Topic 20
High Level Organic Chemistry
IB HL Objective
• 20.1.1 Deduce structural formulas for
compounds containing up to six carbon atoms
with one of the following functional groups:
amine, amide, ester and nitrile.
• 20.1.2 Apply IUPAC rules for naming
compounds containing up to six carbon atoms
with one of the following functional groups:
amine, amide, ester and nitrile.
Amines (R-NH2)
Prefix: aminoOr for the first three carbons:
Suffix: -ylamine
H
R
N
Examples: CH3CH2NH2 ethylamine
CH3CH(NH2)CH2CH2CH3 2-aminopentane
H
Amide (-CONH2)
Suffix: -anamide
Example: CH3CONH2 ethanamide
O
R
C
NH2
Ester (-COOC-)
Because esters are produced from alcohols
heated with carboxylic acids…
The chain from the alcohol provides first part of
the name, and the chain from the carboxylic
acid is given the name of the acid anion.
Example: CH3-CO-O-CH3 methyl ethanoate
Nitriles (R-CN)
Used to be called cyanides.
Suffix: -nitrile added to the hydrocarbon forming
the basis of the acid.
(Sometimes you will see the prefix cyano-)
Example: CH3CH2CN propanenitrile
CH3CN
ethanenitrile
IB HL Objective
• 20.2.1 Explain why the hydroxide ion is a
better nucleophile than water.
20.2.1 Explain why the hydroxide ion is a better
nucleophile than water.
• Nucleophilic: What’s in a name?
• Nucleo—nucleus philic--loving
• So nucleophiles are molecules which are
attracted to more positive charge.
• Nucleophiles will then donate electrons.
20.2.1 Explain why the hydroxide ion is a better nucleophile
than water.
Water vs. Hydroxide
H2O and OH• When a carbon is attached to a more
electronegative atom, it becomes polar.
• Hydroxide is negatively charged, therefore it is
attracted to the partial positive charge on a
carbon.
IB HL Objective
• 20.2.2 Describe and explain how the rate of
nucleophilic substitution in halogenoalkanes
by the hydroxide ion depends on the identity
of the halogen.
• 20.2.3 Describe and explain how the rate of
nucleophilic substitution in halogenoalkanes
by the hydroxide ion depends on whether the
halogenoalkane is primary, secondary or
tertiary.
Objectives 20.2.2 & 20.2.3
• One group look at the polarity differences
between different halogens and carbon bonds.
Predict which halogen would most likely react.
• One group look at the primary, secondary and
tertiary structure with one type of halogen, and
describe how the polarity would differ. Predict
the mechanism for each and which would react
fastest.
• When time is up, you will share your findings
with the other group.
Objectives 20.2.2 & 20.2.3
•
•
•
•
•
Summary
(Different from what you predicted?)
Polarity of the molecule is not as important as the
strength of the carbon-halogen bond.
Iodine is the least polar, but the most reactive,
since the carbon-iodine bond is the weakest.
Reactivity of halogenoalkanes depends on the
ability of the halogen bond to break and leave,
which is why the halogen is referred to as the
leaving group.
Iodine is the best leaving group, and fluorine is the
worst.
You learn best from your mistakes.
Objectives 20.2.2 & 20.2.3
• Primary halogenoalkanes proceed by an SN2
mechanism.
• Tertiary halogenoalkanes proceed by an SN1
mechanism.
• Secondary halogenoalkanes can proceed by either SN1
or SN2.
• Experimentally, the SN1 reactions are found to be faster
than the SN2 reactions.
• This could be due to the activation energy being less to
form a carbocation then the activation energy needed
to form the transition state (five bonds) for SN2.
Objectives 20.2.2 & 20.2.3
•
•
•
•
Review
What would be more likely to react, 1fluorobutane or 1-bromobutane, and why?
What is the order, in increasing rate of reaction,
for the primary, secondary, and tertiary
halogenoalkanes?
A: 1-bromobutane is more likely to react because
of the lower strength of the bromine-carbon
bond.
A: primary>secondary>tertiary
IB HL Objective
• 20.2.4 Describe, using equations, the
substitution reactions of halogenoalkanes
with ammonia and potassium cyanide.
• 20.2.5 Explain the reactions of primary
halogenoalkanes with ammonia and
potassium cyanide in terms of the SN2
mechanism.
Objectives 20.2.4 & 20.2.5
• Reaction with ammonia forms a primary amine.
• Reaction with the cyanide ion forms a nitrile.
• Group 1: Show iodoethane reacting with
ammonia. State the mechanism and show
transfer of electrons using curly arrows. Name
the new molecule.
• Group 2: Show iodoethane reacting with a
cyanide ion. State the mechanism and show the
transfer of electrons using curly arrows. Name
the new molecule.
Objectives 20.2.4 & 20.2.5
H
H C
H
I
C
H
H
SN2 Mechanism
+
NH3
H
C
H
H
H C
H
NH2
Ethanamine or ethylamine
I-
cyanide ion with iodoethane (SN2)
H
CH3
+
C
H
I
CH3
H
CN
CN
H
-
I
-
propanenitrile
SN2
S (substitution)
C
Notice how we increased the length of the carbon
chain. It went from an ethane to a propane.
N(nucleophilic)
2(species reacting
in the slowest step)
IB HL Objective
• 20.2.6 Describe, using equations, the
reduction of nitriles using hydrogen and a
nickel catalyst.
20.2.6 Describe, using equations, the reduction
of nitriles using hydrogen and a nickel catalyst.
• Due to lengthening the carbon chain with the
cyanide ion, we can now make an amine with
an extra carbon by reacting with hydrogen and
a nickel catalyst:
H2/Ni
CH3CH2CN → CH3CH2CH2NH2
IB HL Objective
• 20.6.1 Describe stereoisomers as compounds
with the same structural formula but with
different arrangements of atoms in space.
20.6.1 Describe stereoisomers as compounds with the same
structural formula but with different arrangements of atoms in space.
Isomers
Structural
(Different
Bonding)
Stereo
(Different shape)
Stereoisomers
The order in which they are bonded is the same
(same structural formula), but they have a
different arrangement.
20.6.1 Describe stereoisomers as compounds with the same structural
formula but with different arrangements of atoms in space.
• Two different types of stereoisomers….
• Geometrical isomers
• Optical isomers (enantiomers)
Geometrical isomers occur when bonds are
unable to rotate freely, also known as
restricted rotation.
Optical isomers occur when there are four
different atoms or groups attached to a single
carbon atom. These molecules are often
called asymmetric or chiral.
IB HL Objective
• 20.6.2 Describe and explain geometrical
isomerism in non-cyclic alkenes.
20.6.2 Describe and explain geometrical isomerism in
non-cyclic alkenes.
Molecular Models
1. Two people make but-1-ene.
2. Two people make but-2-ene.
3. One person make 1,2-dichloroethene
Can you make the two but-1-ene look different no
matter which way you turn it?
What about the but-2-ene?
What are the two differences in the models?
20.6.2 Describe and explain geometrical isomerism in
non-cyclic alkenes.
• Same side = cis
• Opposite side = trans
• They exist separately because the double bond
cannot be rotated.
• One bond is a sigma bond and one is a pi bond.
Pi bond is formed by the sideways overlap of the
p orbitals on each carbon atom. Rotation would
involve breaking this pi bond.
IB HL Objective
• 20.6.3 Describe and explain geometrical
isomerism in C3 and C4 cycloalkanes.
• Seunghwan: make a molecular model of cis1,2 dichlorocyclopropane.
• Nibras: make a molecular model of trans-1,2dichlorocyclopropane
IB HL Objective
• 20.6.4 Explain the difference in the physical
and chemical properties of geometrical
isomers.
• Look at study guide for diagrams and
explanations as well.
20.6.4 Explain the difference in the physical and
chemical properties of geometrical isomers.
• Geometric isomers have different physical properties
such as polarity, which gives rise to differences in
boiling point, melting point, solubility, etc.
• For example, the boiling point of cis-1,2-dichloroethene
is 60°C, whereas trans-1,2-dichloroethene is 48°C. This
is because the cis isomer is more polar.
• However, in but-2-ene-1,4-dioic acid, the trans has
higher intermolecular hydrogen bonding between
different molecules, because of the polarity of
carboxylic acid groups. The cis-isomer reacts when
heated to lose water and become a cyclic acid
anhydride.
IB HL Objective
• 20.6.5 Describe and explain optical isomerism
in simple organic molecules.
• Make a model of butan-2-ol
• Make a model of 2-bromobutane
20.6.5 Describe and explain optical isomerism in simple
organic molecules.
Optical Isomers or Enantiomers
• If a carbon atom has four different groups
attached to it, then there are two different
ways in which these groups can be arranged.
• This is known as an asymmetric carbon atom,
or a chiral centre.
IB HL Objective
• 20.6.7 Compare the physical and chemical
properties of enantiomers.
20.6.7 Compare the physical and chemical properties
of enantiomers.
• Enantiomers are so similar, there is very little
difference in physical and chemical properties.
• Chemical differences can arise when enantiomers
interact with other optically active substances.
• Chemical reactions which produce a chiral carbon
often contain equal amounts of the two
enatniomers. This is called a racemic mixture.
• The only difference in the physical properties is
their interaction with polarised light.
• Find some examples in your books/study guides of
where different enantiomers can cause different
effects, and examples of enantiomers that are
non-racemic.
IB HL Objective
• 20.6.5 Outline the use of a polarimeter in
distinguishing between optical isomers.
20.6.5 Outline the use of a polarimeter in distinguishing
between optical isomers.
• Optical isomers rotate the plane of planepolarized light in opposite directions.
• A polarimeter can be used to see how
enantiomers rotate the plane of planepolarized light.
• You need a light, two polarizing lenses, and a
tube in between the lenses to hold the sample
of the enantiomer.
20.6.5 Outline the use of a polarimeter in
distinguishing between optical isomers.
• If the analyzer has to be rotated clockwise, the
enantiomer is dextrorotatory (D), or dextro for
short. From the Latin word meaning “right”.
• If the analyzer has to be rotated counterclockwise, then the enantiomer is laevorotatory
(L), or levo for short. From the Latin word
meaning “left”.
• If both enantiomers are present, they cancel each
other out, and it appears optically inactive. This
indicates a racemic mixture.