MCQ
When $CH_3-CH_2-COOH$ is reduced with $LiAlH_4$, the compound obtained will be
- A$CH_3CH_2COOH$
- B$CH_3CH_2CHO$
- ✓$CH_3CH_2CH_2OH$
- D$H_2C=CH-CH_2-OH$
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$(I)$ $\left[ {{{\left( {P{h_3}P} \right)}_2}PdC{l_2}PdC{l_2}} \right]$
$(II)$ $\left[ {NiBrCl\left( {en} \right)} \right]$
$(III)$ $N{a_4}\left[ {Fe{{\left( {CN} \right)}_5}NOS} \right]$
$(IV)$ $Cr{(CO)_3}{\left( {NO} \right)_2}$
$(I)$ - $(II)$ - $(III)$ - $(IV)$