ALKENES ANSWERS
1.
a)
cis-2-pentene
b) 3-methylcyclopentene
c)
3,5-dimethylcyclopentene
d) 3,3-dimethyl-1-butene
e)
2,4,4-trimethyl-2-pentene
f)
2-methyl-2,5-heptadiene
2.
a)
CH3CH2
H
C
f)
C
CH3
CH2
CH
g)
H3C
H3C
HC
CH
C
CH3
C
CH3
H3C
F
Br
h)
C
c)
i)
CH
C
H
CH2
d)
CH2
F
ClCH2CH2
CH3
C
CH3
C
CH2CH2CH3
CH3CH2
H
H
e)
CCl2
CH3
CH3
CH
CH3
b)
CH2
CH2
CH2CH3
H
H3C
CH
CHCl
C
j)
H3C
C
CHCl
CH3
page 1
ALKENES ANSWERS
3.
a)
cis-3-hexene or (Z)-3-hexene
b) methylidenecyclobutane
c)
ethylidenecyclobutane
d) propylidenecyclobutane
e)
ethenylcyclobutane or vinylcyclobutane
f)
(E)-(1-propenyl)cyclobutane
g) (2-propenyl)cyclobutane or allylcyclobutane
h) (1-methylethenyl)cyclopentane
i)
(Z)-(1-propenyl)cyclohexane
j)
(3E,5Z)-1,3,5-octatriene
CH3
4.
H
a)
H
C
CH2
e)
H3C
C
CH3
CH3
C
CH
CH3
CH
CH3
CH3
H
b)
CH3
C
CH2
f)
CH2
C
CH3
CH3
H
CH
CH
CH
CH2
c)
Br
CH2CH3
C
CH3
g)
CH2CH3
C
H3C
CH2CH2Br
CH2CH3
d)
CH3
CH3
h)
CH3
H3C
page 2
ALKENES ANSWERS
5.
a)
b)
H3C
C
CH2CH3
CH2CH3
6.
H3C
CH3
C
C
H
CHCH2CH3
c)
C
CH3
H
H
CH3CH
H
CH3
CH3CH
CCH3
Markovnikov’s Rule: The more stable, more substituted intermediate carbocation forms.
7.
CH3
CH3
a)
HCl
CH3
CH
C
CH3
CH3
CH2
C
CH2
Cl
b)
CH3
2-chloro-2-methylpentane
I
KI
CH2
CH2
CH3
1-iodo-1-methylcyclohexane
H3PO4
Br
c)
HBr
H3C
+
H3C
1-bromo-3-methylcyclohexane
H3C
Br
1-bromo-4-methylcyclohexane
Equal quantities formed
8.
2-methyl-1-butene
CH3
a)
CH2
CH3
C
CH2
CH3
HBr
CH3
OR
CH3
C
CH
CH3
Br
2-methyl-2-butene
CH3
CH2
C
CH3
CHCH3
b)
ethylidenecyclopentane
HCl
CH2CH3
OR
Cl
CH2CH3
1-ethylcyclopentene
9.
2º C+
CH
H
CH2
KI, H+H2PO4-
H
+
CH
CH2
H
C+ rearrangement
(via methide shift)
3º C+
H
CH2
+
CH2
.. :I:
..
CH2CH3
I
page 3
ALKENES ANSWERS
O
10.
CH2
CH
CHCl
DU = 3
(1 ring and two p bonds)
11.
DU = 4
(1 ring and three p bonds)
CH
NH2
C
DU = 2
(two p bonds)
Halogens count as H’s; O’s are ignored; subtract 1 H for each N present. Compare the formula to a saturated
acyclic HC with generic formula CnH2n+2.
a)
C2HCl3O ~ C2H4 vs. C2H6  2H’s/2 = 1 DU
b) C3H6O2
c)
~
C3H6 vs. C3H8  2H’s/2 = 1 DU
C4H7NO ~ C4H6 vs. C4H10  4H’s/2 = 2 DU
12.
CH2CH3
CH2CH3
CH2CH3
I
KOH in ethanol
+
+
+ H2O
KI
- HI
1-ethylcyclopentene
(major product)
(trisubstituted)
H2SO4
in THF, D
OH
C
H3C
3-ethylcyclopentene
(minor product)
(disubstituted)
CH2
H2C
CH3
- H2O
CH3
C
CH2
+
CH3
C
H3C
CH
CH3
CH3
CH3
2-methyl-2-butene
(major product)
(trisubstituted)
2-methyl-1-butene
(minor product)
(disubstituted)
13.
iodinium cation
(bridged intermediate)
..
:I
..
H
H
..
I:
..
.. - :I:
..
+
:I:
..
:I:
H
+
H
H
H
H
..
:I :
:I : H
..
trans-1,2-diiodocyclopentane
(anti addition)
page 4
ALKENES ANSWERS
14. Zaitzev’s Rule: In elimination reactions, the more highly substituted (more stable) alkene is the major product in
a product mixture.
a vic dihalide
15.
(2,3-dibromobutane)
minor product
.. : Br :
..
CH3 CH3
CH3 CH3
H3C
C
C
..
: Br
..
CH3
H3C
..
Br
.. :
C
C
CH3
H3C
..
Br
..
+
H
CH3 CH3
H3C
C
H + H
:O
CH3
C
CH3
H3C
..
Br
..
C
C
C
C
H
CH3
..
: OH CH3
H3C
CH3
C
- H3O+
CH3 Br
+
CH3
CH3 Br
..
H O
..
..
H O
..
Br
CH3
CH3 Br
a halohydrin
(3-bromo-2-butanol)
major product
When w ater is present in high concentration it competes w ith bromide
in attacking the intermediate so both vic dihalide and halohydrin are formed.
The solubility of Br2 in w ater is only about 4%, so the major product is the
halohydrin as the concentration of w ater far exceeds the conc. of bromide ion.
16.
C
cyclohexane bisulfate
conc.
H
H+HSO4-
H
-
HSO4
H
+
O
H
H
H
H
..
H O
..
O
H
+
H
H
OH
O
H
H
S
H
H
+ H
O
..
H
..
H O
..
H
..
OH
..
+
+
H3O
H
cyclohexanol
H
Bisulfate is a very w eak nucleophile and w ill only bond w ith the carbocation in absence of w ater.
The reaction is exothermic. The alkene dissolves in the conc. acid.
When w ater (a better nucleophile) is present it w ill bond w ith the carbocation instead of bisulfate.
H
17.
CH2
CH
CH2
..
:
X
..
X
CH3
CH3
CH
CH2
CH3
Markovnikov product
..
. Br :
..
18.
CH2
CH
CH2
CH3
Br
CH2
.
CH
antiMarkovnikov product
1-bromobutane
H
..
Br
.. :
Br
H
CH2
CH3
CH2
CH
CH2
CH3
..
+ . Br :
..
Note that the more stable, more substitued free radical forms (consistent w ith Markovnikov's rule).
page 5
ALKENES ANSWERS
19.
H
H
KMnO4 (cold, neutral or alkaline)
(Z)-1,2-cyclopentanediol
(a syn diol)
OH
HO
20.
hot KMnO4
(ketones, carboxylic acids, & CO2 form)
CH2
CH2
CH2
CH2
CH3
H
O3, then Zn in dil. CH3COOH
(only aldehydes and ketones form)
O C
CH2
CH3
C O
CH
CH3C
CH3
H
O
CH2
O C
C O
HO
CH3
CH2
O
O
OH
+
CO2
CH3C
H
C
+
H
H
21.
1. Hg(OAc)2
2. NaBH4
CH3
CH3 OH
H
C
CH2
CH
Markovnikov product
w ithout rearrangement
H
CH3
CH3
C
CH
CH2
dilute
H2SO4
CH3
H
CH3 H
H
C
CH2
CH
Markovnikov product
w ith rearrangement
OH
1. BH3
2. H2O2, OH-
CH3 H
OH
C
CH2
antiMarkovnikov product
CH3
CH
H
page 6
ALKENES ANSWERS
22.
Br
a)
H3C C
KOH
CH2
CH2
in ethanol
CH3
H3C C
CH3
CH3
CH2
CH
1.
BH3
2.
H2O2 , OH
C
CH3
CH2
+
CH3
CH2
CH3
-
CH2
OH
H
CH
C
H2O
CH3
CH2CH3
Hydroboration/oxidation produces
the syn, antiMarkovnikov alcohol.
CH2CH3
cold, neutral, KMnO4
c)
+
KBr
In elimination reactions, the more stable,
more substituted alkene is the major product.
CH3
CH3
b)
CH
C
H2C
CH2CH3
OH
+
CH3
MnO2
The syn, vic diol is produced
OH
H2C C
CH3
CH2CH3
hot, acidic KMnO4
CO2
+
O C
+ Mn+2
CH3
CH2CH3
d)
Br
H
H2C
C
HBr, H2O2
H2C C
CH3
CH2CH3
The antiMarkovnikov product forms
w ith HBr and peroxides.
CH3
H
OH
Antiaddition occurs w ith
formation of the halohydrin.
HOCl
e)
CH3
CH3
Cl
Cl
H
f)
CH3
CH3
CH2
CH
CH3
HCl
C
CH3
CH2
CH
C
CH3
CH2
CH3
CH2
The normal Markovnikov product results.
g)
O
CH3
C
CH3
H
1.
O3
2. Zn , CH3COOH
H
O
C
C
CH3
O
+
O
C
CH3
Ozone oxidizes alkenes to aldehydes and ketones.
page 7
ALKENES ANSWERS
h)
H3C
C
Br
H3C
Br2 , H2O
CH
C
H3C
H3C
HO
i)
CH3
1.
BH3
2.
H2O2 , OH-
Both halohydrin and
vic dihalide are formed
as Br- and H2O compete
as the nucleophile, how ever
halohydrin is by far the major
product ow ing to the low conc.
of Br- ion.
CH
OH
H
Hydroboration/oxidation produces
the syn, antiMarkovnikov alcohol
H
CH3
23.
a)
CH3CH
g)
H3C
ethylene group
methyl group
Cl
b)
H3C
Cl
CH3CH
h)
methyl chloride
Cl
ethylene chloride
c)
H2C
i)
CH2
vinyl group
methylene group
d)
CH
Cl
j)
H2C
CH2
CH
Cl
vinyl chloride
Cl
methylene chloride
e)
k)
CH3CH2
CH2
C
vinylidene group
ethyl group
Cl
f)
CH3CH2
Cl
l)
ethyl chloride
CH2
C
Cl
vinylidene chloride
Note: These are common names. They are commonly used but are not systematic and so must be memorized. Some are
confusing, for example, ‘ethylene’ is the common name for ‘ethene’ (CH2=CH2) but ‘ethylene chloride’ is not an alkene; it is a
saturated alkyl halide. Be careful.
24.
a)
CH2
CH
CH2
allyl group
b)
CH2
CH
CH2
Cl
allyl chloride
page 8
ALKENES ANSWERS
25.
a)
CH2=CCl2
1,1-dichloroethene
no cis-/trans- isomers
c)
CHCl=CHCl
1,2-dichloroethene
H
H
H
C
C
C
Cl
Cl
Cl
cis-1,2-dichloroethene
b) CH2=CHCl
chloroethene
no cis-/trans- isomers
Cl
C
H
trans-1,2-dichloroethene
d) CBrCl=CHCl
1-bromo-1,2-dichloroethene
Br
C
Cl
Br
Cl
H
C
C
H
Cl
cis-1-bromo-1,2-dichloroethene
C
Cl
trans-1-bromo-1,2-dichloroethene
Note that when either one of the doubly-bonded carbons has identical substituents, cis- and trans- isomers are not
possible.
26. Hydrogenation of alkenes occurs rapidly at room temperature and atmospheric pressure if a large surface area of
heterogeneous catalyst such as Ni, Pd and Pt are used.
H2 (g) on Ni
27.
CH
CH2
CH
CH2
1,3-butadiene
28.
CH3 O
C
+
CH3CH C
CH3
H3C
n-butane
1. O3
2. Zn in
dil. CH3COOH
O
a)
CH3CH2CH2CH3
1 atm.
20 ºC
H
H3C
C
H
C
CH3
CH
H3C
CH3
b)
2
CH3
CH2
C
H
O
1. O3
2. Zn in
dil. CH3COOH
CH3
CH2
C
C
H
H
CH2
CH3
page 9
ALKENES ANSWERS
29. Baeyer’s test: If an alkene is present, it will decolorize purple KMnO 4 solution (rendering it colorless) but it
forms a brown precipitate of MnO2. Cold KMnO4 oxidizes alkenes to syn diols. In the process, KMnO4 is
reduced to a brown manganese dioxide precipitate.
Bromine test: Solutions of bromine are orange colored. Bromine is removed from solution as it adds to alkene
double bonds and hence the orange color of the solution disappears.
Sulfuric Acid test: Concentrated H2SO4 (pKa = -5) is a very strong acid-strong enough to protonate alkene p
bonds. The alkene appears to dissolve in the acid as the bisulfate is formed. The reaction is exothermic so the
test tube grows hot.
30.
Br
I
a)
CH3
f)
H
Markovnikov alkyl halide
anti addition vic dihalide
CH3
Br
OH
b)
CH3
g)
anti addition halohydrin
Br
OH
OH
h)
c)
OH
CH3
d)
CH3
i)
O
CH3
O
Markovnikov bisulfate
e)
OH
CH3
H
syn vic diol
Markovnikov alcohol
OSO3H
CH3
H
H
CH3
j)
O
CH3
O
OH
Markovnikov alcohol
H
k)
OH
syn
CH3
antiMarkovnikov
alcohol
H
page 10
ALKENES ANSWERS
31.
CH3
H3C
C
dil. aq. H2SO4
& heat
CH3
HBr
Br
H2C
C
CH3
H3C
CH3
C
CH3
OH
H3C
HOCl
1. BH3 in THF
2. H O , OH2 2
OH
H
H3C
H3C
C
C
CH2Cl
CH2OH
H3C
H3C
Br
32.
a)
OH
Br
- H2O
Br2
in CCl4
conc. H2SO4
&D
b)
H3C
H3C
OH
OH
Cl
- H2O
HOCl
or Cl 2, H2O
H3C
conc. H2SO4
&D
CH3
CH3
c)
I
OH
- HI
CH3
KOH
in ethanol
d)
OH
1. BH3 in THF
2. H O , OH2 2
H3C
CH2
OH
CH3
CH
CH
CH3
CH2
CH
CH3
CH3
- H2O
conc. H2SO4
&D
1. Hg(OAc) 2 in THF
2. NaBH4
CH3
CH2
CH
CH
CH3
page 11
ALKENES ANSWERS
e)
CH
CH3
CH3
CH3
- HBr
Br
H2 on Ni
KOH
in ethanol
f)
CH
CH
25ºC, 1 atm.
CH3
Br
Br
CH3
CH3
CH2
?
CH
CH2
CH3
CH2
CH3
- HBr
KOH
in ethanol
HBr, H2O2
H2C
CH
CH3
g)
OH
OH
OH
- H2O
cold KMnO4
conc. H2SO4
&D
33.
Cl2
CH2
CH
CH
CH2
CH2
CH
Cl
Cl
CH
CH2
Cl
Cl
+
major product
3,4-dichloro-1-butene
CH2
CH
CH
CH2
minor product
1,4-dichloro-2-butene
34.
CH3
a)
isobutylene
H2C
C
2-methylpropene
(isobutylene)
CH3
CH3
b) isoprene
H2C
C
CH
CH2
2-methyl-1,3-butadiene
(isoprene)
page 12
ALKENES ANSWERS
35.
hot alkaline
KMnO4
a)
Oxid. Half Rxn.:
O
+
CO2
+
H2O
+
MnO2
C5H8  C4H6O + CO2 + H2O
 C4H6O + CO2 + H2O
 C4H6O + CO2 + H2O + 8H+
 C4H6O + CO2 + H2O + 8H+ + 8e+ 8OH-  C4H6O + CO2 + H2O + 8H+ + 8e- + 8H2O
8OH-
Balance O’s
Balance H’s
Balance charge
Neutralize H+
C5H8
C5H8
C5H8
C5H8
Cancel Terms
Result
C5H8 + 4H2O + 8OH-  C4H6O + CO2 + H2O + 8e- + 8H2O
4
C5H8 + 8OH-  C4H6O + CO2 + 8e- + 5H2O
+
+
+
+
Red. Half Rxn.:
4H2O
4H2O
4H2O
4H2O
 MnO2
 MnO2 + 2H2O
+ 4H+  MnO2 + 2H2O
+ 4H+ + 3e-  MnO2 + 2H2O
+ 4H+ + 4H2O + 3e-  MnO2 + 2H2O + 4OH4OH+ 4H2O + 3e-  MnO2 + 2H2O + 4OH2
Balance O’s
Balance H’s
Balance charge
Neutralize H+
MnO4MnO4MnO4MnO4MnO4-
Cancel Terms
MnO4-
Result
MnO4- + 2H2O + 3e-  MnO2 + 4OH-
Find Lowest Common Multiple of Electrons Transferred:
(C5H8 + 8OH-  C4H6O + CO2 + 8e- + 5H2O)3 = 3C5H8 + 24OH-  3C4H6O + 3CO2 + 15H2O + 24e(MnO4- + 2H2O + 3e-  MnO2 + 4OH-)8
= 8MnO4- + 16H2O + 24e-  8MnO2 + 32OHCombine Half Reactions and Cancel Like Terms:
3C5H8 + 24OH- + 8MnO4- + 16H2O + 24e-  3C4H6O + 3CO2 + 15H2O + 24e- + 8MnO2 + 32OH3C5H8 + 8MnO4- + H2O  3C4H6O + 3CO2 + 8MnO2 + 8OHb)
CH3
CH3CH
C
H2CrO4
O
CH3COH
CH3
+
O C
CH3
+
Cr+3
CH3
Oxid. Half Rxn: C5H10 + 3H2O  C2H4O2 + C3H6O + 6H+ + 6eRed. Half Rxn: 3e- + 6H+ + H2CrO4  Cr+3 + 4H2O
Balanced Rxn:
C5H10 + 6H+ + 2H2CrO4  C2H4O2 + C3H6O + 2Cr+3 + 5H2O
page 13
ALKENES ANSWERS
36.
O
a)
CH3CH2CH
O
(oxidation)
CH3CH2COH
C3H6O + 2OH-  C3H6O2 + H2O + 2e-
O
b)
CH3CHCH2CH2OH
OH
CH3CCH2COH
(oxidation)
O
C4H10O2 + 6OH-  C4H6O3 + 5H2O + 6e-
c)
MnO4- + 8H+ + 5e- 
d)
HNO3 + H+ + e-  NO2 + H2O
(reduction)
e)
HIO4 + 2H+ + 2e-  HIO3 + H2O
(reduction)
f)
H2O2 + 2OH- + 2e-  2O-2 + 2H2O
(oxidation)
g)
Cl2 + 2e-  2Cl-
(oxidation)
Mn+2 + 4H2O
(reduction)
page 14