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(D) The reactivity of alkanes towards free radical substitution reactions depends on the stability of the intermediate free radical formed during the

Vedclass · Accepted Answer

$(CH_3)_3CH$

Vedclass · Answer

(D) The reactivity of alkanes towards free radical substitution reactions depends on the stability of the intermediate free radical formed during the reaction.In the given alkanes,the stability of the free radical formed by the abstraction of a hydrogen atom follows the order: $3^\circ > 2^\circ > 1^\circ$.$CH_4$ is a primary alkane,$CH_3CH_2CH_3$ has $2^\circ$ hydrogens,$CH_3(CH_2)_2CH_3$ has $2^\circ$ hydrogens,and $(CH_3)_3CH$ (isobutane) has a $3^\circ$ hydrogen.The abstraction of the $3^\circ$ hydrogen from isobutane leads to the formation of a tertiary butyl radical,which is the most stable among the options provided.Therefore,$(CH_3)_3CH$ is the most reactive alkane.

Column $I$	Column $II$
$A$. $(CH_3)_3CH + O_2 \xrightarrow{KMnO_4} (CH_3)_3COH$	$i$. $CH_3COOH + H_2O$
$B$. $2CH_4 + O_2 \xrightarrow[100 \ atm]{Cu/523 \ K} 2CH_3OH$
$C$. $CH_4 + O_2 \xrightarrow[\Delta]{Mo_2O_3} HCHO + H_2O$	$iii$. $2CH_3OH$
$D$. $CH_3-CH_3 + \frac{3}{2}O_2 \xrightarrow{(CH_3COO)_2Mn} CH_3COOH + H_2O$	$iv$. $HCHO + H_2O$

Which of the following is the most reactive alkane?

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