17.1 For all practical purposes, the compound 2,4-cyclohexadien-1-one exists totally in its enol
form. Write the structure of 2,4-cyclohxandien-1-one and of its enol form. What special factor
account for the stability of the enol form?
Answer:
O
OH
Because phenol is a kind of aromatic compounds. It has aromatic ring, so it is much more stable.
17.2 Would you expect optically active ketones such as the following to undergo acid-or
base-catalyzed racemization? Explain your answer.
O
O
C3H7
C3H7
C
C2H5
CC6 H5
C
H3C
(a)
H3C
CH2CC6H5
(b)
H
Answer: (a) It doesn’t have any α-H, therefore, it will not tautomerize to any enol form, no
racemization. (b) It has the α-H, but the sterocenter is the β-carbon and thus enol formation
doesn’t affect it, no racemization.
17.3 When sec-bultyl phenyl ketone is treated with either OD¯ or D3O+ in the presence of D2O, the
ketone undergoes hydrogen-deuterium exchange and produces this compound:
CH3
C2H5
C
D
COC6H5
Write mechanisms that account for this behabior.
Answer: OD- Catalyzed:
C2H5
CH3
O
C
C
H
D
OC2 H5
+
D3O Catalyzed:
C2H5
C6H5
CH3
OD
C
C
C6H5
D
OD
CH3
O-
D
C
C
C6H5
C2H5
CH3
O
C
D
C
OD
C6H5
C2H5
CH3
O
D
O
C
C
C6 H5
D
D
C2H5
CH3
O
D
C
C
C6H5
O
H
H
D
D
C2H5
CH3
OD
C
C
C6H5
D
OD
C2H5
CH3
O
C
D
C
C6H5
17.4 Write a mechanism using sodium ethoxide in ethoxide in ethanol for the epimerization of
cis-decalone to trans-decalone. Draw chair conformational structures that show why
trans-decalone is more stable than cis-decalone.
Answer:
Mechanism:
-O
O
O-
C2H5
HO
O
C2H5
H
H
O
H
H
H
H
or
H
chair conformations:
O
H
H
O
trans form:
H
cis form:
H
17.5 Why do we say that the halogenation of ketones in base is “base promoted” rather than “base
catalyzed”?
Answer:
The reaction is said to be “base promoted” because base is consumed as the reaction takes
place. A catalyst is, by definition, not consumed.
17.6 Addition evidence for the halogenation mechanisms that we just presented comes from the
following facts: (a) Optically active sec-butyl phenyl ketone undergoes acid-catalyzed
racemization at a rate exactly equivalent to the rate at which it undergoes acid-catalyzed
halogenation. (b) sec-Butyl phenyl ketone undergoes acid-catalyzed iodination at the same rate
that it undergoes acid-catalyzed bromination. (c) sec-Butyl phenyl ketone undergoes
base-catalyzed hydron-deuterium exchange at the same rate that it undergoes base-promoted
halogention. Explain how each of these observations supports the mechanisms that we have
presented.
Answer:
(a) From this, we know that the slow step in acid-catalyzed racemization is the same as the
that in acid-catalyzed halogenation.
(b) According to he mechanism given, the slow step for acid-catalyzed iodination is the same
as that for acid-catalyzed bromination. Thus, we would expect both reactions to occur at
the same rate.
(c) The slow step for both reactions is the same.
17.7 (a) Show all steps in the aldol addition that occur when propanal is treated with base. (b)
OH
O
CH3CH2CHCHCH
How can you account for the fact that the product of the aldol addition is
OH
CH3
and not
O
CH3CH2CHCH2CH2CH ?(c) What products would be formed if the reaction mixture were heated?
Answer，
（a）
O
O
O
CH3CH-CH
H
CH3CH
C
H
H
HO
O
O
O
O
CH3CH2CHCHCH
CH3 CH2CH + CH3CHCH
CH3
O
O
CH3CH2 CHCHCH
OH
+
H
OH
O
CH3 CH2CHCHCH
CH3
+
OH
CH3
O
(b) Owing to the mechanism that we get a carbanion, CH3CHCH , we can account for the
product.
(c)
O
CH3 CH2 CH
CCH
CH3
17.8 One industrial process for the synthesis of 1-butanol begins with acetaldehyde. Show how
this synthesis might be carried out.
OH
O
OH
CH3CH
O
O
H
CH3CHCH2CH
H3CHC
CHCH
H2O
H2 \ Ni
CH3CH2CH2 CH2 OH
pressure
17.9 Show how each of the following products could be synthesized from butanal
(a) 2-ethyl-3-hydroxyhexanal
(b) 2-ethyl-2-hexen-1-ol
(c) 2-ethyl-1-hexanol
(d) 2-ethyl-1,3-hexanediol
O
O
(a)
OH
OH
H2 O
H
H
C2 H5
O
(b)
product of (a)
OH
1. LiAlH4
heat
H
H
2. H2O
C2H5
C2H5
OH
(c)
product of (b)
H2 / Pt
H
C2H5
OH
(d)
product of (a)
OH
NaBH4
H
C2H5
17.10 The acid-catalyzed aldol condensation of acetone, also produces some
2,6-dimethyl-2,5-heptadien-4-one. Give a mechanism that explain the formation of this produce.
H+
H+
H+
O
OH
O
O
OH
O
H
CH2
C
H2
H
O
O
OH
OH
H+
O
17.11 Heating acetone with sulfuric acid leads to the
(1,3,5-trimethylbenzene). Propose a mechanism for this reaction.
formation
of
mesitylene
Answer:
H
H3C
OH
O
H
OH
+
OH
O
H
H+
OH2
O
OH
H+
O
OH
H
HO
H+
O
HO
O
H
OH
H2O
H
H2O
H
17.12 Outlined below is a synthesis of a compound used in perfumes, called lily aldehyde. Provide
all of the missing structures.
p-tert-Butylbenzyl alcohol
PCC
CH2Cl2
C11H14O
propanal
OH-
C14H18O
H2, Pd-C
lily aldehyde
(C14H20O)
Answer:
PCC
CH2OH
CHO
CHO
Propanal / OH-
CHO
H2 / Pd
17.13 Show how you could use a crossed aldol reaction to synthesize cinnamaldehyde .Write a
detailed mechanism for the reaction.
O
+
H2C
O
O
OH
+
HC
Ph
H
H
O
Ph
O
H2
C
C
H
H
+
H
O
H
OH
Ph
C
H
O
H
C
O
H
Ph
C
H
C
H
C
H
H
OH-
17.14 When excess formaldehyde in basic solution is treated with acetaldehyde, the following
reaction takes place, write a mechanism that accounts for the formation of the product.
3HCHO+CH3CHO (dilute NaCO3)
(HOCH2)3CCHO
O
O
H2C
HO
H
H2C
H
H
H2
C
H2C
C
H
H
O
O
C
H
O
H
C
H2C
C
OH
C
H
H
HO
H
O
H
O
O
H2C
OH
C
H
C
H2C
H2C
H
OH
H
H
O
C
H
H
HO
O
H
C
H
O
O
C
C
H
OH
H
OH
O
O
HO
O
O
O
H
H2C
H
C
O
H2C
C
OH
H
OH
H
H2C
C
C
C
C
H
OH
OH
OH
OH
17.15 When pseudoionone is treated with BF3 in acetic acid ,ring closure takes place and αand
β-ionone are produced.This is the next step in the vitamin A synthesis.
(a) Write mechanisms that explain the formation ofαandβ-ionone.
H+
O
O
O
H
H
O
O
+
β-Ionone
α-Ionone
(b) β-Ionone is the major product. How can you explain this?
β-Ionone is a conjugated product.
(c) Which would you expect to absorb at longer wavelengths in the UV-visible region? Why?
Theβ-Ionone will absorb at longer wavelengths in the UV-visible region, because it is a
conjugated molecular.
17.16 Assuming that you have available the required aldehyde, ketone and nitroalkanes, show how
you would synthesize each of the following compounds and write a detailed mechanisms for each
reaction.
（a）
CNO2
PhHC
CH3
O
H
CH3CH2NO2
NaOH
Ph
O H
CH3CHNO2
C6H5CHCNO2
NaOH
warm
T.M.
CH3
(b) HOCH2CH2NO2
CH3NO2
O
NaOH
CH2NO2
O
HCH
H
T.M
CH2CHNO2
17.17 a) Write resonance structures for the anion of acetonitrile that account for its being much
more acidic than ethane.
b) Give a step-by-step mechanism for the condensation of benzaldehyde with acetonitrile.
B:
a)
H
H2C
C
H2C
N
O
C
O
H2C
N
H
OH
C
N
CN
b)
PhCH=CHCN
Ph
H
Ph
H
H2C
C
Ph
H
B:
N
17.18 Assuming that dehydration occurs in all instances, write the structures of the two other
products that might have resulted from the aldol cyclization just given. (One of these products will
have a five-membered ring and the other will have a seven-membered ring.)
O
H
O
O
B:
H
H
O
O
O
O
H
O
O
O
CHO
H
O
CHO
O
B:
H
17.19 What starting compound would you use in an aldol cyclization to prepare each of the
following:
O
O
H3 C
CH3
a)
the starting material is:
O
O
O
H
b)
the starting material is:
O
O
O
O
c)
the starting material is:
17.20 What experimental conditions would favor the cyclization process in the intramolecular
aldor reaction over intermolecular condensation?
Answer:
It is necessary for conditions to favor the intramolecular reaction rather than the
intermolecular one. One way to create these conditions is to use very dilute solutions when we
carry out the reaction. When the concentration of the compound to be cyclized is very low, the
probability is greater that one end of a molecular will react with the other end of that same
molecular rather than with a different molecular.
17.21 staring with ketones and aldehydes of your choice,outline a direct aldol synthesis of each of the following
using lithium enolates.
O
H3C
LiN(i-C3H7)2
a
CH
H
CH
O
LDA
O
O
H
CH3CHO
O
OH
O
b
LDA
CH3CH2CHO
O
O
O
O
H2O
O
c
LDA
O
OH
O
OH
C6H5CHO
O
H2O
OH
O
OH
d
O
LDA
O
a
O
CH3CHO
H 2O
17.22
O
O
O
O
O
LDA
O
O
O
O
H2O
O
OH
O
b
OHC
LDA
O
O
O
O
H2O
17.23
(a) Write a reaction involving a lithium enolate for introduction of the methyl group in the
following compound (an intermediate in a synthesis by E. J. Corey cafestol, an
anti-inflammatory agent found in coffee beans).
O
CH 3
O
Answer:
O
O
O
LiN(i-C3 H7 )3
CH3
CH3 I
CH 3
CH 3
CH3
O
O
O
(b) Dienolates can be formed from β-keto esters using two equivalents of LDA. The dienolate
can be alkylated selectively at the more basic of the two enolate carbons. Write a reaction for
synthesis of the following compound using a dienolate and the appropriate alkyl halide.
O
O
OCH3
(H3C)3 Si
Answer:
I
O
O
(H3C)3Si
OCH3
O
O
2
O
(H3C)3Si
O
LDA
OCH3
OCH3
17.24 Starting with 2-methylcyclohexanone, show how you would use α-selenation in a synthesis
of the following compound:
O
H3 C
Answer:
O
H3C
O
LiN(i-C 3H 7 )3
H3C
O
O
C6 H5SeBr H3C
SeC6H5
H2O2
H3C
25℃
17.25 (a) Propose step-by-step mechanisms for both transformations of the Robinson annulation
sequence just shown. (b) Would you expect 2-methyl-1,3-cyclohexanedione to be more or less
acidic than cyclohexanone? Explain you answer.
Solution:
(a) The mechanisms are as follows.
O
O
O
O
HO
H
O
O
O
O
O
O
OH
H
O
O
O
O
H2C
OH
O
O
H
O
O
heating
O
H
OH
O
O
OH
(b) I think 2-methyl-1,3-cyclohexanedione is more acidic than cyclohexanone because its
enolate anion is stabilized by an additional resonance structure.
OH-
O
O
O
O
H
O
O
OHH
O
O
O
O
O
17.26 What product would you expect to obtain from the base-catalyzed Michael reaction (a) of
1,3-diphenyl-2-propen-1-one and acetophenone? (b) of 1,3-diphenyl-2-propen-1-one and
cyclopentadien? Show all steps in each mechanisms.
Answer:
(a)
O
O
+
O
O
(b) The mechanisms are as follows.
H
O
OH
+
O
OH
H
O
OH
product
17.27 When acrolein reacts with hydrazine, the product is a dihydropyrazole:
CHCHO +
CH2
H2N
N
NH2
N
H
Suggest a mechanism that explains this reaction.
Answer:
NH2
H2C
H2C
H2C
H2N
H
H
H
H2N
O
NH2
HN
O
O
NH2
O
HN
NH2
OH
HN
NH
HN
N
17.28 Give structural formulas for the products of the reaction (if one occurs) when propanal is
treated with each of the following reagents:
(a) OH-,H2O
(b) C6H5CHO,OH(c) HCN
(d) NaBH4
(e) HOCH2CH2OH, p-TSOH
(f) Ag2O,OH-,then H3O+
(g) CH3MgI,then H3O+
(h) Ag(NH3)2+OH-, then H3O+
(i) NH2OH
(j)
C6H5HC
P(C6H5)3
(k) C6H5Li, then H3O+
CNa then H3O+
(l) HC
(m) HSCH2CH2SH,HA, then Raney Ni,H2
(n) CH3CH2CHBrCO2Et and Zn, then H3O+
Answer:
h
a
O
H
OH
OH
O
b
O
N
i
OH
C6H5
CHC6H5
j
c
OH
k
N
l
OH
d
e
OH
OH
O
H
m
H
O
f
CH3CH2COOH
n
COOEt
g
OH
OH
Ph
17.29 Give structural formulas for the products of the reaction (if one occurs) when acetone is
treated with each reagent of the preceding problem.
—
(a) OH-, H2O
(h) Ag(NH3)2+OH , then H3O+
(b) C6H5CHO, OH(i) NH2OH
(c) HCN
(j) C6H5CH-P(C6H5)3
(d) NaBH4
(k) C6H5Li, then H3O+
(e) HOCH2CH2OH, p-TsOH
(l) HCCNa, then H3O+
+
(f) Ag2O, OH , then H3O
(m) HSCH2CH2SH, HA, then Raney Ni, H2
+
(g) CH3MgI, then H3O
(n) CH3CH2CHBrCO2Et and Zn, then H3O+
Answer:
(a)
(h)
o
H
C
N
(j)
CN
(c)
OH
O
(i)
HC
(b)
no reaction
C
CHC6H5
CH3
C
H3C
OH
Ph
H
(d)
CH3
C
H3C
(e)
C
(k)
CH3
H2C
OH
O
O
(l)
(m)
(f)
C
HO
OH
no reaction
(n)
(g)
CH3
C
H3C
CH
O
O
CH3
OH
17.30 What products would form when 4-methylbenzaldehyde reacts with each of the following?
(a) CH3CHO, OH(e) Hot KMnO4, OH-, then H3O+
+
(b) CH3CCNa, then H3O
(f)- CH2-P+(C6H5)3
(c) CH3CH2MgBr, then H3O+
(g) CH3COC6H5, OH(d) Cold dilute KMnO4, OH-, then H3O+
(h) BrCH2CO2Et and Zn, then H3O+
Answer:
COOH
O
(e)
(a)
COOH
OH
(b)
C
(f)
CH
(c)
O
(g)
CH
Ph
(h)
O
HOHC
O
COOH
(d)
17.31 Show how each of the following transformations could be accomplished. You may use any
other required reagents.
C6H5CH=CHCOC(CH3)3
(a) CH3CO(CH3 )3
O
H
H
+
O
H
O
O
H
H
O
H
H
OH-
O
O
H2
C
O
(b)
C6 H5CH
C6H5CHO
O
O
H
OH-
O
H2O
EtOH,
heat
O
C6H5CH
O
C6 H5 CH2CH(CH3)NH2
(c) C6H5CHO
CH3
OH-
CH3CH2NO2
EtOH
H3C
NO2
Ph
O
H2,Pt
heat
C6H5CH
CNO2
C6 H5 CH2CH(CH3)NH2
O
O
(4)
H3C
C
CH2
CH2
CH2
CH2
C
CH3
O
CH3CO(CH2)4 COCH3
OH
OH-
CH3
H3C
heat
O
Aluminum isopropoxide
OH
H3 C
CH3 CN
H3CO
OH-
C
H
C
H
CN
H
CH2CN
O
OCH3
EtOH
heat
H3CO
C
H
C
H
CH3
H3C
O
isopropyl alcohol
(5) CH3CN
EtOH
CN
(6)
CH3(CH2)3CH
CH3CH2CH2CH2CH
C(CH2)2CH3
O
CH2OH
O
HC
CH3CH2CH2CH2 CH
OH-
EtOH
H
H3C
heat
CH3(CH2 )3CH
C(CH2)2CH3
O
H
H3C
O
CH2OH
Aluminum isopropoxide
CH3 (CH2)3 CH
isopropyl alcohol
C(CH2)3CH3
O
(7)
CH
O
O
C6H5
O
O
OH-
EtOH
C
H
heat
C6H5
1. HSCH2CH2 SH
C
H
2. Raney Ni
C6H5
17.32 The following reaction illustrates the Robinson annulation reaction (Section 17.9B). Give
mechanism for the steps that occur.
O
C6H5COCH2CH3
+
H2C
C
CH3
Mechanism:
C
CH3
base
C6H5
O
C6H5
O
O
H
C
C
CH3
C6H5
C
C
H
C6H5
CH3
H
C
C
CH3
H
B:
C6H5
C6H5
O
O
O
H
C
C
C
H2C
CH3
C
CH3
CH3
O
C6H5
C6 H5
C6H5
O
O
O
HB
H2
C
H
O
O
O
B:
C6H5
C6 H5
O
C6H5
O
O-
HB
H
O
O
O
C6H5
OH
C6H5
-H2O
H
O
17.33 Write structural formulas for A,B, and C.
HC
CH
(1)NaNH2,liq,NH3
(2)CH 3COCH3
(3)NH3Cl/H2O
A (C5H8O)
Hg2+,H3O+
B (C5H10O2)
H2 O
CH3
HC
Answer: A:
C
C
OH
CH3
H3C
;
B:
C6H5CHO,OH-
O
CH3
C
C
OH
C (C12H14O2)
CH3
;
C
CH
O
CH3
C
C
H
CH3
OH
C:
17.34
The hydrogen atoms of the γcarbon of crotonaldehyde are appreciably acidic (pKa≈20).
γ
β
H3CHC
α
CHCHO
Crotonaldehyde
(a) Write resonance structures that will explain this fact.
(b) Write a mechanism that accounts for the following reaction:
C6H5HC
CHCHO
+
H3CHC
CHCHO
base
C6H5(CH=CH)3 CHO
EtOH
(87%)
A: (a)
O
H3C
C
H
C
H
C
O
H
H3C
C
H
C
H
O
H3C
(b)
C
H
C
H
C
C
H
O
H
H3C
C
H
C
H
C
H
O
H2C
C
H
C
H
O
C
H
H2C
C
H
C
H
C
H
H
B:
O
O
H2C
C
H
C
H
C
H2C
H
H
C
C
H
O
C6H5
C6H5
C
H
C
H
C
H
C
H
C
O
H
CH
C
H
OH
C
H
O
H
+
H2C
C
H
C
H
C
H
O
C
H
C
H
H
C
H
O
-H2O
C6H5
C
H
C
H
CH
C
H
C
H
C
H
C
T.M
H
17.35 What reagents would you use to bring about each step of the following syntheses.
(a)
O
O
O
(1) O 3
NaOH
(2)Zn , H2O
Heat
CH
(b)
O
(1) O3
(2) Zn ,H2O
(c)
O
O
NaOH
Heat
CH 3
CH3
CH3
OH
KMnO4
CHO
CHO
CHO
cold ,dilute
CH3
HIO4
OH
CH3
NaOH
CHO
(d)
O
C6 H5
CH2CC 6H 5
N
NaOH
N
N
N
O
C
CH2CC 6H 5
C6 H5
O
17.36
(a) Infrared spectroscopy gives an easy method for deciding whether the product obtained
from the addition of a Grignard reagent to an α,β-unsaturated ketone is the simple
addition product or the conjugate addition product. Explain. (What peak or peaks would
you look for ?)
(b) How might you follow the rate of the following reaction using UV spectroscopy?
O
O
+
NH2
NH
(a) If the IR spectroscopy has a absorption near 1710cm-1, the product is the conjugate
addition. Otherwise, simple addition.
(b) You use UV to detect the reaction. Because the reactant is conjugated, it has a strong UV
absorption.
17.37 (a) A compound U (C9H10O) gives a negative iodoform test. The IR spectrum of U shows a
strong absorption peak at 1690 cm-1. The 1H NMR spectrum of U gives the following:
Triplet
δ1.2 (3H)
Quartet
δ3.0(2H)
Multiplet
δ7.7 (5H)
What is the structure of U?
(b) A compound V is an isomer of U. Compound V gives a positive iodoform test; its IR spectrum
shows a strong peak at 1705 cm-1. The 1H NMR spectrum of V gives the following:
Singlet
δ2.0 (3H)
Singlet
δ3.5 (2H)
Multiplet
δ7.1(5H)
What is the structure of V?
O
U.
V.
O
17.38 Compound A has the molecular formula C6H12O3 and shows a strong IR absorption peak at
1710 cm-1. When treated with Tollens’ reagent, no reaction occurs; however, if A is treated
first with water containing a drop of sulfuric acid and then treated with Tollens’ reagent, a
silver mirror forms in the test tube. Compound A shows the following 1H NMR spectrum.
Singlet
δ2.1
Doublet
δ2.6
Singlet
δ3.2(6H)
Triplet
δ4.7
Write a structure for A
The answer is
O
OCH3
OCH3
17.39
Treating a solution of cis-1-decalone with base causes an isomerization to take place.
When the system reaches equilibrium, the solution is found to contain about 95%
trans-1-decalone and about 5% cis-1-decalone。Explain this isomerization.
O
H
H
cis-1-Decalone
Answer:
HO
OH
H
OH
O
O
H
H
H
H
And trans-1-decalone is more stable than cis-1-decalone.
H
O
H
H
H
O
trans-1-decalone
cis-1-decalone
17.40 The Witting reaction can be used in the synthesis of
aldehyde, for example.
CH3
O
CH 3O
CCH3
+
CH 3OCH
P(C6H5)3
CH 3 O
CHOCH3
C
(60%)
H 3O
/H 2O
CH3
CH 3O
C
H
CH
O
(85%)
(a)How would you prepare CH 3OCH
second reaction produces an aldehyde
O
CH
from cyclohexanone ?
Answer:
(a)
P(C 6H 5) 3 ?
(b) Show with a mechanism how the
(c) How could you use this method to prepare
OCH3
OCH3
(C6H5 )3P +
CH2Br
(C6H5)3P
CHOCH3
(C6H5)3P
CH
(b)
CH3
CH3O
CH3
OCH3
C
CH
H
O
CH3O
O
C
H2O
H
H
CH3
C
H
H
CH3
CH3O
C
H
O
CH3
CH3
CH3 O
CH
C
H
H
O
OH2
OH
CH3
CH3
CH3 O
C
H
CH3
CH
CH3O
CH
C
H
O
CH
OH
(c)
O
+ CH3OCH
H3O / H2 O
P(C6H5 )3
CHOCH3
CHO
17.41 Aldehydes that have no α hydrogen undergo an intermolecular oxidation-reduction called
the Cannizzaro reaction when they are treated with concentrated base. An example is the
following reaction of benzaldehyde.
2C6H5
CHO
OH
H 2O
C6 H5
CH2 OH + C6H5
CO2
(a) When the reaction is carried out in D2O, the benzyl alcohol that is isolated contains no
deuterium bound to carbon. It is C6H5CH2OD. What dose this suggest about the mechanism for
the reaction? (b) When (CH3)2CHCHO and Ba(OH)2/H2O are heated in a sealed tube the reaction
produces only (CH3)2CHCH2OH and [(CH3)2CHCO2]2Ba. Provide an explanation for the
formation of these products instead of those expected from an aldol reaction.
Solution:
(a) mechanism:
O
O
C
H
OD
OD
O
H
C
H
OD
C
O
C
O
O
OD
C
+
C
H
O
H2
C
OD
+
H
(b) Although an aldol reaction occurs initially, the aldol reaction is reversible. The Cannizzaro
reaction, though slower, is irreversible. Eventually, all the products is in the form of alcohol and
the carboxylic ion.
O
OH
H
C
H
C
H3C
O
O
C
C
H
CH3
C
CH3
H
H3C
H
C
C'
O
C
C
CH3
CH
H3C
CH3
CH3
CHO
H
CH3
CH3
CH
CH3
H3C
OH
O
CH3
O
H3C
CH
C
H3C
H
H3C
OH
H3C
CH
H3C
C
H
C
H
C
CH
CH3
OH
O
O
O
H3C
CH
O
OH
+
H3C
H3C
H3 C
CH
C
CH
H
H3C
H3 C
C
OH
+
CH
H2
C
OH
H3C
H
17.42 When the aldol reaction of acetaldehyde is carried out in D2O, no deuterium is found is the
methyl group of unreacted aldehyde. However, in the aldol reaction of acetone, deuterium is
incorporated in the methyl group of the unreacted acetone. Explain this difference in behavior.
The difference behavior indicates that, for acetaldehyde, the capture of a proton from the solvent
(the reverse pf the reaction by which the enolate ion is formed) occurs much more slowly than the
attack by the enolate ion on another molecular.
When the acetone is used, the equilibrium for the formation of the enolaate ion is unfavorable, but
more importantly, enolate attack on another acetone molecular is disfavored due to the steric
hindrance. Here proton capture (actually deuteron capture) completes very well with the aldol
reaction.
17.43 Shown below is a synthesis of the elm bark beetle pheromone, multistriatin (see Problem
16.46). Give structures for compounds A,B,C, and D.
O
LiAlH4
TsCl
A (C5H10O)
base
CO2H
C (C10 H18O)
RCO3H
D (C10H18O2)
B (C12H16O3S)
Lewis acid
base
O
O
Answer:
A (C5H10O):
OH
O
S
B (C12H16 O3S):
O
CH3
O
O
C (C10H18O):
O
D (C10 H18O2 ): O
17.44 Allowing acetone to react with 2 molar equivalents of benzaldehyde in the presence of KOH
in ethanol leads to the formation of compound X. The
13
C NMR spectrum of X is given in
Fig.17.2 propose a structure for compound X.
Answer:
O
C
H
C
H
C
C
H
C
H
17.45 The following is an example of a reaction sequence developed by Derin C.D’Amico and
Michael E. Jung (UCLA) that results in enantiospecific formation of an aldol addition product by
non-aldol reactions. The sequence includes a Horner-Wadsworth-Emmons reaction, a sharpless
asymmetric epoxidation, and a novel rearrangement that ultimately leads to the aldol-type product.
Propose a mechanism for rearrangement of the epoxy alcohol under the conditions shown to form
the aldol product. [hint: the rearrangement can also be accomplished by preparing a trialkylsilyl
ether form the epoxy alcohol in a separate reaction first, and then treating the resulting silyl ether
with a lewis acid catalyst]
O
H3C
(1) CH3 CHCO2CH3
OH
PO(OCH3)2
H
(2)DIBAL-H
tert-BuOOH
Ti(O-i-Pr)4
D-(-)-diisopropyl tartrate
TBDMSO O
TBDMSOTf
(tert-butyldimethylsilyl triflate).
1.3 equiv.
H
diisoprooylethylamine. 1.35 eqviv.
°
molecular sieves. -42C
H
O
OH
CH3
CH3
Answer:
O
Si
Si
O
CF 3
O
Si
O
O
H
OH
H
CH3
N
O
O
CH3
H3C
H