A) Na + alcohol
B) \[LiAl{{H}_{4}}\] + ether
C) \[{{H}_{2}}\] + Pt
D) Sn + HCl
Correct Answer: B
Solution :
Lithium aluminium hydride \[(LiAl{{H}_{4}})\] is a powerful reducing agent. It reduces acetic acid into ethanol. \[\underset{acetic\text{ }acid}{\mathop{C{{H}_{3}}COOH}}\,\xrightarrow[ether]{LiAl{{H}_{4}}}\underset{ethyl\text{ }alcohol}{\mathop{C{{H}_{3}}C{{H}_{2}}OH}}\,+{{H}_{2}}O\] Na \[+{{C}_{2}}{{H}_{5}}OH\] is used in the reduction of alkyl halides to alkanes. \[R-X+2H\xrightarrow{Na\,+\,{{C}_{2}}{{H}_{5}}OH}RH+HX\] \[{{H}_{2}}+\]Pt is used in the catalytic hydrogenation. \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}\xrightarrow{Pt}C{{H}_{3}}-C{{H}_{3}}\] \[>C=O\xrightarrow{{{H}_{2}}+Pt}>CHOH\] Sn + HCl is used in the reduction of nitroalkanes into primary amine. \[C{{H}_{3}}C{{H}_{2}}N{{O}_{2}}+6\,[H]\xrightarrow{Sn\,\,+\,\,HCl}C{{H}_{3}}C{{H}_{2}}.N{{H}_{2}}\] \[+2{{H}_{2}}O\]
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