Organic Chemistry
A3.1 – Organic Compounds
o
the word “organic” means of, relating to, or
derived from living organisms
o
organic chemistry is defined as the study of
the molecular compounds of carbon

compounds containing carbon-based ions (e.g.
carbonate) are not considered to be organic
compounds because they contain ionic bonds
o
of the over 10 million
chemical compounds
that have been
discovered, at least
90% are carbon-based
molecular compounds
the properties of
organic compounds
are a result of the
covalent bonds
within their
molecules
o
o
recall, covalent
bonds occur when
the valence electrons
of two non-metals
are shared
 carbon
has the highest
number of bonding
electrons of any element

4 bonding electrons
 because
of its high bonding
capacity, carbon atoms
have the ability to form
chains, rings, spheres,
sheets, and tubes of almost
any size
 carbon
can also form
single, double, and triple
covalent bonds
o
Lewis dot diagrams
are useful for the
small and simple
molecules…
H

but are awkward with the
larger, more complex
molecules found in
organic chemistry.
one way to speed up the
process is to expand the
molecular formula into
clusters of carbons
o
e.g. CH3-CH2-CH2- etc…
 to the right are six different ways of
illustrating the compound C5H12:
 1 – the Lewis dot diagram
 2 – complete structural diagram
 3 – condensed structural diagram
 4 – expanded molecular formula
 5 – simplified structural diagram
 6 – line structural diagram

o
in order to manage the enormous number of
different organic compounds, chemists divide
them into families based what makes the
compound special
 types of bonds
single,
 double or
 triple

H
H H
H
H
H
H H
 Functional


groups
groups of atoms with
elements other than
carbon and hydrogen
these additional groups
are responsible for the
unique properties of some
compound groups

Ex. ethanol CH3–CH2 –OH, has
the hydroxyl functional group
(-OH) which gives the
compound its physical and
chemical properties, such as
higher solubility, &
flammability.
hydrocarbons are compounds containing
only carbon and hydrogen atoms
alkanes are hydrocarbons with only
single carbon-to-carbon bonds
the general formula for all alkanes is
CnH2n+2
o
o
o
as the size (and molar mass) of an alkane
increases, the boiling point increases
 bigger molecules will have MORE BONDS, so
MORE ENERGY is needed to break them apart to
change state

All alkane names end in –
ane
The prefix (beginning of the
word) indicates how many
carbons are in the chain
o
o
o
Ex. Meth = one cartbon,
eth = two, prop= 3, etc…
o
notice that the first four (the
smallest ones) are gases,
the next twelve are liquids,
it isn’t until the molecules
get very large (C17H36)that are
big enough to be solid
o
o
name
formula
methane
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
CH4 (g)
C2H6(g)
C3H8(g)
C4H10(g)
C5H12(l)
C6H14(l)
C7H16(l)
C8H18(l)
C9H20(l)
C10H22(l)
- ane
CnH2n+2
Step #1:
o
identify the longest continuous chain of carbon
atoms ( the parent chain) in the structural
diagram
o
Note: they may not be in a straight line, but all must be
connected
Step #2:
o number the carbon atoms
in the parent chain,
starting from the end
closest to a branch.
Name it! The number of
carbons determines what
the prefix is.
o Only singled bonds, means it
ends in –ane
o
o
See page 9 of your data
booklet for the complete
list of alkanes
four carbons
= butane
three carbons
= propane
Step #3:
o
identify any branches and their location, using a
number on the parent chain.
o
o
o
o
one single
carbon branch
on the second
carbon of the
parent chain
branch names always end in –yl
the prefixes (meth = one carbon, eth = two, prop= 3,
etc…) indicates how many carbons are in the branch
if the compound has more than one of the same type
of branch, the number of BOTH locations is listed, and
a prefix is provided
TWO single
carbon branches
BOTH on the
second carbon of
the parent chain
= 2,2-dimethyl ___
= 2-methyl___
Step #4:
o
write the complete IUPAC name, following this format:
o (number of locations) – (branch name)(parent chain)
* one, two carbon branch on carbon 5 = 5-ethyl
10 carbon parent chain = decane
* four, one carbon branches on carbon 3, 4, 4 & 6 = 3,4,4,6-tetra methyl
o
o
where there are more than one type of branch
group, (some with one carbon; methyl, and others
with two carbons; ethyl) the branches are named in
alphabetical order (E-thyl before M-ethyl)
Now put it all together…
5-ethyl -3,4,4,6-tetra methyl decane
o
Name this compound:
o
Step #1 & 2:
the parent chain is
identified and
numbered, beginning
at the end closest to
the first branch.
o 6 carbons = hexane
o
o
Step #3:
o
o
the first branch is
identified as a single
carbon branch (methyl),
on C#2.
methyl on 2 = 2methyl
_________-2-methyl_________
o
Step #3 cont:
o
o
o
3-ethyl__________________
a second branch is identified
as a two carbon
branch(ethyl) on C#3
ethyl on 3 = 3-ethyl
Step #4:
o
Put it all together!
o
remember, where there are
more than one branch
group, the branches are
named in alphabetical order
o
(ethyl before methyl).


to draw alkanes, use the IUPAC name, starting from the
BACK of the compound name:
 Ex. 3-ethyl-2,4-dimethylpentane
Step #1:

identify the name of the parent chain and the
corresponding number of carbon atoms:


pentane = 5 carbon parent chain
draw the skeleton of the parent chain


Step #2:
identify the type of branches and position them
on the appropriate carbons on the parent
chain.
2,4-dimethyl = TWO single carbon
branches on the second and fourth carbon

Draw them in!
3-ethyl =
ONE double
carbon branch
on the third
carbon
o
Step #3:
o
fill-in any unfilled bond sites with hydrogens.
o Don’t add too many! Remember each carbon
has a MAXIMUM bonding capacity of 4!
 Draw
the complete structural diagram
for:
NOT
 ethane
 4-propyloctane
 2,2,4,5-tetramethylhexane

Practice Problems (page
110)


Q. 1-4 (a & c only)
3.1 Summary (page 121)

Q’s 4, 5 (a,c,e 7 g) & 6
Complete the
Alkanes Worksheet
You must submit your
work by the end of class
A3.2 – Alkenes & Alkynes
o
Alkenes and alkynes are hydrocarbons with
double and triple bonds
o hydrocarbons with double bonds are called
alkenes, and with triple bonds are called
alkynes.
o these compounds are named in a similar way to
alkanes, (using prefixes to indicate the number
of carbon atoms), but their names will end in
–ene and –yne.
ethyne
triple bond
name
formula
name
formula
ethene
propene
butene
pentene
hexene
heptene
octene
nonene
decene
C2H4(g)
C3H6(g)
C4H8(g)
C5H10(l)
C6H12(l)
C7H14(l)
C8H16(l)
C9H18(l)
C10H20(l)
ethyne
propyne
butyne
pentyne
hexyne
heptyne
octyne
nonyne
decyne
C2H2(g)
C3H4(g)
C4H5(g)
C5H8(l)
C6H10(l)
C7H12(l)
C8H14(l)
C9H16(l)
C10H18(l)
o recall the general formula for alkanes was CnH2n+2
o the general formula for alkenes is CnH2n
o and for alkynes is CnH2n-2

the type of bond affects the chemical properties,
such as reactivity, and physical properties, such
boiling point, melting point and solubility.
o
a compound will be MORE STABLE if it is larger
o this is because it has more carbon-carbon bonds to
break during the reaction
OR if it DOES NOT contain multiple bonds
o alkenes and alkynes (double or triple bond) are
relatively easy to break
o this is because it has more hydrogen bonds are a
stabilizing force in a molecule
o
o
MORE STABLE compounds
have:
higher melting point/ boiling
point
o are less reactive
o more likely to be found in
solid form
o
recall the phrase
“like dissolves like”
o
o
water is polar, so it
easily dissolves ionic
compounds, acids,
and polar molecular
compounds
o

o
polar solutes
dissolve easily in
polar solvents.
Ex. salt, hydrochloric
acid or ethanol
alkanes, alkenes and alkynes are NON-POLAR, so they
dissolve better in a non-polar solvent (like oil) than a
polar one, such as water.

Rank the following compounds in order of lowest to
highest boiling point
 pentane,
butane, propane
 propene, propane, propyne

Rank the following compounds in order from least to
most reactive
 ethyne,
nonane, pentene
 octane, heptene, octene

Rank the following compounds in order of lowest to highest
boiling point

pentane, butane, propane

MORE STABLE if it is larger = higher boiling point…must rank
smallest to largest

propane (3 C’s) butane (4 C’s) pentane (5C’s)

propene, propane, propyne


MORE STABLE if it contains ALL single bonds = higher boiling
point …must rank from least to most single bonds
propyne (triple bond)  propene (double bond)  propane
(single bonds)

Rank the following compounds in order from least to
most reactive
 ethyne,

nonane, pentene
MORE STABLE if it is larger and/or if it contains ALL
single bonds = less reactive…must rank from
smallest with most double/triple bonds.
 nonane (largest-9 C’s & single bond) 
pentene
(5 C’s &
 ethyne (smallest-2 C’s & triple bond)
 octane, heptene, octene
double bond)
 octane
(largest-8 C’s & single bond)
double bond)
 octene
(8 C’s &
 heptene (smallest-7 C’s & double bond)
o
Naming alkenes and alkynes is similar to naming
alkanes;
o
using prefixes to indicate the number of carbon
atoms and the ending indicates the type of bond
o alkanes end in –ane
o alkenes end in –ene
o alkynes end in -yne
Since the location of a multiple bond affects the
chemical and physical properties of a compound
the name of the compound must specify the
location of the multiple bond
o

pent-1-ene indicates
the double bond is
located between the
1st and 2nd carbons of
the parent chain

pent-2-ene indicates
the double bond is
located between the
2nd and 3rd carbons
of the parent chain
o
the rules for naming
are similar to
alkanes, with three
additional rules:
 Step #1: find the
longest continuous
carbon chain; this is the
parent chain. Use the
number of carbons to
determine which prefix

for alkenes and alkynes
the parent chain must
contain the multiple
bond
five carbons = pentdouble bond = - ene
1
2
3
4
5
 Step #2: to
determine the
location of the
double/triple bond,
the parent chain is
numbered

bond between 2 & 3 = _____-2- ______
for alkenes and
alkynes you start
numbering from the
end closest to the
multiple bond (not
closest to the first
branch as for
alkanes)
 Step #3: identify
any branches and
their location,
using a number on
the parent chain
(just like with
alkanes).
o
o
count from the
SAME side you
started numbering
the bond from
branch names
always end in –yl
Two carbon branch on third
carbon = 3-ethyl _________

Step #4:
o
write the complete IUPAC name, following this format:
o (number of locations)–(branch name)(prefix of carbon
chain-location of bond- suffix of parent chain)
o
3-ethylpent-2-ene
five carbons = pentdouble bond = - ene
bond between 2 & 3
Two carbon branch on third
carbon
 Name
the following compound
five carbons = penttriple bond = - eye
bond between 2 & 3
identify any branches
single carbon branch on fourth carbon
= 4-methyl
 Put it all together



to draw alkenes and alkynes, use the same rules as you did
for drawing alkanes,
use the IUPAC name, starting from the BACK of the compound
name:
 Ex. 2,5-dimethylhept-3-yne
Step #1:

identify the name of the parent chain,

the corresponding number of carbon atoms AND the type/ location of
the bond
2,5-dimethylhept-3-yne


hept- = 7 carbon parent chain,
3-yne is a triple bond on the third carbon

Draw the skeleton
2
1

3
4
5
6
7
Remember the maximum bonding capacity of carbon
is 4… so that will affect the number of hydrogens on
some of the carbons in the parent chain.

Since there are THREE bond in between the C’s and
one, connecting to the rest of the chain there is NO
ROOM for any hydrogens.


Step #2:
identify the type of branches and position them
on the appropriate carbons on the parent
chain.
2,5-dimethylhept-3-yne
2,5-dimethyl = TWO single carbon
branches on the second and fifth carbon

Draw them in!
H
3
2
1
C
H
H
4
H
6
5
C
H
H
7
o
Step #3:
o
fill-in any unfilled bond sites with hydrogens.
o Don’t add too many! Remember each carbon
has a MAXIMUM bonding capacity of 4!
H
2
H C H
H
3
4
5
H C H
H
6
H
H
H
H
H
1
H
2
1
7
H
H C H
H
3
4
5
6
H C H H
H
7
H
o
Isomers are compounds
containing the same
number of carbons and
hydrogens but in different
arrangements
o Isomers have the same
formula, but different
structural diagrams.
C7H 16
-di
C7H 16

Draw the following isomers
 pentane
 2-methylbutane
 2,2-dimethylpropane

What is the chemical formula for each isomer?
o
if an organic compound has carbon-carbon
double bonds, it is said to be unsaturated.
vs.
o
the addition of hydrogen molecules, to an
unsaturated is called hydrogenation and results in a
saturated hydrocarbon
o the term saturated means “full” so think of
alkanes are being “full” of hydrogens, where
alkenes and alkynes ARE NOT
o
o
to become saturated, the double (or triple) bond breaks, and
the hydrogen atoms take the place of the multiple bond
this reaction is called an addition reaction
H
H
o
H
H
there must always be enough hydrogens in an addition
reaction to fill all the unbonded sites.
o
Alkynes have triple bonds so TWO hydrogen molecules (4
hydrogens) are required to fill the bonding sites
+2
o
alkenes and alkynes also tend to react
readily with other small diatomic
molecules such as the halogens
o F2 , Cl2, Br2 , & I2
o Like with a hydrogen
addition reaction, some
of the bonding sites once occupied by multiple
carbon-carbon bonds become filled with a
halogen
Cl Cl
Cl - Cl
o
The terms saturated and
unsaturated are often
used in combination with
“fats”
o
o
animal fats and plant oils
can be a good source of
energy.
fatty acids (aka fats) are
organic molecules made of
 a long chain of carbons
with a -COOH group at
one end
and a methyl group at
the other end
o
Recall that the structure
of a molecule affects it
physical and chemical
properties
oils tend to be unsaturated
fats—(containing double or
triple bonds)
o this causes the molecule to
bend, which is why the
molecules can pack together
as tightly.
o This is why they tend to be
liquids at room temperature
o
Fats (either manmade or from animals)
tend to be saturated fats—(containing
ONLY single bonds)
o this causes the molecule to straight, so
they can pack tightly together.
o This is why they tend to be solids at
room temperature.
o Their solid sate AND the fact that they
are alkanes (containing all single bonds)
mean they are VERY stable
o
o
this means they
have HIGH boiling/
melting points
o
monounsaturated fats –have
one double bond (and two
fewer hydrogens than a
saturated fat)
o
polyunsaturated fats – have
more than one double bond, (so
even fewer hydrogens)
unsaturated fats are better for the
body
o hydrocarbons with double or triple
bonds are more reactive/easy to
break down = less work for our
bodies to break down
o saturated fats are harder for your
body to break down, so they stick
around and cause “body fat”.
o
o
essential fatty acids are fats your body can’t make, and
are found only in certain foods
 “omega-6” fatty acids are found in margarine, processed
foods, corn oil and soybean oil
 while these fats are important for clotting and
inflammation, many people eat too many omega-6s
 “omega-3” fatty acids are found in some fish, flax seeds,
and enriched eggs
 these fats have the opposite effect as omega-6s – they
reduce swelling and slow blood clotting, so a balance
between the two is important
“Cis” and “Trans” refer
to different 3D
arrangements of a
molecule
“Cis” fats are naturally
occurring, but can be
made into “Trans” fats
through a hydrogenation
o
o
o
the addition reaction
involving hydrogen.
o
o
o
Because they are synthetic, trans fats are the
hardest for your body to break down, and are the
worst kinds of fat.
Trans fats are found in many processed foods and
fast foods because
 they are in solid form, so they are easier to work
with than oils
 they are softer and easier to spread that typical
animal fats
 you need less trans fats to have the same effect
on a food, so they qualify for a “low fat” label,
even though the fats are much worse for you
Because food manufacturers are allowed to put up
to 0.5g of trans fats in their food and still label it
as “trans fat free”, be sure to check ingredients
labels for partially hydrogenated oils
Practice Problems (pg 129)
Q. 28 & 29
** CAUTION! Your textbook uses
the old IUPAC naming system,
where the location of the bond
goes BEFORE the parent chain.
You must use the current method:
Ex. A hydrocarbon with a
double bond between
the 1st and 2nd carbon: you
SHOULD write hex-1-ene,
your textbook writes it as
1-hexene


Practice Problems (pg 132)
Q’s 31

3.2 Summary (pg 136)
Q’s 3 & 4, 8 &10
Complete the
Try This Activity
Building Models of
Hydrocarbons
A3.3 &3.4 - Industrial
Processes Involving Organic
Compounds


Use the “Fractional
Distillation” applet, to
complete the Utilizing
Technology: From Petroleum to
Gasoline
Complete the Ch 3.4-Everyday
uses of Hydrocarbons Inquiry
Questions