Target Exam

CUET

Subject

Chemistry

Chapter

Organic: Alcohols, Phenols and Ethers

Question:

Which of the following is an industrial method of preparation of methanol?

Options:

by catalytic reduction of carbon monoxide

by reacting methane with steam in the presence of a catalyst

by reduction of \(HCHO\) with \(LiAlH_4\)

by reduction of \(HCHO\) with aqueous\(NaOH\)

Correct Answer:

by catalytic reduction of carbon monoxide

Explanation:

The correct answer is option 1. by catalytic reduction of carbon monoxide.

Methanol is an organic compound having the functional group alcohol \((-OH)\). Its molecular formula is, \(CH_3OH\). Methanol is a liquid compound at room temperature but it is volatile in nature. Its boiling point is \(64.7^oC\). Commercially, methanol is prepared by catalytic reduction of carbon monoxide. In the process of synthesis of methanol, water gas is used. Water-gas is an equimolar mixture of carbon monoxide, \(CO\) and hydrogen, \(H_2\). The reaction takes place in the presence of zinc oxide, \(ZnO\) and chromium oxide, \(Cr_2O_3\) as the catalyst and at 573K temperature.

The reaction is represented as follows:

Water-gas is converted to methanol at high temperature due to the presence of a catalyst.

Let us look at the other options.

Methane in reaction with steam in the presence of a catalyst gives hydrogen gas, carbon dioxide and carbon monoxide. This is Haber's process for the synthesis of hydrogen.

Reduction of formaldehyde in the presence of \(LiAlH_4\) does yield methanol but this process is not used commercially because formaldehyde has strong tendency to get oxidized in the presence of oxygen in the air to form formic acid and products obtained will be a mixture of methanol and formic acid.

Formaldehyde reacts with strong alkali to undergo Cannizzaro reaction to yield formic acid and methanol.

Therefore, since the catalytic reduction of carbon monoxide yields only one product and it is economically efficient, this process is used in the commercial synthesis of methanol.

Therefore, the answer is option (1).