Question

$RCO - {{OH}}\xrightarrow{{[?]}}{{RC}}{{{H}}_2}{{OH}}$
Here reagent is:
A. $LiAl{H_4}$
B. $NaB{H_4}$
C. Both (A) & (B)
D. $Red\;P/HI$

Answer 1

Hint: RCOOH, that is the carboxylic acid shows different chemical properties due to their H-atom in the carboxylic group and -OH group of carboxylic acid. Apart from this carboxylic acid also show different properties when they react with the carboxylic group as a whole or alkyl group of carboxylic acids

Complete answer:
Here we can see that carboxylic acid is being reduced to alcohol (${\text{RC}}{{\text{H}}_2}{\text{OH}}$). First of all we understood that this is a process of reduction of carboxylic acids to alcohol.

(A) $LiAl{H_4}$: It is a strong reducing agent. It produces hydride ions due to the presence of a polar metal-hydrogen bond. The $Al - H$ is more polar thereby making it a strong reducing agent. The hydride ions in the reducing agent $LiAl{H_4}$, react with carboxylic acid and they attack the lone pair of oxygen and finally remove the double-bonded oxygen and add two hydrogens to give corresponding alcohol.
$RCOOH\xrightarrow{{LiAl{H_4}/ether}}RC{H_2}OH$

(B)$NaB{H_4}$: Sodium borohydride is not a strong reducing agent as Lithium Aluminium hydride is because aluminium is less electronegative than boron and $Al - H$ is more polar thereby making it as a strong reducing agent. Since the carboxylic acid is more electrophilic than the carbonyl in aldehyde or ketone and also there is an acid proton from the carboxylic acid that can react with hydride reagents. Therefore we can say that sodium borohydride $NaB{H_4}$does not reduce carboxylic acid to their corresponding alcohol

(D) Red $P/HI$: The red phosphorus does reduce carboxylic acid but they do not form alcohol whereas they form their respective alkanes.
$RCOOH + HI\xrightarrow{{Red\;P}}\;RC{H_3} + 2{H_2}O + 3{I_2}$
This is the chemical reaction when reacted with red phosphorus and hydroiodic acid.

Therefore, the correct option is A. $LiAlH_4$

Note:
We can also use diborane in ether to reduce carboxylic acid to their respective alcohols.
$RCOOH\xrightarrow{{{B_2}{H_6}/ether}}RC{H_2}OH$
Here, diborane acts as the reducing agent to form alcohol from carboxylic acid.