Chapter 10 10.1 Give IUPAC names for the following alkyl halides:

(a), CH3CH2CH2CH2I Solution: 1-iodobutane

(b), CH3CHCH2CH2ClCH3

Solution: 1-chloro-3-methylbutane

(c),

BrCH2CH2CH2CCH2BrCH3

CH3

Solution: 1,5-Dibromo-2,2-dimethylpentane

(d),

CH3CCH2CH2ClCH3

Cl Solution: 1,3-Dichloro-3-methylbutane

(e), CH3CHCHCH2CH3

I CH2CH2I

Solution: 3-Ethyl-1,4-diiodopentane

(f),

CH3CHCH2CH2CHCH3

Cl

Br

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

Solution:

Cl Cl

(c), 3-Bromo-3-ethylpentane

Solution:

Br

(d), 1,1-Dibromo-4-isopropylcyclohexane

Solution:

Br

Br (e), 4-sec-Butyl-2-chlorononane

Solution:

Cl

(f), 1,1-Dibromo-4-tert-butylcyclohexane

Solution:

Br

Br

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:

Cl2hν

Cl

+ C

Cl

1-chloro-2-methylpentane(chiral)

2-chloro-2-methylpetane

+

C

Cl

3-chloro-2-methylpentane(chiral)

H

+ C

Cl

H

2-chloro-4-methylpentane(chiral)

＋ Cl

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:

Cl2

hν

Cl

+

Cl

6/26 10/26

+

7/26

+

Cl

3/26

Cl

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

NBSlight, CCl4

CH2Br

Major product

Solution:

CH2

NBSlight, CCl4

CH2Br

Major product

H H

BrAllylic position

H

CH2 CH2

Br2

More hindered Less hindered

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

CH3

NBS

H

H

HH

CH3

CH3

CH3

Br

CH3

Br

(b)

H3C C CH

CH3

CH

HC CH3

H

H3C C CH

CH3

CH

HC CH3

H H

H3C CH

CH

CH3

CH

HC CH3H3C C C

H

CH3

CH

HC CH3

H

H3C CHC

CH3

CH

CH

CH3

H

H3C C CH

CH3

HC

HC CH3

H

A B

C D

From the four radicals above, we can get four products drawn below:

A

H3C C CH

CH3

CH

HC CH3

HBr

H3C CH

CH

CH3

CH

HC CH3

Br

H3C C CH

CH3

HC

H2C CH3

Br

C

B

H3C CH

HC

CH3

CH

CH

CH3

D

Br

main

10.8

How do you prepare the following alkyl halides from the corresponding alcohols? (a)

H3C C CH3

Cl

CH3

H3C C CH3

OH

CH3

HCl

EtherH3C C CH3

Cl

CH3

+ H2O

(b)

H3C CH

H2C C

HCH3

Br CH3

H3C CH

H2C C

HCH3

OH CH3

PBr3

EtherH3C C

H

H2C C

HCH3

Br CH3

(c)

BrH2C

H2C

H2C

H2C C

H

CH3

CH3

HOH2C

H2C

H2C

H2C C

H

CH3

CH3PBr3

EtherBr

H2C

H2C

H2C

H2C C

H

CH3

CH3

(d)

H3CH2C C

H

H2C C CH3

CH3

ClCH3

H3CH2C C

H

H2C C CH3

HCl

Ether

CH3

OHCH3

H3CH2C C

H

H2C C CH3

CH3

ClCH3

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 H CH4 + H C C MgBr

(b) CH3MgBr + NH3 CH4 + H2N 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?

?

CH3CHCH2CH3

Br

CH3CHCH2CH3

D

Solution:

CH3CHCH2CH3

Br

CH3CHCH2CH3

D

MgEther

CH3CHCH2CH3

MgBr

D2O

10.11 How would you carry out the following transformations using an organocopper coupling reaction ? More than one step is required in each case.

(a)

?

CH3

Solution:

CH3

NBS

CCl4

Br

Br(CH3)2CuLi

Ether

(b) CH3CH2CH2CH2CH2CH2CH2CH3H3CH2CH2CH2C Br?

Solution:

H3CH2CH2CH2C Br

CH3CH2CH2CH2CH2CH2CH2CH3

2Li

Pentane H3CH2CH2CH2C Li LiBr

H3CH2CH2CH2C Li2 CuIEther

(H3CH2CH2CH2C) Cu+Li-2

H3CH2CH2CH2C Br (H2CH2CH2CH3C) Cu+Li-2

2

Ether

LiBr (H3CH2CH2CH2C) Cu (c)

H3CH2CH2CHC CH2

?

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

Solution:

H3CH2CH2CHC CH21 BH3,THF

2 H2O2,OHH3CH2CH2CH2CH2C OH

H3CH2CH2CH2CH2C OHPB3

H3CH2CH2CH2CH2C Br

H3CH2CH2CH2CH2C Br2Li

PentaneH3CH2CH2CH2CH2C Li+

H3CH2CH2CH2CH2C Li+2 CuIEther

(CH2CH2CH2CH2CH3)2Cu+Li-

LiBr

LiI

H3CH2CH2CH2CH2C Br (CH3CH2CH2CH2CH2)2Cu+Li-

EtherCH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

10.12 Rank each of the following series of compounds in order of increasing oxidation level: (a)

O Cl

Oxidation level order: <

O

=

Cl

<

(b) CH3CN CH3CH2NH2 NH2 CH2CH2NH2 Oxidation level order:

CH3CNCH3CH2NH2 NH2 CH2CH2NH2< < 10.13 Tell whether each of the following reaction is an oxidation, a reduction, or neither.

Explain your answer.

CH3CH2CH

O

NaBH4

H2OCH3CH2CH2OH

1. BH3

2. NaOH, H2O2

OH

(a)

(b)

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) H3C C

H

CH3

CH

Br

CH

Br

H2C C

H

CH3

CH3

Solution: 3,4-dibromo-2,6-dimethyl-heptane

(b) H3CHC CHCH2CHCH3

I

Solution: 5-iodo-2-hexene

(c)

H3C CH2C

CH3

Br

CH

Cl

CH

CH3

CH3

Solution: 2-bromo-4-Chloro-2,5-dimethyl-hexane

(d)

CH2Br

CH

CH2CH2CH3H3CH2C

Solution: 3-bromomethylhexane

(e) ClH2CH2CH2CC CCH2Br Solution: 1-bromo-6-chloro-2-hexyne 10.18 Draw structures corresponding to the following IUPAC names: (a) 2,3-Dichloro-4-methylhexane

Solution:

Cl

Cl

CH3

(b) 4-Bromo-4-ethyl-2-methylhexane

Solution:

Br

(c) 3-Iodo-2,2,4,4-tetramethylpentane

Solution: I

(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:

BrNBS

CCl4 What is wrong with this synthesis plan? What side products would form in addition to the desired product? Solution:

BrNBS

CCl4+

Br

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 NBSCCl4

?

Solution: The products will be

CH3

Br and

CH3

Br

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

Solution: (a)

HCl Cl

(b)

HI I (CH3)2CuLi CH3

(c)

NBS

Br

(d)

1.BH3,THF OH2.H2O2,OH

(e)

HI I

2Li

Pentane

LiI CuI

Ether

Cu Li

(f)

NBS

Br

Base

10.23 Predict the product(s) of the following reactions:

(a)

OHH3CHBr

Ether?

(b) OHSOCl2

?

(c)

NBSCCl4

? (d)

OHPBr3Ether

?

(e) Br

MgEther

A?H2O

B?

(f) Br

LiPentane

A? CuI B?

(g) Br +Ether ?(CH3)2CuLi

Solution: (a)

OHH3CHBr

Ether

H3C Br

+ H2O

(b) OHSOCl2

Cl + SO2 + HCl

(c)

NBSCCl4

Br

+ N

O

O

H

(d)

OHPBr3Ether

3 3

Br+ H3PO3

(e) Br

MgEther

H2O

MgBr

+ Mg(OH)Br

(f)

Br 2LiPentane

CuILi + LiBr (CH3CH2CH2CH2)2CuLi + LiI

(g) Br +Ether(CH3)2CuLi + + LiBrCH3Cu

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?

Solution: Two stereoisomers of the product are formed.

C

CH2CH3

CH2CH2CH3

Br

H3C

(R)-3-bromo-3-methylhexane and

C

CH2CH3

CH2CH2CH3

H3C

Br

(S)-3-bromo-3-methylhexane. They are all chiral and form in a ratio of 1:1.

Conclusions: The radical intermediate is

C

CH2CH3

CH2CH2CH3H3C

. It’s planar and the two sides have equal chance to be attacked. 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)

(R)-2-chloropentane

Cl

(R)(R)

Cl Cl

Cl

(R)(S)

Cl Cl

2,4-Dichloro-pentane 2,4-Dichloro-pentane

Cl+ +Cl2

HCl Cl +

+

Α 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 ofCholomethane

0

100

200300

400

500

600700

800

900

0 2 4 6

Reaction process

Energy

Reaction energy diagram ofBromomethane

0100200300400500600700800900

0 2 4 6

Reaction process

Energy

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:

NBS

Br

Br

Br

Br

Br

Br

major

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:

CCl4

BrNBS

CH2H

368kJ/mol

H

H

H

nearly 464kJ/mol

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:

CH2CH2 CH2

CH2

10.32 Draw resonance structure for the following species:

(a) CH3CH CHCH CHCH CHCH2 (b) (c) CH3C N O

Solution: (a).

(b).

(c).

N O N O

10.33 Rank the compounds in each of the following series in order of increasing oxidation level:

(a)

O O

OH

(b)

NH2 BrCl Br Cl

O

Solution:

a) = <

O

<

O

OH

b) NH2

=Br

<Br Cl<

Cl

O

10.34. Which of the following compounds have the same oxidation level and which have different levels?

OOH

O

1 2 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:

a. CH3CH2OHCrO3

CH3CH

O

Oxidation

b. H2C CHCCH3

O

+ NH3 NH2CH2CH2CCH3

O

Neither

c.

CH3CH2CHCH3

Br

1.Mg

2.H2OCH3CH2CH2CH3

Reduction 10.36 How would you carry out the following syntheses?

(a) Butylcyclohexane from cyclohexene (b) Butylcyclohexane from cyclohexanol (c) Butylcyclohexane from cyclohexane

HBrBr

(CH3CH2CH2CH2)2CuLiIn ether

++ CH3CH2CH2CH2Cu LiBr

OH

PBr3

Ether

Br

(CH3CH2CH2CH2)2CuLiIn ether

++ CH3CH2CH2CH2Cu LiBr

Br

(CH3CH2CH2CH2)2CuLiIn ether

++ CH3CH2CH2CH2Cu LiBr

Heat/hvBr2

10.37 The syntheses shown here are unlikely to occur as written. What is wrong with each? (a)

F 1. MgOH32.

(b)

CH2

CH3

NBSCCL4

CH2

CH3

Br

(c)

F

(CH3)2CuLiEther

CH3

Solution: (a) Organofluorides rarely react with magnesium. So there is no alkyl magnesium fluorides. (b) The reaction reacts as follow:

CH2

CH3

CH2

CH3

NBSCCL4

CH2

CH3 CH3

CH2Br

CH3

CH2Br

is more stable than

CH2

CH3

Br

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

OH OH

MgBr

Mg

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.

Such as BrC

OH

O

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

H Br

OCH3

Br

Draw two possible carbocation intermediates in this electrophilic addition reaction, and explain using resonance why the observed product is formed. Answer:

OCH3

H Br

OCH3 OCH3

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 + (C4H9)3SnX

solution: Initiation step:

(C4H9)3Sn H +(C4H9)3Sn H Propagation steps:

R X

R HX

H(C4H9)3SnH

Termination steps:

R X

+

R H

(C4H9)3Sn (C4H9)3Sn

(C4H9)3Sn

+

+

+

+

+ X

R H

R X

R R

H H

X

++

+

+ X

(C4H9)3Sn H

(C4H9)3Sn R

X H

(C4H9)3Sn Sn(C4H9)3

(C4H9)3Sn X

R R

H H

R HR HR HR HR H

X X

(C4H9)3Sn H

(C4H9)3Sn R

H X 10.42 Identify the reagents a-c in the following scheme:

a

OH

b

Br

c

CH3

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:

RC

OH

O

RC

O

O

RC

O

O

RC

O

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- .