8.1 Give the structure and name of the product that would be obtained from the ionic addition of
IBr to propene.
Answer:
Br
I
I
I
Br
Br
8.3 Provide mechanistic explanations for the following observations：
（a）The addition of hydrogen
chloride to 3-methyl-1-butene produces two products ： 2-chloro-3-methylbutane and
2-chloro-2-methylbutane.
(b) The addition of hydrogen chloride to 3,3-dimethyl-1-butene produces two products:
3-chloro-2,2-dimethylbutane and 2-chloro-2,3-dimethylbutane. (The first explanations for the
course of both of these reactions were provide by F. Whitmore; see Section 7.7A.)
Answer:
(a) step 1:
C+
H+
;
C+
C+
C+
C+
Because the
much more stable then the
step 2:
Cl
C+
+ Cl;
C+
+ Cl+
Me
Cl
(b) step 1:
C+
H+
rearrangement:
C+
C+
C+
C+
much more stable then the
Because the
Step 2:
Cl
C+
+ Cl-
;
C+
+ Cl-
Me
Cl
8.4 In one industrial synthesis of ethanol, ethene is first dissolved in 95% sulfuric acid. In a second
step water is added and the mixture is heated. Outline the reactions involved.
Answer:
H
OSO3H
OH
8.6
Outline all steps in a mechanism showing how 2,3-dimethyl-2-butanol is formed in the
acid-catalyzed hydration of 3,3-dimethyl-1-butene.
Answer:
STEP 1
CH3
H3C
H
C C CH2
H
CH3
+
CH3
H
C C CH3
slow
H O H
H3C
O H
+
H
CH3
STEP2
H3 C
CH3
H
C C
CH 3
H
C C
FAST
CH3
H3C
CH3
CH 3
CH 3
(Rearrangement)
S TE P 3
H3C
H3C
C
H
C
H3C
FAST
+
C H3
O
C H3
H
H3C
H
C
H O
H
C
C H3
C H3
H
STEP4
H3C
H3C
H
C C CH3
+
FAST
O
H
H O CH3
H
H
H3C
H
H
+
C C CH3
H O H
H3C
HO CH3
8.7 The following order of reactivity is observed when the following alkenes are subjected to
acid-catalyzed hydration.
(H3C)2C
CH2 > H3CHC
CH2 > H2C
CH2
Explain this order of reactivity.
Answer: The mechanism of hydration is to form carbocation. For the stability of the carboation is:
(H3C) 2C
So the (H3C) 2C
(H3C)2C
CH3 > H3CHC
CH3 > H2C
CH3
CH3 is the most stable, has the lowest energy, and the
CH2 has the highest reactivity. And the H2C
CH2 has the lowest
reactivity.
8.8 When 2-methylpropene (isobutylene) is dissolved in methanol containing a strong acid, a reaction
takes place to produce tert-butyl methyl ether, CH3OC (CH3)3. Write a mechanism that accounts for
this.
Answer:
H+
C
H2C
+
CH3
H3C
C+
CH3
CH3
CH3
H3C
C+
CH3
O
+
CH3
CH3
H
H
H3C
O+
C
CH3
CH3
CH3
A-
+
H
H3C
O+
C
CH3
O
CH3
H3C
CH3
C
CH3
CH3
CH3
8.9 In section 8.7 you studied a mechanism for the formation of one enantiomer of trans-1,
2-dibromocyclopentane when bromine adds to cyclopentene. Now write a mechanism showing
how the other enantionmer forms.
Br
Br
Br2
+
Br
Br8.10 Outline a mechanism that account for the formation of trans-2-bromocyclopentanol and its
enantiomer when cyclopentene is treated with an aqueous solution of bromine.
H2O
OH
OH
+
Br
Br
Br
8.11 When ethene gas is passed into an aqueous solution containing bromine and sodium chloride,
the products of the reaction are BrCH2CH2Br, BrCH2CH2OH, and BrCH2CH2Cl. Write
mechanisms showing each product is formed;
Answer:
Step1:
Br
+
H2C
CH2
Br
+
_
Br
Step2:
Br
H
+
H
Br
Br
Br
H
Br
+
H
H
H
Br
Cl
Cl
H
H
Br
H
+
Br
OH2
H 2O
H
H
H
H
H
H
Br
H
OH2
+
H
H2 O
Br
OH
+
H3O
H
H
H
8.12 What products would you expect from each of the following reactions?
KOC(CH3)3
CHCl3
(a) trans-2-butene
(b)
Cyclopentene
KOC(CH3)3
CHBr3
CH2I2/Zn(Cu)
(c) cis-2-Butene
diethylether
Answer:
(a)
H3C
H3C
H
H
C
CH3
C
KOC(CH 3)3
C
H
C
CHCl3
H
C
CH3
Cl
+ enantiomer
Cl
(b)
KOC(CH3)3
CHBr3
H
(c)
C
Br2
H
H
H
H
C
H3C
CH2I2/Zn(Cu)
C
H
H3C
CH3
C
C
diethylether
CH3
C
H H
8.13 Starting with cyclohexene and using any other needed reagents, outline a synthesis of
7,7-dibromobicyclo[4.1.0]heptane.
Answer:
KOC(CH3)3
Br
Br
CHBr3
8.14 Treating cyclohexene with 1,1-diiodoethane and a zinc-copper couple leads to two isomeric
products. What are their structures?
Answer:
I
Zn(Cu)
+
CH
CH3
I
H3C
CH3
H
H
8.15 Starting with an alkene, outline syntheses of each of the following:
CH3
OH
OH
CH3
(a)
CH2CH3
OH
OH
(b)
H
OH
H
(c)
OH
Answer:
(a)
CH3
OH
KMnO4
OH-, H2 O
OH
CH3
(b)
H2C
OH
KMnO4
+ enantiomer
OH-, H2 O
OH
(c)
KMnO4
+ enantiomer
OH-, H2 O
OH
8.16 Explain the following facts: (a) Treating (Z)-2-butene with OsO4 in pyridine and then NaHSO3 in
water gives a diol that optically inactive and cannot be resolved. (b) Treating (E)-2-butene with OsO4
and then NaHSO3 gives a diol that optically inactive but can be resolved into enantiomers.
OH
H3C
CH3
CH3
H3C
OsO4
H
H
H
H3C
H
HO
H
meso
OH
NaHSO3
HO
CH3
H
OH
CH3
HO
H3C
H
H
OH
H
OH
CH3
H3C
H
H
CH3
(+)
OsO4
NaHSO3
H
OH
CH3
CH3
H
H
OH
CH3
OH
HO
H
CH3
H3C
H
8.17 Write the structure of the alkenes that would produce the following products when treated
with ozone and then with zinc and acetic acid.
( a ) CH3COCH3 and CH3CH(CH3)CHO
CH3
CH3
H3C
C
C
C
CH3
H
H
( b ) CH3CH2CHO only ( 2 mol produced from 1 mol of alkene )
H2
H2
C
C
CH3
C
C
H3C
H
H
O
and
(c)
HCHO
CH2
8.19 Alkynes A and B have the molecular formula C8H14. Treating either compound A or B with
excess hydrogen in the presence of a metal catalyst leads to the formation of octane. Similar
treatment of a compound C(C8H12) leads to a product with the formula C8H16. Treating alkyne A
with ozone and then acetic acid furnishes a single product CH3CH2CH2CO2H. Treating alkyne C
with ozone and then water produces a single product HO2C(CH2)6CO2H, . Compound B has an
absorption in its IR spectrum at ~3300cm-1, what are compounds A,B and C.
Answer:
A:
B:
C:
8.21 White structural formulas for the products that form when 1-butene reacts with each of the
following reagents:
(a) HI
I
(b) H2, Pt
(c) Dilute H2SO4, warm
OH
(d) Cold concentrated H2SO
O
O
S
O
OH
(e) Cold concentrated H2SO4, then H2O and heat
OH
(f) HBr in the presence of alumina
Br
(g) Br2 in CCl4
Br
Br
(h) Br2 in CCl4, then KI in acetone
(i) Br2 in H2O
Br
OH
(j) HCl in the presence of alumina
Cl
(k) Cold dilute KMnO4, OHOH
OH
(l) O3, then Zn, HOAc
O
CHO ,
(m) OsO4, then NaHSO3/H2O
OH
OH
(n) KmnO4, OH-, heat, then H3O+
HCOOH
O
OH
8.22 White structural formulas for the products that form when cyclohexene reacts with each of
the following reagents:
(o) HI
I
(p) H2, Pt
(q) Dilute H2SO4, warm
OH
(r) Cold concentrated H2SO
O
S
O
OH
(s) Cold concentrated H2SO4, then H2O and heat
OH
(t) HBr in the presence of alumina
Br
(u) Br2 in CCl4
Br
Br
Br
(v) Br2 in CCl4, then KI in acetone
(w) Br2 in H2O
Br
Br
Br
OH
OH
(x) HCl in the presence of alumina
Cl
(y) Cold dilute KMnO4, OHOH
OH
(z) O3, then Zn, HOAc
O
O
(aa) OsO4, then NaHSO3/H2O
OH
OH
(bb) KmnO4, OH-, heat, then H3O+
OH
O
O
OH
8.23 Give the structure of the products that you would expect from the reaction of 1-butyne with:
(a) One molar equivalent of Br2
(b) One molar equivalent of HBr in the presence of alumina
(c) Two molar equivalent of HBr in the presence of alumina
(d) H2 (in excess)/Pt
(e) H2, Ni2B (P-2)
(f) NaNH2 in liquid NH3, then CH3I
(g) NaNH2 in liquid NH3, then (CH3)3CBr
Answer:
(a)
Br
H
Br
(b)
H
H
Br
(c)
Br
Br
(d)
(e)
H
H
H
(f)
(g)
Br
H
Na
H
+
H
8.24 Give the structure of the products you would expect from the reaction (if any) of 2-butyne
with:
(a) One molar equivalent of HBr in the presence of alumina
(b) Two molar equivalents of HBr in the presence of alumina
(c) One molar equivalent of Br2
(d) Two molar equivalents of Br2
(e) H2, Ni2B (P-2)
(f) One molar equivalent of HCl in the presence of alumina
(g) Li/liquid NH3
(h) H2 (in excess)/Pt
(i) Two molar equivalents of H2, Pt
(j) KMnO4, OH-, then H3O+
(k) O3, HOAc
(l) NaNH2, liquid NH3
Answers:
(a)
Br
H3C
C
C
H
(b)
CH3
H3C
H
Br
C
C
H
Br
CH3
(c)
Br
H3C
C
C
CH3
Br
(d)
H3C
Br
Br
C
C
Br
Br
CH3
(e)
H3C
CH3
H
H
(f)
Cl
H3C
C
C
CH3
H
(g)
H3C
H
H
CH3
(h)
H3C
H
H
C
C
H
H
H
H
C
C
H
H
CH3
(i)
H3C
CH3
(j)
O
H3 C
C
OH
(two molar equivalents)
(k)
O
H3 C
C
OH
(two molar equivalents)
(l)
I think they cannot react with each other in this condition.
8.25 show how 1-butyne could be synthesized from each of the follow:
(a) 1-butene
(b) 1-chlorobutane
(c) 1-chloro-1-butene
(d) 1, 1-dichlorobutane
(e) Ethyne and ethylbramide
Answer:
Br2 / CCl4
(a)
Br
3 NaNH2
NH4Cl
Na
Br
t-BuOK
(b)
Cl
then as in (a)
t-BuOH
NH4Cl
2 NaNH2
(c)
Cl
Na
Cl
3 NaNH2
(d)
NH4 Cl
Cl
(e) H
H
Na
NaNH2
CH3CH2 Br
8.26 Starting with 2-methylpropene (isobutylene) and using any other needed reagents, outline a
synthesis of each of the following:
(a) (CH3)3COH (b) (CH3)3CCl (c) (CH3)3CBr (d) (CH3)3CF (e) (CH3)2C(OH)CH2Cl
Answer
(a)
(CH3)2CH=CH2 +H2O
60%H2SO4
(CH3)3COH
(b)
(CH3)2 CH=CH2+HCl
-30oC
(CH3)3CCl
(c)
(CH3)2CH=CH2+HBr
(CH3)3CBr
(CH3)2CH=CH2 +HF
(CH3)3 CF
(d)
(e) (CH3)2CH=CH2 +Cl2+H2O
(CH3)2C(OH)CH2 Cl
8.27 Myrcene, a fragrant compound found in bayberry wax, has the formula C10H16 and is
known not to contain any triple bonds.
(a) What is the index of hydrogen deficiency of myrecene? When treated with excess hydrogen
and a platinum catalyst, myrcene is converted to a compound A with the formular C10H22.
(b) How many rings does myrcene contain?
(c) How many double bonds? Compound Acan be identified as 2,6-dimethyloctane. Ozonolysis
of myrcene followed by treatment with zinc and acetic acid yields 2 mol of formaldehyde
(HCHO), 1 mol of acetone (CHCOCH), and a third compound Bwith the formula C5H6O3.
(d) What is the structure of myrcene?
(e) Of compound B?
Answer:
(a) 3
(b) No
(c) 3
(d)
O
O
O
(e)
8.28 When propene is treated with hydrogen chloride in ethanol, one of the products of the
reaction is ethyl isopropyl ether. Write a plausible mechanism that accounts for the formation of
this product.
Answer:
Step 1:
CH3CH
Step 2:
CH2 + H
Cl
CH3CHCH3 + Cl
CH3CHCH3
CH3 CHCH3 +
CH3CH2OH
OCH2CH3
H
Step 3:
CH3 CHCH3
CH3CHCH3
OCH2CH3
OCH2CH3
H
8.29 When, in separate reactions, 2-methylpropene, propene, and ethene are allowed to react with
HI under the same conditions (i.e., identical concentration and temperature), 2-methylpropene is
found to react fastest and ethene slowest. Provide an explanation for these relative rates.
Answer:
According to Markovnikov rule, in the addition of HI to the alkenes, the hydrogen atom adds to
the carbon atom of the double bond that already has the greater number of hydrogen atoms.
The reactions take place
CH3C
CH2
CH3CCH3
+ HI
CH3
CH3
CH3 CH
CH2
+ I-
CH2
CH2
CH3CHCH3 + I-
+ HI
CH3CH2
+ HI
+ I-
Because the 30 carbocation is the most stable carbocation and the 20 carbocation is more stable
than the 10 carbocation, the free energy in each reaction is 30 carbocation<20carbocation<10
carbocation. As a result, the 2-methylpropene found to react fastest and ethene slowest.
8.30 Farnesene (below) is a compound found in the waxy coating of apples. Give the structure and
IUPAC name of the product formed when farnesene is allowed to react with excess hydrogen in
the presence of a platinum catalyst.
Farnesene
Answer:
H3C
H
(S)-2,6,10-Trimethyl-dodecane
H
CH3
(R)-2,6,10-Trimethyl-dodecane
8.31 Write structural formulas for the products that would be formed when geranial, a component
of lemongrass oil, is treated with ozone and then with zinc and water.
H
O
Answer: The structure formulas for the products are the followings:
O
O
O
O
&
&
O
8.32 Limonene is a compound found in orange oil and lemon oil. When limonene is treated with
excess hydrogen and a platinum catalyst, the product of the reaction is
1-isopropyl-4-methylcyclohexane. When limonene is treated with ozone and then with zinc and
water the products of the reaction are HCHO and the following compound. Write a structural
formula for limonene.
O
O
O
H
Answer:
The structural formula is:
8.33 When 2,2-diphenyl-1-ethanol is treated with aqueous HI, the main product of the reaction is
1-iodo-1,1-diphenylethane. Propose a plausible mechanism that accounts for this product.
I
H
C
H2C
OH
aqueous
Ph + HI
H3C
C
Ph
(main product)
H
C
Ph
Ph
Ph
Mechanism:
H
C
H2 C
OH
Ph
+ H
I
aqueous
H2C
+ I-
O+H2 Ph
Ph
H
H
C
H2C
H 2+ C
Ph
O+H2 Ph
C
Ph
+ H2O
Ph
H
H2+C
C
Ph
H3C
Ph
C+
Ph
Ph
I
H3C
C+
Ph
Ph
+
I-
H3C
C
Ph
Ph
8.34 When 3,3-dimethyl-2-butanol is treated with concentrated HI, a rearrangement takes place.
Which alkyl iodide would you expect from the reaction? (Show the mechanism by which it is
formed.)
Answer:
The main product of the reaction is 3,3-dimethyl-2-iodobutane.
OH2
+
I
8.35 Pheromones are substances secreted by animals that produce a specific behavioral response
in other members of the same species. Pheromones are effective at very low concentrations and
include sex attractants, warning substances, and ”aggregation” compounds. The sex attractant
pheromone of the codling moth has the molecular formula C13H24O. On catalytic hydrogenation
this pheromone absorbs two molar equivalents of hydrogen and is converted to
3-ethyl-7-methyl-1-decanol. On treatment with ozone followed by treatment with zinc and water
the pheromone produce CH3CH2CH2COCH3, CH3CH2COCH2CH2CHO, and OHCCH2OH. (a)
Neglecting the stereochemistry of the double bonds, write general structure for this pheromone. (b)
The double bonds are known to be (2Z, 6E). Write a stereo-chemical formula for the codling moth
sex attractant.
Answer:
(a)
HO
（b）
HO
H
H
8.36.The sex attractant of the common housefly (Musca domestica) is a compound called muscalure.
H
H
(CH2)12
(CH2)7
. Starting with ethyne and any other needed reagents, outline
Muscalure is
a possible synthesis of muscalure.
Solution:
NaNH2
NaNH2
Pb/BaSO4
C
CH3(CH2)6CH2Br
C(CH2)7CH3
H
CH3(CH2)11CH2Br
H
H2
(CH2)12
(CH2)7
HC
C(CH2)7CH3
CH3(CH2)12C
C(CH2)CH3
8.37 Starting with ethyne and 1-bromopentane as your only organic reagents (except for solvents)
and using any needed inorganic compounds, outline a synthesis of the compound shown below.
Answer:
liq.NH3
HC
C
Na NH2
+
H
CH3 CH2CH2CH2CH2 Br +
Na
C
CH3CH2 CH2 CH2CH2C
+
HBr
CH
HC
C
Na
-33 ° C
CH
CH3CH2CH2CH2CH2CBr CH2 +
HBr
Br
Br
8.38 Shown below is the final step in a synthesis of an important perfume constituent, cis-jasmone.
Which reagents would you choose to carry out this last step?
O
O
CH2
C
C
CH2CH3
ANSWER:
The reagents should be (1) Raney Nickel / H2.
8.39 Write stereochemical formulas for all of the products that you would expect from each of the
following reactions. (You may find models helpful.)
(a)
H
HO
OsO4
C
CH3
CH3
CH3
H
H
C
C
OH
+
NaHSO3 ,H2O
C
H
CH2CH3
H3CH2C
OH
H
(b)
C
C
HO
CH2CH3
H
H
CH3
CH3
C
CH3
H
H
C
OsO4
C
H3CH2C
HO
OH
C
+
C
NaHSO3,H2O
H
H
C
OH
CH2CH3
HO
H
CH2CH3
(c)
H
H
CH3
C
CH3
H3C
C
Br2,CCl4
Br
C
+
C
C
H3CH2C
H
Br
H
H
C
Br
CH2CH3
Br
H
CH2CH3
(d)
H
H
CH3
Br
C
H3C
C
Br2,CCl4
Br
C
+
C
C
H
H
CH3
H3CH2C
CH2 CH3
C
Br
Br
CH2CH3
H
H
8.41 When cyclohexene is allowed to react with bromine in an aqueous solution of sodium
chloride, the products are trans-1,2-dibromocyclohexane, trans-2-bromo-cyclohexanol, and
trans-1-bromo-2-chlorocyclohexane. Write a plausible mechanism that explains the formation of
this last product.
Answer:
Nu
Br
Br
Br
H
Br
H
H
+
H
Br
+
Br
H
H
Br
Br
H
H
Br
H
+
Cl
OH
H
Br
Br
+
OH
H
+
Cl
H
8.42 Predict features of their IR spectra that you could use to distinguish between the members of
the following pairs of compounds.
(a) Pentane and 1-pentyne
(b) Pentane and 1-pentene
(c)
(d)
(e)
(f)
(g)
(h)
(i)
1-Pentene and 1-pentyne
Pentane and 1-bromopentane
2-pentyne and 1-pentyne
1-Pentene and 1-pentanol
Pentane and 1-pentanol
1-Bromo-2-pentene and 1-bromopentane
1-Pentanol and 2-penten-1-ol
Answer:
(a) 1-pentyne
C
(
(CHCCH2CH2CH3) has an absorbing peak
C
C
H).
) and about 3300cm-1(
(b)
(CH2CHCH2CH2CH3)
1-pentene
has
an
absorbing
C
C
about
peak
2100-2260
around
cm-1
1620-1680cm-1
H
C
) and about 3010-3095cm-1(
(
).
H
C
C
-1
(c) 1-pentene (CH2CHCH2CH2CH3) has an absorbing peak in 1620-1680cm (
C
and about 3010-3095cm-1(
absorbing
3300cm-1(
peak around
C
H).
H
2100-2260
)
H
), but 1-pentyne (CHCCH2CH2CH3) has an
C
C
) and about
cm-1 (
(d) 1-bromopentane (CH2BrCH2CH2CH2CH3) has an absorption around 690-515 cm-1
C
Br
(
).
(e) 2-Pentyne (CH3CCHCH2CH3) has 2100-2260 cm-1 (
C
H).
(CHCCH2CH2CH3) has 3300cm-1(
C
C
C
(f) 1-Pentene (CH2CHCH2CH2CH3) has 1620-1680cm-1 (
C
3010-3095cm-1(
H
), but 1-pentyne
C
) and about
H
), but 1-pentanol (CH3CH2CH2CH2CH2OH) has a broad absorption
-1
peak 3200-3600cm (OH).
(g) 1-pentanol (CH3CH2CH2CH2CH2OH) has a broad band peak around 3200-3600cm-1(OH).
(h) 1-Bromo-2-pentene (CH2BrCHCHCH2CH3) has a double bond peak around 1620-1680cm-1
C
(
C
), but 1-bromopentane doesn’t have.
C
(i) 2-penten-1-ol has a double bond peak around 1620-1680cm-1 (
C
) , but
1-pentanol doesn’t have.
8.43 The double bond of tetrachloroethene is undetectable in the bromine/carbon tetrachloride test for
unsaturation. Give a plausible explanation for this behavior.
Answer:
Because of the electron-withdrawing nature of chlorine, the density at the double bond is greatly
reduced and attacked by the electrophilic bromine doesn’t occur.
8.44 Three compounds A, B, and C all have the formula C6 H10. All three compounds rapidly
decolorize bromine in CCL4; all three are soluble in cold concentrated sulfuric acid. Compound A
has an absorption in its IR spectrum at about 3300cm-1, but compound B and C do not.
Compounds A and B both yield hexane when they are treat with excess hydrogen in the presence
of a platinum catalyst. Under these condition C absorb only one molar equivalent of hydrogen and
gives a product with the formula C6 H12. When A is oxidized with hot basic potassium
permanganate and the resulting solution acidified, the only organic product that can be isolated is
CH3(CH2)3CO2H. Similar oxidation of B gives only CH3CH2CO2H, and similar treatment of C
gives only HO2C(CH2) 4CO2H. What are structures for A, B, and C?
A:
B:
3-hexyne
C:
8.47 Use your answer to the preceding problem to predict the stereochemistry outcome of the addition
of bromine to maleic acid and to fumaric acid. (a) Which dicarbonoxylic acid would add bromine to
yield a meso compound? (b) Which will yield a racemic form?
OH
O
maleic acid
O
OH
O
fumaric acid
HO
OH
O
meso compound:
COOH
H
COOH
H
Br2
Br
meso compound
H
HOOC
Br
H
COOH
racemic form:
COOH
COOH
H
COOH
Br2
H
Br
Br
H
COOH
Br
H
H
Br
+
H
COOH
COOH
racemic form
8.48 An optically active compound A (assume that it is dextrorotatory) has the molecular formula
C7H11Br. A reacts with hydrogen bromide, in the absence of peroxides, to yield isomeric products,
B and C, with the molecular formula C7H12Br2. Compound B is optically active; C is not.
Treating B with 1 mol of potassium tert-butoxide yields (+)-A. Treating C with 1 mol of
potassium tert-butoxide yields (+)-A. Treating A with potassium tert-butoxide yields D (C7H10).
Subjecting 1 mol of D to ozonolysis followed by treatment with zinc and acetic acid yields 2 mol
of formaldehyde and 1 mol of 1,3-cyclopentanedione.
O
O
1,3-Cyclopentanedione
Propose stereo chemical formulas for A, B, C, and D and outline the reactions involved in these
transformations.
Answer:
CH3
CH3
Br
HBr
Br
(+)-A
CH3
CH3
Br
Br
+
CH2
Br
CH3
CH3
C
B
Br
CH3
Br
t-BuOK
H2 C
CH2
Br
CH3
Br
CH3
(+)-A
(+)-A
CH3
CH3
CH3
CH3
t-BuOK
+
CH2
Br
H2C
Br
Br
Br
C
(+)-A
(-)-A
CH3
1, O3
t-BuOK
H2C
CH2
CH2
O
2, Zn, HOAc O
D
Br
(+)-A
8.49 A naturally occurring antibiotic called mycomycin has the structure shown here. Mycomycin is
optically active. Explain this by writing structures for the enantiomeric forms of mycomycin.
HC
C
C
C
CH
C
CH
CH
CH
CH
CH
CH2COOH
Answer:
The stereo structure for this molecule is
HC
H
C
HC
C
C
C
CH
CH
CH
CH2COOH
C
H
C
The enantiomeric form is
HC
H
C
HC
C
C
C
C
CH
CH
CH
CH2COOH
C
H
8.50 An optically active compound D has the molecular formula C6H10 and shows a peak at about
3300 cm-1 in its IR spectrum. On catalytic hydrogenation D yields E (C6H14). Compound E is
optically inactive and cannot be resolved. Propose structures for D and E.
Answers: D should be
CH3
HC
CH2CH3
C
H
;
E should be
CH3
CH3CH2CHCH2CH3
.
8.51 (a) By analogy with the mechanism of bromine addition to alkenes, draw the likely
three-dimensional structures of A, B, and C.
Reaction of cyclopentene with bromine in water gives A.
Reaction of A with aqueous NaOH (1 equivalent, cold) gives B, C5H8O (no 3590 to 3650 cm-1 infrared
absorption). (See the squalene cyclization discussion for a hint.)
Heating of B in methanol containing a catalytic amount of strong acid gives C, C6H12O2, which does
show 3590 to 3650 cm-1 infrared absorption.
(b) Specify the R or S configuration of the stereocenters in your predicted structures for C. Would it be
formed as a single stereocenter or as a racemate?
(c) How could you experimentally confirm your predictions about the stereochemistry of C?
Answer:
H
A
OH
Br
H
OCH3 H
C
H
S
OH
H
H
&
B
S
HO
Br
O
H
R
OH
OCH3
R
H
(c) C, in contrast to its cis isomers, would exhibit no intramolecular hydrogen bonding, this
would be proven by the absence of infrared absorption in the 3500-3600 cm-1 region when studied as a
very dilute solution in CCl4. C would only show free –OH stretch at about 3625 cm –1.
8.52 Triethylamine, (C2H5)3N, like all amines, has a nitrogen atom that has an unshared pair of
electrons. Dichlorocarbene also has an unshared pair of electrons. Both can be represented as
shown below. Draw the structures of compounds Q, E, and R.
(C2H5)3N
+
CCl2
Q
D (an unstable adduct)
E
H2 O
E
+
R
C2 H4 (by an intramolecular E2 reaction)
(Water effects a replacement that is the reverse of that used to make
gem-dichlorides.)
Answer:
H
H
H
H
H
Cl
N
H
Cl
N
- C2H4
N
N
C
C
C
OH2
Cl
Cl
Cl
H2O
H
H
H
Cl
N
N
N
C
C
C
O
OH
Cl
C
H
O