organic ps chapter 7
TRANSCRIPT
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.
C CCl
Br H
CH2CH2CH3
(a)
C C
H3C
H
CH2CH(CH3)2
CH3
(b)
C C
Cl
I
Br
CH2CH3
(c)
C C
Cl
I
CH3
CH2CH3
(d)
C
H3C
H CH3
C
CH3H
(e)
C
Br
H
Cl
C
H3CH
(f)
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
KOCH2CH3+
Br
CH3CH2OH
major minor
(b) 2-bromo-2,3-dimethylbutane.
BrKOCH2CH3CH3CH2OH
+
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
Br
H HH
HH
B-
Anti periplanar transition state is staggered, therefore, it’s lower energy.
B-
H
Br
H
H
HH
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, trans- bromo-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-
Br
Br
H
EtO-
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.
(b) To
Br
CH3
H
Br
H
H3C
H
H
Br
CH3
HH
H
H
CH3
H
Therefore,
CH3
should be the only product, because the reaction is stereoselective. 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
H3C CH
CH3
O H + H O
H
H
H3C CH
CH3
O H
H
+ H O
H
Step 2
H3C CH
CH3
O H
H
H3C
CH
CH3
+ O H
H
Step 3
H3C
CH
C
H
H H
+ O H
H
H3C
CH
CH2
+ H O
H
H
(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)3CCH2OH , yields 2-methyl-2butene as the major product. Outline a mechanism showing all steps in its formation. Answer: Step 1
H3C C
CH3
CH3
CH2
O H
+ H O+
H
H
H3C C
CH3
CH3
CH2
OH2+
+ H2O
Step 2
H3C C
CH3
CH3
CH2
OH2+
H3C C
CH3
CH3
CH2++ H2O
Step 3
H3C C
CH3
CH3
CH2+H3C C
CH3
CH2
H3C
δ+ δ+
H3C C
CH3
CH2
CH3
+
Step 4
A-
H CH2 C
CH3
CH
H
CH3
+(a)
(b)(a)
(b)
CH2 CH2C
CH3
CH3minor product
CH3 C CH
CH3
CH3 major product
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:
OH OH2
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:
OHH3O+
heat
7.13 Outline all steps in a synthesis of propyne from each of the following: (a) CH3COCH3 (c) CH3CHBrCH2Br (b) CH3CH2CHBr2 (d) CH3CH=CH2
Answer: (a)
O
PCl5
Cl
Cl
NaNH2mineral oilheat H+
(b)
Br
BrNaNH2
Br NaNH2
(c)
NaNH2
Br
NaNH2
BrBr Br
(d)
NaNH2NaNH2
BrBr
Br
Br2
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.
H3C C CH H3C C
CH3
CH3
Br H3C C
CH3
CH3
C CH H3C I
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:
H 3C C
CH3
C
CH3
CH N aN H2+liq.NH 3 H3C C
CH3
CH3
C C:--Na+ + NH 3
Then:
H3C C
CH3
CH3
C C:--Na+ H3C I+nucleophilicsubstitutionSN2
H3C C
CH3
CH3
C C CH3 + NaI
There is another way to synthesize 4,4-dimethyl-2pentyne,
H3C C CH + NaNH2liq.NH3 H3C C C:--Na+ + NH3
H3C C C:--Na+ + H3C C
CH3
CH3
Brnucleophilic
substitutionH3C C
CH3
CH3
C C CH3 + NaBr
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.
H3C C C:-- + H3C C
CH3
CH3
Br E2 H3C C CH + H2C C CH3
CH3
+ Br--
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 (i) (Z)-1-Cyclopropyl-1-pentene (d) (Z)-3-Hexene (j) 5-Cyclobutyl-1-pentene (e) (E)-2-Pentene (k) (R)-4-Chloro-2-pentyne (f) 3,3,3-Tribromopropene (l) (E)-4-Methyl-4-hexen-1-yne Answer: (a) (g)
3-Methylcyclobutene (Z,4R)-4-Methyl-2-hexene (b) (h)
1-Methylcyclopentene Cl
(E,4S)-4-Chloro-2-pentene (c) (i)
2,3-Dimethyl-2-pentene (Z)-1-Cyclopropyl-1-pentene (d) (j)
(Z)-3-Hexene 5-Cyclobutyl-1-pentene (e) (k)
(E)-2-Pentene
Cl(R)-4-Chloro-2-pentyne
(f) (l)
BrBr
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:
(a) 4-Bromo-2-hexene
Br
(S) - 4-Bromo-2-hexene (Z) -
4-Bromo-2-hexene
(b) 3-Chloro-1, 4-hexadiene
Cl
(R) -3-Chloro-1, 4-hexadiene
(Z) - 3-Chloro-1, 4-hexadiene
(c) 2, 4-Dichloro-2-pentene
Cl Cl
(S) - 2,
4-Dichloro-2-pentene (E) - 2,
4-Dichloro-2-pentene
(d) 2-Bromo-4-chloro-2-hexen-5-yne
Cl Br
(R) -
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:
(a) (c)
(e) ClH (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)
Cl
Base 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
Br
Zn, CH3COOH
(-ZnBr2)
(2)
Br
Br
+2NaIacetone
+I2+2NaBr
(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)
ClCl
(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 (k) H3CH2CC CD
(l)
H3C
D
CH3
D
ANSWER: (a)
+ Cl2 Cl + HCl
Cl + CH3MgCl (b)
Cl + CH3CH2MgCl
(c)
+ 2Cl2 Cl Cl + 2HCl
CH3MgClCl Cl + H3C CH3
(d)
H3C CH3 + H2 Pd/BaSO4H3C CH3
(e)
+ CH3CH2MgCl
H3C CH3 + H2
H3C CH3Na/NH3
(f)
Cl CH3CH2CH2MgCl+
(g)
+ Cl2 Cl
CH3CH2CH2MgClCl +
(h)
+ H2Pd/BaSO4
(i)
Na/NH3+ H2
(j)
+ CH3CH2MgCl
(k) Na
Na
Na D2O CD
(l)
D2
Pd/BaSO4 H3C
D D
CH3
7.25 Starting with 1-methylcyclohexene and using any other needed reagents, outline a synthesis of the following deuterium-labeled compound.
D
CH3
H
D
+ D2Ni
D
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:
(a)
C CH3
Br
Br
(1) 3 NaNH2
mineral oil heat(2)H+
C CH
(b)
H2C CH
(1) 3 NaNH2
mineral oil heat(2)H+
C CH
Br
Br
(c)
CH
CH2 Br2
CCl4
C CHCH
Br
CH2Br (1) 3 NaNH2
mineral oil
(2) H+
heat
(d)
C CH3C CHC
Cl
CH3 (1) 3 NaNH2
mineral oil
(2) H+
heat
O
PCl5
0.C
Cl
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
H2C CH
HC CH3
CH3 3-methyl-1-butene
Minor product CH
CH3C CH3
CH3 2-methyl-2-butene (b) Only one product
H2C CH
C CH3
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:
Cl1-Chloro-pentane Pent-1-ene
C2H5ONa
C2H5OH
Cl
C2H5ONa
C2H5OH1-Chloro-3-methyl-butane 3-Methyl-but-1-ene
ClC2H5ONa
C2H5OH
1-Chloro-2,3-dimethyl-butane 2,3-Dimethyl-but-1-ene
CH3
CH3
Cl
C2H5ONa
C2H5OH
4-Methyl-cyclohexene
1-Chloro-4-methyl-cyclohexane
CH3CH3
C2H5ONa
C2H5OH
Cl
1-Chloro-1-methyl-cyclopentane1-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
OH H+OH2 H
OHH+ OH2 H
OH H
OH
H
OH
OH H
H+
H+
H+
(a)
1,2-H shift(b)
(c)
Methyl group transfer(d)
(e)
2-Methyl-2-propanol
3-Methyl-2-butanol
3-Methyl-3-pentanol
2,2-Dimethyl-1-propanol
1,4-Dimethylcyclohexanol
more stable
rearrangementmore stable
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.)
don't(heat base)
Br
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:
CH3
Br
D
HH
H
-OEt
H CH3
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?
CH3
HO
CH3CH
H3CH2C
CH2
H2C
CH
CH3
CH3
Cholesterol (a ) addition reaction
CH3
HO
CH3CH
H3CH2C
CH2
H2C
CH
CH3
CH3
BrBr
CH3
HO
CH3CH
H3CH2C
CH2
H2C
CH
CH3
CH3
Br2
CCl4
( c) debromination reaction
CH3
HO
CH3CH
H3CH2C
CH2
H2C
CH
CH3
CH3
BrBr
CH3
HO
CH3CH
H3CH2C
CH2
H2C
CH
CH3
CH3
Zn, EtOH
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
(c)
C C
H3C
H
CH3
HC C
H3C
H
H
CH3
284kj/mol
5kj/mol
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.
C2H5
CH
H3C
H
CH
CH3
H3C
C2H5
H
H2C C
HCH2
H3C
H
C2H5
H2C C
HCH2
H3C
H
C2H5
H2C C
HCH2
H3C
C2H5
H
H2C
H2C CH3 H3C
H
C2H5
H2C
H2C CH3
1 2
I
J
K
I
J
K
7.45 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
C C
C2H5
CH3C2H5
H
C C
C2H5
CH3H
C2H5
The possible structure for N:
C C
H H
C2H5C2H5H CH3
+ C C
H C2H5H H
C2H5 CH3 7.46 (a) Partial dehydrohalogenation of either (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 dehydrohalogenation of (1R,2S)-1,2-dibromo-1,2-diphenylethane (the meso 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)
H
BrBr
H1R
2R
H
Br
1RH
B -
Br
H
H
BrBr
H1S
2SH
Br
1S
H
B-1
Br
H
(1R,2R)-1,2-dibromo-1,2-diphenylethane
(1S,2S)-1,2-dibromo-1,2-diphenylethane
(Z)-1-bromo-1,2-diphenylethene
(Z)-1-bromo-1,2-diphenylethene
(b)
H
BrBr
H1R
2S
H
Br
1R
H
B -
Br H
(1R,2S)-1,2-dibromo-1,2-diphenylethane
(E)-1-bromo-1,2-diphenylethene
(c)
Br
BrH
H1R
2S
(1R,2S)-1,2-dibromo-1,2-diphenylethane
acetone
I-
H
H
(E)-1,2-diphenylethe
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: H3CH2C
H2C C C
H2C OH
B:
H3CCH2
H2CC C
CH2
OH
H H
C:
H3CH2C
H2C
H2C
HC 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?
CH
O
(a)
(b)C
C
O
O
O
O
CH3
CH3
(c)C
C
O
O
O
O
CH3
CH3
(d)
CH3
H
HC
C
O
O
O
(e)
CN
(f)
CO
OCH3
answer : (a)
andCH
O
(b) (c)
(c)
andCOCH3
O
COCH3
O
and
C
O
C
O
OH3C
O CH3
(d)
andC
C
O
O
O
(e)
and
CN
(f)
andCOCH3
O