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 \u2014 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.
CHAPTER INTEGRATION PROBLEMS
6-29. a. Write a mechanism and fi nal product for the reaction between sodium ethoxide, 
 NaOCH2CH3, and bromoethane, CH3CH2Br, in ethanol solvent, CH3CH2OH.
SOLUTION
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 \u2013 O
2. 
Attack occurs at the carbon attached to bromine in the substrate molecule, CH3 \u2013 CH2Br:
CH3CH2O \ufffd
H3C
H
Br CH3CH2O CH2CH3\ufffdO O}C
H
G
(
Br \ufffd
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).
SOLUTION
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
(i) 
NH2\u161 (ii) Ið\u161\ufffd (iii) Brð\u161\ufffd (iv) OH\u161\ufffd (v) 
Clð\u161\ufffd
 (vi) CNð
SOLUTION
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 \u2014 in part \u2018a\u2019 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 \u201cwhy\u201d in the question. \u201cHow\u201d and \u201cwhy\u201d 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: 
2NH2, 
2OH, and 2CN are too strongly basic for this purpose (thus answering the \u201cwhy not\u201d 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.
SOLUTION
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 \u2014 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.
SOLUTION
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 \u2013 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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