An alkane (molecular weight 72) forms only on | vedclass

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Question

(A) The general formula for an alkane is $C_nH_{2n+2}$.Given molecular weight is $72$.$12n + 1(2n+2) = 72 \implies 14n + 2 = 72 \implies 14n = 70 \imp

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$(CH_3)_4C$

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(A) The general formula for an alkane is $C_nH_{2n+2}$.Given molecular weight is $72$.$12n + 1(2n+2) = 72 \implies 14n + 2 = 72 \implies 14n = 70 \implies n = 5$.So,the molecular formula is $C_5H_{12}$.An alkane that forms only one monochlorinated product must have all its hydrogen atoms equivalent.In $2,2-$dimethylpropane (neopentane),$(CH_3)_4C$,all $12$ hydrogen atoms are equivalent.Therefore,it forms only one monochlorinated product upon reaction with $Cl_2$ in the presence of sunlight.

An alkane (molecular weight $72$) forms only one monochlorinated product. Its formula is

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