JEE MainChemistryModerateOrganic Chemistry

An alcohol (A), 0.22 g of this alcohol librates 56 ml of CH4 at
STP on reaction with CH3MgBr. Write the molecular weight of alcohol
which satisfies these conditions.

Show answer and solution

ROH + {CH}_{3}MgX \rightarrow {CH}_{4} + ROMgX

Let molecular mass of alcohol is M

\frac{56}{22400} = \frac{0.22}{M}

M = \frac{22400 \times 0.22}{56} = 88gm

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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