Pooja kulhade last edited by
The triphenylacetic acid is dehydrated in the presence of concentrated sulfuric acid to give an acylium ion, namely triphenylacetylium cation, which is stabilized by resonance.
Then, the triphenylacetylium cation undergoes decarbonylation (removal of carbon monoxide) to give triphenylmethyl cation, which is stabilized by the three phenyl groups.
The oxygen in methanol acts as a nucleophile to attack the carbocation, and is deprotonated to give our desired product triphenylmethyl methyl ether.
It is noted that this reaction should be done in a well-ventilated area because of the evolution of toxic carbon monoxide gas.
Option c is only the correct option.
As the first step of the reaction is dehydration to remove the water molecule from the given compound and then further formation of tryphenyl acetone. From that the lone pair of 'O' will attack on the electron deficient carbon of methanol and form displace hydrogen ion which will further form water by reacting with hydroxyl group liberated from methanol. And the reaction at the end will form a product in a form of triphenyl methyl ether.
Therefore option C must be the correct option