alkene
TRANSCRIPT
8.1 Give the structure and name of the product that would be obtained from the ionic addition of IBr to propene. Answer:
I BrI
Br
Br
I
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:
H+ C+
;
C+
C+
Because the
C+
much more stable then the
C+
step 2:
C+
+ Cl-
Cl
;
C+ + Cl+
Me Cl
(b) step 1:
H+ C+
rearrangement:
C+
C+
Because the
C+
much more stable then the
C+
Step 2:
C+
+ Cl-
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:
H3C CCH3
CH3
CH
CH2 + H O HH
slowH3C C
CH3
CH3HC CH3 + O H
H
STEP 1
STEP2
H3C CCH3
CH3HC CH3 H3C C
CH3HC CH3
CH3
FAST
(Rearrangement) STEP3
H3C CH3C
HC C H3
CH3
OHH
+FAST
H3C C
H3CHC C H3
C H3OHH
STEP4
H3C C
H3CHC CH3
CH3OHH
OHH
+FAST
H3C CH3C
HC CH3
CH3HO
+H O H
H
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:
CH3(H3C)2C > H3CHC CH3 > H2C CH3
So the CH3(H3C)2C is the most stable, has the lowest energy, and the
(H3C)2C 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:
H2C C CH3
CH3
H+ +C+H3C
CH3
CH3
C+H3C
CH3
CH3+
O CH3
H CH3C
CH3
CH3
O+H CH3
CH3C
CH3
CH3
O+H CH3+A-
CH3C
CH3
CH3
O 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.
Br2Br+
Br-
BrBr
8.10 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.
Br
H2O
+OH
Br Br
OH
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:
H2C CH2 Br Br+ _
Br+
Step2:
Br+
Br
Br
H H
Br
HH
Br+
ClCl
H H
Br
HH
Br+
H2O
OH2
H H
Br
HH
OH2
H H
Br
HH
H2OOH
H H
Br
HH
++ H3O
8.12 What products would you expect from each of the following reactions?
(a) trans-2-butene KOC(CH3)3
CHCl3
(b)
Cyclopentene KOC(CH3)3CHBr3
(c) cis-2-Butene CH2I2/Zn(Cu)diethylether
Answer: (a)
KOC(CH3)3C C
CH3
HH3C
HCHCl3
C C
C
H CH3H3C H
ClCl + enantiomer (b)
(c)
KOC(CH3)3CHBr3
H HCBr2
CH2I2/Zn(Cu)diethylether
C C
CH3
HH
H3C
C C
C
H3C CH3H
HH
H
8.13 Starting with cyclohexene and using any other needed reagents, outline a synthesis of
7,7-dibromobicyclo[4.1.0]heptane. Answer:
Br
Br
KOC(CH3)3
CHBr3
8.14 Treating cyclohexene with 1,1-diiodoethane and a zinc-copper couple leads to two isomeric products. What are their structures? Answer:
+ CH
I
I
CH3
Zn(Cu)
CH3
H
H3C
H 8.15 Starting with an alkene, outline syntheses of each of the following:
(a)
CH3
CH3
OH
OH
CH2CH3
H
OH
OH
(b)
OHH
OH
(c)
Answer: (a)
CH3
CH3
OH
OH
KMnO4
OH-, H2O
(b)
H2COH
OH
KMnO4
OH-, H2O+ enantiomer
(c)
KMnO4
OH-, H2O
OH
+ enantiomer
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.
CH3
HH
H3C
OsO4
NaHSO3 OH
H3C H
OH
H3C H
CH3
HO H
CH3
HO H
H
CH3H
H3C
OsO4
NaHSO3
OH
H CH3
OH
H3C H
OH
H3C H
OH
H CH3
CH3
HO H
CH3
H OH
CH3
H OH
CH3
HO H
( + )
meso
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
H3C C CH
CH3
CH
CH3
CH3
( b ) CH3CH2CHO only ( 2 mol produced from 1 mol of alkene )
H3CH2C C
HCH
H2C CH3
( c )
O
and 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
S
OH
OO
(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
OH
Br
(j) HCl in the presence of alumina
Cl (k) Cold dilute KMnO4, OH-
OH
OH
(l) O3, then Zn, HOAc
CHO, O
(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
S
OH
OO
(s) Cold concentrated H2SO4, then H2O and heat
OH
(t) HBr in the presence of alumina
Br
(u) Br2 in CCl4
Br
Br
Br
Br
(v) Br2 in CCl4, then KI in acetone
(w) Br2 in H2O
Br
OH
Br
OH
(x) HCl in the presence of alumina
Cl
(y) Cold dilute KMnO4, OH-
OH
OH
(z) O3, then Zn, HOAc
O
O
(aa) OsO4, then NaHSO3/H2O
OH
OH
(bb) KmnO4, OH-, heat, then H3O+
OO
OH
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)
H
Br
Br (b)
Br
H
H
(c)
Br
Br (d)
(e)
H
H
H
(f)
(g)
HNa
Br
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)
C CH3C CH3
H
Br
(b)
C CH3C CH3
H Br
H Br
(c)
C CH3C CH3
Br
Br
(d)
C CH3C CH3
Br
Br
Br
Br
(e)
CH3H3C
H H
(f)
C CH3C CH3
H
Cl
(g)
HH3C
H CH3 (h)
C CH3C CH3
H
H
H
H (i)
C CH3C CH3
H
H
H
H
(j)
H3C C
O
OH (two molar equivalents) (k)
H3C C
O
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:
(a)
(b)
Br2 / CCl4
Br
Br3 NaNH2
Na
NH4Cl
Clt-BuOK
t-BuOHthen as in (a)
(c)
(d)
(e)
Cl2 NaNH2
Na
NH4Cl
Cl
Cl
3 NaNH2Na
NH4Cl
HHNaNH2 CH3CH2Br
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+H2O60%H2SO4
(CH3)3COH (b)
(CH3)2CH=CH2+HCl (CH3)3CCl
-30oC
(c)
(CH3)2CH=CH2+HBr (CH3)3CBr (d)
(CH3)2CH=CH2+HF (CH3)3CF
(e) (CH3)2CH=CH2+Cl2+H2O (CH3)2C(OH)CH2Cl 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)
(e) O O
O
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 CH2 + H Cl CH3CHCH3 + Cl
Step 2:
CH3CHCH3 + CH3CH2OHCH3CHCH3
OCH2CH3H
Step 3:
CH3CHCH3
OCH2CH3
CH3CHCH3
OCH2CH3H
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
CH3
+ HI CH3CCH3
CH3
+ I-
CH3CH CH2 + HI CH3CHCH3 + I-
CH2 CH2 + HI CH3CH2 + 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
OO
OO
& &
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.
H
O O
O
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.
H2C
OH
HC
Ph
Ph + HI aqueous H3C C
Ph
Ph
I
(main product)
Mechanism:
H2C
OH
HC
Ph
Ph + H aqueousI H2C
O+H2
HC Ph
Ph
+ I-
H2C
O+H2
HC Ph
Ph
H2+C C Ph
Ph
H
+ H2O
H2+C C Ph
Ph
H
H3C C+ Ph
Ph
H3C C Ph
Ph
H3C C+ Ph
Ph
+ I-
I
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.
Muscalure is
(CH2)12
HH
(CH2)7
. Starting with ethyne and any other needed reagents, outline
a possible synthesis of muscalure. Solution:
NaNH2 CH3(CH2)6CH2BrHC C(CH2)7CH3
NaNH2C C(CH2)7CH3
CH3(CH2)11CH2BrCH3(CH2)12C C(CH2)CH3
Pb/BaSO4
H2
(CH2)12
HH
(CH2)7
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:
HC C H + Na NH2
liq.NH3
-33° CHC C Na
CH3CH2CH2CH2CH2 Br + C CHNa
CH3CH2CH2CH2CH2C CH + HBr 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
CH2 C C CH2CH3
O
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)
C
C
H CH3
H CH2CH3
OsO4
NaHSO3,H2O
C
C
HO HCH3
HH3CH2C OH
+C
CHO CH2CH3
H OH
H
CH3
(b)
OsO4
NaHSO3,H2O
C
C
HO HCH3
CH2CH3
H OH
+C
C
H CH3
H3CH2C H
C
C
OHHCH3
CH2CH3
HHO
(c)
Br2,CCl4 C
C
Br CH3
H
CH2CH3
H Br
+C
C
H CH3
H3CH2C H
C
C
BrH3CH
CH2CH3
HBr
(d)
Br2,CCl4 C
C
Br CH3
H
HH3CH2C Br
+C
C
H CH3
H CH2CH3
C
C
BrH3CH
HCH2CH3Br
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:
Br Br Br +
Nu Br
H
Br
H
OH
H
Br
H +
Cl
H
Br
H
+ +
H
Br
H
Br
H
Br
H
OH +
H
Br
H
Cl
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) 1-Pentene and 1-pentyne (d) Pentane and 1-bromopentane (e) 2-pentyne and 1-pentyne (f) 1-Pentene and 1-pentanol (g) Pentane and 1-pentanol (h) 1-Bromo-2-pentene and 1-bromopentane (i) 1-Pentanol and 2-penten-1-ol Answer: (a) 1-pentyne (CHCCH2CH2CH3) has an absorbing peak about 2100-2260 cm-1
( C C ) and about 3300cm-1( C H). (b) 1-pentene (CH2CHCH2CH2CH3) has an absorbing peak around 1620-1680cm-1
(C C
) and about 3010-3095cm-1(
C H
H ).
(c) 1-pentene (CH2CHCH2CH2CH3) has an absorbing peak in 1620-1680cm-1 (C C
)
and about 3010-3095cm-1(
C H
H ), but 1-pentyne (CHCCH2CH2CH3) has an
absorbing peak around 2100-2260 cm-1 ( C C ) and about 3300cm-1( C H).
(d) 1-bromopentane (CH2BrCH2CH2CH2CH3) has an absorption around 690-515 cm-1
(
C Br
).
(e) 2-Pentyne (CH3CCHCH2CH3) has 2100-2260 cm-1 ( C C ), but 1-pentyne (CHCCH2CH2CH3) has 3300cm-1( C H).
(f) 1-Pentene (CH2CHCH2CH2CH3) has 1620-1680cm-1 (C C
) and about
3010-3095cm-1(
C H
H ), but 1-pentanol (CH3CH2CH2CH2CH2OH) has a broad absorption
peak 3200-3600cm-1(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.
(i) 2-penten-1-ol has a double bond peak around 1620-1680cm-1 (C 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?
maleic acid
O
OHO
OH
fumaric acid
O
OH
O
HO
meso compound:
H COOH
HOOC H
Br2H Br
COOH
COOH
BrHmeso compound
racemic form:
H Br
COOH
COOH
HBr
H COOH
H COOH
Br2
Br H
COOH
COOH
BrH+
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 O1,3-Cyclopentanedione
Propose stereo chemical formulas for A, B, C, and D and outline the reactions involved in these transformations. Answer:
CH3
Br
CH2
CH3
Br
Br
CH3
B
CH3
Br
CH3
Br
+
C
CH3
Br
Br
CH3
HBr
t-BuOK
CH3
Br
CH2
Br
CH3
H2C
(+)-A (+)-A
(+)-A
CH3
Br
CH3
Br
C
t-BuOKCH3
Br
CH2
CH3
Br
H2C+
(+)-A (-)-A
CH3
Br
CH2
(+)-A
t-BuOKCH2H2C
D
1, O3
2, Zn, HOAc OO
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 C C C
C C C
HC CH CH CH CH2COOHH
H
The enantiomeric form is
HC C C C
C C C
HC CH CH CH CH2COOHH
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
C HCH2CH3HC ;
E should be
CH3CH2CHCH2CH3
CH3
. 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:
A
OHH
Br H &
HHO
H Br
B O
C
HOCH3
OHH
OCH3H
HOH
RRSS
(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 D (an unstable adduct)
E + C2H4 (by an intramolecular E2 reaction)Q
E RH2O(Water effects a replacement that is the reverse of that used to makegem-dichlorides.)
Answer:
N C
Cl
Cl
H
HH
N C
Cl
ClH
- C2H4N C
Cl
H
H2O
N C
OH2
ClH
N C
OH
ClH
N C
O
H
H
N C
O
H