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Organic Chemistry
Notes
SL Chemistry
Intro
1. Organic chemistry is the study of compounds containing carbon.
2. Hydrocarbons are compounds of C and H. The simplest hydrocarbon is methane, CH4,
which is used for heating and in gas stoves for cooking.
3. Saturated hydrocarbons have only single bonds. (Saturated means “full.” Saturated
hydrocarbons are “full” of hydrogen.)
4. Unsaturated hydrocarbons have double or triple bonds.
10.1.1 Define the features of a homologous series
1. differ by a CH2 group.
2. Members can be described by a general formula.
(a) A homologous series is a “family” of compounds. Example: alkanes are
hydrocarbons with only single bonds.
(b) The first two alkanes are CH4 and C2H6. CH4 plus CH2 is C2H6, so they differ by
a CH2 group.
(c) The general formula is CnH2n+2
10.1.2 Predict and explain the trends in boiling points of members of a homologous series
(a) As the number of carbons increase, boiling points increase. C4H10 has a higher
boiling point than C3H8.
(b) Why? Higher molecular mass causes stronger Van der Waal’s forces, which
make the boiling point higher. The increase is larger at first (from methane to
ethane) because there is a larger increase in chain length.
10.1.3 Distinguish between empirical, molecular and structural formulas
Three types of structural formulas:
1. Full Structural formula: Shows all bonds as a line.
H
H
H
H
H
C
C
C
C
H
H
H
H
H
2. CH3CH2CH2CH3 – condensed structural formula.
3. Stereochemical formula – a representation of the 3-D structure. These formulas
allow you to identify stereoisomers
a. The lines show bonds in the plane of the page
b. The dark triangle represents a bond coming out of the page
c. The dashed line represents a bond going out of the back of the page
Br
Cl
Cl
O
H3C
H
C
H
H
H
I am trying to show a that three atoms of the tetrahedral make a
plane, and the other two stick out of the page – either behind or in front.
2. C4H10 – molecular formula, gives only the number of atoms in the molecule
3. C2H5 – empirical formula, gives only the ratio of elements
Example
a. Here is the structural formula for a chemical.
H
H
H
C
H
H
C
C
H
H
H
C
C
C
H
O
H
H
H
H
H
b. How do you write the condensed structural formula of the above
molecule? Condensed structural formula (CH3)2CHCH2CH(OH)CH3 –
When there are ( ), the group inside is attached to the carbon before it.
10.1.4 Describe structural isomers as compounds with the same molecular formula but with
different arrangement of atoms.
Structural isomers are compounds that have the same molecular formula, but different
arrangement of atoms.
 Example: butane and 2-methyl propane both have the molecular formula C4H10.
 Properties – Isomers have different physical and chemical properties. The more
branches there are on a compound, the lower its boiling point.
1. carbon most often bonds to the following
a. Hydrogen – makes 1 bond
b. Nitrogen – makes 3 bonds, one lone pair
c. Oxygen – makes 2 bonds, two lone pairs
d. Group 7 Halogens: make 1 bond, three lone pairs.
2. When determining organic formulas – follow the rules above so that you don’t have to
worry about counting electrons to determine the Lewis structure
Naming:
 IUPAC names have three parts: stem prefix, suffix, side chain prefix.
a. Number of carbons in main chain given by stem prefix (meth, eth, etc.)
b. Functional group given by suffix (-ane, -ene, etc.)
c. Side chain/substituent group: prefix (methyl-, ethyl-, etc.)
#C
1
2
3
4
5
6
Stem
prefix
MethEthPropButPentHex-
The suffix for each functional group.
Homologous series
Suffix
Alkane
-ane
Alkene
-ene
Alcohol
Organic acid / Carboxylic
acid
-anol
-anoic acid
Aldehyde
-anal
Ketone
-anone
Functional
group
Single bonds
C=C
-OH
-CHO
-CO
-COOH
Side chains have a prefix
Side chain
Prefix
-CH3
Methyl
-CH2CH3
Ethyl
-CH2CH2CH3
Propyl
-F, -Cl, -Br, -I
Fluoro, chloro, bromo, iodo
-NH2
Amino
-OH
hydroxy
Prefix when more than one substituent group is present
Number of
Prefix
Substituent
Groups
1
-2
Di
3
tri
4
tetra
10.1.5 Deduce structural formulas for the isomers of non-cyclic alkanes.
1. When drawing isomers, first draw the straight chain alkane.
2. Next, shorten the main chain by one carbon and try moving it to another carbon in the
chain.
3. Repeat this to get different structures. To make sure you didn’t draw the same
structure twice, name each structure. If you get the same name, it is the same
structure.
10.1.6 Apply IUPAC rules for naming the isomers of non-cyclic alkanes.
1. Naming alkanes. The following hydrocarbon’s name is “2-methylbutane.”
a. Number carbons on the main chain so that the substituent group gets the lowest
number. Do you number from the left or right? In this case, If we numbered from the
left, the substituent group would be number 3. 2 is smaller than 3, so we number
from the right.
b. Name the substituent group with the number of the main chain it is on and the prefix
for the number of carbons in the substituent group. The substituent group is called
“2-methyl” because meth means one carbon and 2 means it is on the second carbon
of the main chain.
c. Give the main chain a prefix for the number of carbons.
10.1.7 Deduce structural formulas for the isomers of the straight-chain alkenes up to C6.
10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6.
10.1.9 Deduce structural formulas for alcohols, aldehydes, ketones, carboxylic acids and
halogenoalkanes.
10.1.10 Name alcohols, aldehydes, ketones, carboxylic acids and halogenoalkanes
Naming Functional Groups
Functional Groups that have a prefix
1. Halogenoalkanes
a. Name the longest carbon chain
b. Give a prefix for the halogen:
i. Flouro
ii. Chloro
iii. Bromo
iv. Iodo
c. Number the location of the halogen so that it has lowest # possible
d. If more than one halogen, use di/tri prefix
Functional Groups that have a different ending than -ane
2. Alkenes
a. Find longest carbon chain
b. Add –ene for a double bond
c. Number the location of the double bonds so it has lowest # possible
d. Don’t need a number for ethene and propene because 2-ethene is the same as
1-ethene.
3. Alcohols
a. Name longest chain
b. Use ending –anol
c. Number carbons to figure out what carbon the –OH group is on
d. Number can go at the front or right before ending. The latter is often used
when there is more than one functional group on a molecule
OH
OH
propan-1-ol
or: 1-propanol
propan-2-ol
or: 2-propanol
OH
pentan-3-ol
or: 3-pentanol
i. Classification
1. primary 1 - attached to 1 R group (at the end of a carbon chain)
2. secondary 2 - attached to 2 R groups (at a branch of a carbon
chain)
3. tertiary 3 - attached to 3 R groups
R
R
H
R C O
H
H
H
C
primary
secondary
R
O
R
H
C
O
R
H
tertiary
4. Aldehydes/Ketones
a. Name longest chain
b. If carbonyl (C=O) is
i. on the end: Use ending –anal
1. no number needed unless higher priority functional group is on
molecule
ii. in the middle: Use ending -anone
1. Number carbons to figure out what carbon the C=O group is on
2. Number can go at the front or right before ending. The latter is
often used when there is more than one functional group on a
molecule
O
O
H3C
CH 3
H
2-propanone
propanal
ethanal
Ketone
Aldehyde
5. Carboxylic Acids
a. Name longest chain
b. Use ending –anoic acid
O
O
H3C
OH
ethanoic acid
OH
propanoic acid
1. Alkenes are hydrocarbons with a functional group of one double bond.
10.1.11 Identify the following functional groups when present in structural formulas: amino,
benzene ring and esters.
1. Benzene has a unique structure that you may see as part of larger organic
structures
H
H
C
H
C
C
C
C
H
C
H
H
2. Ester functional group R-COO-R’
3. Amino functional group: R-NH2-R’
10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and
halogenoalkanes.
1. A primary carbon is attached to one functional group and 2 or 3 hydrogens.
CH3CH2OH (ethanol) has a primary carbon.
2. A secondary carbon is attached to one functional group and two alkyl groups.
(Alkyl group is a chain with a carbon).
3. A tertiary carbon is attached to one functional group and three alkyl groups.
10.1.13 Discuss the volatility and the solubility in water of compounds containing the
functional groups listed in 10.1.9.
1. Volatility is a measure of tendancy of a substance to change to the gaseous state. High
volatility = low boiling point.
a. Weak intermolecular forces = high volatility. So you have to evaluate 1. the type of
intermolecular forces and 2. the strength of the intermolecular forces.
① Strength of intermolecular forces
Intermolecular Force Type
Bonding Type
Hydrogen bonding
H bonded to N, O or F
Dipole-dipole interactions
Polar
Van der Waals forces
Nonpolar
Strength of forces from strongest to weakest: Hydrogen bonding, dopole-dipole
interactions, Van der Waals’ forces.
② More polar bonds = stronger dipole-dipole/hydrogen bonds.
Example: discuss the volatility of ethanoic acid and ethanol. Answer: Ethanol is
probably more volatile because it has weaker intermolecular forces. Both ethanol
and ethanoic acid have hydrogen bonding, but ethanoic acid has two oxygen
atoms bonded, which makes the dipole-dipole interactions stronger.
③ Higher molecular mass, stronger Van Der Waals.
Example: which chemical has a lower boiling point, pentanol or hexanol?
Pentanol has a lower molecular mass, so it would have a lower boiling point.
④ More branches (substituent groups) = weaker Van der Waal’s forces because the
molecules do not pack as closely.
Example: the chemical on the right would be more volatile because the Van der
Waal’s forces will be weaker. They have the same molecular mass, but the
brached molecule will not be as closely packed.
Volitility of functional groups:
Most volatile
least volatile
Alkane, halogenoalkane, aldehyde, ketone, alcohol, carboxylic acid
Increasing strength of intermolecular forces 
Increasing boiling point 
Solubility
1. “Like dissolves like”
a. Polar chemicals are soluble in polar solvents
b. Nonpolar chemicals are soluble in nonpolar solvents.
c. Water is a polar molecule, so polar substances are soluble and nonpolar substances
are not.
Group
Polarity
Alkanes
CnH2n+2
Alkenes
CnH2n
Haloalkanes
R–X
Alcohols
R- OH
Non-polar
Soluble in
Water
No
Non-polar
No
Slightly Polar –
non-polar
Polar bond capable
of H-bonds
Slightly
Aldehydes
R-CHO
Polar carbonyl
group (C=O)
Yes,
decreases w/
length of
carbon chain
Yes,
decreases w/
O
R
H-bond can form
between carbonyl
and water
length of
carbon chain
Polar carbonyl
group (C=O)
H-bond can form
between carbonyl
and water
Yes,
decreases w/
length of
carbon chain
Polar, capable of
hydrogen bonds
Yes decreases
w/ length of
carbon chain
H
Ketones
R-CO-R
O
R
R
Carboxylic
Acids
R-COOH
O
R
OH
Combustion 10.2.2 and 10.4.1
Combustion of hydrocarbons: burning of hydrocarbon in presence of oxygen
gas producing heat, carbon dioxide and water. ALWAYS EXOTHERMIC!
1. complete combustion occurs when there is sufficient oxygen available
a. CxHy + O2  CO2 +H2O
b. CxHyOz + O2  CO2 +H2O
c. complete combustion = blue flame
2. incomplete combustion occurs when there is insufficient oxygen
available
a. CxHy + O2  CO +H2O
b. CxHy + O2  C + H2O
c. incomplete combustion = smoky, yellow flame
d. The black soot is carbon and the yellow flame comes from glowing carbon
atoms
e. CO is toxic for humans b/c it binds to hemoglobin in blood like oxygen – but
unlike oxygen, it is never released and can cause suffocation if enough sites on
the hemoglobin molecule are blocked.
f. Particulates in the air from unburned hydrocarbons can also impair breathing
ability.
3. How to balance combustion reactions
a. Balance the carbons
b. Balance the hydrogens
c. If you need an odd number of oxygens, double all coefficients
Example
1. _____ C4H10 + _____O2  __4___CO2 + _____H2O
2. _____ C4H10 + _____O2  __4___CO2 + ___5__H2O
3. Need 13 oxygens – so give O2, coefficient of 13 and double all other coefficients
___2__ C4H10 + __13___O2  __8___CO2 + __10___H2O
Organic Reactions
1. 3 basic reaction types
a. Addition Reactions
i. Hydrogenation
1. saturating an unsaturated carbon chain
2. alkene to alkane
ii. Hydration: alkene to alcohol
iii. Halogenation: alkene to halogenoalakane
iv. Polymerization – Alkene to polymer
b. Substitution Reactions (like single or double replacement reactions where one
atom/ion/functional group is replaced by another)
i. Alkane to halogenoalkane
ii. Halogenoalkanes to alcohols: SN1, SN2
2. Basic process of organic reactions is through attraction of positively and negatively
charged parts of molecules
a. Organic has special names for positively and negatively charged parts of a
molecule




Electrophiles
“loves electrons” = attracted to
negative charge
maybe positively charged or
have deficit of electrons b/c
atom is attached to very
electronegative atom
carbon of carbonyl group
acids



Nucleophiles
“loves nuclei” = attracted to
positive charge
often negatively charged or
lone pairs
high electronegativity




alkenes
Hydroxide –OH
Chloride –Cl
Ammonia – NH3

b. many organic reactions happen through the attraction of electrophiles for
nucleophiles
c. in reaction mechanisms, curly arrows show how electrons move – generally
electrons from nucleophile move to electrophile
3. Low reactivity: Alkanes have low reactivity because they have nonpolar CC and CH
bonds. But they can react.
4. High reactivity:
a. Alkenes have pi bonds in which electrons are easily reached.
b. Other functional groups have highly electronegative atoms like O, N or
halogens
5. The table below gives the characteristic reactions for several functional groups. The
name of the rxn is given followed by what it makes in ( )
Functional
Addition
Substitution
Group
Alkane
Halogenation
(haloalkanes)

Alkene




Hydrohalogenation
(mono-haloalkanes,)
Hydration (alcohols)
Halogenation (dihaloalkanes)
Hydrogenation
(alkanes)
Oxidation (-OH,
C=O, COOH)
Alcohol
Oxidation
(aldehyde, ketone,
COOH)
-OH
Halogenoalkanes
*All do combustion reactions, remember.
Reactions (10.2.3-4, 10.3.1-3, 10.4.2-3, 10.5.1-2)
1. Halogenation of alkane
a. Alkane + halogen gas  haloalkane
b. Need ultraviolet light for rxn to occur
c. Depending on time and amount of reactants, more than one halogen can added
to the alkane
d. Also see Nucleophilic Substitution Notes
H
H
H
+
Cl Cl
H
h
Cl
H
H
H
methane
chloromethane
h
H
Cl
Cl
H
h
H
dichloromethane
Cl
Cl
h
Cl
Cl
chloroform
also: trichloromethane
Cl
Cl
Cl
tetrachloromethane
also: carbontetrachloride
e. Reaction occurs through homolytic fission to form a free radical
i. Free radical is an element or molecule with an unpaired electron
ii. Homolytic fission
1. Fission means splitting apart
2. Homolytic means the bond is split in half – each side takes 1
electron and 2 free radicals are formed
iii. Formation of free radicals often results in chain reactions – reaction
keeps occurring until all reactant is used up. See polymerization notes
form mechanism.
2. Hydrohalogenation
a. Alkene + acid halide  monohaloalkane
b. Halide ion adds to larger side (more substituted side of alkene)
i. Hydrohalogenation of ethene
H
H
H
H
+
H Cl
H Cl
H
H
H H
ii. Hydrohalogenation of 1-propene: notice that the chlorine adds to the
larger side of the alkene.
H
H
H
CH3
+
H Cl
H Cl
H
H
H CH3
3. Hydration
a. Alkene + water in acidic solution  alcohol
b. Acid acts as catalyst in rxn
c. –OH group adds to larger side (more substituted side) of alkene
d. Uses: hydration is used for commercial manufacture of ethanol
i. Hydration of ethene
H H
H
H
acid
H
H
+
H
O
H
H
O
H
H
H
ii. Hydration of 1-propene
H
H
H
CH3
acid
+
H
O
H H
H
H
O H
H CH3
4. Halogenation
a. Alkene + halogen gas  1,2-dihaloalkane
b. Diatomic gas has two atoms – both add to opposite sides of the double bond
(and opposite sides of the molecule)
c. Uses: Chlorine + ethane  1,2-dichloroethane: used as starting material for
PVC
d. Uses: Br2 dissolved in dichloromethane is used to distinguish between alkenes
and alkanes. If reddish-brown color of Br2 disappears when added to unknown,
the unknown has alkenes in it.
H
H
Cl Cl
Cl
Cl
+
H
H
H
H
H H
5. Hydrogenation
a. Alkene + Hydrogen gas (with catalyst)  alkane
b. Also called reduction (carbon is reduced in this reaction but is also reduced in
many of the reactions above)
c. Heterogeneous Catalyst: Ni (rxn occurs on a metal surface)
d. Uses: unsaturated vegetable oils are saturated to produce saturated fats (more
solid at room temp than unsaturated) for margarines
H
H
H
H
PtO2/Pd
+
H H
H H
H
H
H H
6. Oxidation of Alcohols: 10.4.2, 10.4.3
a. Primary Alcohol + oxidizing agent  Aldehyde  (longer time) Carboxylic acid
i. Distill to get aldehyde: heat up the solution and collect the vapor
ii. Heat under reflux to get carboxylic acid: heat under reflux. To heat
under reflux you need to attached a condensing tube to your flask so
that vapors condense before they can escape.
Distill to get aldehyde
Reflux to get COOH
water out
b. C
aldehyde vapor
o
m
p
l
e
boling solution
t
e
condensing tube
condensing tube
water in
water in
water out
vapor condenses
and falls back
into flask
condensed
aldehyde out
boling solution
c. Primary Alcohol
i. Primary alcohol + oxidizing agent  carboxylic acid
acid
O
K
Cr
O
+ 2 2 7
OH
propan-1-ol
OH
propanoic acid
ii. Partial Oxidation primary alcohol + oxidizing agent  aldehyde
acid
O
K
Cr
O
+
OH
2
2 7
propan-1-ol
H
1-propanal
d. Secondary alcohol
Secondary alcohol + oxidizing agent  ketone
acid
+ K2Cr2O7
OH
butan-2-ol
O
butan-2-one
e. Tertiary alcohols do not oxidize because the C-C bond would need to break,
which takes much more energy than breaking a C-H bond.
Substitution of Halogenoalkanes
1. Substitution reactions are like single replacement reactions – one ion or atom takes
the place of a functional group on a molecule
2. 2 Kinds of mechanisms
a. SN1 (Substitution (S), Nucleophilic (N), Unimolecular (1))
i. Unimolecular = 1 (Rate only depends on concentration of 1 reactant)
ii. Occurs with tertiary and secondary haloalkanes – more likely with tertiary
because ion formed is more stable.
iii. Takes place in 2 steps
iv. Results in mixture of optical isomers
v. Mechanism
1. SLOW (Rate determining step) Bromine spontaneously leaves
with electrons from the bond.
a. This makes a carbocation and bromide ion.
b. This is the rate determining step
2. FAST: Nucleophile attaches to carbon intermediate that is formed
H
C
H3CH2C
H CH3
CH3
Br
Rate
determining
step: spotaneous
dissociation of
leaving group
C
+
+
H3CH2C
Transition State:
Formation of
Carbocation
OHBr-
HO C
Very fast step:
reaction of
nucelophile and
carbocation
CH3
H
CH2CH3
b. SN2:
i. Bimolecular (Rate depends on conc. of 2 reactants)
ii. Occurs with primary and secondary haloalkanes – faster with primary
because reaction is more hindered with secondary haloalkanes.
iii. Takes place in one step
iv. Mechanism:
1. Negatively charged hydroxide ion attatches to carbon with
halogen.
2. The halogen handles more negative charge better and leaves the
carbon as a halide ion.
H
OH- +
H3CH2C
CH3
C
H CH3
Br
HO
C
Br
HO C
H3CH2C
Transition State:
As OH- attaches,
Br- leaves
Works Cited
Some notes are modified from the following:
Wiseman. “IB Chemistry.” M. Wisemen’s Page. Accessed Apr. 20, 2012.
http://www.mwiseman.com/courses/chem_ib/
CH3
H
+
CH2CH3
Br-