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प्रश्न
Which would undergo SN1 reaction faster in the following pair and why?
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उत्तर १
Carbocations are the intermediates in the SN1 reaction. Greater the stability of the carbocations, more easily will the product be formed and hence faster will be the rate of the reaction. Because the stability of the carbocations decreases in the order: 3° carbocation > 2° carbocation > 1° carbocation > CH3+.
Therefore, the reactivity of alkyl halides towards SN1 reactions decreases in the same order, i.e. 3° alkyl halides > 2° alkyl halides > 1° alkyl halides > methyl halides.
The two structures are
Bromoethane is a primary alkyl halide which forms a 1° carbocation intermediate in the SN1 reaction. The other compound is 2-bromo-2-methylpropane which is a tertiary alkyl halide which forms a tertiary carbocation intermediate in the SN1 reaction. Hence, 2-bromo-2-methylpropane undergoes an SN1 reaction faster than bromoethane
उत्तर २
A tertiary alkyl halide tends to undergo the SN1 mechanism because it can form a tertiary carbocation, which is stabilised by the three alkyl groups attached to it. As alkyl groups are electron donating, they allow the positive charge in the carbocation to be delocalised by the induction effect. Hence, out of the given pairs, (CH3)3C-Br would undergo SN1 reaction faster than CH3-CH2-Br.
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\[\ce{{n}BuBr + KCN ->[EtOH-H2O] {n}BuCN}\]
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(i)
(ii)
CH3CH2CH2Cl
(iii)
\[\begin{array}{cc}
\ce{CH3}\phantom{..}\\
|\phantom{....}\\
\ce{CH3-CH-CH2Cl}
\end{array}\]
(iv)
\[\begin{array}{cc}
\phantom{..}\ce{H}\\
\phantom{..}|\\
\ce{CH3-C-Cl}\\
\phantom{..}|\\
\phantom{.....}\ce{C2H5}
\end{array}\]
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(i) Both the compounds form same product on treatment with alcoholic KOH.
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(iii) Both the compounds form same product on reduction.
(iv) Both the compounds are optically active.
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