Solution Found!
Each of the two carbocations in Solved 6-2 can also react
Chapter 8, Problem 35P(choose chapter or problem)
Each of the two carbocations in Solved Problem 6-2 can also react with ethanol to give a sub-
stitution product. Give the structures of the two substitution products formed in this reaction.
Eliminations can also take place under second-order conditions with a strong base present. As an example, consider the reaction of tert-butyl bromide with methoxide ion in methanol. This is a second-order reaction because methoxide ion is a strong base as well as a strong nucleophile. It attacks the alkyl halide faster than the halide can ionize to give a first-order reaction. No substitution product (methyl tert-butyl ether) is observed, however. The \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is blocked because the tertiary alkyl halide is too hindered. The observed product is 2-methylpropene, resulting from elimination of \(\text {HBr}\) and formation of a double bond.
\(\text { Rate }=k_{r}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]\left[^{-} \mathrm{OCH}_{3}\right]\)
The rate of this elimination is proportional to the concentrations of both the alkyl halide and the base, giving a second-order rate equation. This is a bimolecular process, with both the base and the alkyl halide participating in the transition state, so this mechanism is abbreviated \(\text {E2}\) for Elimination, bimolecular.
\(\mathrm{E} 2 \text { rate }=k_{r}[\mathrm{RX}]\left[\mathrm{B}:^{-}\right]\)
In the \(\text {E2}\) reaction just shown, methoxide reacts as a base rather than as a nucleophile. Most strong nucleophiles are also strong bases, and elimination commonly results when a strong base nucleophile is used with a poor \(\mathrm{S}_{\mathrm{N}} 2\) substrate such as a \(3^{\circ}\) or hindered \(2^{\circ}\) alkyl halide. Instead of attacking the back side of the hindered electrophilic carbon, methoxide abstracts a proton from one of the methyl groups. This reaction takes place in one step, with bromide leaving as the base abstracts a proton.
In the general mechanism of the \(\text {E2}\) reaction, a strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group. As the base abstracts a proton, a double bond forms and the leaving group leaves. Like the \(\mathrm{S}_{\mathrm{N}} 2\) reaction, the \(\text {E2}\) is a concerted reaction in which bonds break and new bonds form at the same time, in a single step.
Reactivity of the Substrate in the \(\text {E2}\) The order of reactivity of alkyl halides toward E2 dehydrohalogenation is found to be
\(3^{\circ}>2^{\circ}>1^{\circ}\)
Equation Transcription:
Text Transcription:
S_{N}2
HBr
S_{N}2
CH_{3}O
E2
CH_{3}
CH_{3}
Br
E2
CH_{3}-O-H
CH_{3}
C-C
CH_{3}
Rate =k_{r}[(CH_{3})_{3}C-Br][^{-}OCH_{3}]
E2
E2 rate=k_{r}[RX] [B:^{-}]
E2
S_{N}2
3^o
2^o
E2
S_{N}2
E2
E2
3^o>2^o>1^o
Questions & Answers
QUESTION:
Each of the two carbocations in Solved Problem 6-2 can also react with ethanol to give a sub-
stitution product. Give the structures of the two substitution products formed in this reaction.
Eliminations can also take place under second-order conditions with a strong base present. As an example, consider the reaction of tert-butyl bromide with methoxide ion in methanol. This is a second-order reaction because methoxide ion is a strong base as well as a strong nucleophile. It attacks the alkyl halide faster than the halide can ionize to give a first-order reaction. No substitution product (methyl tert-butyl ether) is observed, however. The \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is blocked because the tertiary alkyl halide is too hindered. The observed product is 2-methylpropene, resulting from elimination of \(\text {HBr}\) and formation of a double bond.
\(\text { Rate }=k_{r}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]\left[^{-} \mathrm{OCH}_{3}\right]\)
The rate of this elimination is proportional to the concentrations of both the alkyl halide and the base, giving a second-order rate equation. This is a bimolecular process, with both the base and the alkyl halide participating in the transition state, so this mechanism is abbreviated \(\text {E2}\) for Elimination, bimolecular.
\(\mathrm{E} 2 \text { rate }=k_{r}[\mathrm{RX}]\left[\mathrm{B}:^{-}\right]\)
In the \(\text {E2}\) reaction just shown, methoxide reacts as a base rather than as a nucleophile. Most strong nucleophiles are also strong bases, and elimination commonly results when a strong base nucleophile is used with a poor \(\mathrm{S}_{\mathrm{N}} 2\) substrate such as a \(3^{\circ}\) or hindered \(2^{\circ}\) alkyl halide. Instead of attacking the back side of the hindered electrophilic carbon, methoxide abstracts a proton from one of the methyl groups. This reaction takes place in one step, with bromide leaving as the base abstracts a proton.
In the general mechanism of the \(\text {E2}\) reaction, a strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group. As the base abstracts a proton, a double bond forms and the leaving group leaves. Like the \(\mathrm{S}_{\mathrm{N}} 2\) reaction, the \(\text {E2}\) is a concerted reaction in which bonds break and new bonds form at the same time, in a single step.
Reactivity of the Substrate in the \(\text {E2}\) The order of reactivity of alkyl halides toward E2 dehydrohalogenation is found to be
\(3^{\circ}>2^{\circ}>1^{\circ}\)
Equation Transcription:
Text Transcription:
S_{N}2
HBr
S_{N}2
CH_{3}O
E2
CH_{3}
CH_{3}
Br
E2
CH_{3}-O-H
CH_{3}
C-C
CH_{3}
Rate =k_{r}[(CH_{3})_{3}C-Br][^{-}OCH_{3}]
E2
E2 rate=k_{r}[RX] [B:^{-}]
E2
S_{N}2
3^o
2^o
E2
S_{N}2
E2
E2
3^o>2^o>1^o
ANSWER:
Solution 35P
Rearrange carbonation (tertiary) reacts with