Name: Oxymercuration–Demercuration Reaction
Uploaded: 2019-07-19
Duration: 6 min 52 s
Description: In an oxymercuration–demercuration reaction, an alkene is treated with mercury(II) acetate, and the product is treated with sodium borohydride.

Question

0.44 g

of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP

112 {cm}^{3}

of methane, with PCC the same alcohol forms a carbonyl compound that answers silver mirror test. Identify the compound.

Accepted Answer

$0.44 g$ of a monohydric alcohol reacts with methyl magnesium iodide in ether by the following reaction and liberates $112 {cm}^{3}$ methane at STP,

$R - OH + {CH}_{3} MgI \to {CH}_{4} + MgI (OR) (1 mole) (1 mole)$

We know that at STP, the volume of any gas is $22400 {cm}^{3}$ .

And at STP,

$the number of moles of methane = \frac{volume of methane}{volume of gas at STP} = \frac{112 {cm}^{3}}{22400 {cm}^{3}} = 0.005 mole$

$0.005 mole$ of methane corresponds to $0.005 mole$ of alcohol.

In terms of amount of alcohol,

$number of moles = \frac{Mass of alcohol}{Molar mass of alcohol} 0.005 mole = \frac{0.44 g}{Molar mass of alcohol} Molar mass of alcohol = \frac{0.44 g}{0.005 mole} = 88 {gmol}^{- 1}$

As we know that the general formula of alcohol is $C_{n} H_{2 n + 1} OH$ .

$∴ n (12) + 2 n + 1 (1) + 16 + 1 = 88 14 n + 18 = 88 14 n = 70 n = 5$

The formula of alcohol is $C_{5} H_{11} OH$ .

Neopentyl alcohol reacts with $PCC$ to form neopentyl aldehyde which shows positive silver mirror test. Therefore, the compound is neopentyl alcohol.

$0.44 g$ of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP $112 {cm}^{3}$ of methane, with PCC the same alcohol forms a carbonyl compound that answers silver mirror test. Identify the compound.

Important Questions on Hydroxy Compounds and Ethers

Learn and Prepare for any exam you want !

0.44 g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 cm3 of methane, with PCC the same alcohol forms a carbonyl compound that answers silver mirror test. Identify the compound.

Important Questions on Hydroxy Compounds and Ethers

Learn and Prepare for any exam you want !

$0.44 g$ of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP $112 {cm}^{3}$ of methane, with PCC the same alcohol forms a carbonyl compound that answers silver mirror test. Identify the compound.