Download organic synthesis handout

Organic Synthesis
How do we choose between reactions?
Br
How can we form this?
CH CH2 CH3
CH
Substitution?
No; the pi bond will break
before Br contributes to a CH2 CH CH2 CH3
substitution reaction
Br
We could use hydrogenation C C CH CH
2
3
of 1-bromo-1-butyne
(provided only one of the Br
pi bonds break)
CH2 CH CH2 CH3
Performing an elimination
on 1-bromo-2-butanol or
1-bromo-1-butanol
would work best.
Br
OH
CH CH2 CH2 CH3
OH
Best choice: only one
product possible.
Notice that their may be more than one way to
form a particular compound
The handout is an oversimplification …
There are thousands of reaction mechanisms,
many of which are very specific
Considerations when choosing include …
Yield (how much product forms)
Potential for multiple products
Ease of separating contaminating structures
Type of isomer desired
Economics: e.g. cost of starting materials
Answers: 1, 2
Answers: 3
1) An addition reaction involves breaking a
double bond and adding two parts of a
molecule across the bond
• Examples: 1) halogenation, 2) hydrogenation,
3) hydrolysis, 4) addition polymerization
• The ‘opposite’ of breaking the bond is forming
a double bond - i.e. elimination
2) Condensation is a reaction that involves the
production of water.
• Examples: 1) elimination, 2) esterification,
3) condensation polymerization
• oxidation is not an example – the H2O comes
from H2SO4, not from an organic molecule
a) 1,2-dichlorocyclopentane is best formed via
halogenation of cyclopentene. Substitution
using Cl2 would work, but would not be
efficient because many other products would
form (e.g. 1,2,4-trichlorocyclopentane, etc.)
b) H H H
d) 1-butene can be formed from the elimination
reaction involving 1-butanol. Using 2-butanol
could result in either 1-butene or 2-butene
and therefore is a less desirable choice.
H H H
H C C C C C C C C H + H2
H H H
H H H
H H H H H H H H
Room temp. H C C C C C C C C H
H H H H H H H H
c) 2,2,3,3-tetrabromopentane can be formed
via substitution using Br2. It is more efficiently
formed via the halogenation of 2-pentyne
H
H H
H
H H
Room temp.
Cl
b) The only way to prepare octane from 4octyne is via hydrogenation of both pi bonds.
(see top of next page for diagram)
HO H H H
H H
catalyst
H C C C C H
H C C C C H
heat
H H H H
H H H H + H2O
e) Propanoic acid is best formed from the
oxidation of propanal
H H O
+ Br2
H C C C C C H
Cl
+ Cl2
H Br Br H H
Room
temp. H C C C C C H
H Br Br H H
f) Ethanol is most easily formed from the
hydrolysis of ethene.
H OH
acid
H C C H + H2O
H C C H
catalyst
H H
H H
g) Ethyl propanoate is an ester, formed via
esterification (ethanol plus propanoic acid)
O
H C C C H
H H O
+ Cr2(SO4)3
H C C C OH
H H + K2Cr2O7
+ K2SO4
H H
+ H2SO4
+ H2SO4
Formation of a polyamide
O
O
O
HO
NH
HO O
H2 N
OH
NH
O
NH
H3C
OH + HO
H+
heat
O
CH3
NH2
+ H2O
H3C
O
CH3
A polyamide “backbone” forms with R groups
coming off. This protein is built with amino acids.