C HAPTER 9 HW S OLUTIONS : A LCOHOLS + E THERS
ALCOHOL + ETHER NOMENCLATURE
1. Give the IUPAC name for each compound. Include cis/trans or R/S stereochemistry if necessary.
OH
OH
HO
Structure
CH3
Name
3-methyl-1-heptanol
1-t-butylcyclopentanol
HO
OH
Structure
(2R,3S)-3-methyl-2-pentanol
Cl
OH
Name
4,5-dimethyl-3-decanol
(1S,2S)2-chlorocycloheptanol
trans-4-(3methylbutyl)cyclohexanol
(can’t use R,S because it’s achiral)
2. Give the IUPAC or common name for each compound. Include cis/trans or R/S stereochemistry if needed.
O
O
Structure
Name
OCH2CH2CH3
hexyl isopropyl ether
or 1-isopropoxyhexane
cyclopentyl propyl ether
or propoxycyclopentane
dicyclohexyl ether
(no R,S- not chiral)
O
Structure
O
O
Name
(R)-4-cyclopropoxyoctane
OH
(1S,4R)4-t-butoxycyclooctanol
O
1,6-diethoxy-2-methylheptane
Page 1
REVIEW OF ALCOHOL SYNTHESES
3. Provide the starting alkyl halide and reagents needed in order to produce each alcohol through a substitution
reaction.
NaOH
CH3X
a.
S N2
X
H2O
X
b.
NaOH
c.
CH3OH
OH
S N1
H2O
X
d.
OH
S N2
OH
S N1
WILLIAMSON ETHER SYNTHESIS
4. What is the purpose of the sodium hydride (NaH) in the following reaction?
a. NaH
CH3OH
b. CH3I
CH3OCH3
H– (hydride) is a strong base and removes the H from the alcohol (acid-base).
This converts CH3OH (poor nucleophile) into CH3O– (good nucleophile), so
the reaction is faster.
5. Give the curved arrow mechanism for the following reactions.
a.
a. NaH
CH3OH
CH3OCH3
b. CH3I
CH3O
H
H
acid-base
H3C
O
CH3
H3C
I
O
CH3
S N2
a. NaH
b.
b.
OH
O
I
H
O
H
O
I
O
a. NaH
c.
CH3OH
b.
Br
CH3OH +
Br
CH3O
H
H
acid-base
H3C
O
H
E2 because 2o
or 3o RX
+ CH3OH
Page 2
6. Give the major organic product for the following reactions.
OH
a.
a. NaH
O
OH
d.
CH3
b. CH3Br
I
O
a. NaH
O
e.
b. CH3CH2CH2Br
CH3OH
b.
OH
a. NaH
b.
c.
a. NaH
a. NaH
b. Cl
OH
OH
Br
a. NaH
OH
f.
b.
b.
E2
O
Cl
7. Using the Williamson Ether Synthesis, show a synthetic route (complete with reagents) that efficiently produces
each ether below.
OCH3
a.
a. NaH
OH
OCH3
b. CH3Br
a. NaH
CH3OH
b.
O
HO
Both methods work.
OCH3
Br
b.
O
a. NaH
b. CH3CH2Br
Below doesn't produce the ether well because E2 is the main pathway with a 2o RX.
CH3CH2OH
a. NaH
2o RX.
Br
OCH2CH2CH3
OH
c.
OCH2CH2CH3
a. NaH
b.
OH
E2
CH3CH3O + Br
b.
I
a. NaH
b.
I
This doesn't work well:
E2 occurs instead.
8. Provide the reagents needed to complete each reaction.
a.
CH3CH2OH
Cl
S N1
OCH2CH3
a. NaH
b.
OH
b. CH3CH2X
Williamson Ether Synth.
OCH2CH3
Page 3
INTRAMOLECULAR REACTIONS
9. Give the curved arrow mechanism for the following reaction.
Cl
NaH
O
OH
Cl
Cl
O
O
H
H
acid-base
O
Intramolecular S N2
10. Give the major organic product of each reaction.
a. Cl
5
NaH
3
1
O
5
b.
OH
1
Br
HO 1
3
2
4
NaH
O
4
1
3
3
2
DEHYDRATION REACTIONS
11. Give the curved arrow mechanism for each reaction. Include the Lewis structure of the acid in your mechanism.
a.
con. H2SO4
OH
heat
O
O
H
OH
O
O
S
H
H
O
O
S
O
OH
E1
H
con. H2SO4
b.
heat
OH
H
O
OH
H
O
OH
c.
OH
O
S
HSO4
OH2
OH
con. H3PO4
heat
O
OH
H
O
P
OH
OH
H
OH2
O
P
O
OH
OH
E2
1 o dont form carbocations
Page 4
12. For the following reaction,
OH
heat
a. Draw the curved arrow mechanism.
O
OH
H
O
con. H2SO4
O
S
H2O
OH
HSO4
H
Step 2
Step 1
Step 3
b. Use the mechanism to identify two reasons why “acid” is a catalyst in the dehydration reaction.
Note: “acid” is any strongly acidic source, such as H2SO4, H3O+, or ROH2+.
• Acid accelerates the reaction by turning a bad leaving group (OH–) into a good leaving group (OH2).
This makes the catalytic pathway have a lower activation barrier.
• The acid is not consumed in the reaction. It is used in step 1, but is regenerated in step 3.
c. Draw the energy diagram.
Alkene is higher
energy than
alcohol reactant.
E
OH
H 2O
Step 1 Acid-base
reaction is favorable
(check pKas)
13. Draw all probable dehydration products for these reactions, including stereoisomers. Then decide which should
be the major product and briefly explain your answer.
con. H2SO 4
a.
heat
OH
most substituted (di) and trans has
fewer repulsions, so lowest energy.
OH
b.
con. H3PO4
heat
Trisubsituted is lower
energy than disub.
OH
con. H2SO 4
c.
No other possibility.
Δ
con. H3PO4
d.
OH
heat
Trisub lower E than disub. Major
put bulkiest groups opposite. (In
truth all 4 trisub are similar E.)
Page 5
HYDRIDE AND ALKYL SHIFTS
14. Draw the intermediate formed after each mechanistic step.
H
H
a.
H
H
C
C
H
H
C
C
C
H
C
H
H
H
3
4
5
H
H
H H
C
C
C
C
H
H
H
H
H
4
3
H
H3C
H3C
2
1
5
6
CH3
CH3
b.
CH3
4
2
H3C
c.
H
H
1
d.
5
1
4
3
6
8
6
6
5
2
3
2
7
7
1
8
15. What is a possible driving force (or reason) for the rearrangement in:
a. Problem 14a?
A secondary carbocation is converted into a tertiary carbocation through the rearrangement, which is more
stabilized by hyperconjugation. The motivation for the shift may be to lower the energy of the carbocation.
b. Problem 14c?
A four-membered ring is converted into a five-membered ring through the rearrangement, which has less
ring strain. The motivation for the shift may be to relieve the ring strain (lower the energy of the system).
16. Give the curved arrow mechanism for each reaction. Include the Lewis structure of the acid.
Reaction
con. H2SO4
a.
heat
OH
H
O O
S
O
OH
H
H-shift
OH2
OH
H
H2O
H
or HSO4
b.
OH
O O
H
S
O
OH
con. H2SO 4
Δ
H
OH2
another
H-shift
H-shift
H
HSO4
H
H
H
Page 6
con. H3PO4
c.
heat
O
H
P
O
OH
OH
OH
CH3 methyl
shift
OH2
d.
CH3
H
H2PO4
OH
con. H2SO 4
Δ
O O
H
S
O
OH
HSO4
H
alkyl
OH2
=
shift
(ring expansion)
17. Identify which of the reactions W-Z would synthesize 3,3-dimethylcyclopentene most efficiently (with the
fewest competing products). Then explain why the other routes would be less efficient.
con. H2SO 4
W
OH
heat
Br
KOC(CH3)3
+
b
X
a
BEST
+
minor
major
Br
KOC(CH3)3
Y
OH
Z
con. H2SO 4
heat
+
likely
Reaction Y would most efficiently
produce 3,3-dimethylcyclopentene as
there are no significant alternative
products. There is only one type of betahydrogen, so the E2 reaction could make
only one possible product. (Also SN2
does not compete much when using such
a bulky base.)
In Reaction W, two different betahydrogens are present, leading to a likely
mixture of the products shown. This
would split the yield, lessening the
quantity of 3,3-dimethylcyclopentene
obtained. Also this mechanism proceeds
through a carbocation, so rearrangements
could further divert the yield.
In Reaction X, a bulky base would more likely remove a beta-hydrogen from position “a,” creating the wrong
alkene isomer as the major product. Removal of a beta-hydrogen from position “a” is still possible, but the
intended alkene would be a minor product.
In Reaction Z, the carbocation generated in this mechanism is neighboring a
quaternary center, so rearrangements involving a methyl shift are very likely,
producing the wrong product.
3,3-dimethylcyclopentene
Page 7
ALCOHOL REACTION WITH HX
18. Give the curved arrow mechanism for each reaction.
Reaction
a.
HBr
OH
Br
OH
OH2
H
b.
Br
Br
S N2
CH3
HI
CH3
Br
I
OH
CH3
H
OH
CH3
H
CH3
I
hydride
shift
I
CH3
CH3
I
SN1
OH2
c.
Cl
HCl
OH
ethyl
shift
Cl
Cl
OH2
(same as above)
PBr 3 , SOCl 2 AND TOSYLATE REACTIONS
19. Fill in the boxes with the organic product from each reaction.
NaH
OH H
CH3I
O
TsCl
OH H
py.
OCH3
KN3
OTs
N3
N
N
N
SOCl2
OH H
KN3
Cl
N3
Page 8
COMBINED ALCOHOL REACTIONS
20. Give the major organic product for each reaction. Consider plausible rearrangements.
O
Cl S
O
HCl
a.
OH
Cl
OH
HBr
OH
Br
PBr3
d.
a. TsCl, py.
N3
b. NaN3
1˚ don’t form carbocations, so no rearrangement occurs
Br
OCH2CH3
b. KOCH2CH3
OH
k.
OH
CH3 OTs
py.
a. SOCl2
j.
b.
HO
CH3 OH
I
HI
c.
i.
OH
CN
Br
a. PBr3
l.
b. NaCN
Double SN2
a. TsCl, py.
OH
OH
m.
Cl
SOCl2
e.
b. O
O
O
ONa
Single SN2
f.
OH
TsCl
CH3OH
CH3OTs
py.
HBr
n.
D
o.
NaH
g.
Br
CH3
CH3
OH
OH
D
HCl
Cl
O
1˚ alcohol goes through SN2, so inversion occurs
HO
h.
PBr3
Br
Page 9
21. The synthesis of the product shown in reaction L (call it product Q) is best achieved by this method. Explain
why methods M-O will not effectively synthesize product Q.
CH3
CH3
Best synthesis of Q: Double SN2 gives the right
stereochemistry of the SH group (PBr3 inverts the
center, SH inverts the center again).
a. PBr3
(L)
b. NaSH
OH
SH
Q
CH3
Reaction (M) doesn’t work: OH– is not a good leaving group for SN2
reactions.
NaSH
(M)
OH
CH3
Reaction (N) is not efficient: the reaction goes through a flat
carbocation intermediate, so attack of H2S gives both isomers (with
SH out and SH back). Carbocations also often rearrange, and a
significant product might be with the SH on the same carbon as the
methyl group. E1 may also compete.
H2SO 4
(N)
H2S
OH
CH3
a. TsCl, py.
(O)
Reaction (O) doesn’t product Q: single SN2 produces the wrong
isomer.
b. NaSH
OH
EPOXIDE REACTIONS
22. Give the curved arrow mechanism for the following reactions.
O
a.
OH
NaOH
OH
H2O
O
H
O
O
OH
H
OH
OH
OH
OH
H+
O
b.
CH3CH2OH
OCH2CH3
H
H+
O
OH
OH
O
CH3CH2OH
CH3CH2OH
H
O
OCH2CH3
CH2CH3
O
KOCH2CH3
c.
O
CH3CH2O
DMF
Cl
O
O
CH3CH2O
CH3CH2O
CH3CH2O
Cl
O
Cl
Page 10
23. Concerning the following two reactions:
a
SH
O
b
NaSH
CH3
H2O
OH
O
OH
HCl
CH3
CH3
CH3
Cl
a. Explain the regioselectivity of the reactions.
The first reaction involves a negatively charged nucleophile (good Nu) so reacts SN2-style at the least
hindered side of the epoxide, at site “a.”
The second reaction involves protonation of the epoxide before addition of the
nucleophile (Cl–). Since the LG is better now, the LG has already partially left (SN1like). There are partial positive charges on the two carbons of the epoxide, but the
more substituted side (3˚vs. 2˚) stabilizes a partial carbocation better. Thus there is a
greater δ+ on the more hindered side, which is where the nucleophile attacks.
δ+ H
O
δ+
δ+
CH3
b. Explain the stereoselectivity of the reactions.
Both reactions occur with inversion of the reacting carbon, SN2-style, with backside attack.
Even though the protonated epoxide in the second reaction has a partial carbocation, it is not a full
carbocation. The reaction mechanism is really in between SN1 (strong δ+ on the C) and SN2 (backside
attack).
24. Give the major organic product for each reaction. Indicate if a racemic mixture is formed.
OH
O
CH3CH2OK
a.
O
CH3CH2OH
O
e.
H+
b.
H2O
CH3
Br
O
CH3OH
OH
O
HBr
c.
KOH
f.
racemic
CH3 OCH3
OH
O
O
(CH3)2CHOH
CH3OH
CH3
OH
OH
O
H+
I
HI
g.
OH
O
d.
OH
NaCN
H2O
h.
CN
O
a. HC
C
b. H2O
OH
C
CH
Page 11
COMBINED ALCOHOL + EPOXIDE REACTIONS
25. Give the major organic product for these reactions. Consider plausible rearrangements and indicate if a racemic
mixture is formed.
OH
OCH2CH3
a. NaH
a.
CH2CH3 b. CH3CH2I
O
CH2CH3
OH
O
H2O
Br
OH
NaOH
b.
HO
HBr
e.
OH
I
HI
f.
racemic
H
OTs
NaI
g.
OH
con. H2SO 4
c.
or
or
OTs
heat
I
CH3
CH3
H3C
PBr3
CH3O
h.
CH3O
H3C
CH3
Br
OH
d.
I
H3C
H3C
O
KCN
H2O
CH3
OH
CN
Page 12