Question
Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using $$NaOH.$$ The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :
A.
2, 2, 2 - Trichloroethanol
B.
Trichloromethanol
C.
2, 2, 2 - Trichloropropanol
D.
Chloroform
Answer :
2, 2, 2 - Trichloroethanol
Solution :
$$CC{l_3}CHO + NaOH \to \mathop {CC{l_3}C{H_2}OH + CC{l_3}COONa}\limits_{{\text{2, 2,2 }} - \,\,{\text{trichloroethanol}}} $$
In Cannizzaro’s reaction the compounds which do not contain $$\alpha $$ - hydrogen atoms undergo oxidation and reduction simultaneously i.e undergo disproportion ation and form one molecule of sodium salt of carboxylic acid as oxidation product and one molecule of alcohol as reduction product.