Download Practice Sheet Answers

Alcohols, Ethers, and Epoxides 9–15
9.38 Use the directions from Answer 9.2.
a.
(CH3)2CHCH2CH2CH2OH
[1]
[2]
H
CH3
C CH2CH2CH2OH
H
CH3
1
CH3
[3]
4-ethyl-6-methyl-3-heptanol
[3]
4-ethyl-5-methyl-3-octanol
[3]
cis-1,4-cyclohexanediol
[3]
3,3-dimethylcyclohexanol
[3]
(2R,3R)-2,3-butanediol
[3]
5-methyl-2,3,4-heptanetriol
[3]
3-isopropylcyclopentanol
4-methyl
5 carbons = pentanol
(CH3)2CHCH2CH(CH2CH3)CH(OH)CH2CH3
[1]
H
CH3
[2]
H OH
C CH2 C CH
CH2CH3
H
CH3
CH2CH3
CH3
c.
[1]
C CH2 C CH
6-methyl
4-ethyl
5-methyl
[2]
OH
8 carbons = octanol
HO
OH
[1]
[2]
1
HO
OH
cis
[2]
HO
HO
6 carbons = cyclohexanol
[1]
HO H
4-ethyl
3
4
cyclohexanediol
e.
CH2CH3
CH2CH3
OH
d.
[1]
3
H OH
CH3
7 carbons = heptanol
f.
4-methyl-1-pentanol
C CH2CH2CH2OH
CH3
b.
[3]
3,3-dimethyl
[2]
HO H
3
2
HO H
HO H
2R, 3R
4 carbons = butanediol
g.
[2]
OH
OH
[1]
4 3 2
OH
h.
OH
OH
7 carbons = heptanetriol
[1]
5-methyl
[2]
HO
CH(CH3)2
OH
1
HO
CH(CH3)2
5 carbons = cyclopentanol
3-isopropyl
Chapter 9–16
9.39 Use the rules from Answers 9.4 and 9.5.
a.
d.
O
O
1,2-epoxy-2-methylhexane
or 1-butyl-1-methyloxirane
or 2-methylhexene oxide
common name: dicyclohexyl ether
4,4-dimethyl
b.
CH2CH3
e.
OCH2CH2CH3
longest chain =
heptane
substituent =
3-propoxy
IUPAC name: 4,4-dimethyl-3-propoxyheptane
c.
epoxy
O
2
5 carbons =
cyclopentane
IUPAC name: 1,2-epoxy-1-ethylcyclopentane
CH3
O
f.
common name: ethyl isobutyl ether
CH3
CH3
C O C CH3
CH3
CH3
IUPAC name: 1-ethoxy-2-methylpropane
tert-butyl tert-butyl
common name: di-tert-butyl ether
9.40 Use the directions from Answer 9.3.
a. 4-ethyl-3-heptanol
3
1
e. 3-chloro-1,2-propanediol
4
HO
2
3
Cl
OH
OH
1
OH
b. trans-2-methylcyclohexanol
OH
f. diisobutyl ether
O
or
CH3
HO
CH3 g. 1,2-epoxy-1,3,3-trimethylcyclohexane
1
3
O
c. 2,3,3-trimethyl-2-butanol
3
2
d. 6-sec-butyl-7,7-diethyl-4-decanol
h. 1-ethoxy-3-ethylheptane
OH
4
6
7
O 1
3
Alcohols, Ethers, and Epoxides 9–17
9.41
Eight constitutional isomers of molecular formula C5H12O containing an OH group:
OH 1-pentanol
1 OH 3-methyl-1-butanol
3
1 OH 2-methyl-1-butanol
OH
2-pentanol
OH
2 1
2,2-dimethyl-1-propanol
2-methyl-2-butanol
2
OH
3-methyl-2-butanol
OH
OH
3-pentanol
2
9.42 Use the boiling point rules from Answer 9.6.
CH3CH2OCH3
(CH3)2CHOH
CH3CH2CH2OH
ether
no hydrogen bonding
lowest bp
2° alcohol
hydrogen bonding
intermediate bp
1° alcohol
hydrogen bonding
highest bp
a.
b.
CH3CH2CH2CH2CH2CH3
no OH group
lowest water solubility
CH3CH2CH2CH2CH2CH2OH
HOCH2CH2CH2CH2CH2CH2OH
one OH group
intermediate water solubility
two OH groups
highest water solubility
9.43
a. CH3CH2CH2OH
b. CH3CH2CH2OH
c. CH3CH2CH2OH
H2SO4
NaH
HCl
CH3CH CH2
+ H2O
CH3CH2CH2O Na+
CH3CH2CH2Cl
+ H2
+ H2O
ZnCl2
d. CH3CH2CH2OH
e. CH3CH2CH2OH
f. CH3CH2CH2OH
HBr
SOCl2
pyridine
PBr3
CH3CH2CH2Br + H2O
CH3CH2CH2Cl
CH3CH2CH2Br
g. CH3CH2CH2OH
TsCl
pyridine
CH3CH2CH2OTs
h. CH3CH2CH2OH
[1] NaH
CH3CH2CH2O Na+
i.
CH3CH2CH2OH
[1] TsCl
CH3CH2CH2OTs
[2] CH3CH2Br
[2] NaSH
CH3CH2CH2OCH2CH3
CH3CH2CH2SH
Chapter 9–18
9.44
a.
OH
NaH
b.
OH
NaCl
c.
OH
HBr
d.
OH
HCl
e.
OH
H2SO4
f.
OH
NaHCO3
Br
g.
OH
[1] NaH
Cl
h.
OH
POCl3
pyridine
O
Na+
+ H2
N.R.
+ H2O
N.R.
O
[2] CH3CH2Br
O CH2CH3
9.45 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major
product.
TsOH
a.
OH
tetrasubstituted
major product
b.
CH2CH3
OH
c.
OH
disubstituted
CH2CH3
CHCH3
TsOH
TsOH
trisubstituted disubstituted
major product
d.
CH3CH2CH2CH2OH
TsOH
CH3CH2CH CH2
OH
TsOH
CH3CH2CH CHC(CH3)3
e.
disubstituted
tetrasubstituted
major product
disubstituted
Two products formed
by carbocation rearrangement
9.46 The more stable alkene is the major product.
OH
H2SO4
trans and disubstituted
major product
monosubstituted
cis and disubstituted
Alcohols, Ethers, and Epoxides 9–19
9.47 OTs is a good leaving group and will easily be replaced by a nucleophile. Draw the products by
substituting the nucleophile in the reagent for OTs in the starting material.
a. CH3CH2CH2CH2
OTs
b. CH3CH2CH2CH2
OTs
c. CH3CH2CH2CH2
OTs
d. CH3CH2CH2CH2
OTs
CH3SH
SN2
NaOCH2CH3
SN2
NaOH
SN2
K+
–
CH3CH2CH2CH2
SCH3 + HOTs
CH3CH2CH2CH2
OCH2CH3
CH3CH2CH2CH2
OH
OC(CH3)3
+ Na+
+ Na+
–
–
OTs
OTs
+ –
CH3CH2CH CH2 + (CH3)3COH + K OTs
E2
9.48
HBr
a.
+
Br
HO H
OH
b.
H D
HCl
ZnCl2
H
Cl
1° Alcohol will undergo SN2.
inversion
H D
SOCl2
c.
HO H
pyridine
H Cl
TsCl
pyridine
d.
H Br
2° Alcohol will undergo SN1.
racemization
SOCl2 always implies SN2.
inversion
KI
HO H
TsO H
SN2
inversion
H
I
Configuration is maintained.
C–O bond is not broken.
9.49
NaH
CH3I
A=
B=
O H
a.
TsCl
pyridine
HO H
PBr3
CH3O H
CH3O
C=
TsO H
E=
D=
H OCH3
CH3O
F=
H Br
CH3O H
b. B and D are enantiomers.
c. B and F are identical.
9.50 Acid-catalyzed dehydration follows an E1 mechanism for 2o and 3o ROH with an added step to
make a good leaving group. The three steps are:
[1] Protonate the oxygen to make a good leaving group.
[2] Break the C−O bond to form a carbocation.
[3] Remove a β hydrogen to form the π bond.
Chapter 9–20
OH
a.
CH3 H OSO3H
+
overall
reaction
The steps:
CH3
CH3
CH2
+
+ H2O
H
O H
CH3
CH3
H
+ HSO4
HSO4
2° carbocation
and
+ HSO4
H
CH3
+ H2SO4
CH3
+ H2O
H
H
1,2-H shift
CH2
CH2
H
3° carbocation
and
2° carbocation
+ HSO4
H
CH3
CH3
CH3
b.
+ H2O
overall
reaction
OH
+ H2SO4
CH3
H OSO3H
CH3
+ H2SO4
CH3
The steps:
CH3
CH3
CH3
CH3 + H O
2
+ HSO4
1,2-CH3 shift
CH3
+ H2SO4
CH3
H
O H
H
CH3
2° carbocation
CH3
+ HSO4
3° carbocation
9.51 To draw the mechanisms:
[1] Protonate the oxygen to make a good leaving group.
[2] Break the C−O bond to form a carbocation.
[3] Look for possible rearrangements to make a more stable carbocation.
[4] Remove a β hydrogen to form the π bond.
Dark and light circle are meant to show where the carbons in the starting material appear in the product.
OH
H OSO3H
O H
+ H2O
H
+ HSO4–
2o carbocation
H
–
+ HSO4
o
3 carbocation
+ H2SO4
Alcohols, Ethers, and Epoxides 9–21
9.52
HO
H
Br
HBr
a.
H
H
S
HO
H
b.
R
H
PBr3
2° alcohol
SN1 = racemization
Br
PBr3 follows
SN2 = inversion
Br
R
HO
H
Cl
HCl
c.
H
H
R
S
HO
H
H
SOCl2
d.
pyridine
2° alcohol
SN1 = racemization
Cl
SOCl2 follows
SN2 = inversion
Cl
R
9.53
OH
3-methyl-2-butanol
[1] HBr
1,2-H shift
H
Br–
[2] –H2O
3° carbocation
2° carbocation
2-methyl-1-propanol
H
HBr
HO
H2O
Br
Br
H Br
Br
The 2° alcohol reacts by
an SN1 mechanism to
form a carbocation
which rearranges.
The 1° alcohol reacts with HBr
by an SN2 mechanism.
no carbocation intermediate =
no rearrangement possible
SN2
no carbocation
9.54 Conversion of a 1° alcohol into a 1° alkyl chloride occurs by an SN2 mechanism. SN2 mechanisms
occur more readily in polar aprotic solvents by making the nucleophile stronger. No added ZnCl2 is
necessary.
HCl
R OH
R Cl
HMPA
polar aprotic solvent
This makes Cl− a better nucleophile.
Chapter 9–22
9.55
H2SO4 , NaBr
CH3CH2CH2CH2OH
CH3CH2CH2CH2Br
overall reaction
CH3CH2CH CH2
CH3CH2CH2CH2OCH2CH2CH2CH3
Step [1] for all products: Formation a good leaving group
CH3CH2CH2CH2
H OSO3H
OH
CH3CH2CH2CH2 O H
+ HSO4
H
Formation of CH3CH2CH2CH2Br:
O H
CH3CH2CH2CH2
CH3CH2CH2CH2Br + H2O
H
Na+ Br–
Formation of CH3CH2CH=CH2:
CH3CH2CH CH2 O H
CH3CH2CH CH2
+ H2O
H
H
HSO4–
Ether forms (from the protonated alcohol):
CH3CH2CH2CH2
OH2
CH3CH2CH2CH2 O CH2CH2CH2CH3 + H2O
H
HSO4
CH3CH2CH2CH2 O H
CH3CH2CH2CH2 O CH2CH2CH2CH3
H2SO4
9.56
TsCl
OH
OTs
pyridine
C CCH2OR
C CCH2OR
9.57
H
OTs
CH3CH2
a.
H
OTs
=
H
CH2CH3
H
H
H
H
2 anti H's
2 elimination products
This conformer reacts.
H
H
H
H
H
OTs
CH3CH2
axial OTs group
OTs
CH3CH2
H
CH3CH2
H
H
H
H
H
=
=
elimination CH3CH2
–H, OTs
two axial β hydrogens
two possible products
CH2CH3
major product
trisubstituted
CH2CH3
disubstituted
Alcohols, Ethers, and Epoxides 9–23
H
OTs
H
CH3CH2
b.
=
H
CH2CH3
H
OTs
H
H
CH3CH2
TsO
H
axial OTs group
H
only 1 axial H
1 elimination product
This conformer reacts.
CH3CH2
only one β axial H
only one product
H
H
H
H
elimination
–H, OTs
OTs
CH3CH2
H
=
H
H
CH2CH3
only product
9.58
a.
c.
O
O
CH3CH2OCH2CH2CH3
O
Br
2° halide
CH3CH2O + BrCH2CH2CH3
O
Br
1° halide
1° halide
less hindered RX
preferred path
OCH2CH2CH3
b.
CH3CH2OCH2CH2CH3
CH3CH2Br + OCH2CH2CH3
1° halide
Neither path preferred.
OCH2CH2CH3
Br
O
OCH2CH2CH3
+ BrCH2CH2CH3
1° halide
2° halide
less hindered RX
preferred path
9.59 A tertiary halide is too hindered and an aryl halide too unreactive to undergo a Willamson ether
synthesis.
Two possible starting materials:
CH3
O C CH3
Br
CH3
O C CH3
CH3
CH3
aryl halide
unreactive in SN2
CH3
Br
C CH3
CH3
3° alkyl halide
too sterically
hindered for SN2
O
Chapter 9–24
9.60
a. (CH3)3COCH2CH2CH3
b.
HBr
HBr
O
2
c.
H2O
(CH3)3CBr + BrCH2CH2CH3
HBr
OCH3
CH3Br
H2O
Br
H2O
Br
9.61
overall reaction
CH3
a.
O
I
CH3
H
+ H 2O
I
I
The steps:
CH3
O
CH3
I
+ I
H
H
+ I
CH3
O
H
CH3
O
CH3
CH3
H
b.
O
Cl
Na+ H–
H
H
I
O
Cl
CH3
O
I
CH3
H
I
+ H2 + NaCl
+ H2 + Na+
O
9.62
O
CH3
R O H
Na+ H–
O
R O
+
+ Na + H2
O S OCH3
R OCH3
(dimethyl sulfate)
+
OSO3CH3
+
H2
Dimethyl sulfate is a reactive methylating agent because –OSO3CH3 is a
very good leaving group; it is a resonance-stabilized, weak conjugate base.
The conjugate acid of –OSO3CH3 is HOSO3CH3, which is a strong acid, similar in acidity to H2SO4.
9.63
O
a. H
C
O
C
H
H
H
HBr
BrCH2CH2OH
d. H
C
H
O
b. H
C
C
H
H
H
O
c. H
H
C
C
H2O
H2SO4
[2] H2O
[1] HC C–
H
H
O
HOCH2CH2OH
e. H
C
H
H
O
CH3CH2OCH2CH2OH
f. H C
H
HC C CH2 CH2OH
[2] H2O
[1] −OH
C
H
[1] CH3CH2O−
H
H
C
C
HOCH2CH2OH
[2] H2O
[1] CH3S−
H
H
[2] H2O
CH3SCH2CH2OH
Alcohols, Ethers, and Epoxides 9–25
9.64
O
a.
H
CH3
CH3
H
b.
Br
CH3CH2O H
O
CH3
OH
[1] CH3CH2O− Na+
O
OH
HBr
c.
CH3 C C H
H2SO4
CH3
O
CH3 OH
CH3CH2OH
OH
d.
[2] H2O
CN
[1] NaCN
[2] H2O
OCH2CH3
9.65
CH3
Cl
H
C
a.
CH3
Cl
H
C
C
H O
H
CH3
H
CH3 C
C
O
H
CH3
Cl
CH3
C
b.
CH3
H
C
C
H O
H
CH3
O
Cl
CH3
H C
C
H
The 2 CH3 groups are anti in the
starting material, making them
trans in the product.
CH3
H
The 2 CH3 groups are gauche in the
starting material, making them
cis in the product.
O
C 4 H8 O
Na+ H−
H
C
CH3
H
CH3
C
O
C 4 H8 O
Na+ H−
9.66
a.
KOC(CH3)3
OTs
b.
OH
*
Bulky base favors E2.
HBr
H CH3
c.
H CH3
O
[1] −CN
CH3CH2 C C H
CH2CH3
H
CH3CH2
H
[2] H2O
Keep the stereochemistry at the stereogenic
center [*] the same here since no bond is
broken to it.
C
C
CN
H
CH3CH2
CH2CH3
H
H
C
C
CH2CH3
CN
H
H
OH
e.
+
KCN
(CH3)3C
HO
OH
OTs
d.
SN2
inversion
PBr3
CH3
f.
Br
*
CN
(CH3)3C
Br
SN2
inversion
CH3
OH
H D
TsCl
pyridine
OTs
H D
CH3CO2
CH3CO2
H D
identical
Chapter 9–26
g.
O
h.
OH
Br
Br
OH
HBr
O
[2] H2O
OH
OCH3
OH
[1] NaOCH3
OH
OCH3
O
NaH
OCH2CH3
CH3CH2I
i.
CH2CH3
CH3
j.
CH3CH2
CH2CH3
CH2CH3
C O CH3
CH3
HI
CH3CH2
C
CH3
I
+
I
CH3
+ H2O
CH3
9.67
a.
OH
b.
OH
c.
HBr
or
PBr3
Br
HCl
or
SOCl2
Cl
[1] Na+ H−
[2] CH3CH2Cl
O
OH
O
[1] TsCl/pyridine
d.
OH
OTs
[2] N3−
N3
Make OH a good leaving
group (use TsCl), then add N3−.
9.68
a.
OH
HCl
or
SOCl2
b.
OH
H2SO4
c.
OH [1] Na+ H−
d.
OH [1] TsCl/pyridine
Cl
O
[2] CH3Cl
OTs [2] –CN
OCH3
CN
Make OH a good
leaving group (use TsCl),
then add −CN.
Alcohols, Ethers, and Epoxides 9–27
9.69
Br
(a) = HBr
(b) = KOC(CH3)3
or PBr3
or other
strong base
OH
OTs (d) = KOC(CH )
3 3
(c) =
(e) =
or other
strong base
TsCl/pyridine
H2SO4 or
TsOH
Br
(f) = KOC(CH3)3
NBS
or other
strong base
HOCl
OH
(g) = NaH
OH
(h) =
O
+ enantiomer
−
OH/H2O
Cl
OH
9.70
Net Reaction: Nu replaces OH.
(C6H5)3P
[ROY]
ROH
O
R Nu
O
CH3CH2OCN NCOCH2CH3
Not isolated.
(DEAD)
O
(C6H5)3P
a.
OH
DEAD
CH3COOH
O
C
CH3
b.
OH
(C6H5)3P
DEAD
CH3SH
SCH3
Chapter 9–28
9.71
Cl
Cl
(CH3)3C
(CH3)3C
OH
A
Cl
OH
(CH3)3C
OH
B
C
Cl
(CH3)3C
Cl
OH
(CH3)3C
Cl
(CH3)3C
OH
OH
(CH3)3C
rapid reaction
This isomer can have the OH and Cl in the trans
diaxial positions, while the larger group C(CH3)3
can be in the favorable equatorial position.
no reaction
This isomer has the OH and Cl in a cis position,
prohibiting epoxide formation. The OH must be
able to approach from the backside.
OH
Cl
OH is in a favorable arrangement
for backside attack of the
nucleophile on the leaving group.
intermediate reactivity
This isomer can have the OH and Cl in the trans
diaxial positions. But the larger group C(CH3)3
must be in the unfavorable axial position, making
this reaction slower.
Cl
(CH3)3C
O
9.72
OH OH
CH3
C
C CH3
H OSO3H
OH
OH2 OH
CH3
C
CH3 CH3
CH3
C CH3
C
1,2-CH3
CH3
CH3
C
shift
CH3 CH3
CH3 CH3
pinacol
C CH3
CH3
+ H2O
+ HSO4
CH3
CH3
C
CH3
O
CH3 O H
C
CH3
CH3
C
CH3
pinacolone
9.73
O
O H
I
O
H
H
O
Na+ –OH
C
I
O
O
O
C
O
O
H
C
H
H
O
OH
C
CH3
C
H
I
O
H
H
O
C
CH3
HSO4
Alcohols, Ethers, and Epoxides 9–29
9.74 You must draw the product that places the nucleophile and leaving group (O−) trans and diaxial.
a. (CH3)3C
This bond was above the six-membered ring
in the epoxide and it stays above it in the product.
O
=
OH (axial)
O
O
(CH3 )3C
H2O
(CH3)3C
OCH3
(CH3)3C
OCH3
=
H
H
(CH3)3C
OH
OCH3 (axial)
attack from below
OCH3
only
b. (CH3)3C
O
=
(CH3)3C
O
attack from above
OCH3 (axial)
OCH3
OCH3
H2O
(CH3)3C
(CH3)3C
O
This bond was below the six-membered ring
in the epoxide and it stays below it in the product.
H
H
= (CH3)3C
OCH3
OH (axial)
OH
only
The nucleophile must always approach by backside attack; i.e. if the epoxide is drawn "up" it must attack
from below. Even though both ends of the epoxide are equally substituted, nucleophilic attack occurs at
only one C–O bond, the one that gives trans diaxial products, as drawn.