An alkane (Y) requires 8 moles of oxygen for | vedclass

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(D) $1$. The general formula for the combustion of an alkane is $C_nH_{2n+2} + (\frac{3n+1}{2})O_2 \rightarrow nCO_2 + (n+1)H_2O$.$2$. Given that the

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$4$

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(D) $1$. The general formula for the combustion of an alkane is $C_nH_{2n+2} + (\frac{3n+1}{2})O_2 ightarrow nCO_2 + (n+1)H_2O$.$2$. Given that the moles of oxygen required is $8$, we set $\frac{3n+1}{2} = 8$, which gives $3n+1 = 16$, so $3n = 15$, and $n = 5$.$3$. The alkane is pentane $(C_5H_{12})$. For an alkane to yield only one monochlorinated product upon reaction with $Cl_2/h
u$, all hydrogen atoms must be equivalent. This structure is neopentane ($2,2$-dimethylpropane).$4$. In neopentane, there are $4$ methyl groups attached to a central quaternary carbon atom. Each methyl group contains a primary carbon atom. Therefore, there are $4$ primary carbon atoms.

An alkane $(Y)$ requires $8$ moles of oxygen for complete combustion and on chlorination with $Cl_2/h\nu$, $(Y)$ gives only one monochlorinated product $(Z)$. The total number of primary carbon atoms in $(Y)$ is . . . . . . .

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