Chapter 6
6.1 Calculate the degree of the unsaturation in the following hydrocarbons:
(a) C8H14;
(b) C5H6
(c) C12H20
(d) C20H32
(e) C40H56
Solution:
(a)△=2
(b)△=3
(c)△=3
(d)△=5
(e)△=13
6.2 Calculate the degree of the unsaturation in the following formula, and the draw as many structures
as you can for each:
(a) C4H5;
(b) C4H6;
(c) C3H4;
Solution:
(a)△=1
H2
C
H2C
C
H
H3C
C
H
C
H
H2C
C
CH3
CH3
CH3
CH3
(b)△=2
H2C
C
C
H
CH3
H2C
C
H
C
H
CH2
HC
C
H2
C
CH3
H3C
C
C
CH3
(c)△=2
HC
C
CH3
H2 C
C
CH2
6.3 Calculate the degree of unsaturation in the following formulas:
(1) C6H5N 5
(2) C6H5NO2 5
(3) C8H9Cl3
3
(4) C9H10Br2
1
(5) C10H12N2O3 6
(6) C20H32ClN 5
6.4 Give the IUPAC name for the following compounds:
(1)
3,4,4-trimethyl-1-pentene
(2)
3-methyl-3-hexene
(3)
4,7-dimethyl-2,5-octadiene
6.5 Draw structures corresponding to the following IUPAC names:
(a)2-Methyl-1,5-hexadiene
(b)3-Ethyl-2,2-dimethyl-3-heptene
(c)2,3,3-Trimethyl-1,4,6-octatriene (d)3,4-diisopropyl-2,5-dimethyl-3-hexene
(e)4-tert-Butyl-2-methylheptane
Solution:
CH3
H3C
H2
C
C
H2C
H2
C
C
H
CH2
CH3
(a)
C
C
CH3
CH2
H2
C
C
H
CH3
(b)
H3 C
CH3
CH
CH3
H 2C
(c)
C
C
CH3
CH3
H3C
C
H
C
H
C
H
CH3
C
H
CH3
H
C
C
CH3
(d)
CH3
C
C
H
CH3
CH
H3 C
CH3
CH3
H3C
H2
C
C
H
H3C
H
C
H2
C
C
CH3
CH3
CH3
(e)
6.6 Name the following cycloalkenes:
CH3
CH3
CH3
CH3 (b)
(a)
Solution:
CH(CH3)2
(c)
(a) 1,2-Dimethyl-cyclohexene
(b) 4,4-Dimethyl-cycloheptene
(c) 3-Isopropyl-cyclopentene
6.7 Which of the following compounds can exist as pairs of cis-trans isomers? Draw each cis-trans pair,
and indicate the geometry of each isomer.
CH2
(b) (H3C)2C
(a) H3CHC
CHCH3
(d) (H3C)2C
(c) H3CH2CHC
(e) ClHC
(f) BrHC
CHCl
CHCH3
C(CH3)CH2CH3
CHCl
Solution:
(c)
H3CH2C
C
cis:
H
H3CH2C
CH3
H
C
C
H
trans:
H
C
CH3
(e)
Cl
Cl
C
Cl
H
C
cis: H
(f)
Br
C
C
H
trans: H
Cl
Cl
Br
H
C
cis: H
C
C
trans:
H
C
H
Cl
6.8 Name the following alkenes, including the cis or trans designation:
H
CH3
H3C
CH
CH
(a)
C
CH3
C
H3C
H3C
C
H
HC
H
CH2
(b)
CH2
C
H
H3C
Solution:
(a) cis-4,5-Dimethyl-2-hexene
(b) trans-6-Methyl-3-heptene
6.9 Which member in each of the following sets has higher priority?
(a) –H or –Br
(d) –NH2 or –OH
(b) –Cl or –Br
(e) –CH2OH or –CH3
(c) –CH3 or –CH2CH3
(f) –CH2OH or –CH=O
Solution:
(a) –Br
(b) –Br
(c) –CH2CH3
(d) –OH
(e) –CH2OH
(f) –CH=O
6.10 Rank the following sets of substituents in order of Cahn-Ingold-Prelog priorities.
(a) –CH3, -OH, -H, –Cl
(b) –CH3, –CH2CH3, -CH=CH2
(c) –CO2H, -CH2OH, -C≡N, -CH2NH2
(d) –CH2CH3, -C≡CH, -C≡N, -CH2OCH3
Solution: (From High to low)
(a) –Cl, -OH, –CH3, -H
(b) -CH=CH2, –CH2CH3, –CH3
(c) –CO2H, -CH2OH, -C≡N, -CH2NH2
(d) -CH2OCH3, -C≡N, -C≡CH, –CH2CH3
CH3
6.11 Assign E or Z configuration to the following alkenes:
Solution:
H3C
CH2OH
C
Cl
C
C
H3CH2C
CH2CH3
Cl
C
H3CO
CH2CH2CH3
E
Z
H
H3 C
CO2H
C
CN
C
C
C
H3 C
CH2NH2
CH2 OH
Z
E
6.12 Assign stereochemistry to the following alkene, and convert the drawing into a skeletal structure:
Solution:
COOCH3
CH2OH
Z
6.13
Name the following alkenes, and tell which compounds in each of the following pairs are more
stable:
(a)
H2C
CHCH2CH3
CH3
or
H2C
1-butene
CCH3
2-methyl-1-propene
2-methyl-1-propene is more stable because it is more substituents.
(b)
H
H
H
C
CH2CH2CH3
or
C
H3C
H3C
CH2CH2 CH3
H
trans-2-hexene
cis-2-hexene
trans-2-hexene is more stable because it is trans isomer.
(c)
CH3
CH3
C
C
or
3-methyl-1-cyclohexene
1-methyl-1-cyclohexene
1-methyl-1-cyclohexene is more stable because the carbon has only sp2-sp2 bonds while the other
has sp2-sp3 bonds.
Predict the products of the following reactions:
6.14
+
HCl
(a)
Cl
H3C
H2C
C
(b)
CH3
CH
Br
HBr
CH3
H3C
I
(c)
CH3CH2CH2CH
CH2
H3PO4
KI
Br
CH2
CH3
+
HBr
(d)
6.15 What alkenes would you start with to prepare the following alkyl halides?
(a) Bromocyclopentane
(b) 1-Ethyl-1-iodocyclohexane
(c)
Br
CH3CH2 CHCH2CH2CH3
Solution:
(a) Cyclopentene
(b) 1-Ethylcyclohexene
CHCH2CH3
(c) H3CH2CHC
6.16 Show the structures of carboations you would expect in the following reactions:
(a)
CH3
CH3
H3CH2CC
?
+ BrH
CHCHCH3
(b)
CHCH3
+
HI
?
Solution:
(a)
CH3
H3CH2CC
CH3
CH2CH2CH3
(b)
CH2CH3 +
6.17 Draw a skeletal structure of the following carbocation. Identify it as primary, secondary, or tertiary,
and identify the hydrogen atoms that are involved in hyperconjugation in the conformation shown.
CH3
CH3
H
C
C
H
CH3
Solution:
It is secondary carbocation.
Only the methyl-group C—H that is parallel to the carbocation p orbital can show
hyperconjugation.
6.18 What about the second step in the electrophilic addition of HCl to an alkene—the reaction of
chloride ion with the carbocation intermediate? Is this step exergonic or endergonic? Does the
transition state for this second step resemble the reactant (carbocation) or product (alkyl chloride)?
Make a rough drawing of what the transition-state structure might look like.
Solution: This step is exergonic. The transition state for this second step resemble carbocation.
Br
H
R
C
C
R
R
R
transition state
Intermediate
Product
6.19 On treatment with HBr, vinylcyclohexane undergoes addition and rearrangement to yield
1-bromo-1-ethylcyclohexane. Using curved arrows, propose a mechanism to account for this
result.
Solution:
CH3
+
H
Br
+
C
Br
H
vinylcyclohexane
CH3
C
CH2CH3
+
Br
Br
6.20 Name the following alkenes, and convert each drawing into a skeletal structure.
(a)
(b)
2,4,5-trimethyl-2-hexene
1-ethyl-3,3-dimethyl-1-cyclohexene
6.21. Assign stereochemistry (E or Z) to each of the following alkenes, and convert each drawing into a
skeletal structure:
(a)
(b)
SOLUTION:
(a)
(b)
Cl
O
Z
E
O
O
HO
6.22. The following carbocation is an intermediate in the electrophilic addition reaction of HCl with
two different alkenes. Identify both, and tell which C
H bonds in the carbocation are
aligned for maximum hyperconjugation with the vacant p orbital on the positively charged
carbon.
SOLUTION:
1
2
3
6.23 Calculate the degree of unsaturation in the following formulas, and draw five possible for each:
(a) C10H16
(b)
C8H8O
(c) C7H10Cl2
(d) C10H16O2 (e) C5H9NO2 (f) C8H10ClNO
Solution”
a)
It has 3 unsaturation degrees.
Possible structures:
b) It has 5 unsaturation degrees.
Possible structures:
O
OH
O
OH
OH
c)
It has 2 unsaturation degrees.
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Possible structures:
Cl
d) It has 3 unsaturation degrees.
OH
OH
OH
OH
OH
OH
HO
Possible structures:
e)
HO
it has 2 unsaturation degrees.
OH
OH
OH
OH
N
OH
Possible structures:
f)
OH
N
H
OH
N
OH
OH
N
H
H
OH
HO
OH
N
H
H
it has 4 unsaturation degrees.
OH
OH
OH
N
N
N
H
H
H
OH
OH
Possible structures:
N
N
H
H
6.24 A compound of formula C10H14 undergoes catalytic hydrogenation but absorbs only 2 molar
equivalents of hydrogen. How many rings does the compound have?
Solution:
It has 4 unsaturation degrees. Two of them are rings.
6.25 A compound of formula C12H13N contains two rings. How many molar equivalents of hydrogen
does it absorb if all the remaining unsaturations are double bonds?
Solution:
C12H13N=C12H12
Degree of Unsaturation:
(12 × 2 + 2 − 12)
=7
2
It can absorb 7-2=5 molar equivalents of hydrogen.
6.26 Name the following alkenes:
H3 C
H
CHCH2CH3
C
(a)
H3 C
C
H
CH2CH3
CH3
CH3CHCH2CH2CH
CH3
C
(b)
C
H
H
CH2CH3
(c)
H2C
CCH2CH3
H
CH3
H2C
CH3
C
C
CHCHCH
(d)
H
CH3
H
H3C
C
C
(e)
H3CH2CH2C
(f)
H2C
C
H
C
CH3
C
CH3
CHCH3
Solution:
(a).trans-4-Methyl-2-hexene or (E)-4-Methyl-2-hexene
(b).cis-4-Ethyl-7-methyl-2-octene or (Z)-4-Ethyl-7-methyl-2-octene
(c).2-Ethyl-1-butene
(d).trans-3,4-Dimethyl-1,5-heptadiene or (E)-3,4-Dimethyl-1,5-heptadiene
(e).(2-cis,4-trans)-4,5-Dimethyl-2,4-octadiene or (2Z,4E)-4,5-Dimethyl-2,4-octadiene
(f).1,2-Butadiene
6.27 Ocimene is a triene found in the essential oils of many plants. What is its IUPAC name, including
stereochemistry?
Solution:
(3E)3,7-dimethyl-octa-1,3,6-triene
6.28 α-Farnesene is a constituent of the natural wax found on apples. What is its IUPAC
name, including stereochemistry?
Solution:
α-Farnesene
(3E,6E)3,7,11-trimethyl-dodeca-1,3,6,10-tetraene
6.29 Draw structures corresponding to the following systematic names:
(a) (4E)-2,4-Dimethyl-1,4-hexadiene
H2
C
H
H
C
C
C
C
H
H3C
Solution: H3C
CH3
(b) cis-3,3-Dimethyl-4-propyl-1,5-octadiene
H3C
CH3
H
H
H
C
C
C
C2H5
CH
C
C
H
C3H7 H
Solution:
(c) 4-Methyl-1,2-pentadiene
CH3
H
C
C
CH
H
C
H
Solution:
CH3
(d) (3E,5Z)-2,6-Dimethyl-1,3,5,7-octatetraene.
H
H
H
C
C
C
CH3
C
C
C
C
H
Solution: H
H
C
H
H
CH3
(e) 3-Butyl-2-heptene
C4H9
H
C4H9
Solution: H3C
(f) trans-2,2,5,5-Tetramethyl-3-hexene
H3C
CH3
H
C
H3C
C
C
C
Solution:
6.30
H3C
Menthene,
CH3
H
CH3
a
hydrocarbon
found
in
mint
plants,
has
the
systematic
name
1-isopropyl-4-methylcyclohexhene. Draw its structure.
Solution:
6.31 Draw and name the 6 pentene isomers, C5H10, including E, Z isomers.
1-Pentene
(Z)-2-Pentene
(E)-2-Pentene
2-M ethyl-1-butene
2-M ethyl-2-butene
3-M ethyl-1-butene
6.32 Draw and name the 17 hexene isomers, C6H12, including E, Z isomers.
1-Hexene
(E)-2-Hexene
(Z)-2-Hexene
(E)-3-Hexene
(Z)-3-Hexene
2-M ethyl-1-pantene
2-M ethyl-2-pantene
(E)-4-M ethyl-2-pantene
4-M ethyl-1-pantene
3-M ethyl-1-pantene
(E)-3-M ethyl-2-pantene
(Z)-3-M ethyl-2-pantene
2,3-Dim ethyl-1-butene
(Z)-4-M ethyl-2-pantene
2-Ethyl-1-butene
3,3-Dim ethyl-1-butene
2,3-Dim ethyl-2-butene
6.33 trans-2-Butene is more stable than cis-2-butene by only 4 KJ/mol, but trans-2, 2, 5,
5-tetramethyl-3-hexene is more stable than cis-2, 2, 5, 5-tetrameethyl-3-hexene by 39 KJ/mol.
Explain.
ΔΗohydrog
Alkene
(KJ/mol)
(kcal/mol)
cis-2-butene
-119.7
-28.6
trans-2-Butene
-115.5
-27.6
cis-2, 2, 5, 5-tetrameethyl-3-hexene
-151.5
-36.2
trans-2,2,5,5-tetrameethyl-3-hexene
-112.6
-26.9
Solution: Between cis-2-butene and trans-2-Butene, the cis-2-butene has steric strain, so the
trans-2-Butene is more stable. But the steric strain is not large, the difference between them
is not very large. And the cis-2, 2, 5, 5-tetrameethyl-3-hexene has large steric strain, so the
difference
between
cis-2,
2,
5,
5-tetrameethyl-3-hexene
and
trans-2,
2,
5,
5-tetrameethyl-3-hexene is larger.
6.34 Cyclodecene can exist in both cis and trans forms, but cyclohexene cannot. Explain.(Making
molecular models is helpful.)
Solution: the cyclodecene can be
But the cyclohexene can only be:
6.35 Normally, a trans alkene is more stable than its cis isomer. trans-Cyclooctene, however, is less
stable than cis-cyclooctene by 38.5 kJ/mol. Explain.
Build models of the two cyclooctenes and notice that the large amount of torsional strain in
trns-cyclooctene relative to cis-cyclooctene. This strain causes the trans isomer ot be of higher
energy.
6.36 Trans-Cyclooctene is less stable than cis-cyclooctene by 38.5 kJ/mol, but trans-cyclononene is less
stable than cis-cyclononene by only 12.2 kJ/mol. Explain.
trans-Cyclononene has more carbon than trans-Cyclooctene, and the ring is bigger, so the
steric strain is smaller.
6.37 Allene(1.2-propadiene), H 2C
C
CH 2 ,has two adjacent double bonds. What kind of
hybridization must the central carbon have? Sketch the bonding π orbitals in allene. What shape
do you predict for allene.
Solution:
The central carbon is sp hybridized.
The bonding πorbitals in allene is showed as follow:
H
H
H
C
C
H
H
C
H
H
C
C
C
H
The structure of allene is showed as follow:
H
H
H
C
C
C
H
6.38 The heat of hydrogenation for allene(Problem 6.37) to yield propaneis -298 kJ/mol, and the heat
of hydrogenation for a typical monosubsituted alkene such as propene is -126kJ/mol. Is allene
more or less stable then you might except for diene? Explain.
Solution:
Less stable. Because allene give out more heat than diene in hydrogenation which means that it
contains more energy.
6.39 Predict the major product in each of the following reactions:
(a)
CH3
CH3CH2CH
CH3
+
CCH2CH3
HCl
CH3CH2CH2 CCH2CH3
Cl
(b)
+
HBr
Br
1-Ethylcyclopentene
(c)
CH3
+
HI
CH3
CH3CCH2 CCH2 CH3
CH3
I
2,2,4-Trimethyl-3-hexene
(d)
Cl
+
1,6-Heptadiene
Cl
2HCl
(e)
CH3
CH3
+
HBr
6.40 Predict the major product from addition of HBr to each of following alkenes:
(a)
Br
CH3
CH2
+
Br
HBr
(b)
Br
+
HBr
(c)
CH3
H3CHC
HBr
+
CHCHCH3
Br
6.41 Rank the following sets of substituents in order of priority according to the Cahn-Ingold-Prelog
sequence rules:
CH3,
(a)
OCH3,
C O 2H,
Solution: 1.
H.
CH3, 4.
COOH
COOH , 4.
H,
OCH3, 2.
Solution: 1.
I
Br, 3.
I, 2.
Solution: 1.
OH ,
(b)
(c)
H,
Br,
OH ,
3.
CO2CH3,
CO2CH3, 2.
CH3
CH 2 OH ,
C O 2 H 3.
H.
CH3.
CH 2 OH , 4.
O
CH3,
(d)
CH2CH3,
CH 2CH 2OH ,
CCH3
O
C
H
(e)
C
H
Solution:
,
CH 2Br ,
Solution: 1.
(f)
CH2
1.
CH2
CN,
2.
,
CH2OCH3 ,
CN, 3.
CH2CH3,
2.
CH3.
CH2CH3, 4.
CH 2CH 2OH , 3.
CCH3, 2.
Solution: 1.
CH 2Br
CH 2NH 2,
C
H
CH 2NH 2, 4.
CH2OCH3,
CH 2 OH ,
CH2CH3.
6.42 Assign E or Z configuration to each of the following alkenes:
3.
CH2
.
CH 2 OH
C
H
CH2
,
4.
HOH2C
CH3
HOOC
H
Z
Z
C
(a)
C
H3 C
C
(b)
H
NC
CH3
Cl
OCH3
H3CO2C
HC
E
C
(c)
C
H3CH2C
C
C
(d)
CH2OH
Z
CH2
C
HO2C
CH2CH3
6.43 Name the following cycloalkenes:
CH3
(b)
（a）
(c)
(e)
(d)
(f)
Solution: (a) 3-Methyl-1-cyclohexene
(b) 2,3-Dimethyl-1-cyclopentene
(c) 1-Ethyl-1,3-cyclobutadiene
(d) 1,2-Dimethyl-1,4-cyclohexadiene
(e) 5-Methyl-1,3-cyclohexadiene
(g) 1,5- cyclooctadiene
6.44 Which of the following E,Z designations are correct, and which are incorrect?
H3C
H
CO2H
C
C
Z
Br
NC
C
H
Z
(c)
CH2NHCH3
CH3
C
(d)
(H3C)2NH2C
E
HOH2C
Br
C
(e) H
CH2CH(CH3) 2
E
(b)
CH2NH2
C
(f)
H3COH2C
C
CH2CH3
CO2H
C
C
Z
CH2
C
H3C
H
(a)
CH2CH
C
E
C
COCH3
Solution: (a) correct
(b) Correct
(c) incorrect, it should be E
(d) correct
(e) Doesn’t show E / Z isomerism.
(f) correct
6.45 tert-Butyl esters [RCO2C(CH3)3] are converted into carboxylic acids (RCO2H) by reaction with
trifluoroacetic acid, a reaction useful in protein synthesis (see Section 26.10). Assign E,Z
designation to the double bonds of both reactant and product in the following scheme, and explain
why there is an apparent change of double-bond stereochemistry:
O
O
C OCH3
C OCH3
H
H
CF3CO2H
C C
C C
C OH
C OC(CH3 )3
H3C
H3C
O
O
O
Solution: The reactant is Z designation, while the product is E designation. Because
receives higher priority than
C OC(CH3)3
O
O
C OCH3 , while
C OCH3 receives higher priority than
O
O
O
C OH, and in the reaction
C OH substitute for
C OC(CH3)3.
6.46 Use the bond dissociation energies in Table 5.4 to calculate ΔH0 for the reaction of ethylene with
HCl, HBr, and HI. Which reaction is most favorable?
H2C
CH2
+
235 KJ / mol
H2C
CH2
CH2
235 KJ / mol
H
Cl
+
H
H
Br
H
H2
C
420 KJ / mol
366 KJ / mol
+
H2
C
420 KJ / mol
432 KJ / mol
235 KJ / mol
H2C
H
H
I
H2
C
420 KJ / mol
298 KJ / mol
H2
C
Cl
H o = -91 KJ / mol
338 KJ / mol
H2
C
Br
H o = -104 KJ / mol
285 KJ / mol
H2
C
I
H o = -109 KJ / mol
222 KJ / mol
6.47: Each of the following carboncations can rearrange to a more stable ion. Propose structure for the
likely rearranges for the likely rearrangement product.
(a)
Solution: (a)
(b)
(c)
(b)
(c)
6.48 Addition of HCl to 1-isopropyl-cyclohexane yields a rearranged product. Propose a mechanism,
showing the structures of the intermediates and using curved arrows to indicate electron flow in
each step.
Cl
ClH
+
Solution:
H
Cl
H
(E)
+
Cl
+
Cl
Cl
+
6.49
Addition
of
HCl
to
1-isopropenyl-1-methylcyclopentane
yields
1-chloro-1,
2,
2-trimethylcyclohexane. Propose a mechanism, showing the structures of the intermediates and
using curved arrows to indicate electron flow in each step.
Cl
CH3
+
H
Cl
CH3
CH3
CH3
Solution:
CH3
CH3
CH3
+
CH3
H
Cl
CH3
Cl
CH3
CH3
CH3
Cl
6.50 Vinylcyclopropane reacts with HBr to yield a rearranged alkyl bromide. Follow the flow of
electrons as represented by the curved arrows, show the structure of the carbocation intermediate
in brackets, and show the structure of the final product.
H
Br
Br
?
?
Solution:
H
Br
Br
Br
6.51: Calculate the degree of unsaturation in each of the following formulas.
Solution:
C27H46O
D=5
(b) C14H9Cl5
D=8
(c)
C20H34O5
D=4
(d)
C8H10N4O2 D=6
(e)
C21H28O5
D=8
(f)
C17H23 NO3
D=7
(a)
6.52: Is the rearrangement exergonic or endergonic? Draw the transition state.
Solution: It is a exergonic.
State
H3 C
H
H3C
CH2
6.53 Draw a reaction energy diagram for the addition of HBr to 1-pentene. Let one curve on your
diagram show the formation of 1-bromopentane product and another curve on the same diagram
show the formation of 2-bromopentane product. Label the postions for all reactants, intermediates,
and products. Which curve has higher-energy carbocation intermediate? Which curve has the
higher-energy first transition state?
Solution:
H2
C
Br
H2
C
C
H2
CH3
C
H2
Br
H3C
C
H
H2
C
H2
C
CH3
The vertical axial presents energy, and the horizontal one shows the progress of the reaction.
The red one stands for 2-bromopentane, and the blue one stands for 1-bromopentane.
The blue one has higher-energy cabocation intermediate and the higher-energy first transition state.
6.54 Make sketches of the transition state structures involved in the reaction of HBr with 1-pentene
(Problem 6.53). Tell whether each structure resembles reactant or product.
Solution:
The first one is for forming 1-bromopentane, and the second one is for forming 2-bromopentane.
δ
Br
Br
δ
H
H
δ
C
C
H
H
R
δ
H
H
δ
C
C
δ
Br
Br
H
R
H
Each structure resembles carbocation.
6.55 Aromatic compounds such as benzene react with alkyl chlorides in the presence of AlCl3 catalyst
to yield alkylbenzenes. The reaction occurs through a carbocation intermediate, formed by
reaction of the alkyl chloride with AlCl3 (R－Cl+ AlCl3
R+＋AlCl4-). How can you
explain the observation that reaction of benzene with 1-chloropropane yields isopropylbenzene as
the major product?
AlCl3
H3CH2CH2C Cl
Solution:
H
AlCl3
H3CH2CH2C Cl
AlCl4
H C C CH3
H H
H3C CH
CH3
H
AlCl4
AlCl3
HCl
6.56 Alkenes can be converted into alcohols by acid-catalyzed addition of water. Assuming that
Markonikov’s rule is valid, predict the major product from each of the following alkenes.
CH3
(a)
H3CH2CC CH2
CH2
(b)
CH3
(c)
H3C C CH2CH CH2
H
Solution:
CH3
CH3
H3CH2CC CH2
H2O
H3CH2C C CH3
(a)
OH
CH2
H 2O
OH
(b)
CH3
H3 C C CH2CH CH2
H
(c)
H 2O
H
(H3C)2H2C C CH3
OH
6.57 Reaction of 2,3-dimethyl-1-butene with HBr leads to an alkyl bromide C6H13Br. On treatment of
this alkyl bromide with KOH in methanol, elimination of HBr occurs and a hydrocarbon that is
isomeric with the starting alkene is formed. What is the structure of this hydrocarbon, and how do
you think is formed from the alkyl bormide?
Solution:
Br
H
H2C
CH3
CH3
C
C
H
CH3
H3C
CH3
CH3
C
C
H
CH3
H3C
Br
CH3
CH3
C
C
Br
H
CH3
OH
H3C
CH3
CH3
C
C
CH3