Question:
The correct order of increasing reactivity of the following alkyl halides, CH$_3$CH$_2$CH(Br)CH$_3$ (I), CH$_3$CH$_2$CH$_2$Br (II), (CH$_3$)$_2$CClCH$_2$CH$_3$ (III) and CH$_3$CH$_2$CH$_2$Cl (IV) towards S$_N$2 displacement is
The correct order of increasing reactivity of the following alkyl halides, CH$_3$CH$_2$CH(Br)CH$_3$ (I), CH$_3$CH$_2$CH$_2$Br (II), (CH$_3$)$_2$CClCH$_2$CH$_3$ (III) and CH$_3$CH$_2$CH$_2$Cl (IV) towards S$_N$2 displacement is
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S$_N$2 rule:
Primary > Secondary >> Tertiary
Br > Cl > F (leaving group ability)
Updated On: May 1, 2026
- \( I < II < III < IV \)
- \( III < I < IV < II \)
- \( III < I < II < IV \)
- \( II < IV < I < III \)
- \( I < III < II < IV \)
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The Correct Option is B
Solution and Explanation
Concept: S$_N$2 reaction depends on:
• Steric hindrance (less hindrance → faster)
• Nature of leaving group (Br better than Cl)
• Structure: Primary > Secondary >> Tertiary Analysis of compounds:
• (III) (CH$_3$)$_2$CClCH$_2$CH$_3$ → highly hindered (tertiary-like) → slowest
• (I) Secondary bromide → moderate
• (IV) Primary chloride → better than secondary but Cl is poor LG
• (II) Primary bromide → least hindered + good LG → fastest Final order: \[ III < I < IV < II \]
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