Question

$0.40g$ of an organic compound $\left(A\right)$, (Molecular formula $C_5{H}_{8}O$) reacts with x mole of ${CH}_{3}MgBr$ to liberate $224mL$ of a gas at STP. With excess of $H_2$, $\left(A\right)$ gives pentan-1-ol. The correct structure of $\left(A\right)$ is:

• A. ${CH}_3-C\equiv C-{CH}_2-{CH}_2-OH$
• B. ${CH}_3-{CH}_2-C\equiv C-{CH}_2-OH$
• C. $H-C\equiv C-{CH}_2-{CH}_2-{CH}_2-OH$
• D. $H-C\equiv C-C{H}_2-\underset{OH}{\underset{|}{C}H}-C{H}_3$

Answer 1

The correct option is C $H-C\equiv C-{CH}_2-{CH}_2-{CH}_2-OH$

Molecular weight of A= $12\times 5+8\times 1+16=84$

$\therefore 0.40g$ of A= $\frac{0.4}{84}=0.0047\approx 0.005$ mole

$224ml$ of a gas at STP$\equiv \frac{224}{22400}$mole= $0.01$ mole

Therefore, $0.005$ mole of A give $0.01$ mole i.e. $1$ equivalent of A gives $2$ equivalent of gas.

$C{H}_{3}MgBr\stackrel{H^{+}}{\to }C{H}_4\uparrow$

So, A should have $2$ equivalent of acidic hydrogen.

$\underset{C}{\stackrel{\delta -}{H}-\underset{sp}{\stackrel{\delta +}{C}}\equiv C-C{H}_2-C{H}_2-C{H}_2-\stackrel{\delta +}{O}-\stackrel{\delta -}{H}}\stackrel{H_2}{\to }\underset{pentan-1-ol}{H_{3}CC{H}_{2}C{H}_{2}C{H}_{2}C{H}_{2}OH}$

Here $C$ has $2$ acidic hydrogens and also $C$ gives $pentan-1-ol$ with excess of $H_2$ i.e. reduction of alkyne group of $C$.