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प्रश्न
Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
\[\begin{array}{cc}\ce{CH3CH2CHCH3}\\
\phantom{.....}|\\
\phantom{......}\ce{Br}\\
\end{array}\] or \[\begin{array}{cc}\phantom{.......}\ce{CH3}\\
\phantom{...}|\\
\ce{H3C - C - Br}\\
\phantom{...}|\\
\phantom{.......}\ce{CH3}\\
\end{array}\]
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उत्तर
The SN2 process involves a transition state with both an incoming nucleophile and a leaving group surrounding the carbon atom. Five atoms are simultaneously bonded together. A transition state requires minimal steric hindrance. Hence, 1° alkyl halides are the most reactive to SN2, followed by 2° and 3°.
1° RX > 2° RX > 3° RX
Based on the above order, the more reactive alkyl halide is:
\[\begin{array}{cc}\ce{CH3CH2CHCH3}\\
\phantom{.....}|\\
\phantom{......}\ce{Br}\\
\end{array}\]
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| (I) |  |
| (II) |  |
| (III) |  |
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| Column I | Column II | |
| (i) |  |
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| (ii) | \[\begin{array}{cc} \ce{CH3 - CH = CH2 + HBr -> CH3 - CH - CH3}\\ \phantom{............................}|\phantom{}\\ \phantom{.............................}\ce{Br}\phantom{} \end{array}\] |
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| (iii) |  |
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| (iv) |  |
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| (v) | \[\begin{array}{cc} \ce{CH3 CH2 CH CH3 ->[alc.KOH] CH3 CH = CH CH3}\\ \phantom{}|\phantom{..........................}\\ \phantom{}\ce{Br}\phantom{........................} \end{array}\] |
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(i) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5OH] \underset{}{(CH3)2CH - CH2OC2H5 + HBr}}\\
\end{array}\]
(ii) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5O-] \underset{}{(CH3)2CH - CH2OC2H5 + Br-}}\\
\end{array}\]
The mechanisms of reactions (i) and (ii) are respectively:
