Arrange the following compounds in increasing order of reactivity by SN2 mechanism.
A. CH3CH2CH2CH2Br
B. (CH3)2CHCH2Br
C. CH3CH2CH2CH2I
D. (CH3)3CBr
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → D, B, A, C
In an SN2 reaction the nucleophile attacks the carbon from the side opposite the leaving group, in one step. Two factors decide the rate:
Steric hindrance: bulky groups on or near the carbon bearing the halogen block the attack. So the rate falls from 1° to 2° to 3° halides, and branching at the next (β) carbon also slows a primary halide.
Leaving group: for the same alkyl group, the weaker C–X bond breaks more easily: R–I > R–Br > R–Cl.
Matching the given compounds:
(D) \( \ce{(CH3)3CBr} \): 2-bromo-2-methylpropane (3°, most hindered)
(B) \( \ce{(CH3)2CHCH2Br} \): 1-bromo-2-methylpropane (1°, but branched at the β-carbon)
(A) \( \ce{CH3CH2CH2CH2Br} \): 1-bromobutane (1°, unbranched)
(C) \( \ce{CH3CH2CH2CH2I} \): 1-iodobutane (1°, unbranched, best leaving group)
Shortcut: The tertiary bromide (D) is the slowest and the iodide (C) is the fastest. Only Option (4) starts with (D) and ends with (C).