The least reactive alkane towards free-radical substitution reaction is:

_{4}

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(CH

_{3})_{3}

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_{3}

_{3}

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_{3}

_{2}

_{3}

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Solution

The correct option is A CH $_{4}$
Methane $C H_{4}$ is the least reactive alkane towards free radical substitution reaction. More number of alkyl groups are present, higher will be the stability of the free radical intermediate and higher will be the reactivity. Thus, higher alkanes are more reactive towards free radical substitution reaction.

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Highly reactive compound towards $S N^{2}$ reaction is

C H_{3} - C H_{3}, C H_{3} - C H_{2} - C H_{3}, (C H_{3})_{2} C H - C H_{3}, C H_{3} - C H_{2} - C H (C H_{3})_{2} .

C H_{3} C H_{2} C H O

C H_{3} C O C H_{3}

(1) C H_{3} C H = C H_{2} + H B R - ---------- \to C H_{3} C H_{2} C H_{2} - B r

(2) C H_{3} C H = C H_{2} + C I_{2} - --------- \to C H_{3} C H (C l) C H_{3}

(3) C H_{3} C H = C H_{2} + H C I - ---------- \to C l - C H_{2} C H = C H_{2}

(4) C H_{3} C H_{3} + C l_{2} h r - \to C I - C H_{2} C H = C H_{2}

B r_{2}

C C l_{4}

K M n O_{4}

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