A hydrocarbon A (V.D. = 36) forms only one mo | vedclass

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(B) The molecular weight of hydrocarbon $A = 2 	imes V.D. = 2 	imes 36 = 72$. For an alkane with formula $C_nH_{2n+2}$,$12n + 2n + 2 = 72$ $\Rightar

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neo-pentane

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(B) The molecular weight of hydrocarbon $A = 2 	imes V.D. = 2 	imes 36 = 72$. For an alkane with formula $C_nH_{2n+2}$,$12n + 2n + 2 = 72$ $\Rightarrow 14n = 70$ $\Rightarrow n = 5$. The molecular formula is $C_5H_{12}$. Among the isomers of $C_5H_{12}$,neo-pentane $(CH_3-C(CH_3)_2-CH_3)$ has all $12$ hydrogen atoms equivalent. Therefore,it forms only one monochloro substitution product.

$A$ hydrocarbon $A$ $(V.D. = 36)$ forms only one monochloro substitution product. $A$ will be:

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