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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, \[\begin{array}{cc}
\ce{CH3CH2CHCH3}\\
\phantom{...}|\\
\phantom{....}\ce{Br}\
\end{array}\] is more reactive.
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संबंधित प्रश्न
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\[\ce{CH3CH2CH2CH3 -> CH3CH2CH2CH2Cl + CH3CH2CHClCH3}\]
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(i) The given reaction follows SN2 mechanism.
(ii) (b) and (d) have opposite configuration.
(iii) (b) and (d) have same configuration.
(iv) The given reaction follows SN1 mechanism.
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| Column I | Column II | |
| (i) |  |
(a) Nucleophilic aromatic substitution |
| (ii) | \[\begin{array}{cc} \ce{CH3 - CH = CH2 + HBr -> CH3 - CH - CH3}\\ \phantom{............................}|\phantom{}\\ \phantom{.............................}\ce{Br}\phantom{} \end{array}\] |
(b) Electrophilic aromatic substitution |
| (iii) |  |
(c) Saytzeff elimination |
| (iv) |  |
(d) Electrophilic addition |
| (v) | \[\begin{array}{cc} \ce{CH3 CH2 CH CH3 ->[alc.KOH] CH3 CH = CH CH3}\\ \phantom{}|\phantom{..........................}\\ \phantom{}\ce{Br}\phantom{........................} \end{array}\] |
(e) Nucleophilic substitution (SN1) |
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