Alkyl Halides - Result Question 21
####21. In $S _N 2$ reactions, the correct order of reactivity for the following compounds $CH _3 Cl, CH _3 CH _2 Cl,\left(CH _3\right) _2 CHCl$ and $\left(CH _3\right) _3 CCl$ is
(2014 Main)
(a) $CH _3 Cl>\left(CH _3\right) _2 CHCl>CH _3 CH _2 Cl>\left(CH _3\right) _3 CCl$
(b) $CH _3 Cl>CH _3 CH _2 Cl>\left(CH _3\right) _2 CHCl>\left(CH _3\right) _3 CCl$
(c) $CH _3 CH _2 Cl>CH _3 Cl>\left(CH _3\right) _2 CHCl>\left(CH _3\right) _3 CCl$
(d) $\left(CH _3\right) _2 CHCl>CH _3 CH _2 Cl>CH _3 Cl>\left(CH _3\right) _3 CCl$
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Answer:
Correct Answer: 21. (b)
Solution:
- Steric hindrance (crowding) is the basis of $S _N 2$ reaction, by using which we can arrange the reactant in correct order of their reactivity towards $S _N 2$ reaction.
$$ \text { Rate of } S _N 2 \propto \frac{1}{\text { Steric crowding of ‘C’ }} $$
As steric hinderance (crowding) increases, rate of $S _N 2$ reaction decreases.
| Primary halides | > | Secondary halides | > | Tertiary halides |
|---|---|---|---|---|
| $(1^{\circ})$ | $(2^{\circ})$ | $(3^{\circ})$ |
Note The order of reactivity towards $S _N 2$ reaction for alkyl halides is
