Vollhardt  Capítulo 7 (Haloalcanos 2)
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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 \u201cno reaction.\u201d
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
CH3
CH3OH CH3O\ufffdNa\ufffd, CH3OH
A
CH3CH2CHBr ?
CH3
A
?
(b) (CH3CH)2N
\ufffdLi\ufffd, (CH3CH)2NH
CH3
Nitromethane
A
CH3
A
I I
??
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 C h a p t e r 7 275
THE BIG PICTURE
We have completed our study of haloalkanes by describing three new reaction pathways \u2014 SN1, 
E1, and E2 \u2014 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 \u2013 oxygen 
single bonds, comparing them with polar carbon \u2013 halogen bonds. The alcohols will be fi rst; 
they add the feature of a polar O \u2013 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.
CHAPTER INTEGRATION PROBLEMS
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.
CH3
CH3
Cl
CH3CHCH3
0
%
\u221e
~
NaOCH2CH3, CH3CH2OH
SOLUTION
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 \u2013 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:
Cl
CH3
CH3 Cl\ufffd \ufffd(CH3)2CH(CH3)2CH
CH3CH2O
CH3
H
CH3CH2OH
CH3
0%
\u161 \u161\u161
\u161 \u161\u161\u161
\u161\u161\u161 \ufffd
\ufffd\u161\u161
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
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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 \u2014 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.
SOLUTION
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 \u2013 Br bond to give the carbocation, halide ion attacks 
at carbon to give the fi nal product directly. The very polar nitromethane