Identify A and B in the following reaction $\begin{array}{l} {\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_3}\xleftarrow{B}{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}\xrightarrow{A}{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{A}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{B}} \end{array}$

${\rm{LiAl}}{{\rm{H}}_4}{\rm{HI}} + {\rm{red}}\;{\rm{ P}}$

${\rm{HI}} + {\rm{red}}\;{\rm{ PLiAl}}{{\rm{H}}_4}$

${\rm{Ni}}/\Delta {\rm{LiAl}}{{\rm{H}}_4}$

${\rm{Pd}} - {\rm{BaS}}{{\rm{O}}_4}{\rm{Zn}} + {\rm{HCl}}$

JEE CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

1 Medium

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Understanding +4 | -1

Identify A and B in the following reaction

\begin{array}{l} \mathrm{C}{\mathrm{H}_3} - \mathrm{C}{\mathrm{H}_3}\xleftarrow{B}\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\xrightarrow{A}\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{A}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{B} \end{array}

View answer and explanation

Correct option: C

\mathrm{LiAl}{\mathrm{H}_4}\mathrm{HI} + \mathrm{red}\;\mathrm{ P}

Option C (

\mathrm{LiAl}{\mathrm{H}_4}\mathrm{HI} + \mathrm{red}\;\mathrm{ P}

) is correct because

LiAlH_4

reduces carboxylic acid to alcohol, and HI/red P reduces it to alkane.