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0.34 g of hydrocarbon when heated with methyl magnesium bromide gives 112 ml of $C H_{4}$ at STP. Possible structure of hydrocarbon is

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$C H_{3} C H_{2} - \underset{\underset{C H_{3}}{|}}{C} H - C \equiv C H$

$C H_{3} C H_{2} C \equiv C - C H_{3}$

$C H_{3} C H_{2} C H_{2} C \equiv C H$

$C H_{3} C H_{2} C H_{2} C H_{2} C \equiv C H$

answer is A.

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Detailed Solution

112ml of CH₄ at STP is obtained from 0.34g 22400ml of CH₄ at STP is obtained from $\frac{0.34}{112} \times 22400 = 68 g m o l^{- 1}$ So alkyne is $C_{5} H_{8} = 5 \times 12 + 8 = 68 g m o l^{- 1}$ . So correct ans is (a)

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