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

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to these details enables us to understand why the reaction occurs and what
makes it easier or more diffi cult — in short, how changes in structure affect function. With
this understanding comes the ability to predict: to extrapolate by analogy from examples
of reactions we have seen to reactions we have not seen. Rational prediction is an essential
component of scientifi c reasoning, which we illustrate often throughout this book, beginning
in the very next chapter. There, we shall expand our study of haloalkanes and examine
additional processes they can undergo. Our continued focus will be the effect that structural
changes have on chemical behavior.

6-29. a. Write a mechanism and fi nal product for the reaction between sodium ethoxide,
 NaOCH2CH3, and bromoethane, CH3CH2Br, in ethanol solvent, CH3CH2OH.

The mechanism is backside attack in which the nucleophilic atom of the reagent attacks the atom of
the substrate that contains the leaving group (Section 6-5). We begin by identifying each of these
components. The nucleophilic atom is the negatively charged oxygen atom in ethoxide ion, CH3CH2 – O

Attack occurs at the carbon attached to bromine in the substrate molecule, CH3 – CH2Br:





Br �

The products are bromide ion and ethoxyethane, CH3CH2 O

. . CH2CH3, an ether.

b. How would the preceding reaction be affected by each of the following changes?
1. Replace bromoethane with fl uoroethane.
2. Replace bromoethane with bromomethane.
3. Replace sodium ethoxide with sodium ethanethiolate, NaSCH2CH3.
4. Replace ethanol with dimethylformamide (DMF).

1. Table 6-4 tells us that fl uoride is a stronger base than bromide and, therefore, a poorer leaving

group. The reaction would still take place but would be very much slower. (The actual rate decrease
is of the order of 104.)

2. The carbon containing the leaving group in bromomethane is less sterically hindered than that in
bromoethane, so the rate of reaction would increase (Section 6-9). The product of the reaction
would be CH3OCH2CH3, methoxyethane.

3. Both ethoxide and ethanethiolate are negatively charged. Oxygen in ethoxide is more basic than
sulfur in ethanethiolate (Table 6-4), but the sulfur atom in ethanethiolate is larger, more polariz-
able, and less tightly solvated in the hydrogen-bonding ethanol solvent (compare Figure 6-6). We
know that strong bases are good nucleophiles, but base strength is outweighed by the increased
polarizability and reduced solvation of the larger atoms within the same column of the periodic
table (Section 6-9). Ethanethiolate reacts hundreds of times as fast, giving as a product
CH3CH2SCH2CH3, an example of a sulfi de (Section 9-10).

4. Conversion from a protic, hydrogen-bonding solvent to a polar, aprotic one accelerates the reaction
enormously by reducing solvation of the negatively charged oxygen atom (compare Table 6-6).
See Problem 57 for a similar exercise.

Stefanie Schore, providing a visual
demonstration of relative SN2 reac-
tivity. The three test tubes contain,
from left to right, solutions of
1-bromobutane, 2-bromopropane,
and 2-bromo-2-methylpropane in
acetone, respectively. Addition of
a few drops of NaI solution to each
causes immediate formation of
NaBr (white precipitate) from the
primary bromoalkane (left), slow
NaBr precipitation only after warm-
ing from the secondary substrate
(center), and no NaBr formation
at all from the tertiary halide even
 after extended heating (right).

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 C h a p t e r 6 245

6-30. a. Which of the following compounds would be expected to react in an SN2 manner at a rea-
sonable rate with sodium azide, NaN3, in ethanol? Which will not? Why not?

I m p o r t a n t C o n c e p t s

NH2š (ii) Iðš� (iii) Brðš� (iv) OHš� (v)


 (vi) CNð

As we did in Chapter Integration Problem 3-14, let us apply the WHIP approach to break down the
process of solving this problem. Again, this strategy will be described in detail in the Interlude that
follows Chapter 11.

What is the problem asking? This may be obvious — in part ‘a’ one merely has to identify which of
the compounds shown reacts with azide in ethanol via an SN2 process. However, there is a bit more
to it, and the clue is the presence of the word “why” in the question. “How” and “why” questions
invariably require a closer look at the situation, usually from a mechanistic perspective. It will be
necessary to consider fi ner details of the SN2 mechanism in light of the structures of each of the
substrate molecules.

How to begin? Characterize each substrate in the context of the SN2 process. Does it contain a viable
leaving group? To what kind of carbon atom is the potential leaving group attached? Are other rel-
evant structural features present?

Information needed? Does each of these six molecules contain a good leaving group? If necessary,
look in Section 6-7 for guidance: To be a good leaving group, a species must be a weak base. Next,
can you tell if the leaving group is attached to a primary, secondary, or tertiary carbon atom? See
their defi nitions in Section 2-5. Anything else? Section 6-9 tells you what to look for: steric hindrance
in the substrate that may obstruct the approach of the nucleophile.

Proceed. We identify fi rst the molecules with good leaving groups. Referring to Table 6-4, we see that,
as a general rule, only species that are the conjugate bases of strong acids (i.e., with pKa values , 0)
qualify. So, (i), (iv), and (vi) will not undergo SN2 displacement. They lack good leaving groups:

2OH, and 2CN are too strongly basic for this purpose (thus answering the “why not” for these three).
Substrate (ii) contains a good leaving group, but the reaction site is a tertiary carbon and sterically
incapable of following the SN2 mechanism. That leaves substrates (iii) and (v), both of which are
primary haloalkanes with minimal steric hindrance around the site of displacement. Both will trans-
form readily by the SN2 mechanism.

b. Compare the rates at which the SN2 reactions of substrates (iii) and (v) with azide will proceed.

Following the WHIP protocol, we look for any differences between the two substrates that might be
signifi cant in the context of the SN2 mechanism. Both are comparable in steric bulk with respect to
backside displacement: Both possess branching only at the remote g-carbon, sterically without adverse
consequences. That leaves as the only reasonable deciding factor the identity of the leaving group
itself. Bromide is better than chloride in this respect (HBr is a stronger acid than HCl — Table 6-4)
and is more readily displaced. Therefore we expect the rate of reaction of (iii) to be greater. Problem
56 requires similar reasoning.

c. Name substrates (iii) and (v) according to the IUPAC system.

Review Sections 2-5 and 4-1 if necessary.

(iii) 1-Bromo-3-methylbutane (v) (2-Chloroethyl)cyclopentane

Important Concepts
 1. A haloalkane, commonly termed an alkyl halide, consists of an alkyl group and a halogen.

 2. The physical properties of the haloalkanes are strongly affected by the polarization of the C – X
bond and the polarizability of X.

 3. Reagents bearing lone electron pairs are called nucleophilic when they attack positively polarized
centers (other than protons). The latter are called electrophilic. When such a reaction leads to
displacement of a substituent, it is a nucleophilic substitution. The group being displaced by the
nucleophile is the leaving group.

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