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 \u2013 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
\ufffdðð
 1 : I
..
. .:
2 uy 
H
2-Iodobutane
I
CH3CCH2CH3
A
A
ðð\ufffd
 1 :Br
..
. . :
2
4. 
H
CH3CCH2I
CH3
A
A
1-Iodo-2-methylpropane
ð\u161\ufffd 1 :N\u201aC:
2 uy 
H
CH3CCH2C N
CH3
A
A
3-Methylbutanenitrile
q ð 1 : I
..
. .:
2
5. 
Br
Bromocyclohexane
Br\ufffd
\ufffd
ð
 1 CH3S
..
. .:
2 uy 
SCH3
Methylthiocyclohexane
\ufffd
\ufffd
 1 :Br
..
. . :
2
6. CH3CH2 I
..
. .: 1 :NH3 uy 
H
\ufffd
Ethylammonium
iodide
H
CH3CH2NH
A
A
 1 : I
..
. .:
2
 Indoethane
7. CH3Br
..
. . : 1 :P(CH3)3 uy 
\ufffd
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 \u2013 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 \u201cWhat substances must react by nucleophilic substitution to give the desired product?\u201d 
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
\u2022 Example 5 from Table 6-3 gives us a model for a reaction that forms a sulfur compound with 
two C\u2013S 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:
\ufffdOSH3C CH3CH2Br
Delete C\u2013S bond
Attach any suitable
halide leaving group
S
H3C CH2
CH3 O O\u161ð\ufffd\u161\ufffd
\u2022 We fi nish by writing the preparation in the forward direction, the way we would actually carry 
it out:
\ufffd \ufffdCH3S CH3SCH2CH3CH3CH2Br\ufffd \ufffd\u161\ufffd \u161\ufffd\ufffd
\ufffd
ð ð Br\ufffd
\ufffd
ð ð
\u2022 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:
\ufffd \ufffdCH3I CH3SCH2CH3SCH2CH3\ufffd \ufffd\u161\ufffd \u161\ufffd\ufffd
\ufffd
ð ð I\ufffd
\ufffd
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 \u201cElectron-Pushing\u201d 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 \u2014 nucleophiles are attracted to electrophiles \u2014 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\ufffd \ufffd
\ufffd
\u161\ufffd \u161\ufffd
\ufffd
Cl \ufffd\u161\ufffd
\ufffd\ufffd
O E HH H
H
\ufffdO
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\u2013Cl 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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