Trichloroacetaldehyde was subjected to Cannizzaro's reaction by using $$\text{NaOH}$$. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is:

Cannizzaro reaction: Non‐enolizable aldehydes (i.e. aldehydes that possess no $$\alpha$$-hydrogen) undergo self-oxidation-reduction in concentrated alkali. For every two molecules of such an aldehyde:
• one molecule is oxidised to a carboxylate anion,
• the other molecule is reduced to the corresponding alcohol.

The given aldehyde is trichloroacetaldehyde (commonly called chloral): $$CCl_3CHO$$.
Its $$\alpha$$-carbon already carries three $$Cl$$ atoms, so there is no $$\alpha$$-hydrogen. Hence it satisfies the condition for the Cannizzaro reaction.

Let us apply the reaction with $$NaOH$$.

Step 1 Nucleophilic addition: $$OH^-$$ attacks the carbonyl carbon giving an alkoxide intermediate $$CCl_3CH(OH)O^-$$.

Step 2 Hydride transfer: A hydride ion ($$H^-$$) migrates from this alkoxide to the carbonyl carbon of a second aldehyde molecule. Thus one molecule gets reduced while the other is oxidised.

Products obtained:

• Oxidised product: $$CCl_3COO^-$$ combines with $$Na^+$$ to give sodium trichloroacetate.
• Reduced product: $$CCl_3CH_2OH$$, named 2, 2, 2-trichloroethanol.

Therefore the “other compound” present in the mixture is 2, 2, 2-trichloroethanol.

Checking the options:
Option A Trichloromethanol $$\;(CCl_3OH)$$ - wrong carbon skeleton.
Option B 2, 2, 2-Trichloropropanol $$\;(CCl_3CH_2CH_2OH)$$ - one extra $$CH_2$$ group.
Option C Chloroform $$\;(CHCl_3)$$ - not an alcohol.
Option D 2, 2, 2-Trichloroethanol $$\;(CCl_3CH_2OH)$$ - matches the Cannizzaro reduction product.

Hence, the correct answer is:
Option D which is: 2, 2, 2-Trichloroethanol.