Question

When acetaldehyde is reacted with $$LiAl{H}_4$$, What is the product formed?
A.$$C{H}_{3}COOH$$
B.$$C{H}_{3}C{H}_{2}OH$$
C.$$C{H}_{3}OH$$
D.$$HCOOH$$

Answer 1

Hint: The molecular formula of acetaldehyde is $$C{H}_{3}CHO$$. Lithium aluminium hydride $$(LiAl{H}_4)$$ is a reducing agent which facilitates the reduction of the reacting substrate. Reduction describes the addition of hydrogen.

Complete answer:
The molecular formula of acetaldehyde is $$C{H}_{3}CHO$$. Lithium aluminium hydride $$(LiAl{H}_4)$$ is a reducing agent which facilitates the reduction of the reacting substrate. Reduction describes the addition of hydrogen. Thus, the reduction of aldehyde in presence of $$LiAl{H}_4$$ lead to the formation of alcohol. This is illustrated in the following reaction:
$$C{H}_{3}CHO+LiAl{H}_4\to C{H}_{3}C{H}_{2}OH$$
The general reaction mechanism for the above reduction is given as:

From the above mechanism, we conclude that the reduction of aldehyde takes place via hydride transfer. Upon reduction, the number of carbon atoms in the moiety remains the same. Therefore, option (B) is correct. In contrast, option (A) contains $$C{H}_{3}COOH$$ which is the oxidised form of acetaldehyde thus this option is incorrect. Option (C) contains $$C{H}_{3}OH$$ which contains less number of the carbon atoms than acetaldehyde. Thus, option (A) is incorrect. Option (D) contains $$HCOOH$$ which is the oxidised form of formaldehyde instead of acetaldehyde. Thus, option (D) is incorrect.
From the above discussion, we conclude that option (B) is correct.

Note:
It is important to note that acetaldehyde undergoes reduction when reacted with $$LiAl{H}_4$$. This reduction takes place via hydride transfer. The final product will be $$C{H}_{3}C{H}_{2}OH$$ and hence option (B) is correct.