7.1 Using the (E)-(Z) designation [and in parts (e) and (f) the (R-S) designation as well], give
IUPAC names for each of the following.
(a)
C
(b)
C
Cl
Br
(c)
C
C
C
H
CH2CH3
Cl
Br
(f)
C
C
I
CH3
H3 C
(e)
CH3
Cl
CH2 CH3
H
C
CH3
(d)
C
I
C
H
CH2CH2CH3
Cl
CH2CH(CH3)2
H3C
H
Br
C
C
H
CH3
H3C
H
Answer:
(a) (Z)-1-Bromo-1-chloro-1-pentene
(b) (Z)-3,5-Dimethyl-2-hexene
(c) (E)-2-Bromo-1-chloro-1-iodo-1-butene
(d) (Z)-1-Chloro-1-iodo-2-methyl-1-butene
(e) (2Z,4S)-3,4-Dimethyl-2-hexene
(f) (1Z,3R)-1-Bromo-2-chloro-3-methyl-1-hexene
7.2 Heats of hydrogenation of three alkenes are as follows:
2-methyl-1-butene (-119kJ/mol)
3-methyl-1-butene (-127kJ/mol)
2-methyl-2-butene (-113kJ/mol)
(a) Write the structure of each alkene and classify it as to whether its doubly bonded atoms are
monosubstituted, disubstituted, trisubstituted, or tetrasubstituted. (b) Write the structure of the
product formed when each alkene is hydrogenated. (c) Can heats of hydrogenation be used to
relate stabilities of these three alkene? (d) If so, what is the predicted order of stability? If not,
why not? (e) What other alkene isomers are possible for these alkenes? Write their structures.
(f) What data would be necessary to relate the stabilities of all these isomers?
Answer:
(a)
2-methyl-1-butene
disubstituted
3-methyl-1-butene
monosubstituted
2-methyl-2-butene
trisubstituted
(b)
2-methyl-butane
(c) Yes. Because the products are the same.
(d) stability: 3-methyl-1-butene<2-methyl-1-butene<2-methyl-2-butene
(e)
1-pentene
2-pentene
(f) heats of combustion and heats of hydrogenation.
7.3 Predict the more stable alkene of each pair. (a) 2-Methyl-2-pentene or
2,3-dimethyl-2-butene, (b) cis-3-hexene or trans-3-hexene, (c) 1-hexene or cis-3-hexene, and
(d) trans-2-hexene or 2-methyl-2-pentene.
Answer:
(a) 2,3-dimethyl-2-butene , (b) trans-3-hexene , (c) cis-3-hexene , (d) 2-methyl-2-pentene
are more stable alkenes of each pair.
7.4 Consider the pairs of alkenes, for which pairs could you use heats of hydrogenation to
determine their relative stabilities? For which pairs would you be required to use heats of
combustion? (a) 2-methyl-2-pentene or 2,3-dimethyl-2-butene, (b) cis-3-hexene or
trans-3-hexene, (c) 1-hexene or cis-3-hexene, (d) trans-2-hexene or 2-methyl-2-pentene
Answer:
To both (b) and (c), we could use heats of hydrogenation to analysis their relative stability, to
(a) and (d), we need to use heats of combustion to analysis their relative stabilisty.
7.5 List the alkenes that would be formed when each of the following alkyl halides is
subjected to dehydrohalogenation with potassium ethoxide and use Zaitsev’s rule to predict
the major product of each reaction: (a) 2-bromo-3-methylbutane and (b)
2-bromo-2,3-dimethylbutane.
Answer:
(a) 2-bromo-3-methylbutane
Br
KOCH2CH3
CH3CH2OH
+
major
minor
(b) 2-bromo-2,3-dimethylbutane.
Br
KOCH2CH3
CH3CH2 OH
+
major
minor
7.6 Consider a simple molecule such as ethyl bromide and show with Newman projection
formulas how the anti periplanar state would be favored over the syn periplanar one.
Answer
B-
H
H
H
H
H
Br
Anti periplanar transition state is staggered, therefore, it’s lower energy.
B-
Br H
H
H
H
H
Syn periplanar transition state is eclipsed, therefore, it has higher energy.
So, the anti periplanar state would be favored over the syn periplanar one.
7.7 When cis-1-bromo-4-tert-butylcyclohexane is treated with sodium ethoxide in ethanol, it
reacts rapidly; the product is 4-tert-butylcyclohexene, under the same condition, transbromo-4-tert-butylcyclohexane reacts very slowly. Write conformational structures and
explain the difference in reactivity of these cis-trans isomers.
Answer:
Br
EtONa / EtOH
H
EtO-
EtOH
Br
Br
7.8 (a) When cis-1-bromo-2-methylcyclohexane undergoes an E2 reaction, two products
(cycloalkenes) are formed. What are these two cycloalkenes, and which would you expect to
be the major product? Write conformational structures showing how each is formed. (b) When
trans-1-bromo-2methylcyclohexane reacts in an E2 reaction, only one cycloalkene is formed.
What is this product? Write conformational structures showing why it is the only product.
Answer:
(a)
Br
+
CH3
CH3
is the major product.
Br
(b)
CH3
To
H
Br
H
H
H Br
H
H
H
H
H3C
H
CH3
CH3
CH3
Therefore,
stereoselective.
should be the only product, because the reaction is
7.9 Dehydration of 2-propanol occurs in 75% H2SO4 at 100oC. (a) Using curved arrows write
all steps in a mechanism for the dehydration. (b) Explain the essential role performed in
alcohol dehydrations by the acid catalyst. (Hint: Consider what would have happen if no acid
were present.)
Answer:
(a) Step1
H
CH3
H3C
C
H
O
+
H
H
H3C
O
CH3
H
C
H
O
H
+
H
H
H
Step 2
CH3
H
C
H
O
CH3
CH
H3C
H
O
+
H
H
H3C
Step 3
H
H
C
H
CH2
H
+
CH
O
H
+
CH
H
H3C
H
O
H
H3C
(b) In step1, an acid-base reaction, a proton is rapidly transferred from the acid to one of
the unshared electron pairs of the alcohol. So step1 will not be occurred without a strong acid;
and so do the following steps.
7.10 Acid-catalyzed dehydration of neopentyl alcohol, (CH3 )3CCH2 OH , yields
2-methyl-2butene as the major product. Outline a mechanism showing all steps in its
formation.
Answer:
Step 1
CH3
H3C
H
C
CH2
+ H
O+
CH3
O
H
H
CH3
H 3C
C
CH2
CH3
OH2
+
Step 2
CH3
H 3C
C
CH3
CH3
CH2
OH2
+
H3 C
O
C
CH3
CH2 +
+
H2 O
+
H2 O
Step 3
H3
C
CH3
H3 C
C
H 3C
CH2
+
CH3
δ+
C
δ
CH3
+
+
CH2
H 3C
CH3
C
CH2
CH3
Step 4
(b)
A-
(a)
H
(a)
CH2
+
H
CH2
C
CH
CH3
CH3
C
H2
C
CH3
minor product
CH3
(b)
CH3
C
C
H
CH3
major product
CH3
7.11 Acid-catalyzed dehydration of either 2-methyl-1-butanol or 3-methyl-1-butanol gives
2-methyl-2-butene as the major product. Write plausible mechanisms that explain these
results.
Because the stability of carbocations is tertiary>secondary>primary, the mechanisms of
the results can be explain in the following way:
OH2
OH
OH
OH2
7.12 When the compound called isoborneol is heated with 50% sulfuric acid, the product of
the reaction is the compound called camphene and not bornylene as one might expect. Using
models to assist you, write a step-by-step mechanism showing how camphene is formed.
Solution:
Mechanism:
OH
H3O+
heat
7.13 Outline all steps in a synthesis of propyne from each of the following:
(c) CH3CHBrCH2Br
(a) CH3COCH3
(d) CH3CH=CH2
(b) CH3CH2CHBr2
Answer:
(a)
Cl
O
PCl5
NaNH2
mineral oil
heat H+
Cl
(b)
Br
Br
NaNH2
NaNH2
Br
(c)
Br
Br
Br
Br
NaNH2
NaNH2
(d)
Br
Br
Br2
Br
NaNH2
NaNH2
Br
7.14 Predict the products of the following acid-base reactions. If the equilibrium would not
result in the formation of appreciable amounts products, you should so indicate. In each case
label the stronger acid, the stronger base, the weaker acid, and the weaker base.
(a) CH3CH=CH2 + NaNH2 →
(b) CH3C≡CH + NaNH2 →
(c) CH3CH2CH3 + NaNH2 →
(d) CH3C≡C:- + CH3CH2OH →
(e) CH3C≡C:- + NH4Cl →
Answer:
(a) CH3CH=CH2 (Weaker acid) + NaNH2 (Weaker base) → CH3CH=CH:- (Stronger base)
+ Na+ + NH3 (Stronger acid) (no appreciable amounts products)
(b) CH3C≡CH (Stronger acid) + NaNH2 (Stronger base) → CH3C≡C:- (Weaker base) +
Na+ + NH3 (Weaker acid)
(c) CH3CH2CH3 (Weaker acid) + NaNH2 (Weaker base) → CH3CH2CH2:- (Stronger base)
+ Na+ + NH3 (Stronger acid) (no appreciable amounts products)
(d) CH3C≡C:- (Stronger base) + CH3CH2OH (Stronger acid) → CH3C≡CH (Weaker acid)
+ CH3CH2O:- (Weaker base)
(e) CH3C≡C:- (Weaker base) + NH4Cl (Weaker acid)→ CH3C≡CH (Stronger acid)+ NH2:(Stronger base) + HCl (no appreciable amounts products)
7.15 Your goal is to synthesize 4,4-dimethyl-2-pentyne. You have a choice of beginning with
any of the following reagents.
CH3
CH3
H3C
C
H3C
CH
C
Br
H3C
C
C
CH
H3C
I
CH3
CH3
Assume that you also have available sodium amide and liquid ammonia. Outline the best
synthesis of the required compound.
Answer:
The best way is that:
CH3
CH3
H 3C
C
C
CH
liq.NH 3
+ N aN H 2
H3C
C
C
C : -- N a +
+ NH 3
CH3
CH3
Then:
CH3
H3C
C
CH3
CH3
C
C:--Na+
+ H3C
I
nucleophilic
substitution
SN2
H3C
C
CH3
C
C
CH3
+ NaI
There is another way to synthesize 4,4-dimethyl-2pentyne,
H3C
C
CH
liq.NH3
+ NaNH2
H3C
C
C:--Na+
nucleophilic
+ H3 C
C
+ NH3
CH3
CH3
H3C
C:--Na+
C
H3C
Br
substitution
C
C
C
CH3
+ NaBr
CH3
CH3
But this synthesis fails when the alkynide ion acts as a base rather than as a nucelophile, and
the major result is an E2 elimination. The products of the elimination are alkene and the
alkyne from which the sodium alkynide was originally formed.
CH3
H3C
C
C:--
+ H3 C
C
CH3
Br
E2
H3C
C
CH
+ H2C
C
CH3
+
Br--
CH3
7.16 What is the index of hydrogen deficiency of 2-hexene?
(a) Of methylcyclopentane?
(b) Does the index of hydrogen deficiency reveal anything about the location of the double
bond in the chain?
(c) About the size of the ring?
(d) What is the index of hydrogen deficiency of 2-hexyne?
(e) In general terms, what structural possibilities exist for a compound with the molecular
formula C10H16?
Answers:
(a) 1
(b) 1
(c) No, it doesn’t.
(d) No, either.
(e) 2
(f) The index of hydrogen deficiency of C10H16 equals 3. So there must be six structural
possibilities for a compound.
(1) It has three double bonds.
(2) It has one triple bond and one double bond.
(3) It has two double bonds and a ring.
(4) It has one triple bond and a ring.
(5) It has one double bond and two rings.
(6) It has three rings. (Perhaps this possibility cannot exist very stably, but I think we
can consider it as well.)
7.17 Zingiberene, a fragrant compound isolated from ginger, has the molecular formula
C15H24 and is known not to contain any triple bonds. (a) What is the index of hydrogen
deficiency of zingiberene? (b) When zingiberene is subject to catalytic hydrogenation using
an excess of hydrogen, 1 mol zingiberene absorbs 3 mol of hydrogen and produce a
compound with the formula C15H30. How many double bonds does a molecule of zingiberene
have? (c) How many rings?
Answer: (a) Index of hydrogen deficiency = 4
(b) It has 3 double bonds.
(c) It has 1 ring.
7.18 Each of the following names is incorrect. Give the correct name and explain your
reasoning.
(a) trans-3-Pentene
(d) 4-Methylcyclobutene
(b) 1,1-Dimethylethene
(e) (Z)-3-Chloro-2-butene
(c) 2-Methylcyclohexene
(f) 5,6-Dichlorocyclohexene
Answer:
(a) thans-2-pentene
Because IUPAC rules that number the main chain beginning with the end of the chain
nearer the double bond when alkenes are named.
(b) 2-Methyl-1-propene
Because IUPAC rules that locate the longest continuous chain of carbon atoms when
alkenes are named
(c) 1-Methylcyclohexene
Because IUPAC rules that number the longest chain beginning with the end of the chain
nearer the substituent when alkenes are named
(d) 3-Methylcyclobutene
Because IUPAC rules that number the longest chain beginning with the end of the chain
nearer the substituent when alkenes are named
(e) (Z)-2-chloro-2-butene
Because IUPAC rules that number the longest chain beginning with the end of the chain
nearer the substituent when alkenes are named
(f) 1,2-Dichlorocyclohexene
Because IUPAC rules that number the longest chain beginning with the end of the chain
nearer the substituent when alkenes are named
7.19 Write a structural formula for each of the following:
(a) 3-Methylcyclobutene
(g) (Z,4R)-4-Methyl-2-hexene
(b) 1-Methylcyclopentene
(h) (E,4S)-4-Chloro-2-pentene
(c) 2,3-Dimethyl-2-pentene
(d) (Z)-3-Hexene
(e) (E)-2-Pentene
(f) 3,3,3-Tribromopropene
Answer:
(a)
(i) (Z)-1-Cyclopropyl-1-pentene
(j) 5-Cyclobutyl-1-pentene
(k) (R)-4-Chloro-2-pentyne
(l) (E)-4-Methyl-4-hexen-1-yne
(g)
3-Methylcyclobutene
(b)
(Z,4R)-4-Methyl-2-hexene
(h)
Cl
1-Methylcyclopentene
(E,4S)-4-Chloro-2-pentene
(c)
(i)
2,3-Dimethyl-2-pentene
(d)
(Z)-1-Cyclopropyl-1-pentene
(j)
(Z)-3-Hexene
(e)
5-Cyclobutyl-1-pentene
(k)
Cl
(E)-2-Pentene
(f)
(R)-4-Chloro-2-pentyne
(l)
Br
Br
Br
3,3,3-Tribromopropene
(E)-4-Methyl-4-hexen-1-yne
7.20 Write three-dimensional formulas for and give names using (R-S) and (E)-(Z)
designations for the isomers of:
(a) 4-Bromo-2-hexene
(b) 3-Chloro-1, 4-hexadiene
(c) 2, 4-Dichloro-2-pentene
(d) 2-Bromo-4-chloro-2-hexen-5-yne
Answer:
Br
(S) - 4-Bromo-2-hexene
(a) 4-Bromo-2-hexene
(Z)
-
4-Bromo-2-hexene
Cl
(b) 3-Chloro-1, 4-hexadiene
4-hexadiene
(R) -3-Chloro-1,
(Z)
-
3-Chloro-1,
4-hexadiene
Cl
Cl
(S) - 2,
(c) 2, 4-Dichloro-2-pentene
4-Dichloro-2-pentene
(E)
-
2,
4-Dichloro-2-pentene
Cl
(d)
Br
(R)
2-Bromo-4-chloro-2-hexen-5-yne
2-Bromo-4-chloro-2-hexen-5-yne
(E)
-2-Bromo-4-chloro-2-hexen-5-yne
7.21 Give the IUPAC names for each of the following:
(c)
(a)
(e)
Cl
H
(b)
Cl
-
(d)
(f)
Answer:
(a) 3,5-Dimethyl-2-hexene
(b) 4-Chloro-3-methyl-cyclopentene
(c) 6-Methyl-3-heptyne
(d) 1-sec-Butyl-2-methyl-cyclohexene
(e) R-3-chloro-hept-4-en-1yne
(f) 6-Methylene-undecane
7.22 Outline a synthesis of propene from each of the following：
（a） Propyl chloride
（b） Isopropyl chloride
（c） Propyl alcohol
（d） Isopropyl alcohol
（e） 1，2-Dibromopropane
（f） Propyne
Answer：
（a）
Base
Cl
heat
(-HCl)
（b）
Cl
C2H5O
Na
C2H5OH
heat
or
Cl
Base
heat
(-HCl)
（c）
OH
H+, heat
(-HOH)
（d）
OH
H+, heat
(-HOH)
（e）
（1）
Br
Zn, CH3COOH
(-ZnBr2)
Br
（2）
Br
+2NaI
acetone
+I2+2NaBr
Br
（f）
H2,Pd/CaCO3
(Lindlar's catalyst)
quinoline
(syn addition)
7.23 Outline a synthesis of cyclopentene from each of the following:
(a) Bromocyclopentane
(b) 1,2-Dichlorocyclopentane
(c) Cyclopentanol
Answer:
(a)
Br
(b)
Cl
Cl
(c)
OH
7.24 Starting with ethyne, outline syntheses of each of the following. You may use any other
needed reagents, and you need not show the synthesis of compounds prepared in earlier parts of
this problem.
(a) Propyne
(b) 1-Butyne
(c) 2-Butyne
(d) cis-2-Butyne
(e) trans-2-Butyne
(f) 1-Pretyne
(g) 2-Hexyne
(h) (Z)-2-Hexene
(i) (E)-2-Hexene
(j) 3-Hexyne
CD
(k) H3CH2CC
H3C
D
CH3
(l)
D
ANSWER:
(a)
+ Cl2
+
Cl
Cl
+
CH3MgCl
(b)
Cl
+
CH3CH2MgCl
(c)
+ 2Cl2
Cl
Cl + 2HCl
HCl
Cl + CH3MgCl
Cl
H 3C
CH 3
(d)
H3C
CH3
H2
+
Pd/BaSO4
H3C
CH3
H3C
CH3
(e)
CH3CH2MgCl
+
H3C
CH3
Na/NH3
H2
+
(f)
Cl
+
CH3CH2CH2MgCl
(g)
+
Cl2
Cl
Cl
+ CH CH CH MgCl
3
2
2
(h)
+
H2
+
H2
Pd/BaSO4
(i)
(j)
Na/NH3
+
CH3CH2MgCl
(k)
Na
Na
Na
D2 O
CD
(l)
Pd/BaSO4
H3C
CH3
D2
D
D
7.25 Starting with 1-methylcyclohexene and using any other needed reagents, outline a
synthesis of the following deuterium-labeled compound.
D
CH3
D
H
D
Ni
+ D2
CH3
D
H
7.27 Outline a synthesis of phenylethyne from each of the following:
(a) 1,1-Dibromo-1-phenylethane
(b) 1,1-Dibromo-2-phenylethane
(c) phenylethene (styrene)
(d) Acetophenone (C6H5COCH3)
Answer:
Br
C
(1) 3 NaNH2
C
Br
(a)
CH3
mineral oil
(2)H+
heat
CH
Br
H2
C
C
(1) 3 NaNH2
CH
CH
Br
mineral oil
(2)H+
heat
(b)
(c)
C
H
CH2
Br2
CCl 4
CH
CH2Br
(1) 3 NaNH 2
mineral oil
Br
C
CH
C
CH
heat
(2) H +
(d)
Cl
O
C
CH3
PCl5
0 .C
C
CH3
(1) 3 NaNH 2
mineral oil
Cl
heat
(2) H+
7.29 Match the following heats of combustion (3375 kJ mol-1, 3369 kJ mol-1, 3365 kJ mol-1,
3361 kJ mol-1, 3355 kJ mol-1) with the following alkenes: cis-2-pentene, trans-2-pentene,
2-methyl-2-butene, 1-pentene, 2 -methyi-1-butene.
Answer:
cis-2-pentene′combustion is 3369 kJ mol-1
trans-2-pentene′combustion is 3361 kJ mol-1
2-methyl-2-butene′combustion is 3355 kJ mol-1
1-pentene′combustion is 3375 kJ mol-1
2-methyl-1-butene′combustion is 3365 kJ mol-1
7.30 Without consulting tables, arrange the following compounds in order of decreasing
acidity.
Pentane
1-pentene
1-pentyne 1-pentanol
Answer: pentane <1-pentene<1-pentyne<1-pentanol
7.31 Write structural formulas for all the products that would be obtained when each of the
following alkyl halides is heated with sodium ethoxide in ethanol. When more than one
product results you should indicate which would be the major product and which would be
the minor product(s). You may neglect cis-trans isomerism of the products when answering
this question.
(a) 2-Bromo-3-methylbutane
(b) 3-Bromo-2, 2-dimethylbutane
(c) 3-Bromo-3-methylpentane
(d) 1-Bromo-1-methylcyclohexane
(e) Cis-1-Bromo-2-ethylcyclohexane
(f) trans-1-Bromo-2-ethylcyclohexane
Answer:
(a) Structural formulas for all the products of (a).
1.
2.
3.
4.
OEt
EtO
The second structure is major product. The third and forth structure just very little.
(b) Structural formulas for all the products of (b).
1.
2.
3.
OEt
OEt
The first structure is major product.
(c) Structural formulas for all the products of (c).
1.
2.
OEt
The first structure is major product.
(d) Structural formulas for all the products of (d).
1.
2.
3.
OEt
The first structure is major product.
(e) Structural formulas for all the products of (e).
1.
2.
3.
4.
OEt
OEt
The first structure is major product.
(f) Structural formulas for all the products of (f)
1.
2.
3.
4.
OEt
OEt
The second structure is major product.
7.32 Write structural formulas for all the products that would be obtained when each of the
following alkyl halides is heated with potassium tert-butoxide in tert-butyl alcohol. When
more than one product results you should indicate which would be the major product and
which would be the minor product(s). You may neglect cis-trans isomerism of the products
when answering this question.
(a) 2-Bromo-3-methylbutane
(b) 4-Bromo-2,2-dimethylbutane
(c) 3-Bromo-3-methylpentane
(d) 4-Bromo-2,2-dimethylpentane
(e) 1-Bromo-1-methylcyclohexane
Answer:
According to the Hofmann rule, when potassium tert-butoxide reacts with alkyl halides as an
E2 reaction, the products are mainly the less substituted alkenes rather than the more
substituted alkene which is more stable. Therefore the solution to this problem is as follows:
(a) Major product
H 2C
C
H
H
C
CH3
CH3
3-methyl-1-butene
Minor product
H3C
C
H
C
CH3
2-methyl-2-butene
CH3
(b)
Only one product
CH3
H2C
C
H
C
CH3
CH3
3, 3-dimethyl-1-butene
(c)
Major product
3-Methylene-pentane
Minor product
3-methyl-2-pentene
(d)
Major product
2, 2-dimethyl-4-pentene
Minor product
2, 2-dimethyl-3-pentene
(e)
Main product
1-methylene-cyclohexane
Minor product
1-Methyl-cyclohexene
7.33 Starting with an appropriate alkyl halide and base, outline syntheses that would yield
each of the following alkenes as the major (or only) product.
(a) 1-Pentene
(b) 3-Methyl-1-butene
(c) 2,3-Dimethyl-1-butene
(d) 4-Methylcyclohexene
(e) 1-Methylcyclopentene
Answer:
C2H5ONa
Cl
C2H5OH
1-Chloro-pentane
Pent-1-ene
C2H5ONa
C2H5OH
Cl
1-Chloro-3-methyl-butane
3-Methyl-but-1-ene
C2H5ONa
Cl
C2H5OH
2,3-Dimethyl-but-1-ene
1-Chloro-2,3-dimethyl-butane
CH3
CH3
C2H5ONa
C2H5OH
4-Methyl-cyclohexene
Cl
1-Chloro-4-methylcyclohexane
CH3
CH3
Cl
C2H5ONa
C2H5OH
1-Chloro-1-methyl-cyclopentane
1-Methyl-cyclopentene
7.34 Arrange the following alcohols in order of their reactivity toward acid-catalyzed dehydration
(with the most reactive first):
1-pentanol
2-methyl-2-butanol
3-methyl-2-butanol
Answer:
2-methyl-2-butanol ﹥3-methyl-2-butanol ﹥1-pentanol
Because the ease with which alcohols undergo dehydration is in the following order:
(R)3COH﹥(R)2CHOH﹥RCH2OH
7.35 Give the products that would be formed when each of the following alcohols is subjected
to acid-catalyzed dehydration. If more than one product would be formed, designate the
alkene that would be the major product. (Neglect cis-trans isomerism).
(a) 2-Methyl-2-propanol
(b) 3-Methyl-2-butanol
(c) 3-Methyl-3-pentanol
(d) 2,2-Dimethyl-1-propanol
(e) 1,4-Dimethylcyclohexanol
H+
OH
(a)
OH2
H
2-Methyl-2-propanol
OH
OH2
H+
(b)
1,2-H shift
3-Methyl-2-butanol
(c)
OH
H
more stable
H
H+
3-Methyl-3-pentanol
(d)
Methyl group
transfer
H+
OH
H
rearrangement
more stable
2,2-Dimethyl-1-propanol
OH
H+
(e)
OH
H
1,4-Dimethylcyclohexanol
7.36
1- Bromobicicyclo [2’2’1]heptane does not undergo elimination (below) when heated with
base. Explain this failure to react.(construction of models may help.)
Br
don't
(heat base)
和双键相连的 6 个原子需在同一平面内，原料如消去，产物双键两端的 6 个原子分占
在两个相互垂直的平面内，能量很高，不稳定。且形成π键的条件是两原子的 P 轨道肩并
肩进行组合，而产物中的两个 P 轨道互相成 90 度角，不能重叠，即不能形成π键。故不会
有如上反应。
7.37 When the deuterium-labeled compound shown below is subjected to
dehydrohalogenation using sodium ethoxide in ethanol, the only alkene product in
3-methyl-cyclihexene. Provide an explanation for the result.
Answer:
H
Br
CH3
H
H
H
CH3
D
-
OEt
7.39 Cholesterol is an important steroid found in nearly all body tissues; it is also the major
component of gallstones. Impure cholesterol can be obtained from gallstones by extracting
them with an organic solvent. The crude cholesterol thus obtained can be purified by (a )
treatment with Br2 in CCl4 (b ) careful crystallization of the product, and (c) treatment of the
latter with zinc in ethanol .What reactions are involved in this procedure?
H2
C
H3 C
CH3
CH
H2
C
C
H2
CH3
CH
CH3
CH3
HO
(a ) addition reaction
Cholesterol
H2
C
H3C
CH
H2
C
C
H2
CH3
CH3
H2
C
H3C
CH
CH3
CH
H2
C
C
H2
CH3
CH3
CH3
CH
CH3
CH3
Br2
CCl4
HO
HO
Br
Br
( c) debromination reaction
H2
C
H3C
CH
CH3
H2
C
C
H2
CH3
CH
CH
CH3
CH3
H2
C
C
H2
CH3
CH
CH3
CH3
Zn, EtOH
HO
H2
C
H3C
CH3
HO
Br
Br
7.40 Caryophyllene, a compound found in oil of cloves, has the molecular formula C15H24 and
has no triple bonds. Reaction of caryophyllene with an excess of hydrogen in the presence of
a platinum catalyst produces a compound with the formula C15H28. How many (a) double
bonds, and (b) rings does a molecule of caryophyllene have?
Answer:
The index of hydrogen deficiency of C15H24 are 4, and the index of hydrogen deficiency of
C15H28 is 2. The index of hydrogen deficiency of a double bond is 1, which is the same as
that of a ring. The ring will not be destroyed by the addition of hydrogen with platinum acting
as catalyst.
So there should be two double bonds and two rings in the molecule of caryophyllene.
7.41 Squalene, an important intermediate in the biosynthesis of steroids, has the molecular
formula C30H50 and has no triple bonds. (a) What is the index of hydrogen deficiency of
squalene? (b) Squalene undergoes catalytic hydrogenation to yield a compound with the
molecular formula C30H62. How many double bonds does a molecule of squalene have? How
many rings?
Answer:
(a) 2*30+2=62
62-50=12
12/2=6, so the index of hydrogen deficiency of the
molecular are 6.
(b) The ring will not be destroyed by adding hydrogen. And the index of hydrogen
deficiency of C30H62 is 0. So there is no ring in the molecular. The index of hydrogen
deficiency of a double bond is 1. So there are 6 double bonds.
7.42 Consider the interconversion of cis-2-butene and trans-2-butene. (a) What is the value of
△ Ho for the reaction, cis-2-butene →trans-2-butene? (b) Assume △ Ho ≈ △ Go. What
minimum value of △G+ would you expect for this reaction? (c) Sketch a free-energy
diagram for the reaction and label △Go and △G+.
Answer: (a) 5 KJ / mol.
(b) 284 KJ / mol
284kj/mol
H3C
CH3
C
C
H
H3C
5kj/mol
H
H
C
(c)
C
H
CH3
7.44 Compound I and J both have the molecular formula C7H14. Compounds I and J are both
optically active and both rotate plane-polarized light in the same direction. On catalytic
hydrogenation I and J yield the same compound K (C7H16). Compound K is optically active.
Propose possible structures for I, J, and K.
1
2
H
C2H5
H3C
I
C
H
C
H
CH3
H2
C
C
H
CH2
J
CH2
H
H2
C
H2
C
CH3
K
H
7.45
C
H
C2H5
C2H5
H3C
H2
C
H3C
H
K
CH2
H
C2H5
J
C
H
C2H5
H
H3C
H2
C
H3C
I
H2
C
H3C
H2
C
CH3
C2H5
Compounds Land M both have the molecular formula C7H14. Compounds L and M are
optically inactive , are nonresolvable, and are diastereomers of each other . Catalytic hydrogenation of
either L or M yields N. Compound N is optically inactive but can be resolved into separate
enantiomers . Propose possible structures for L ,M, and N.
The answer:
The possible structures for L, M
C2H5
C
C
C
C
C2H5
C2H5
C2H5
H
H
CH3
CH3
The possible structure for N:
C
H
7.46
(a)
Partial
H
C2H5
CH3
C 2 H5
H
C
C2H5
H
H
C
+
H
dehydrohalogenation
C2H5
of either
C
CH3
(1R,2R)-1,2-dibromo-1,2-diphenylethane
or
(1S,2S)-1,2-dibromo-1,2-diphenylethane enantiomers (or a race mate of the two) produces
(Z)-1-bromo-1,2-diphenylethene as the product, whereas (b) partial
(1R,2S)-1,2-dibromo-1,2-diphenylethane
(the
meso
dehydrohalogenation of
compound)
gives
only
(E)-1-bromo-1,2-diphenylethene. (c) Treating (1R,2S)-1,2-dibromo-1,2-diphenylethane with sodium
iodide in acetone produces only (E)-1,2-diphenylethe. Explain these results.
Answer:
(a)
B
H
-
H
1R
H
1R
H
2R
Br
Br
Br
(1R,2R)-1,2-dibromo-1,2-diphenylethane
H
(Z)-1-bromo-1,2-diphenylethene
Br
-1
B
H
H
1S
H
2S
Br
1S
Br
Br
(1S,2S)-1,2-dibromo-1,2-diphenylethane
Br
(Z)-1-bromo-1,2-diphenylethene
H
(b)
H
B
H
-
H
1R
H
1R
2S
H
Br
Br
Br
(1R,2S)-1,2-dibromo-1,2-diphenylethane
(E)-1-bromo-1,2-diphenylethene
Br
H
(c)
I
-
Br
H
H
1R
2S
H
acetone
Br
H
(E)-1,2-diphenylethe
(1R,2S)-1,2-dibromo-1,2-diphenylethane
Problem 7.48
Predict the structures of compounds A, B, and C:
A is an unbranched C6 alkyne that is also a primary alcohol.
B is obtained from A by use of hydrogen and nickel boride catalyst or dissolving metal reduction.
C is formed from B on treatment with aqueous acid at room temperature. This compound C has no
infrared absorption in either the 1620 to 1680 cm-1 or the 3590 to 3650 cm-1 region. It has an index
of hydrogen deficiency of one and has one stereocenter but forms as the race mate.
Answer:
A:
B:
H 3C
H2
C
H2
C
C
C
H2
C
OH
H3C
OH
CH2
CH2
H2C
C
C
H
H
C:
H 3C
H2
C
H2
C
H2
C
H
C
CH2
O
7.49 what is the index of hydrogen deficiency for:
(a) C7H10O2
(b) C5H4N4
Answer:
(a) the index is 3
(b) the index is 6
13.24 which diene and dienophile would you employ in a synthesis of each of the following?
O
CH3
O
H
CH
C
(a)
(d)
O
C
H
O
O
CN
C
O
CH3
(e)
(b)
C
O
CH3
O
O
(c)
C
O
CH3
C
O
CH3
(f)
C
O
O
answer :
(a)
O
CH
and
(b)
(c)
OCH3
O
COCH3
and
COCH3
O
(c)
O
C
and
H3C
O
C
O
(d)
O
C
and
O
C
O
(e)
CN
and
(f)
O
CH3
and
COCH3
O