Reaction with Cl2 and Br2
C l2
Cl
Cl
B r2
Br
Br
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4
Stereochemistry
Reaction provides the trans product
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5
Explanation
Not a carbocation intermediate as shown
Bromonium ion intermediate forms
H
H
Br +
Br
H
H
Br
Br-
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6
2
Biological Halogenation
Marine organisms
Haloperoxidase
H2O2 oxidizes Cl- or Br- to X+
Cl
Br
Br
Cl
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7
Reaction with X2 in H2O
Cl2 in water yields HOHO-Cl (hypochlorous acid)
Br2 in water yields HOHO-Br (hypobromous
(hypobromous acid)
Br2/H2O
OH
Br
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8
Hydration of Alkenes
Alkenes react with water to give alcohols
Require high temperatures and pressures
H2O
CH3CH2OH
Does not work well in the laboratory
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9
3
Biological Hydration of Alkenes
O
O
O-
-O
O
fumarase
fumarate
OH
-O
O
maleate
Relatively rare reaction
Cellular constraints are not present.
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10
Cellular Constraints
Solvent is water
Narrow pH range
Fixed temperature
Limited elemental choice
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11
Laboratory Hydration of Alkenes
Oxymercuration
Mercuric Acetate in THF
Markovnikov Product
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12
4
Laboratory Hydration of Alkenes
Hydroboration
NonNon-Markovnikov Product
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13
Mechanism of Hydroboration
Borane is a Lewis acid
Alkene is Lewis base
Transition state involves
anionic development on B
The components of BH3 add
across C=C
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14
Reduction and Oxidation
Carbon always has 4 bonds
Oxidation changes are more difficult to see
Reduction:
Increase in H content
Decrease in O content
Oxidation:
Decrease in H content
Increase in O content
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15
5
Reduction of Alkenes: Hydrogenation
Addition of H2
Requires Pt or Pd catalyst (or NR)
Heterogeneous Reaction
Process is not in solution
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16
Mechanism of Catalytic Hydrogenation
Heterogeneous – reaction between phases
Addition of HH-H is syn
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17
Biological Reductions
Rare Reaction
Uses NADPH as reducing agent
NH2
N
O
O
N
H
HO
H
OH
O
P
O-
O
O
O
P O
N
N
OHO
O
N
-2
OPO2
NH2
Nicotinamide Adenine Dinucleotide Phosphate
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18
6
Oxidation of Alkenes: Epoxides
mcpba
O
CH2Cl2
H
O
OOH
peroxide
mcpba =
Cl
Reaction with a peracid
Epoxide or oxirane
Cyclic ether
Fall, 2007
19
Epoxide Preparation
From Halohydrin
Br2/H2O
OH
base
O
Br
bromohydrin
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20
Biological Epoxidation
Present in variety of processes
Does not involve peracids
Peroxides formed by reaction with O2
Very selective reaction (see Figure 7.8)
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21
7
Hydroxylation of Alkenes
Diol formation
H3O+
OH
O
OH
Laboratory and Biological Reaction
Biological process useful for detoxification
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22
Laboratory Hydroxylation
Reaction with osmium tetroxide
Stereochemistry of addition is syn (product is cis)
cis)
Product is a 1,21,2-dialcohol or diol (also called a glycol)
glycol)
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23
Reaction with Carbenes
H2C:
The carbene functional group
Carbenes are electrically neutral with six electrons in the outer
shell
They add symmetrically to double bonds giving cyclopropanes
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24
8
Formation of Dichlorocarbene
Base removes proton
from chloroform
Stabilized carbanion
remains
Unimolecular
Elimination of Clgives electron
deficient species,
dichlorocarbene
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25
Reaction of Dichlorocarbene
Addition of dichlorocarbene is stereospecific cis
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26
Simmons-Smith Reaction
Equivalent of addition of CH2:
Reaction of diiodomethane with zinczinc-copper alloy produces
a carbenoid species
Forms cyclopropanes by cycloaddition
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27
9
Radical Reactions
Mechanism of addition of HBr was hotly debated
NonNon-Markovnikov product was observed
Peroxides form readily in starting material
HBr
Br
HBr
Br
On occasion
Br
+
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28
Radical Reactions - HBr
If reaction is done with HBr/peroxides
HBr/peroxides
Get the nonnon-Markovnikov product
HBr/peroxides
Br
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29
Radical Reactions: Polymer Formation
Polymer – a very large molecule made of
repeating units of smaller molecules
(monomers)
Biological Polymers
Starch
Cellulose
Protein
Nucleic Acid
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10
Polymers
Alkene polymerization
Initiator used generally is a radical
n
r e p e a t in g
u n it
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31
Mechanism
Initiation
Propagation
Termination
See page 241 in text for details
High reactivity of radicals limits usefulness
Not true in biological chemistry
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32
Biological Radical Reactions
Enzyme permits a single substrate at a time
at the active site
Greater control over reactivity
Radical reactions are common
Example given on page 244 for biosynthesis
of the PGAs
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33
11
Dienes
Contain two double bonds
NonNon-conjugated
Conjugated
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34
Common Feature in Nature
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35
Conjugation
Absorption of visible light produces color
Conjugated hydrocarbon with many double
bonds are polyenes
Lycopene - red color in tomatoes
Carrotene – orange color
Extended conjugation in ketones (enones)
enones)
found in hormones such as progesterone
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36
12
Conjugated Dienes
Chemistry is slightly different
More stable than nonnon-conjugated dienes
Heat of hydrogenation
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37
Greater Stability
Why?
Orbital Picture of alkene bonding
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38
Fall, 2007
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13
Orbital picture of conjugated diene
Electrons are delocalized (spread(spread-out) over
the entire pi framework
Impact upon the chemistry
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40
Fall, 2007
41
Reactions
With HBr
Br
(71%)
HBr
H
Br
(29%)
H
Why?
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42
14
Mechanism
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43
Allylic Cation
Fall, 2007
44
Some Data
X
1,2 product
Nucleophile
Bromide
Chloride
H
H
HX
1,2 Product
71%
30%
X
1,4 product
1,4 Product
29%
70%
If HBr is added at 0 oC we see the above data.
If the reaction is done at 40 oC, we see 30% of the 1,2
product and 70% of the 1,4 product.
How do we explain these results?
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45
15
A B+C
B forms faster than C
Energy of activation is lower for B than C
C is more stable than B
Constructing reaction energy diagram
energy
B
A
C
reaction progress
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46
Thermodynamic Control
Transition state leading to more stable species is
higher in energy, therefore, it is easier to get to the
less stable product
Reaction is reversable
At high temperatures, sufficient E for both
reactions to occur
A
B (fast) and A
C (slower)
or B
A
C
We see more stable product dominate.
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47
Kinetic Control
At low temperatures
Reaction is not reversable
Equilibrium is not reached
Insufficient energy for A to C
Sufficient energy for A to B
Less stable product dominates.
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48
16
Reactions of Alkynes
Alkynes are rare in biological chemistry
Chemistry is similar to alkenes
Generally less reactive than alkenes
Reactions can be stopped at alkene stage
using one equivalent of the reagent
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49
Reactions with HX
Regiochemistry is Markovnikov
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50
Reactions with X2
Initial addition gives trans intermediate
Product with excess reagent is tetratetra-halide
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17
Reactions with H2
Reduction using Pd or Pt does not stop
Alkene is more reactive than alkyne
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52
Reactions with H2
Lindler’s catalyst is poisoned
Not as reactive
Stops at ciscis-alkene
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Reduction using dissolving metals
Anhydrous ammonia (NH3) is a liquid below -33 ºC
Alkali metals dissolve in liquid ammonia
Provide a solution of e- in NH3
Alkynes are reduced to trans alkenes with sodium or lithium in
liquid ammonia
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18
Hydration of Alkynes
Hydration (Hg+2) of terminal alkynes provides methyl ketones
Hydration (BH3) of terminal alkynes provides aldehydes
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55
Alkyne Acidity: Acetylide Anion
Terminal alkynes are weak Brø
Brønsted acids
pKa is approximately 25
alkenes and alkanes are much less acidic
Reaction of strong anhydrous bases with a
terminal acetylene produces an acetylide ion
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Alkylation of Acetylide Anions
Acetylide ions are nucleophiles
Acetylide ions are bases
React with a primary alkyl halides
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19
PROBLEMS ON ALKENE CHEMISTRY
©2004 OCHeM.com
1. Answer the following questions concerning the three alkenes shown below. Your answer should
use words and illustrations.
1
2
3
CH3
A
4
5
O
6
CH3
B
CF3
C
a) Which alkene would be most reactive with H+ ? Why?
b) Which alkene would be least reactive with H+ ? Why?
c) Which alkene would be most reactive under conditions of catalytic hydrogenation? Why?
d) Will all three alkenes undergo Markovnikov addition of HBr? If not, which alkene or alkenes
won’t and why?
1
PROBLEMS ON ALKENE CHEMISTRY
©2004 OCHeM.com
2. Provide a mechanism for the following reaction. Then, predict the reaction energy diagram for the
entire reaction in the space provided. In your diagram, you must indicate the ΔH of the reaction,
label the activation energy (Ea) of the rate-determining step, and clearly identify all intermediates
and products of the reaction.
H Br
Energy
Br
Reaction Coordinate
2
PROBLEMS ON ALKENE CHEMISTRY
©2004 OCHeM.com
3. What starting materials & reagents are needed to produce the following compound?
Br
Br
4. Draw the major product of each of the following reactions. Be sure to include stereochemistry in
your answers where appropriate.
a)
CH3
Br 2
b)
H Cl
CH3
CH3
c)
1) Hg(OAc) 2 / H2 O
CH3
2) NaBH4
1) BH3 •THF
2) H2O 2 , NaOH
d)
Cl 2
CH3 CH2 OH
3
Alkene Reactions
• Addition Reactions - only one product
• 1. Hydrogenation (H2):
Alkene + Hydrogen --> Alkane
• 2. Halogenation (X2):
Alkene + diatomic halogen --> Dihaloalkane
• 3. Halgenation (HX):
Alkene + HX --> Haloalkane
• 4. Hydration (HOH):
Alkene + HOH --> alcohol
• 5. Polymerization:
Many alkenes add together into a long alkane chain
Alkene Reactions
1. Hydrogenation Reaction
• Alkene + Hydrogen --> Alkane
• Hydrogen molecule adds to carbons with double bond.
H3C
H
C
CH2
+
H2
H3C
H3C
H
C
C
H
H
H
H
H2
C
CH3
H
H3C
C
H
H
C
H
H
1
Alkene Reactions
Alkene Reactions
2. Halogenation Reaction
• Alkene + Halogen --> Dihaloalkane
• Halogen molecule adds to carbons with double bond.
Br
H3C
H
C
CH2
+
CH CH2
Br2
H3C
H3C
H
C
C
H
Br
Br
Br
H
H
H3C
C
Br
H
C
H
Br
2
Alkene Reactions
Unsymmetrical Addition Reactions
• Markovnikov’s Rule
• For double bonds that have unequal numbers of
hydrogen atoms attached.
• For unsymmetrical reactants such as HX and HOH
• The hydrogen of the reactant goes to the carbon of
the double bond that already has the most hydrogen
atoms. “the rich get richer”
• The -X or the -OH go the other carbon (the one with
the least amount of hydrogen
3. Halogenation Reaction
• Alkene + HX --> haloalkane
• HCl molecule adds to carbons with double bond.
H3C
H3C
H
C
CH CH3
+ HCl
Cl
CH2
H3C
H
C
C
H
H
H
Cl
H
Mark's Rule
H3C
C
Cl
H
C
H
H
3
Alkene Reactions
4. Hydration Reaction
• Alkene + HOH --> alcohol
• HOH molecule adds to carbons with double bond.
H3C
H3C
CH CH3
H
C
CH2
+
HOH
H3C
H
C
HO
C
H
H
HO
H
H
H3C
C
OH
H
C
H
H
Mark's Rule
Alkene Reactions
4
5. Polymerization Reaction
• Alkene + Alkene + Alkene --> long chain of carbons
• Double bonds convert to single bonds
H
H H
H H
H
C
C
H
C
H
H
C
H
C
H
C
H H
H
H
H
H
C
C
H
H
H
.C C.
H
.C
H
H
H
H H
C
C
H H
.C
C.
H
C.
H H
H
H
C
C
H H
H
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
Alkene Reactions
What is the structure for
Alkene Reactions
5
Alkene Reactions
Alkene Reactions
6
1. In the reaction shown below, only one product is formed. Why?
OH
+
H2SO4
H2O
2. Consider the following reaction.
H3C
C
H3C
Br
H
H
Br2
C
CCl4
H
H
H3C
CH3
Br
a. What is the IUPAC name for the starting material in this reaction? What is the
stereochemistry of the starting material (if there is any)?
b. What is the IUPAC name for the product in this reaction?
3. Draw the major organic product formed (showing stereochemistry where applicable) for the
reaction of the following alkene under each of the reaction conditions listed below.
CH3
CH3
C
CH3
a)
HBr
HBr
b)
peroxides
H2O
c)
H+
Cl2
CCl4
Br2
H2O
1. Hg(OAc)2, H2O
2. NaBH4
CH
CH2
4. Draw the major organic product formed (showing stereochemistry where applicable) for the
reaction of the following cycloalkene under each of the reaction conditions listed below.
CH3
d)
e)
f)
g)
HBr
HBr
peroxides
H2O
H+
Cl2
CCl4
h)
i)
Br2
H2O
1. Hg(OAc)2, H2O
2. NaBH4
1. BH3 THF
2. H2O2/NaOH
1. OsO4
2. H2O2
1. O3
2. (CH3)2S
H2
metal catalyst
Copyright, Arizona State University
Introduction to Synthesis (more...)
Alkenes II
1 Additional Reactions of Alkenes
• here are a series of mainly oxidation reactions that we need to complete our survey of alkene chemistry
• some are not "pretty", i.e. with complex mechanisms, some that are not fully known...
• Some of the stuff here you will just have to "know", i.e. here is some of the "bad bit" of Organic 1.1 Formation of cis-Diols (more)
• TWO sets of reagents will accomplish this reaction
The reaction
OH
cold KMnO4/-OH/H2O
OR
syn-addition
(same side)
OH
OsO4 / H2O2
The mechanisms
O
O
O O
Mn
Mn
O
O
O
HO
OH
O
OH
+ MnO2
syn-addition!
don't need to know!!
-
OH
–
–
O O
HO Mn
O
O
aqueous
workup
–
–
O O–
HO Mn OH
O O
HO Mn
–
O
O
–
-
O
–
O
O
OH
radical
reduction
O–
+ MnO2(OH)2
addition/elimination mechanism
• although this is obviously complex, the important part is that the MnO4 ion starts the reaction by adding to both
ends of the alkene at the same side, which is why a cis-diol must be formed
• note that in mechanisms involving metal atoms, the metal has enough electrons and empty orbitals to give and
take electrons on its own, almost at will (almost like cheating to an organic chemist!!)
O
O
Os
O
O
O O
Os
O
H2O2
O
HO
OH
+ OsO4
catalyst
• syn-addition! regenerated!
H2O2
don't need to know.....
Alkenes II
1
Copyright, Arizona State University
why TWO reagents?
KMnO4 - inexpensive, used for large scale reactions, variable yields
OsO4 - expensive, good yields, used in small scale syntheses
• This illustrates the principle that in general there will always be more than one reagent to accomplish any
transformation even if we only discuss one in this course
Examples
cold KMnO4/-OH/H2O
H3C
CH3
C C
H
H
OsO4 / H2O2
H3C
H
C C
H
CH3
OsO4 / H2O2
H
H
HO
OH
H3C CH3
H C C H
HO
OH
cis-diol
meso- diol
OH
H3C
H
C C
H
CH
HO (±) 3
OH
=
HO
(±)
1.2 Formation of Epoxides and trans-Diols (more)
New reagent
O
Cl
O
R C O OH
peroxy acid
R O O R
peroxide
O
C O OH
R C OH
carboxylic acid
meta-chloroperbenzoic acid (MCPBA)
Mechanism
O
C
C
O
C
R
C
C
H O
O
epoxide
+
O
C
R
O
H
carboxylic acid
• concerted mechanism - all bonds made and broken at the same time
• no chance for bond rotation "in the middle" - stereospecific reaction!
Examples
MCPBA
cis-alkene
O
cis-epoxide formed
• reaction is STEREOSPECIFIC
Synthesis of a trans-Diol
H3C
CH3
C C
cis-alkene
H
CH3 OH
H
(±)
C C
CH3
HO
H
ANTI-addition
Alkenes II
MCPBA
H
LA/BA
H CH3 OH
C C
CH3
H O
H
H
O
H3C C C CH3
H
LB/BB H
H3O+
H
O
H
H O LA/BA
H
LA
H3C C C CH3
H
H backside attack!
O
O
H H LB/BB
H H LB
2
Copyright, Arizona State University
+
• when H3O is a reagent, this means aqueous acid (e.g. HCl or H2SO4 in water)
• the intermediate is an oxonium ion (onium means more than usual valence, in this case 3 for oxygen), compare
with bromonium etc.,
• this reaction sequence makes a trans-diol as opposed to the cis-diols we saw above, so........
1. MCPBA
trans-diol
H
2. H3O
OsO4
H2O2
+
cis-diol
H
OH
H
HO
OH
meso compound!
HO
H
(±) racemic mixture
1.3 Oxidative Cleavage of Alkenes: Ozonolysis (more)
New reagent
ozone
=
O3
=
O
O
O
O
O
O
• has separated charges and more non-bonding electrons, much more reactive than molecular oxygen
Mechanism
O
O
O
H3C
H
C C
H3C
CH3
O
O
O
H3C C C H
H3C
CH3
moloxide
O
H3C C
CH3
H3C
H3C
C
O
C
O O
ozonide
O
O
C H
H3C
H
C O + O C CH
3
H3C
ketone aldehyde
H3C
Me2S
H
CH3
S
(reducing
agent)
Me
Me
+
S O
Me
Me
don't need to know!!
• the ozonide is the primary product, but it is never isolated
• in the presence of Me2S it is reduced to (in this case) are an aldehyde and ketone
• if Me 2S is replaced by the oxidizing hydrogen peroxide, different products result.....
H3C
H
C O + O C CH
3
H3C
ketone aldehyde
1. O3
2. Me2S
H3C
H
C C
H3C
CH3
H3C
1. O3
2. H2O2
Alkenes II
H3C
3
C O +
ketone
O
C
OH
CH3
carboxylic acid
Copyright, Arizona State University
Examples
O
1. O3
1. O3
OH
HO
2. H2O2
O
O
H
H
2. Me2S
O
H
1. O3
O
O
2. Me2S
O
O
2 Synthesis of Alkenes
2.1 From Alkyl Halides (seen before, review)
• i.e. reactions of alkyl halides that have alkenes as the products
• in useful reactions we want avoid carbocations, thus we want to do E2 elimination
Br
H3C
base
CH
H3C
CH3
CH
CH2
Which base to use to ensure elimination versus SN2???? Remember......
• E2 reaction is favored for 3Y halides
Y
• For 2 halides, E2 can be forced over SN2 by using a bulky base, see below
Y
• For 3 halides, a bulky base is not necessary, the product will be the Saytzeff product
Y
• For 3 halides, a bulky base will give the least substituted alkene, for steric reasons
Examples of bulky bases
O
NH
diisopropylamine (i-Pr2NH)
–
t-butoxide (t-BuO )
N
dimethylpyridine
Example Reactions
Br
–+
t-BuO K
acetone
• 2Y halide, use bulky base to ensure no SN2, get Saytzeff alkene product
Br
+–
Na OMe
acetone
Alkenes II
4
Copyright, Arizona State University
• 3Y halide, use NON-bulky base (no SN2 not possible), get Saytzeff alkene product
t-BuO– +K
acetone
Br
• 3Y halide, use BULKY base and get Non-Saytzeff (Hoffmann) alkene product
Bottom line
Y
–
• for 2 halides, use t-BuO to ensure no SN2 and to obtain Saytzeff product
Y
–
–
• for 3 halides, use CH3O to obtain Saytzeff product and t-BuO to obtain Hoffman product
2.2 From Alcohols (E1 and E2 elimination in a new context)
The reaction
H
C
OH
C
conc. H2SO4
C
heat
C
+
H2O
• note a special kind of SOLVENT EFFECT here! In an aqueous medium, acid catalyzes water ADDITION to the
alkene to make an alcohol. In conc. sulfuric acid medium, the acid helps to REMOVE water from an alcohol to
make an alkene (the sulfuric acid DEHYDRATES the alcohol)
Mechanism: you already know it - either an E1 or an E2 elimination!
–
• in the mechanism, H2O is the leaving group, OH is a poor leaving group (this is an important general principle
that we will return to again later....)
H
C
OH
C
X
does not happen!
+
H
C
C
OH
good nucleophile but
poor leaving group
H
H
C
OH2
C
H
C
C
H2O
poor nucleophile but
good leaving group
• in general, small neutral molecules such as water make excellent leaving groups, since they tend to contain low
energy electrons
Example
LB/BB OH
H OSO2OH
LA/BA
conc. H2SO4
major
heat
OH2
H
E1
LA/BA
OSO2OH
LB/BB
• 3Y and 2Y alcohols almost certainly E1 mechanism
• carbocation intermediates means rearrangements
• the sulfuric acid is the initial acid, the bisulfate anion is a likely base to deprotonate, recovering the acid catalyst
Alkenes II
5
Copyright, Arizona State University
Example
H OSO2OH LA/BA
LB/BB
OH
conc. H2SO4
heat
LA/BA H
LA/BA
OSO2OH LB/BB
OH2
H OSO2OH
OSO2OH LB/BB
H
E2
H
LA/BA
LB/BB
• Almost certainly an E2 mechanism with a 1Y alcohol, but the first alkene product must get protonated, it is in
conc. sulfuric acid after all, resulting in overall formation of a more substituted alkene
• the final product is the SAME as if the mechanism was E1 followed by carbocation rearrangement
3 Synthesis of Alkyl Halides
3.1 From Alkanes (seen before, review)
Br2
Br
h
Br2
Br
h
• actually, this is a pretty poor reaction since it is unselective, but it is the only one we have for alkanes!
Y
• use only if all hydrogens are identical (first example) or there is an obvious 3 hydrogen (second example)
3.2 From Alkenes (seen before, review)
Br
NBS
h
• we already learned that NBS is the best reagent to use for allylic bromination, don't use Br2
• to be safe, you CAN USE NBS and light for ALL RADICAL BROMINATIONS, including ALKANES
HBr
Br
HBr
ROOR
Br
• Markivnokov and anti-Markovnikov addition to an alkene also forms an alkyl halide
Alkenes II
6
Copyright, Arizona State University
4 Summary of Reactions (more)
• all of them so far!! (non-bonding electrons are not shown in this summary for simplicity)
Br
Br2, h
Br
NBS, h
Br
+–
Na
(±)
CN
SN2
CN
acetone
(and analogues)
Ph
Ph
Br
+–
E2
Na O-t-Bu
DMF
Ph
Ph
conc. H2SO4
OH
heat
Br
HBr
(±)
CCl4
OH
H2O
H2SO4
(±)
rearranged!
1. Hg(OAc)2/H2O
OH
(±)
2. NaBH4
H
1. Hg(OAc)2/EtOH
2. NaBH4
OEt
Br
Br2
CCl4
Br
Br2
CH3OH
Alkenes II
D
2. OH/H2O2
7
(±)
(±)
OCH3
1. BH3.THF
–
Br
(±)
D
H
OH
(±)
Copyright, Arizona State University
HBr
ROOR
Br
H
H2
Pd/C
H
OH
–
dil. KMnO4/ OH
or
OsO4/H2O2
(±)
MCPBA
OH
O
1. MCPBA
OH
+
2. H3O
1. O3
HO
O
O
(±)
OH
2. H2O2
1. O3
2. Me2S
O
O
H
• looks like a lot, actually not (there will be more next semester!!)
• do not attempt to memorize these, to learn them WORK THE PROBLEMS. After you have worked the problems
and understood the mechanisms you will realize that you actually know most of them or can work them out. After
working the problems you might like to test yourself on the reactions above to make sure that you have all the
details about reagents and conditions correct
Alkenes II
8
Copyright, Arizona State University
Copyright, Arizona State University
Alkenes II PROBLEMS
Reaction Practice
Provide the missing major organic product(s), reagents/conditions or reactant structure
as required.
O
+
1
(standard)
O
OH
2
(standard)
O
HO
O
OH
conc. H2SO4
Answer
Correct Y / N
heat
3
(standard)
Answer
Correct Y / N
Answer
Correct Y / N
1. HBr
2. NaOH
1. O3
4
(standard)
Answer
Correct Y / N
2. Me2S
OH
OH
OH
5
(challenging)
(ignore
stereochemistry)
Answer
Correct Y / N
O
OH
6
(challenging)
O
Answer
Correct Y / N
REMEMBER, doing problems is the ONLY WAY that you will learn organic chemistry. Keep track of exactly what
problems you get correct and which incorrect. Keep coming back until you get them ALL correct.
Alkenes II
9
Copyright, Arizona State University
5 Synthesis
• a large part of organic chemistry involves building larger or more complex molecules from smaller ones using a
designed sequence of reactions, i.e. chemical synthesis
• this involves putting a series of reactions together in sequence
• here we will look at some simple example to practice this, later we will learn a more systematic method for
solving complex synthesis problems, retrosynthesis
• to do these problems you need to know the reactions, and PRACTICE, practice, practice, practice......
Example Problems: make the molecule on the right from the one on the left. this can not be done in one reaction.
Give reagents and conditions and the intermediate molecules at each step.
???
+–
NBS, h
Na OCH3
acetone
Br
• brominate to get a functional group onto the alkene, then E2, which is the standard way to make an alkene
???
Br
NBS, h
NBS, h
Br Na+ –OMe
acetone
• brominate to get a functional group onto the alkene, then E2, which is the standard way to make an alkene, then
brominate again (note that I changed this example from the one that was originally in he notes, next semester you
will see why)
???
Br
need to add –Br and –OCH3
(±)
OCH3
Br2
NBS, h
CH3OH
Br t-BuO– +K
DMF
• need to add Br AND -OMe, we know how to do that from an ALKENE, thus make the alkene first as usual
???
HBr
X
NBS, h
HBr
Br
NBS, h
X
ROOR
–+
Br
Alkenes II
t-BuO K
DMF
10
Copyright, Arizona State University
• need to add Br at a position that is not possible by direct bromination, the obvious way is by addition of HBr to
an alkene Anti-Markovnikov, so first, make an alkene as usual
???
NBS, h
Br
OCH3
–+
H3CO K
HBr
–+
t-BuO K
DMF
DMF
Br
ROOR
• can't add -OMe to an alkane, so we need to a a functional group, a LEAVING group at that carbon, once we
recognize this then the strategy is same as previous problem
OH
OH
???
conc. H2SO4, heat
H2SO4/H2O
• we have two ways to make an alcohol, SN2 or water addition to an alkene, the best thing to "do" with the starting
alcohol is make an alkene, which decides for us which alcohol synthesis method to use
Alkenes II
11
Copyright, Arizona State University
Copyright, Arizona State University
Alkenes II PROBLEMS
Cumulative Problems
Provide a synthesis of the target molecule on the right from the starting molecule on the
left. This cannot be done in one reaction. Give reagents, conditions and intermediate
molecules at each step. Do not show any mechanisms or transient intermediates
1
(easier)
Answer
Correct Y / N
CN
2
(easier)
3
(easier)
OH
Answer
Correct Y / N
Answer
Correct Y / N
Br
4
(easier)
Answer
Correct Y / N
OH
Br
OCH3
5
(easier)
Answer
Correct Y / N
6
(standard)
Answer
Correct Y / N
OH
7
(standard)
Answer
Correct Y / N
Br
OH
8
(standard)
O
9
(standard)
O
H
H
10
(standard)
OH
Answer
Correct Y / N
Answer
Correct Y / N
Answer
Correct Y / N
HO
See also the RETROSYNTHESIS Web Pages
REMEMBER, doing problems is the ONLY WAY that you will learn organic chemistry. Keep track of exactly what
problems you get correct and which incorrect. Keep coming back until you get them ALL correct.
Alkenes II
12
Copyright, Arizona State University
ALKENES AND ALKYNES – REACTIONS
A STUDENT WHO HAS MASTERED THE MATERIAL IN THIS SECTION SHOULD BE
ABLE TO:
1.
Given the starting materials and reaction conditions, predict the products of the following
reactions of alkenes and alkynes.
Markovnikov addition of acids to alkenes and alkynes, including the acid-catalyzed addition
of water (hydration). Rearrangement is possible in the additions to alkenes;
tautomerization occurs in the hydration of alkynes.
Anti-Markovnikov addition of HBr.
Addition of halogens (and halohydrin formation). These reactions are anti additions. In
halohydrin formation, the OH goes to the more substituted carbon.
Epoxidation, and the hydrolysis of the resulting epoxides to glycols (overall anti addition).
Glycol formation using either KMnO4 (cold) or OsO4 (these are syn additions).
Degradation of alkenes and alkynes using either ozonolysis or KMnO4 (hot)
Addition of carbenes to alkenes
2.
Using any of the above reactions, propose syntheses of compounds that can be made using
alkenes as starting materials or intermediates. As always, synthesis problems may require
any reaction that you have studied in the course so far.
3.
Propose mechanisms, and predict and explain experimental results using your knowledge of
mechanism. Important reactions include:
Markovnikov additions (which proceed by protonation of the alkene to give a carbocation).
Addition of bromine and bromohydrin formation (by formation of the bromonium ion).
Hydrolysis of epoxides (under acid conditions the oxygen is protonated first).
4.
Identify unknown alkenes and alkynes when given either the products of ozonolysis or of
KMnO 4 degradation. You will not be called on to distinguish between cis and trans
isomers; these methods do not give that information. (Further information can be obtained
from the index of hydrogen deficiency).
5.
Use the results of simple chemical tests in identifying unknown compounds. Data from IR
spectra may also be used. Important tests include:
Solubility in concentrated sulfuric acid (compounds containing only alkanes, halogens, and
aromatic rings do not dissolve)
Bromine in carbon tetrachloride (alkenes and alkynes give an immediate reaction,
decolorizing the reddish-brown bromine solution)
Potassium permanganate (cold, dilute) (alkenes and alkynes give an immediate brown
precipitate; other compounds leave the permanganate solution purple)
Silver nitrate in alcohol solution (gives a white precipitate with alkyl chlorides, a tan ppt.
with alkyl bromides, a brown ppt. with alkyl iodides, no reaction when the halogen is
attached to an sp2 carbon, and no reaction with other materials)
Silver nitrate in ammonia (gives a white ppt. with terminal alkynes)
96
A STUDENT WHO HAS MASTERED THE OBJECTIVES ON THE PREVIOUS PAGE
SHOULD BE ABLE TO SOLVE THE FOLLOWING PROBLEMS AND RELATED ONES:
1.1
Predict the major organic product or products of each of the following reactions.
a) (CH3 )2 CHCH=CH2 + HCl ----->
no peroxides
b) CH3 CH2 CH2 CH=CH 2 + HBr ------------------>
peroxides
c) CH3 CH2 CH2 CH=CH 2 + HBr --------------->
d)
(cold, dilute)
+ KMnO4 ----------------->
e)
KOC(CH 3 )3
---------------->
CHCl3
f)
CCl4
+ Br2 -------->
g)
+ Cl 2 + H2 O ----->
OH , H2 O, heat H3 O
+ KMnO4 --------------------> --------->
h)
COOH
HOOC
i)
C
H
j)
C
CCl4
+ Br2 -------->
H
CH2 I2 /Zn(Cu)
------------------>
ether
97
1.2
For each of the following compounds, compare the reaction products from the addition of
HBr in the presence of peroxides with the addition of HBr in the absence of peroxides.
CH3
|
a) CH3 —CH—CH=CH2
H
b)
CH=CH 2
2.
Propose a synthesis of each of these compounds, from the given starting material and any
needed inorganic reagent and/or solvent.
CH3
a)
CH3
from
Br
b) CH3 CH2 CHBr2 from CH3 C≡CH
OH
c)
from
OH
OH
d)
Br
from
OH
98
2.2
Propose a synthesis of each of these compounds from the given starting materials(s) and
any needed inorganic reagents or solvents.
CH3
CH3
a)
from
Br
b) CH3 CH2 CH2 CH2 Br from HC≡CH and CH3 CH2 I
c) (CH3 )3 CBr from (CH3 )2 CHCH 2 OH
OH
d)
from
OH
CH3
e)
C
H
H
CH3
from
C
CH3
C
H
CH3
C
H
f) (CH3 )2 CHCH 2 CH2 CH3 from (CH3 )2 C=CH 2 and H-C≡CNa
99
3.
Propose a mechanism for each of the reactions shown.
OH
H3 O
a)
O + H 2 O ---------->
OH
b) (CH3 )2 CHCH=CH2 + HCl -----> (CH 3 )2 CClCH2 CH3
CH2 CH3
CH=CH 2
c)
H3 O
+ H 2 O ---------->
OH
d) CH3 CH2 CH=CH 2 + Br2 + I– -----> CH3 CH2 CH—CH 2 + Br–
|
|
I
Br
HO
CH2
H3 O
+ H 2 O ---------->
e)
CH3
CH2
f)
CH3
H3 O (trace)
------------------->
100
4.1
Identify each of these unknowns from the information given.
O
O
O3
Zn
||
||
a) C6 H12 ----> ------> CH3 C-CH3 + CH3 CH2 C-H
H2 O
O
||
CH3 CH2 C-H
b) C9 H16
O3
Zn
----> ------>
H2 O
c) C6 H10
O
O
KMnO 4 , OH
H3 O
||
||
-------------------> ---------> HO-C-CH2 CH2 CH2 C-CH3
heat
O +
O
KMnO 4 , OH
H3 O
d) C10 H16 -------------------> --------->
heat
4.2
Predict the products of the following reactions.
O3
Zn
a)
----> ------>
H2 O
CH2
b)
CH=CHCH3
5.1
CH2 CH2 CH2 COOH
H3 O
KMnO 4 , OH
--------->
------------------->
heat
Match each set of test results with one of the compounds shown.
OH
CH3 CH2 CH2 CH2 Br
A
B
C
D
a) Decolorizes Br2 in CCl4
Soluble in H2 SO 4
No reaction with AgNO3 in ammonia
b) No reaction with cold dilute KMnO4
No reaction with AgNO3 in alcohol
Soluble in conc. H2 SO 4
101
CH3 CH2 CH2 C
E
C-H
5.1
Unknowns, continued. The possibilities are:
OH
CH3 CH2 CH2 CH2 Br
A
B
C
D
c) Gives a brown precipitate with cold dilute KMnO4
Gives a white precipitate with AgNO3 in ammonia
Soluble in conc. H2 SO 4
d) No reaction with cold dilute KMnO4
No reaction with alcoholic AgNO3
Insoluble in conc. H2 SO 4
5.2
Describe simple chemical tests that can distinguish between:
a)
b)
c)
d)
e)
cyclohexene and cyclohexyl bromide
1-hexene and 1-hexyne
tert-butyl alcohol and tert-butyl bromide
pentane and 1-pentene
ethanol and 2-pentyne
102
CH3 CH2 CH2 C
E
C-H
ANSWERS TO THE PROBLEMS:
1.1
Predict the major organic product or products of each of the following reactions.
a) (CH3 )2 CHCH=CH2 + HCl -----> (CH 3 )2 CClCH2 CH3
no peroxides
b) CH3 CH2 CH2 CH=CH 2 + HBr ------------------> CH3 CH2 CH2 CHBrCH 3
peroxides
c) CH3 CH2 CH2 CH=CH 2 + HBr ---------------> CH3 CH2 CH2 CH2 CH2 Br
(cold, dilute)
+ KMnO4 ----------------->
d)
OH
OH
KOC(CH 3 )3
---------------->
CHCl3
e)
f)
+ Br2
Cl
Cl
Br
CCl4
-------->
+ enantiomer
Br
Cl
g)
+ Cl 2 + H2 O ----->
+ enantiomer + HCl
OH
COOH
OH , H2 O, heat H3 O
+ KMnO4 --------------------> --------->
h)
COOH
COOH
HOOC
i)
C
H
j)
C
+ Br2
CCl4
-------->
H
COOH
HOOC
C
Br H
CH2 I2 /Zn(Cu)
------------------>
ether
103
C
Br H
+ enantiomer
1.2
Compare the addition of HBr in the presence and in the absence of peroxides to each of the
following compounds. (Hint: Predict the products.)
CH3
|
a) CH3 —CH—CH=CH2
HBr, no peroxides
HBr, peroxides
(CH3 )2 CBr—CH2 CH3
(CH3 )2 CH—CH 2 CH2 Br
Note that both of these problems involve rearrangements in the absence of peroxides.
Br
HBr, no peroxides
H
CH2 CH3
b)
CH=CH 2
H
HBr, peroxides
CH2 CH2 Br
2.1
Synthesis problems. These answers are in shorthand (not balanced equation) form.
a)
CH3
CH3
HBr, peroxides
-------------------->
Br
2 HBr, peroxides
b) CH3 C≡CH -----------------------> CH3 CH2 CHBr2
c)
C6 H5 CO3 H
---------------->
H2 O, H3 O
O ----------------->
OH
OH
d)
Br KOH, heat
--------------->
C6 H5 CO3 H
H2 O, H3 O
----------------> ----------------->
OH
OH
(Last two steps as in c) above)
2.2
More synthesis:
Br
a)
CH3 Br2 , light
------------->
KOH
HBr
CH3 --------> ------------->
peroxides
heat
104
CH3
Br
NaNH 2
CH3 CH2 I
b) HC≡CH -----------> HC≡CNa --------------> HC≡CCH2 CH3 and then
H2 , Ni2 B
HBr
HC≡CCH2 CH3 ------------> H2 C=CHCH 2 CH3 ------------> CH3 CH2 CH2 CH2 Br
peroxides
H2 SO 4 , heat
HBr, no peroxides
c) (CH3 )2 CHCH 2 OH ----------------> (CH3 )2 CH=CH 2 ----------------------> (CH3 )3 CBr
Br2
------>
light
d)
CH3
e)
C
H
CH3
C
H
Br KOH
------>
heat
C6 H5 CO3 H H3 O
---------------> --------->
H2 O
Br2
NaNH 2
-------> CH3 CHBrCHBrCH 3 ---------> CH3 C
CCl4
CH3 C
CCH3
CH3
Li, NH3
----------->
C
H
H
C
CH3
HBr, peroxides
f) (CH3 )2 C=CH 2 ---------------------> (CH3 )2 CHCH 2 Br, then:
H-C≡CNa
(CH3 )2 CHCH 2 Br ---------------> (CH3 )2 CHCH 2 C≡CH, then:
2H2 , Pt
(CH3 )2 CHCH 2 C≡CH ----------> (CH3 )2 CHCH 2 CH2 CH3
105
CCH3
OH
OH
then
3)
Mechanisms. Note that balanced equations are used throughout.
OH
H3 O
a)
O + H 2 O ---------->
OH
O + H3O
----->
O
H + H2 O
OH
O H + H2 O ----->
OH 2
OH
OH
+ H2 O ----->
+ H3O
OH 2
OH
b) (CH3 )2 CHCH=CH2 + HCl -----> (CH 3 )2 CClCH2 CH3
(CH3 )2 CHCH=CH2 + HCl -----> (CH 3 )2 CHCHCH 3 + Cl
H
|
(CH3 )2 C—CHCH 3 -----> (CH3 )2 C—CH2 CH3
(CH3 )2 C—CH2 CH3 + Cl -----> (CH 3 )2 CClCH2 CH3
CH=CH 2
c)
CH2 CH3
H3 O
+ H 2 O ---------->
OH
CHCH 3
CH=CH 2
+ H3 O
+ H2O
----->
CHCH 3
CH2 CH3
----->
H
CH2 CH3
+ H 2 O ----->
CH2 CH3
OH 2
CH2 CH3
OH 2
CH2 CH3
+ H 2 O ----->
106
OH
+ H3 O
3)
Mechanisms, continued. Note again that balanced equations are used throughout.
d) CH3 CH2 CH=CH 2 + Br2 + I– -----> CH3 CH2 CH—CH 2 + Br–
|
|
I
Br
CH3 CH2 CH=CH 2 + Br—Br -----> CH 3 CH2 CH—CH 2
Br
CH3 CH2 CH—CH 2 + I
-----> CH3 CH2 CH—CH 2
I
Br
HO
CH2
CH3
H3 O
+ H 2 O ---------->
e)
CH2
CH3
+ H3 O
---------->
+ H2O
H3 C OH 2
CH3
+ H 2 O ------>
OH 2
H3 C
HO CH3
+ H3 O
+ H 2 O ------>
CH3
CH2
f)
H3 O (trace)
------------------->
CH3
CH2
+ H3 O
----->
CH3
+ H2 O
CH3
H
+ H2 O ----->
+ H3 O
107
Br
+ Br
4.1
The unknowns are:
a) CH3 CH2 CH=C(CH 3 )2
CH3
CHCH 2 CH3
d)
c)
4.2
b)
The products are:
O3
Zn
----> ------>
H2 O
a)
O
CHO
H3 O
KMnO 4 , OH
-------------------> --------->
heat
CH=CHCH3
CH2
b)
O
O
||
||
+ HCCH2 CH
O
+ CH3 COOH
COOH
+ CO2
5.1
The unknowns are: a) B; b) A; c) E; d) C
5.2
Descriptions of simple chemical tests that can distinguish between the pairs. (Only one
answer is required.)
a) Cyclohexene gives no reaction with alcoholic AgNO3 , while cyclohexyl bromide gives a
white precipitate. Cyclohexene reacts with bromine in carbon tetrachloride and with
cold dilute potassium permanganate, and dissolves in concentrated sulfuric acid;
cyclohexyl bromide does none of these things.
b) 1-Hexene gives no reaction with AgNO3 in ammonia, while 1-hexyne gives a white
precipitate.
c) tert-Butyl alcohol dissolves in concentrated sulfuric acid, while tert-butyl bromide does
not. The bromide gives a white precipitate with AgNO3 in alcohol, while the tert-butyl
alcohol does not react.
d) 1-Pentene decolorizes bromine in carbon tetrachloride, gives a brown precipitate with
cold dilute potassium permanganate, and dissolves in conc. sulfuric acid. Pentane does
not react with any of these reagents.
e) 2-Pentyne reacts with bromine in carbon tetrachloride and with cold dilute potassium
permanganate; ethanol does neither of these things.
108
Name ___________________________________________________ Eighth Drill Test (Sample)
Organic Chemistry 2210 DR
Answer All Questions
1)
Predict the major organic product or products the following reaction.
CH3 CH2 CH2 CH=CH 2 + C6 H5 CO3 H ------>
2)
Propose a synthesis of each of the following compounds from the given starting material
and any needed inorganic reagents or solvents.
OH
a)
from
OH
b)
Br
Br
from
c) (CH3 )3 COH from (CH 3 )2 CHCH 2 OH
3)
Propose a mechanism for each of the reactions shown.
H3 O+
a) CH3 CH2 CH=CH 2 + H2 O -------> CH3 CH2 CHOHCH3
Br
b)
+ Br2 + Cl ----> Br +
Cl
4)
Give the structures of these unknowns from the information given.
O
O3 Zn, H2 O
||
a) C8 H16 ----> -----------> 2 CH3 CH2 CH2 C-H
KMnO 4 , base, heat acid
b) C8 H14 ------------------------> ------>
5)
O + CH3 CO2 H
Which of these compounds gives a peak in the IR spectrum near 1640 cm-1 , is soluble in
conc. H2 SO 4 , reacts with cold alkaline KMnO4 , and gives no reaction with either silver
nitrate in ethanol or Ag(NH3 )2 +?
A. CH3 CH=CH 2
B. CH3 C≡C-H
C. CH 3 CH2 CH3
109
D. CH 3 CH2 OH
E. CH 3 CH2 Cl
Name _________________________________________________ Eighth Drill Test (Sample B)
Organic Chemistry 2210 DR
Answer All Questions
1)
Predict the major organic product of each of the following reactions.
H2 SO 4
a) CH3 CH=CH 2 + H2 O ---------->
b)
2)
peroxides
CH3 + HBr ------------->
Propose a synthesis of each. You may use any needed inorganic reagents and solvents.
a) (CH3 )3 CCl from (CH3 )3 COH
b) bicyclo[3.1.0]hexane from cyclopentene
c) trans-1,2-cyclohexanediol from bromocyclohexane
3)
Propose a mechanism for each of the reactions shown. For problem a), first complete the
equation by predicting the product.
H3 O
a)
O + H 2 O ---------->
b) CH3 CH=CH 2 + Br2 + Cl
4)
Which of the compounds shown is soluble in conc. H2 SO 4 , reacts with cold alkaline
KMnO 4 , and gives no reaction with either AgNO3 in alcohol or Ag(NH3 )2 +?
A. CH3 CH2 Br
5)
------> CH3 CHClCH2 Br + Br
B. CH3 CH2 OH
C. CH 3 CH=CH 2
D. CH 3 C≡C-H
Give the structures of the unknowns from the information given.
O
KMnO 4 , base, heat acid
||
a) C7 H12 ------------------------> ------> HO
O
O
|
|
||
O3 Zn, H2 O
b) C7 H14 ----> -----------> H
+
110
E. CH3 CH2 CH3
O
||
OH
Type and definition of reaction:
1. Substitution (pg. 1025)
The replacement of one atom (or
group) by another atom (or group).
Note substitution is the only way to
add a halogen to an alkane.
Example
propane + Br2
Reaction equation, structural formulas, names, conditions:
H
H
(H)Br (H)Br H
H
catalyst
H
C
C
C
H
H
H
H
+
Br
H
Br
heat, (pressure)
C
C
C
H
H
H
H
+
Br(H)
Product: 2-bromopropane or 1-bromopropane,
also could be 1,2- or 1,3- or 1,1- or 2,2-dibromopropane)
2. Halogenation (pg. 1023)
The addition of a halogen (group VII
element) to a multiple bond (the
halogen atoms add across a pi bond).
propene + Cl2
H3C
H3C
CH
CH2
Cl
+
CH
Cl room temp.
CH2
Cl
Cl
1,2-dichloropropane
3. Hydrogenation (pg. 1023)
The addition of H2 to a multiple bond
(across a pi bond).
(Hydrogenation, also known as
saturation, is used commercially in the
production of margarine).
1-butene + H2
4. Hydrolysis (pg. 1023)
The addition of a water molecule to a
double bond.
propene + H2O
H3C H2C
catalyst
CH
CH2
+
H
H3C H2C
H
CH
heat, pressure
CH2
H
H
butane
H3C
acid catalyst
CH
H3C
CH
CH2 + H2O
CH2
(HO)H
OH(H)
2-propanol or 1-propanol
5. Elimination (pg.1027)
The loss of a small molecule from a
larger molecule. When the small
molecule is H2O the process is also
referred to as “condensation” or
“dehydration”.
ethanol
H
H
catalyst
H
C
C
H
OH
H
C
heat
H
H
C
H
H
+
H
O
H
ethene
water
H
Type and definition of reaction:
6. Esterification (pg. 1030)
The condensation reaction that joins
an alcohol with a carboxylic acid to
produce an ester.
Example
producing
isopentyl acetate
Reaction equation, structural formulas, names, conditions:
H3C
CH
+
CH2
H3C
CH2 OH
O
CH
C
heat
CH3
HO
3-methyl-1-butanol
(isopentyl alcohol)
H3C
H+
O
ethanoic acid
(acetic acid)
CH2
H3C
+
C
CH2 O
H2O
CH3
3-methylbutyl ethanoate water
(isopentyl acetate)
7. Polymerization
The conversion of “monomers” into “polymers”. I.e. the joining of small molecules to form large molecules with repeating units.
a) addition (pg. 1023)
eth(yl)ene
monomer
catalytic process
H 2C
CH2
...
CH2
CH2
CH2
CH2
CH2
CH2
n
CH2
CH2
...
polyeth(yl)ene
b) condensation (pg. 1031)
8. Oxidation (pg. 1028)
(More) oxygen atoms are placed on an
organic molecule.
General Reaction
(for Dacron
replace R1 with
-CH2-CH2- and R2
with benzene.
ethanal + K2Cr2O7
O
HO
R1
+
OH
HO
C
O
R2
C
O
OH
...
R1 O
diol
dicarboxylic acid
(ethylene glycol) (terephthalic acid)
H
C
O
R2
C
polyester
(Dacron)
O
H
O
C
C
O
R1
...
+ H2O
Cr2(SO4)3
K2Cr2O7
H
C
H
C
H +
H
H2SO4
room temp.
O
H
H
+
K2SO4
H2O
ethanoic acid
1. Define addition reaction. Which of these reactions are addition reactions? Which reaction is the opposite of an addition reaction?
2. Define condensation reaction. Which of these reactions are condensation reactions?
3. Based on these reactions how could you make the following chemicals (draw reaction, showing reactants, products and conditions. Also
indicate the type of reaction): 1,2-dichlorocyclopentane, octane (using 4-octyne), 2,2,3,3-tetrabromopentane, 1-butene, propanoic acid,
ethanol, ethyl propanoate.
Unit 5 Review: Hydrocarbons
1. Briefly define or explain the significance of the following terms:
a) acetylene
h) aromatic
b) acyclic
i) bond energy
c) addition reaction
j) calorimeter
d) aliphatic
k) cyclic
e) alkane
l) endothermic
f) alkene
m) exothermic
g) alkyne
n) fractionation
o)
p)
q)
r)
s)
t)
u)
monomer
organic chemistry
petroleum
polymer
saturated hydrocarbon
unsaturated hydrocarbon
Wöhler
2. What general formula describes a) alkanes, b) alkenes, c) alkynes, d) cycloalkanes?
3. Draw the structural diagram for benzene.
4. For 3-methylpentane draw the
a) complete structural diagram, b) condensed structural diagram, and c) line structural diagram
5. a) Write a balanced equation for the complete combustion of hexane.
b) Write two possible balanced equations for the incomplete combustion of hexane.
6. Draw structures for these compounds:
a) 2-methylbutane
b) 4-propyl-3-heptene
c) 5-ethyl-4,4,5-trimethyldecane
d) cis-1,3-dimethylcyclohexane
Name these structures:
CH3
f
e
CH3 CH CH2 CH2 C CH3
CH3
CH
H 2C
CH 2
CH 2
CH3
7. Identify each pair as structural isomers, geometric isomers, or neither e
a) cyclopentane, pentane
b) 1,1-dichloroethene, trans-1,2-dichloroethene
c) cis-1,2-dichlorocyclopentane, trans-1,2-dichlorocyclopentane
d) cis-1,2-dichlorocyclopentane, trans-1,3-dichlorocyclopentane
CH 3
CH
HC
CH2
Br
Br
Cl
C
C
C
H
H
f
H
Cl
H
C
Cl
H
C
Cl
H
C
Cl
C
H
H
C
Cl
8. Using Br2(aq), how can you easily distinguish between ethane, ethene, and ethyne?
9. Using a table of bond energies calculate the heat of reaction when excess Br2 reacts with
a) 1 mole of ethene, b) 1 mole of ethyne.
10. Differentiate the following terms with respect to definition, symbol, and units:
a) heat capacity, b) specific heat capacity, c) heat of reaction, d) specific heat, e) molar heat of reaction
11. A forensics lab receives a small 0.16 g sample of metal. To identify the metal they heat it using exactly
3.0 J of energy. The temperature rises from 20°C to 98°C. What is the unknown metal (see pg. 568)?
12. 50.0 grams of butane is placed in a calorimeter. The 350 grams of water in the calorimeter rose from
19.7°C to 21.2°C. a) Calculate the heat released by the butane. b) Calculate the molar heat of reaction.
13. There is something wrong with each of the following names. Identify the error in each case (often the
correct name can be determined by trying to draw the structure and then renaming it).
a) 5-methyl-3-hexyne
c) 1,2-dichlorocyclobutane
e) 2-ethyl-2-methylhexane
b) 3,3-dimethyl-3-hexyne
d) 3-methyloctene
f) 2,3-dimethyl-4-ethylnonane