Vollhardt  Capítulo 6 (Haloalcanos)
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Vollhardt Capítulo 6 (Haloalcanos)

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This substitution is a general synthetic method for converting a methyl or
primary haloalkane into an alcohol.

A variation of this transformation is reaction 2. Methoxide ion reacts with iodoethane
to give methoxyethane, an example of the synthesis of an ether (Section 9-6).

In reactions 1 and 2, the species attacking the haloalkane is an anionic oxygen nucleo-
phile. Reaction 3 shows that a halide ion may function not only as a leaving group, but
also as a nucleophile.

Reaction 4 depicts a carbon nucleophile, cyanide (often supplied as sodium cyanide,
Na12CN), and leads to the formation of a new carbon – carbon bond, an important means
of modifying molecular structure.

Table 6-3 The Diversity of Nucleophilic Substitution

Reaction Leaving
number Substrate Nucleophile Product group

1. CH3Cl
..

. . : 1 HO
..

. . :
2 uy CH3OH

..
. . 1 :Cl

..
. . :

2

 Chloromethane Methanol

2. CH3CH2 I
..

. .: 1 CH3O
..

. . :
2 uy CH3CH2OCH

..
. . 3 1 : I

..
. .:

2

 Iodoethane Methoxyethane

3.

H

Br

CH3CCH2CH3
A

A

2-Bromobutane
�ðð

 1 : I
..

. .:
2 uy

H

2-Iodobutane

I

CH3CCH2CH3
A

A
ðð�

 1 :Br
..

. . :
2

4.

H

CH3CCH2I

CH3

A

A

1-Iodo-2-methylpropane

ðš� 1 :N‚C:
2 uy

H

CH3CCH2C N

CH3

A

A

3-Methylbutanenitrile

q ð 1 : I
..

. .:
2

5.
Br

Bromocyclohexane

Br�
�

ð
 1 CH3S

..
. .:

2 uy
SCH3

Methylthiocyclohexane

�
�

 1 :Br
..

. . :
2

6. CH3CH2 I
..

. .: 1 :NH3 uy

H
�

Ethylammonium
iodide

H

CH3CH2NH
A

A
 1 : I

..
. .:

2

 Indoethane

7. CH3Br
..

. . : 1 :P(CH3)3 uy
�

Tetramethylphosphonium
bromide

CH3PCH3

CH3

CH3

A

A
 1 :Br

..
. . :

2

 Bromomethane

Note: Remember that nucleophiles are red, electrophiles are blue, and leaving groups are green.
Anionic nucleophiles give neutral products (Reactions 1 – 5). Neutral nucleophiles give salts as products
(Reactions 6 and 7).

6 - 2 N u c l e o p h i l i c S u b s t i t u t i o n

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220 C h a p t e r 6 P r o p e r t i e s a n d R e a c t i o n s o f H a l o a l k a n e s

Reaction 5 shows the sulfur analog of reaction 2, demonstrating that nucleophiles in the
same column of the periodic table react similarly to give analogous products. This conclu-
sion is also borne out by reactions 6 and 7. However, the nucleophiles in these two reactions
are neutral, and the expulsion of the negatively charged leaving group results in a cationic
species, an ammonium or phosphonium salt, respectively.

All of the nucleophiles shown in Table 6-3 are quite reactive, but not all for the same
reasons. Some are reactive because they are strongly basic (HO2, CH3O

2). Others are weak
bases (I2) whose nucleophilicity derives from other characteristics. Notice that, in each
example, the leaving group is a halide ion. Halides are unusual in that they may serve as
leaving groups as well as nucleophiles (therefore making reaction 3 reversible). However,
the same is not true of some of the other nucleophiles in Table 6-3 (in particular, the strong
bases); the equilibria of their reactions lie strongly in the direction shown. These topics are
addressed in Sections 6-7 and 6-8, as are factors that affect the reversibility of displacement
reactions. First, however, we shall examine the mechanism of nucleophilic substitution.

Exercise 6-1

What are the substitution products of the reaction of 1-bromobutane with (a) : I
..

. .:
2; (b) CH3CH2 O

..
. .:

2;
(c) N3

2; (d) :As(CH3)3; (e) (CH3)2 Se
..

. . ?

Exercise 6-2

Working with the Concepts: Planning a Synthesis

Suggest starting materials for the preparation (synthesis) of CH3CH2SCH3.
Strategy
The question does not specify a method for preparing this molecule, but it makes sense to use our
newest reaction, nucleophilic substitution. A powerful method for designing synthetic preparations
involves working backward from the structure of the target molecule and is called retrosynthetic
analysis. We demonstrate the idea here and will return to it in Section 8-9. Begin by rephrasing the
question as “What substances must react by nucleophilic substitution to give the desired product?”
Write the structure in full, so as to see clearly all the bonds it contains, and identify one that might
be formed in the course of nucleophilic substitution.
Solution
• Example 5 from Table 6-3 gives us a model for a reaction that forms a sulfur compound with
two C–S bonds. Proceed in the same way, even though the problem does not tell us which halide
leaving group to displace by the sulfur nucleophile. We may choose any one that will work,
namely, chloride, bromide, or iodide:

�OSH3C CH3CH2Br

Delete C–S bond

Attach any suitable
halide leaving group

S
H3C CH2

CH3 O Ošð�š�

• We fi nish by writing the preparation in the forward direction, the way we would actually carry
it out:

� �CH3S CH3SCH2CH3CH3CH2Br� �š� š��
�

ð ð Br�
�

ð ð

• Notice that we could have just as easily conducted our reverse analysis by deleting the bond
between the sulfur and the methyl carbon, rather than the ethyl. That would give us a second,
equally correct method of preparation:

� �CH3I CH3SCH2CH3SCH2CH3� �š� š��
�

ð ð I�
�

ðð

As before, the choice of halide leaving group is immaterial.

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 C h a p t e r 6 221

In Summary Nucleophilic substitution is a fairly general reaction for primary and second-
ary haloalkanes. The halide functions as the leaving group, and several types of nucleophilic
atoms enter into the process.

6-3 Reaction Mechanisms Involving Polar Functional Groups:
Using “Electron-Pushing” Arrows

In our consideration in Chapter 3 of radical halogenation, we found that a knowledge of its
mechanism was helpful in explaining the experimental characteristics of the process. The
same is true for nucleophilic substitution and, indeed, virtually every chemical process that
we encounter. Nucleophilic substitution is an example of a polar reaction: It includes
charged species and polarized bonds. Recall (Chapter 2) that an understanding of electro-
statics is essential if we are to comprehend how such processes take place. Opposite charges
attract — nucleophiles are attracted to electrophiles — and this principle provides us with a
basis for understanding the mechanisms of polar organic reactions. In this section, we
expand the concept of electron fl ow and review the conventional methods for illustrating
polar reaction mechanisms by moving electrons from electron-rich to electron-poor sites.

Curved arrows depict the movement of electrons
As we learned in Section 2-2, acid-base processes require electron movement. Let us briefl y
reexamine the Brønsted-Lowry process in which the acid HCl donates a proton to a mole-
cule of water in aqueous solution:

Depiction of a Brønsted-Lowry Acid-Base Reaction by Using Curved Arrows

A
O O OH

This electron pair
becomes shared

This electron pair is
completely displaced

H H Cl� �

�

š� š�

�

Cl �š�

��

O E HH H

H

�O

Notice that the arrow starting at the lone pair on oxygen and ending at the hydrogen of
HCl does not imply that the lone electron pair departs from oxygen completely; it just
becomes a shared pair between that oxygen atom and the atom to which the arrow points.
In contrast, however, the arrow beginning at the H–Cl bond and pointing toward the
 chlorine atom does signify heterolytic cleavage of the bond; that electron pair becomes
separated from hydrogen and ends up entirely on the chloride ion.

Exercise 6-3

Try It Yourself

Suggest starting materials for
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