Chapter 10
10.1 Give IUPAC names for the following alkyl halides:
(a),
CH3CH2CH2CH2I
Solution: 1-iodobutane
CH3
(b), CH3CHCH2CH2Cl
Solution: 1-chloro-3-methylbutane
CH3
BrCH2CH2CH2 CCH2Br
CH3
(c),
Solution: 1,5-Dibromo-2,2-dimethylpentane
CH3
CH3CCH2CH2Cl
Cl
(d),
Solution: 1,3-Dichloro-3-methylbutane
I CH2CH2I
(e), CH3CHCHCH2CH3
Solution: 3-Ethyl-1,4-diiodopentane
Br
CH3CHCH2CH2CHCH3
Cl
(f),
Solution: 2-Bromo-5-chlorohexane
10.2 Draw structures corresponding to the following IUPAC names:
(a), 2-Chloro-3,3-dimethylhexane
Solution:
Cl
(b), 3,3-Dichloro-2-methylhexane
Cl
Cl
Solution:
(c), 3-Bromo-3-ethylpentane
Br
Solution:
(d), 1,1-Dibromo-4-isopropylcyclohexane
Br
Br
Solution:
(e), 4-sec-Butyl-2-chlorononane
Cl
Solution:
(f), 1,1-Dibromo-4-tert-butylcyclohexane
Br
Br
Solution:
10.3 Draw and name all monochloro products you would expect to obtain from radical chlorination of
2-methylpentane. Which, if any, are chiral?
Solution:
hν Cl2
Cl
Cl
Cl
H
C
+
C
+
+
C
＋
Cl
H
Cl
1-chloro-2-methylpentane
(chiral)
2-chloro-2-methylpetane
3-chloro-2-methylpentane
(chiral)
2-chloro-4-methylpentane
(chiral)
1-chloro-4-methylpentane
10.4 Taking the relative reactivities of 1°,2°,3°hydrogen atoms into account, what products
would you expect to obtian from monochlorination of 2-methylbutane? What would the
approximate percentage of each product be?
Solution:
Cl
Cl
Cl2
+
hν
10/26
6/26
Cl
+
+
Cl
7/26
3/26
10.5 Draw three resonance forms for the cyclohexadienyl radical.
Cyclohexadienyl radical
Solution:
10.6 The major product of the reaction of methylenecyclohexane with N-bromosuccinimide is
1-(bromomethyl)cyclohexene. Explain.
CH2
CH2Br
NBS
light, CCl4
Major product
Solution:
Br
Allylic position
Less hindered
More hindered
H
H
H
CH2
CH2
CH2
NBS
light, CCl4
Br2
CH2Br
Major product
One more reason is the relative stability of alkene.
10.7 What products would you expect from the reaction of the following alkenes with NBS? If
more than one product is formed , show the structures of all.
(a)
CH3
H
CH3
H
CH3
CH3
NBS
H
Br
H
CH3
CH3
Br
(b)
CH3
H3C
C
C
H
H
C
C
H
CH3
H
CH3
H3C
C
C
H
C
H
H
H
C
CH3
H
B
A
CH3
H3C
C
CH3
C
H
C
H
H
C
CH3
H3C
C
H
C
H
C
H
H
C
CH3
H
C
D
CH3
H3C
C
C
H
H
C
H
C
H3C
CH3
CH3
C
H
C
C
H
C
H
CH3
H
C
CH3
H
H
From the four radicals above, we can get four products drawn below:
A
main
H3C
CH3
C
B
C
H
C
H
Br
H
C
CH3
H3C
CH3
C
H
C
H
C
H
H
Br
C
CH3
H3C
C
D
C
H
H
C
Br
10.8
H2
C
CH3
H3C
CH3
C
H
H
C
Br
C
H
C
H
CH3
How do you prepare the following alkyl halides from the corresponding alcohols?
(a)
Cl
H3C
C
CH3
CH3
Cl
OH
C
HCl
r
e
h
t
E
H3C
CH3
H3C
C
CH3
+
H2O
CH3
CH3
(b)
Br
H3C
H2
C
C
H
OH
C
H
C
H
CH3
Br
CH3
H2
C
PBr3
C
H
r
e
h
t
E
H3C
CH3
CH3
H3C
C
H
CH3
H2
C
C
H
CH3
(c)
CH3
Br
H2
C
H2
C
H2
C
H2
C
C
H
CH3
CH3
(d)
H2
C
H2
C
H2
C
H2
C
C
H
CH3
PBr3
r
e
h
t
E
HO
CH3
Br
H2
C
H2
C
H2
C
H2
C
C
H
CH3
CH3
H2
C
H3C
Cl
H2
C
C
H
C
CH3
CH3
CH3
H2
C
C
H
CH3
C
HCl
r
e
h
t
E
H2
C
H3C
OH
CH3
H3C
H2
C
C
H
Cl
H2
C
CH3
C
CH3
10.9
Just how strong a base would you expect a Grignard reagents to be? Look at Table 8.1, and the
predict whether the following reactions will occur as written. (The pKa of NH3 is 35)
(a) CH3MgBr + H
C
C
CH4 + H
H
CH4 + H2N
(b) CH3MgBr + NH3
C
C
MgBr
MgBr
Solution: The Grignard reagent can be thought of as the magnesium salt of a hydrocarbon acid.
Because hydrocarbons are very weak (pKa’s in the range 44 to 60), it is very strong base.
All the two reactions can occur.
10.10
How might you replace a halogen substituent by a deuterium atom if you wanted to prepare a
deuterated compound?
Br
D
?
CH3CHCH2CH3
Solution:
CH3 CHCH2CH3
Br
MgBr
CH3CHCH2CH3
Mg
Ether
CH3CHCH2CH3
D
D2O
CH3CHCH2CH3
10.11
How would you carry out the following transformations using an organocopper coupling reaction ?
More than one step is required in each case.
?
CH3
(a)
Solution:
CH3
NBS
CCl4
Br
Ether
(CH 3)2CuLi
Br
CH3
?
(b) H3CH2CH2CH2C
CH3CH2CH2CH2CH2CH2CH2CH3
Br
Solution:
2Li
H3CH2CH2CH2C Br
2 H3CH2CH2CH2C
Li
Pentane
H3CH2CH2CH2C Li
Ether
CuI
LiBr
(H3CH2CH2CH2C)2 Cu+Li-
H3CH2CH2CH2C Br 2(H2CH2CH2CH3C)2 Cu+Li-
Ether
CH3CH2CH2CH2CH2CH2CH2CH3
(H3CH2CH2CH2C)
LiBr
Cu
(c)
?
H3CH2CH2CHC CH2
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
Solution:
H3CH2CH2CHC CH2
1 BH 3,THF
2 H 2O2,OH
H3CH2CH2CH2CH2C OH
PB3
H3CH2CH2CH2CH2C OH
2Li
H3CH2CH2CH2CH2C Br
H3CH2CH2CH2CH2C Br
Pentane
H3CH2CH2CH2CH2C
2 H3CH2CH2CH2CH2C
Li+
CuI
Li+
LiBr
Ether
(CH2CH2CH2CH2CH3)2Cu+Li-
LiI
(CH3CH2CH2CH2CH2)2Cu+Li-
H3CH2CH2CH2CH2C Br
Ether
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
10.12
Rank each of the following series of compounds in order of increasing oxidation level:
(a)
Cl
O
O
<
Oxidation level order:
(b)
CH3CN
CH3CH2NH2
Cl
=
<
NH2 CH2CH2NH2
Oxidation level order:
CH3CH2NH2
<
NH2 CH2CH2NH2
<
CH3CN
10.13
Tell whether each of the following reaction is an oxidation, a reduction, or neither.
Explain your answer.
O
(a)
CH3CH2CH
NaBH4
H2O
CH3CH2CH2OH
OH
1. BH3
(b)
2. NaOH, H2 O2
Solution: (a) Because of breaking one C-O, so it is reduction.
(b) Because of forming one C-O and forming one C-H, so it is neither reduction
nor oxidation.
10.14
Give an IUPAC name for each of the following alkyl halides (yellow-green=Cl):
(a)
cis-1-Chloro-3-methyl-cyclohexane
(b)
4-chloro-2-methyl-2-heptene
10.17
Name the following alkyl halides according to IUPAC rules:
(a)
H3 C
CH3
Br
Br
C
H
C
H
C
H
CH3
H2
C
C
H
CH3
Solution: 3,4-dibromo-2,6-dimethyl-heptane
I
(b) H3CHC
CHCH2CHCH3
Solution: 5-iodo-2-hexene
Br
H3 C
H2
C
C
Cl
CH3
C
H
C
H
CH3
CH3
(c)
Solution: 2-bromo-4-Chloro-2,5-dimethyl-hexane
CH2Br
(d)
H3CH2C
C
H
CH2CH2CH3
Solution: 3-bromomethylhexane
CCH2Br
(e) ClH2CH2CH2CC
Solution: 1-bromo-6-chloro-2-hexyne
10.18
Draw structures corresponding to the following IUPAC names:
(a) 2,3-Dichloro-4-methylhexane
Cl
Solution:
CH3
Cl
(b) 4-Bromo-4-ethyl-2-methylhexane
Br
Solution:
(c) 3-Iodo-2,2,4,4-tetramethylpentane
I
Solution:
(d) cis-1-bromo-2-ethylcyclopentane
Solution:
Br
CH2CH3
H
H
10.19:
Draw and name the monochlorination products you might obtain by radical chlorination of
2-methylbutane. Which of the products are chiral? Are any of the products optically active?
Solution: As there are four kinds of hydrogen atoms, so there should be four kinds of products, witch
are
(a):1-chloro-2-methylbutane,
(b):2-chloro-2-methylbutane,
(c):3-chloro-2-methylbutane
and
(d):4-chloro-2-methylbutane.
The products (a) and (c) are chiral. And the products (a) and (c) are optically active.
10.20:
A chemist requires a large amount of 1-bromo-2-pentene as starting material for a synthesis and
decides to carry out an NBS allylic bromination reaction:
NBS
Br
CCl4
What is wrong with this synthesis plan? What side products would form in addition to the desired
product?
Solution:
Br
NBS
Br
CCl4
+
Mixture will be obtained.
10.21
What product(s) would you expect from the reaction of 1-methylcyclohexene with NBS? Would you
use this reaction as part of a synthesis?
CH3
NBS
CCl4
?
CH3
Br
CH3
Solution: The products will be
Br
and
As a part of synthesis I will not use this reaction. Because from this reaction I will get two
mixed products. But usually in synthesis we only need one single product.
10.22
How would you prepare the following compounds, starting with cyclopentene and any other reagents
needed?
(a) Chlorocyclopentane
(b) Methylcyclopentane
(c) 3- Bromocyclopentene
(d) Cyclopentanol
(e) Cyclopentylcyclopentane
(f) 1,3-Cyclopentadiene
HCl
Cl
Solution: (a)
HI
I
CH3
(CH3)2CuLi
(b)
Br
NBS
(c)
1.BH3,THF
(d)
OH
2.H2O2,OH
(e)
HI
I
I 2Li
Li
CuI
Ether
Pentane
Br
NBS
(f)
10.23
Predict the product(s) of the following reactions:
Base
Cu
Li
H3C
OH
HBr
(a)
Ether
(c)
NBS
CCl4
?
OH
(b)
SOCl2
?
OH
Mg
Ether
(e)
PBr3
?
H2O
A?
?
Ether
(d)
B?
Br
Br
(f)
(CH3)2CuLi
Br +
H3C OH
HBr
(g)
CuI
Li
A?
Pentane
B?
Ether
?
H3 C
Br
+ H2O
Ether
Solution: (a)
OH
(b)
SOCl2
+ HCl
Cl + SO2
Br
O
NBS
CCl4
+
N H
O
(c)
OH
PBr3
3
Ether
(d)
Br
Mg
Ether
(e)
Br
+ H3PO3
3
H2O
+ Mg(OH)Br
MgBr
(f)
Br
2Li
Pentane
(g)
Li
+ LiBr
CuI
(CH3)2CuLi
Br +
(CH3CH2CH2CH2)2 CuLi + LiI
Ether
+ CH3Cu + LiBr
10.24
(S)-3-Methylhexane undergoes radical bromination to yield optically inactive 3-bromo-3-methylhexane
as the major product. Is the product chiral? What conclusions can you draw about the radical
intermediate?
CH2CH3
Br
H3C
C
CH2CH2CH3
Solution: Two stereoisomers of the product are formed. (R)-3-bromo-3-methylhexane and
CH2CH3
C
H3C
CH2CH2CH3
Br
(S)-3-bromo-3-methylhexane. They are all chiral and form in a ratio of 1:1.
CH2CH3
C
H3C
CH2CH2CH3.
Conclusions: The radical intermediate is
and the two sides have equal chance to be attacked.
It’s planar
10.25
Assume that you have carried out a radical chlorination reaction on (R)-2-chloropentane and have
isolated (in low yield) 2,4-dichloropentane. How many stereoisomers of the product are formed and in
what ratio? Are any of the isomers optically active? (See Problem 10.24)
Cl
Cl
+
Cl
+
HCl
Cl2
Cl
+
(R)-2-chloropentane
Cl
Cl
(S)
(R)
2,4-Dichloro-pentane
Α
+
Cl
Cl
(R)
(R)
2,4-Dichloro-pentane
Β
Solution:
Two stereoisomers of the product are formedin 1 : 1 ratio. The first one (A) is a meso compound,
so it is optically inactive. The second one (B) is optically active.
10.26
Calculate ΔH for the reactions of Cl
and Br
with CH4 , and then draw a reaction energy
diagram showing both processes. Which reaction is likely to be faster?
Solution:
Product bonds formed
Reactant bonds broken
C-Cl
D=351kJ/mol
C-H
D=438kJ/mol
H-Cl
D=432kJ/mol
Cl-Cl
D=243kJ/mol
Total
D=783kJ/mol
Total
D=681kJ/mol
ΔH＝681－783＝－102kJ/mol
Product bonds formed
Reactant bonds broken
C-Br
D=293kJ/mol
C-H
D=438kJ/mol
H-Br
D=366kJ/mol
Br-Br
D=193kJ/mol
Total
D=659kJ/mol
Total
D=631kJ/mol
ΔH＝631-659＝－28kJ/mol
The reaction energy diagram:
Reaction energy diagram of
Bromomethane
900
800
700
600
500
400
300
200
100
0
Energy
Energy
Reaction energy diagram of
Cholomethane
0
2
4
6
Reaction process
900
800
700
600
500
400
300
200
100
0
0
2
4
Reaction process
6
The reaction between Cholorine radical and methane is much faster.
10.27
Use the bond dissociation energies listed in Table 5.3 on page 154 to calculate △H0 for the reactions
of Cl. and Br. with a secondary hydrogen atom of propane. Which reaction would you expect to
be more selective?
For Cl. △H0=401-339=62kj/mol
For Br. △H0=401-274=127kj/mol
It is clear that Cl. has higher reactivity while Br. has lower reactivity; thus according to
“lower reactivity higher selectivity”, the reaction of Br. with a secondary hydrogen atom
of propane is more selective.
10.28
What product(s) would you expect from the reaction of 1, 4-hexdiene with NBS? What is the structure
of the most stable radical intermediate?
Solution: The reaction will be following:
Br
major
Br
NBS
Br
Br
Br
Br
That’s because of relative stability of conjugated diene.
10.29
Alkyl benzenes such as toluence (methylbenzene) react with NBS to give products in which bromine
substitution has occurred at the position next to the aromatic ring (the benzylic position). Explain,
based on the bond dissociation energies in Table 5.3
The reason is show bellow:
Br
NBS
CCl 4
H
H
368kJ/mol
CH2
H
nearly 464kJ/mol
H
The carbon hydrogen bond of the methyl has the lowest bond dissociation energy, so this bond is
most likely to be broken and yields the most stable radical.
10.30:
Draw resonance structures for the benzyl radical, C6H5CH2 , the intermediate produced in the NBS
bromination reaction of toluene (Problem 10.29)
Solution:
CH2
CH2
CH2
CH2
10.32
Draw resonance structure for the following species:
(a) CH3CH
CHCH
CHCH
CHCH2 (b)
(c)
CH3C
N
N
O
Solution:
(a).
(b).
(c).
N
O
10.33
Rank the compounds in each of the following series in order of increasing oxidation level:
O
(a)
O
OH
O
NH2
Br
Br
Cl
(b)
Cl
O
Solution:
O
O
a)
=
<
<
OH
Cl
NH2
Br
=
b)
<Br
Cl <
O
10.34.
Which of the following compounds have the same oxidation level and which have different levels?
O
O
OH
2
1
3
O
4
5
Solution: compound 1, 2, 4 have the same oxidation level and compound 3， 5 has different
oxidation level.
10.35
Tell whether each of the following reactions is an oxidation or reduction:
O
CrO3
a. CH3CH2 OH
CH3CH
Oxidation
O
O
b. H2C
CHCCH3
+
NH3
NH2CH2CH2 CCH3
Neither
CH3CH2CHCH3
c.
Br
1.Mg
2.H2O
CH3CH2CH2CH3
Reduction
10.36
How would you carry out the following syntheses?
(a) Butylcyclohexane from cyclohexene
(b) Butylcyclohexane from cyclohexanol
(c) Butylcyclohexane from cyclohexane
Br
HBr
(CH3CH2CH2CH2)2CuLi
In ether
+
+ CH3CH2CH2CH2Cu
OH
LiBr
Br
(CH3CH2CH2CH2)2CuLi
PBr3
In ether
Ether
+ CH3CH2CH2CH2Cu
+
LiBr
Br
Br2
Heat/hv
(CH3CH2CH2CH2)2CuLi
In ether
+ CH3CH2 CH2 CH2Cu
+
LiBr
10.37
The syntheses shown here are unlikely to occur as written. What is wrong with each?
(a)
F
1. Mg
2.OH3
(b)
Br
CH2
CH2
NBS
CCL4
CH3
CH3
(c)
F
CH3
(CH3)2CuLi
Ether
Solution:
(a) Organofluorides rarely react with magnesium. So there is no alkyl magnesium fluorides.
(b) The reaction reacts as follow:
CH2
CH2
CH2
CH2Br
CH3
CH3
CH3
NBS
CCL4
CH3
Br
CH2Br
CH2
CH3 is more stable than
CH3 because of Zaitsev’s rule.
(c)
Gilman reagents couldn’t react with fluorides because of high density of electron on fluorine atom.
So this reaction could not occur.
10.38
Why do you suppose it’s not possible to prepare a Gringard reagent from a bromo alcohol such as
4-bromo-1-pentanol?
Br
MgBr
OH
Mg
OH
Give another example of a molecule that is unlikely to form a Grignard reagent.
Solution: If 4-bromo-1-pentanol can form a Gringard reagent, it will nucleophilic attack the active
atom H in –OH, so it’s not possible to prepare a Gringard reagent.
O
C
Such as Br
OH which has active H is unlikely to
form a Grignard reagent.
10.39
Addition of HBr to a double bond with an ether substiutent occurs regiospecifically to give a
product in which the Br and OR are bonded to the same carbon:
OCH3
OCH3
H
Br
Br
Draw two possible carbocation intermediates in this electrophilic addition reaction, and
explain using resonance why the observed product is formed.
Answer:
OCH3
OCH3
H
OCH3
Br
OCH3
more stable form
10.40
Phenols, compounds that have an –OH group bonded to a benzene ring, are relatively acidic because
their anions are stabilized by resonance. Draw resonance structures for the phenoxide ion.
O
phenoxide ion
Solution:
O
O
O
O
O
10.41
Alkyl halides can be reduced to alkanes by a radical reaction with tributyltin hydride, (C4H9)3SnH ,
in the presence of light (hv):
R
+
X
(C4H9 )3SnH
R
+
H
(C4 H9 )3SnX
solution:
Initiation step:
(C4H9)3Sn
H
(C4H9)3Sn
+
H
Propagation steps:
(C4H9 )3SnH
H
R
X
X
R
H
Termination steps:
(C4 H9)3 Sn
Sn(C4H9)3
(C4H9 )3Sn
+
(C4H9)3 Sn
+
X
R
+
H
R
+
X
R
R
+
R
R
R
H
+
H
H
H
X
+
X
X
X
(C4H9)3Sn
+
H
(C4H9)3Sn
H
(C4H9)3Sn
+
R
(C4H9)3 Sn
R
+
H
(C4H9)3Sn
(C4H9)3Sn
R
X
H
R
X
10.42 Identify the reagents a-c in the following scheme:
X
H
X
H
OH
CH3
Br
c
b
a
Solution: a, 1) BH3, 2) OH-, H2O2; b, PBr3; c, 1) Li, 2) CuI, 3) CH3I
10.43
Tertiary alkyl halides, R3CX, undergo spontaneous dissociation to yield a carboncation,
R3C+.Which do you think reacts faster,(CH3)3CBr,or H2C=CHC(CH3)2Br? Explain
Solutions: H2C=CHC(CH3)2Br .Because of double bond’s conjugate effect. So the carboncation is
more stable .And the △G is smaller .And as a result the reaction is faster.
10.44
Carboxylic acids (RCO2H; pKa≈5) are approximately 1011 times more acidic than alcohols (ROH;
pKa≈16). In other words, a carboxylate ion (RCO2-) is more stable than an alkoxide ion(RO-).
Explain, using resonance.
Solution:
O
O
H
C
R
C
O
R
O
O
C
R
O
C
O
R
O
The carboxylate ion has two resonance forms, which is rather than the alkoxide
ion,only one
Form. Thus the RCO2-is much stable than RO- .