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Vollhardt Capítulo 7 (Haloalcanos 2)

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a p t e r 7 F u r t h e r R e a c t i o n s o f H a l o a l k a n e s

hindered strong bases, such as potassium tert-butoxide, are employed. In these cases, the
SN2 pathway is slowed down suffi ciently for steric reasons to allow the E2 mechanism to
take over. Another way of reducing substitution is to introduce branching. However, even
in these cases, good nucleophiles still furnish predominantly substitution products. Only
strong bases, such as alkoxides, RO2, or amides, R2N

2, tend to react by elimination.

Primary (and methyl) haloalkanes react so exceedingly slowly with poor nucleophiles
that for practical purposes we consider the combination to give “no reaction.”

Secondary Haloalkanes. Secondary alkyl systems undergo, depending on conditions, both
eliminations and substitutions by either possible pathway: uni- or bimolecular. Good nucleo-
philes favor SN2, strong bases result in E2, and weakly nucleophilic polar media give mainly
SN1 and E1.

Tertiary Haloalkanes. Tertiary systems eliminate (E2) with concentrated strong base and
are substituted in nonbasic media (SN1). Bimolecular substitution is not observed, but elim-
ination by E1 accompanies SN1.

Exercise 7-18

Write the structure of the major organic product of the reaction of 1-bromopropane with (a) NaCN
in acetone; (b) NaOCH3 in CH3OH; (c) (CH3)3COK in (CH3)3COH.

Exercise 7-19

Write the structure of the major organic product of the reaction of 1-bromo-2-methylpropane with
(a) NaI in acetone; (b) NaOCH2CH3 in CH3CH2OH.

Exercise 7-20

Write the structure of the major organic product of the reaction of 2-bromopropane with
(a) CH3CH2OH; (b) NaSCH3 in CH3CH2OH; (c) NaOCH2CH3 in CH3CH2OH.

Exercise 7-21

Write the structure of the major organic product of the reaction of (a) 2-bromo-2-methylbutane
with water in acetone; (b) 3-chloro-3-ethylpentane with NaOCH3 in CH3OH.

Exercise 7-22

Predict which reaction in each of the following pairs will have a higher E2 : E1 product ratio and
explain why.

(a) CH3CH2CHBr





(b) (CH3CH)2N
�Li�, (CH3CH)2NH












 C h a p t e r 7 275

We have completed our study of haloalkanes by describing three new reaction pathways — SN1,
E1, and E2 — that are available to this class of compounds, in addition to the SN2 process.
We see now that details can make a big difference! Small changes in substrate structure
can completely alter the course followed by the reaction of a haloalkane with a nucleophile.
Similarly, correctly identifying a reaction product may require that we be able to assign
both nucleophile and base strength to a reagent.

The fi nal two sections in this chapter showed how the interplay of these factors affects
the outcome of a reaction. However, learning to solve problems of this type correctly
requires the use of accurate information. Therefore, we make the following study
 suggestion: Practice solving problems fi rst with your textbook and your class notes open,
so that you can fi nd the correct, applicable information as you are trying to devise a
solution. Remember, much of the problem solving you will be doing throughout this
course will require extrapolation from examples presented to you in various contexts. If
you are comfortable studying in small groups, propose reaction problems to one another,
and analyze their components in order to determine what process or processes are likely
to occur.

Where do we go from here? We next examine compounds containing polar carbon – oxygen
single bonds, comparing them with polar carbon – halogen bonds. The alcohols will be fi rst;
they add the feature of a polar O – H bond as well. Over the next two chapters we shall see
that some of the conversions of these and related compounds follow very similar mecha-
nistic pathways, as do the reactions of haloalkanes. Such parallels between the structures
of functional groups and the mechanisms of their reactions provide a useful framework for
comprehending the extensive body of information in organic chemistry.

7-23. Consider the reaction shown below. Will it proceed by substitution or by elimination? What factors
determine the most likely mechanism? Write the expected product and the mechanism by which it forms.








The fact that the compound is cyclic does not change how we approach the problem. The substrate
is secondary because the leaving group is attached to a secondary carbon. The nucleophile (ethoxide
ion, derived from NaOCH2CH3) is a good one, but, like hydroxide and methoxide (see Table 6-4), it
is also a strong base. Based on the evaluation criteria in Sec tions 7-8 and 7-9, the combination of a
secondary substrate and a strongly basic nucleophile will favor elimination by the E2 pathway (Table 7-4).
This mechanism has specifi c requirements with regard to the relative geometric arrangement of the
leaving group and the hydrogen being removed by the base: They must possess an anti conformation
with respect to the carbon – carbon bond between them (Section 7-7). In order to prepare a reasonable
structural representation of the molecule, we must make use of what we learned in Chapter 4 (Sec tion 4-4)
about conformations of substituted cyclohexanes. The best available shape is the chair shown below,
in which two of the substituents, including the bulky 1-methylethyl (isopropyl) group, are equatorial.
Note that the leaving group is axial, and one of the neighboring carbons possesses an axial hydrogen,
positioned in the required orientation anti to the chlorine:



CH3 Cl� �(CH3)2CH(CH3)2CH







š šš

š ššš

ššš �


C h a p t e r I n t e g r a t i o n P r o b l e m s



276 C h a p t e r 7 F u r t h e r R e a c t i o n s o f H a l o a l k a n e s

We draw curved arrows appropriate to the one-step E2 mechanism, arriving at the product structure
shown, a cyclic alkene, with the double bond located in the indicated position. The methyl group on
the alkene carbon moves into the plane of the double bond (recall from Section 1-8 that alkene carbons
exhibit trigonal planar geometry — sp2 hybridization). Notice that the remaining two alkyl groups,
which started cis, remain cis with respect to each other, because no chemical change occurred to the
ring carbons on which they were located.
 Avoid the very common error of removing a hydrogen from the wrong carbon atom in an E2
process. In the correct mechanism the hydrogen comes off a carbon atom adjacent to that bearing the
leaving group; never from the same carbon.

7-24. a. 2-Bromo-2-methylpropane (tert-butyl bromide) reacts readily in nitromethane with chloride
and iodide ions.

1. Write the structures of the substitution products, and write the complete mechanism by which
one of them is formed.

2. Assume equal concentrations of all reactants, and predict the relative rates of these two reactions.

3. Which reaction will give more elimination? Write its mechanism.

1. We begin by analyzing the participating species and then recognizing just what kind of reaction

is likely to take place. The substrate is a haloalkane with a good leaving group attached to a
tertiary carbon. According to Table 7-4, displacement by the SN2 mechanism is not an option,
but SN1, E1, and E2 processes are possibilities. Chloride and iodide are good nucleophiles and
weak bases, suggesting that substitution should predominate to give (CH3)3CCl and (CH3)3CI as
the products, respectively (Section 7-8). The mechanism (SN1) is as shown in Section 7-2 except
that, subsequent to initial ionization of the C – Br bond to give the carbocation, halide ion attacks
at carbon to give the fi nal product directly. The very polar nitromethane