A) \[B{{H}_{3}}/THF\] and\[{{H}_{3}}{{O}^{+}}\]
B) \[NaB{{H}_{4}}\]
C) \[Na/EtOH\]
D) \[{{H}_{2}}/catalyst\]
Correct Answer: A
Solution :
\[{{H}_{2}}/\]catalyst, \[NaB{{H}_{4}}\]and \[Na/{{C}_{2}}{{H}_{5}}OH\]etc., are weak reducing agent, hence these reduce aldehyde, ketone, alkene etc., groups but these cannot reduce carboxylic acid. Hence, -COOH group is reduced to \[-C{{H}_{2}}OH\]group by the reaction of \[B{{H}_{3}}/THF\]and \[{{H}_{3}}{{O}^{+}}.\] \[\underset{\text{propanoic}\,\text{acid}}{\mathop{{{C}_{2}}{{H}_{5}}COOH}}\,\xrightarrow[(ii)\,{{H}_{3}}{{O}^{+}}]{(i)\,B{{H}_{3}}/THF}\underset{\text{propyl}\,\,\text{alcohol}}{\mathop{{{C}_{2}}{{H}_{5}}.C{{H}_{2}}OH}}\,\]
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