Download OrganicChem10 RxPaths SOLUTIONS (2014)

Chemistry 11 (HL)
Unit 6 / IB Topics 10.6 and 20.5
Organic Chemistry 10: Reaction Pathways
Chemists in a variety of industries need to design reactions to
make certain organic compounds. Many organic compounds are
derived from crude oil products – alkanes or alkenes.
The individual reactions you have looked can be linked together to
make a particular product from a particular reactant, via a reaction
pathway.
PRACTICE PROBLEMS:
1.
Which of the following would be a suitable intermediate in the conversion of methane to
methanol?
a)
CH3Cl
b)
CH2Cl2
c)
CHCl3
d)
CCl4
2.
An alkene can be converted to a ketone via
a)
an aldehyde.
b)
an alcohol.
c)
a halogenoalkane.
d)
an alkane.
3.
Which one of the following cannot be produced by the oxidation of an alcohol?
a)
an aldehyde
b)
a ketone
c)
a carboxylic acid
d)
an alkane
4.
Which two compounds can react to form an amide?
a)
alcohol and carboxylic acid
b)
c)
halogenoalkane and amine
d)
amine and an alcohol
amine and carboxylic acid
5.
Which of the following reactions does not require a catalyst?
a)
a nitrile reacting with hydrogen
b)
an alcohol reacting with carboxylic acid
c)
a halogenoalkane reacting with dilure aqueous sodium hydroxide
d)
an alkene reacting with water
6.
OPTION A (preferred because it has fewer steps than option B):
1. but-2-ene to butan-2-ol (addition reaction with water)
CH3CHCHCH3 + H2O  CH3CH2CH(OH)CH3
conditions: heat (with steam)
2. butan-2-ol to butanone (oxidation reaction)
CH3CH2CH(OH)CH3 + [O]  CH3CH2COCH3
conditions: acidified K2Cr2O7
heat under reflux
Chemistry 11 (HL)
Unit 6 / IB Topics 10.6 and 20.5
OPTION B:
1. but-2-ene to 2-bromobutane (other halogen is acceptable) (addition reaction with HBr)
CH3CHCHCH3 + HBr  CH3CH2CHBrCH3
2.
conditions: room temperature
2-bromobutane to butan-2-ol (nucleophilic substitution)
–
CH3CH2CHBrCH3 + OH  CH3CH2CH(OH)CH3 + Br
–
–
conditions: aqueous OH
warm temperature
mechanism: SN2
3. butan-2-ol to butanone (oxidation reaction)
CH3CH2CH(OH)CH3 + [O]  CH3CH2COCH3
7.
conditions: acidified K2Cr2O7
heat under reflux
A: ethanol to ethanoic acid (oxidation reaction): Heat with acidified K2Cr2O7 under reflux.
CH3CH2OH + [O]  CH3COOH
ethanol
ethanoic acid
B: ethanoic acid + ethanol (condensation reaction): H2SO4 as the catalyst; heat
CH3COOH + HOCH2CH3  CH3COOCH2CH3 + H2O
H H
HO C C H

H H
ethanol
8.
ethanoic acid
ethyl ethanoate
This is definitely a CHALLENGE QUESTION. You are only responsible for two step reaction
pathways.
A. ethane to bromoethane (other halogen is acceptable) (free radical substitution)
CH3CH3 + Br2  CH3CH2Br + HBr
conditions: uv light
mechanism: free radical
B. bromoethane to ethanamine (nucleophilic substitution - SN2)
CH3CH2Br + NH3  CH3CH2NH2 + HBr
conditions: concentrated NH3
high pressure
mechanism: SN2
Chemistry 11 (HL)
Unit 6 / IB Topics 10.6 and 20.5
C. bromoethane to ethanol (nucleophilic substitution - SN2)
–
CH3CH2Br + OH  CH3CH2OH + Br
–
conditions: aqueous OH
warm temp
mechanism: SN2
–
D. ethanol to ethanoic acid (oxidation)
CH3CH2OH + [O]  CH3COOH
conditions: acidified K2Cr2O7
heat under reflux
E. ethanoic acid (from D) + ethanamine (from B) to N-ethylethanamide (condensation)
CH3COOH + H2NCH2CH3  CH3CONCH2CH3 + H2O
H H
H
N C C H
H
H H
+

9.
a)
isomers of P
H H H H
H C
H H OH H
C C OH
C
H H
H H
H C
H H
butan-1-ol
H
i)
H H
H CH3 H
C C OH
H C
H H
C C H
H OH H
2-methylpropan-1-ol
b)
C C H
butan-2-ol
H CH3 H
H C
C
butan-2-ol is oxidized to butanone
2-methylpropan-2-ol
H H O H
H C
C
H H
ii)
+
H /Cr2O7
2
C C H
H
heat (under reflux)
(NOTE: Secondary alcohols can ONLY be oxidized to ketones. This can be
done under distillation or reflux conditions. Distillation would give a purified
product whereas reflux would increase the yield. For primary alcohols, the
condition DOES influence the type of product.)
c)
2-methylpropan-2-ol (tertiary alcohols cannot be oxidized)
Chemistry 11 (HL)
d)
Unit 6 / IB Topics 10.6 and 20.5
Two of the isomers of P may be oxidized to produce carboxylic acids.
+
2
i)
H /Cr2O7
heat (under reflux)
ii)
Name and draw the two acid isomers.
H H H
H C
C
H H
C C
H
H CH3
O
H C
OH
H
butanoic acid
e)
C C
H
O
OH
2-methylpropanoic acid
Compound P can be formed by two different types of nucleophilic substitution
mechanisms
i)
The replacement of a halogen atom bonded to a carbon atom by an electronrich species.
Note: You cannot use the terms “substitution” or “nucleophile” in your answer
because these terms are in the question.
–
–
ii)
C4H9Br + OH  C4H9OH + Br
iii)
2-methylpropan-2-ol is formed by an SN1 mechanism.
(SN1 mechanisms involve tertiary halogenoalkanes. Tertiary halogenoalkanes
form tertiary alcohols.)
Q = 2-methyl-2-bromopropane
mechanism:
Chemistry 11 (HL)
iv)
Unit 6 / IB Topics 10.6 and 20.5
Identify the isomers of P that were formed by an SN2 reaction. Draw and name
the starting materials.
butan-1-ol and 2-methylpropan-1-ol (SN2 mechanisms involve primary
halogenoalkanes. Primary halogenoalkanes produce primary alcohols.)
starting materials:
1-bromobutane will produce butan-1-ol
H H H H
H C
C
H H
C C Br
H H
1-bromo-2-methylpropane will produce 2-methylpropan-1-ol
H CH3 H
H C
H
C C Br
H H