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\[{{H}_{3}}{{O}^{+}}\]. \[\underset{propanoic\text{ }acid}{\mathop{{{C}_{2}}{{H}_{5}}COOH}}\,\xrightarrow[(II){{H}_{3}}{{O}^{+}}]{(I)B{{H}_{3}}/THF}\underset{propyl\text{ }alcohol}{\mathop{{{C}_{2}}{{H}_{5}}.C{{H}_{2}}OH}}\,\]
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