Question

2.2 g of an alcohol (A) when treated with $C{H}_3$-MgI liberates 560 mL of $C{H}_4$ at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having the same number of the carbon atom.
The molecular mass of (A) is?

• A. 74
• B. 88
• C. 60
• D. 102

Answer 1

The correct option is B 88

$ROH+C{H}_3-MgI\to C{H}_4+ROMgI$
Number of moles of $C{H}_4$ = number of moles of alcohol
$\frac{560}{22400}=\frac{2.2}{M}=M=88$ [where M = molecular weight]
Thus molecular formula of alcohol A is $C_5{H}_{11}OH$.