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

ROH+CH3MgX→CH4+ROMgXLet molecular mass of alcohol is M5622400=0.22MM=22400×0.2256=88gm

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An alcohol (A), 0.22g of this alcohol liberates 56 ml of $C H_{4}$ at STP on reaction with CH₃MgBr. Write the molecular weight of alcohol which satisfy these conditions.

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answer is 88.

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$R O H + C H_{3} M g X \to C H_{4} + R O M g X$

Let molecular mass of alcohol is M

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

$M = \frac{22400 \times 0.22}{56} = 88 g m$

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