0.40 , g of an organic compound (A) with mole | vedclass

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(C) Molar mass of $C_5H_8O = 84 \, g/mol$. Moles of $(A) = \frac{0.40}{84} \approx 0.00476 \, mol$. Moles of gas $(CH_4)$ liberated at $STP = \frac{22

Vedclass · Accepted Answer

$HC \equiv C-CH_2-CH_2-CH_2-OH$

Vedclass · Answer

(C) Molar mass of $C_5H_8O = 84 \, g/mol$. Moles of $(A) = \frac{0.40}{84} \approx 0.00476 \, mol$. Moles of gas $(CH_4)$ liberated at $STP = \frac{224}{22400} = 0.01 \, mol$. Number of active hydrogens $= \frac{0.01}{0.00476} \approx 2$. Since hydrogenation of $(A)$ with excess $H_2$ gives pentan-$1$-ol $(CH_3-CH_2-CH_2-CH_2-CH_2-OH)$,$(A)$ must have a straight $5$-carbon chain with a terminal alkyne group $(-C \equiv CH)$ and a primary alcohol group $(-OH)$. Therefore,the structure is $HC \equiv C-CH_2-CH_2-CH_2-OH$.

$0.40 \, g$ of an organic compound $(A)$ with molecular formula $C_5H_8O$ reacts with $x$ moles of $CH_3MgBr$ to liberate $224 \, mL$ of a gas at $STP$. Upon hydrogenation with excess $H_2$,$(A)$ yields pentan-$1$-ol. The correct structure of $(A)$ is:

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