Question

$375mg$ of an alcohol reacts with required amount of methyl magnesium bromide and releases $140mL$ of methane gas at STP. The alcohol is

• A. Ethanol
• B. $n$-butanol
• C. Methanol
• D. $n$-propanol
• E. Phenol

Answer 1

Answer: (D)

$n$-propanol

The required reaction is

$\underset{375mg}{ROH}+C{H}_{3}MgBr\to \underset{140mlatSTP}{C{H}_4}+Mg\left(OR\right)Br$

Moles of $C{H}_4\left(n\right)$ at STP $=\frac{140}{22400}=0.00625$ moles

Here, $C{H}_3-$ of $C{H}_{3}MgBr$ combines with $H$-atom of $R-OH$ to give $C{H}_4$.

$\therefore n\left(C{H}_4\right)=n(R-OH)$

Alos, $n=\dfrac { W }{ M\left( molar\quad mass \right) } $

$0.00625=\frac{375\times {10}^{-3}}{M}$

$\therefore M=\frac{375}{0.00625\times 1000}=60$

Also, general formula for alcohol $=C_n{H}_{2n+1}\cdot OH$

Thus, $12n+2n+1+17=60$ $(\because OH=16+1=17)$

$\therefore 14n=60-18=42$

$\therefore n=3$

Hence, alcohol is $n$-propanol $\left(C_3{H}_{7}OH\right)$.