When compound $X$ is oxidised by acidified potassium dichromate, compound $Y$ is formed. Compound $Y$ on…
When compound $X$ is oxidised by acidified potassium dichromate, compound $Y$ is formed. Compound $Y$ on reduction with $\mathrm{LiAlH}_4$ gives $X . X$ and $Y$ respectively are :
When ethyl alcohol is oxidised by acidified potassium dichromate, $\mathrm{CH}_3 \mathrm{COOH}(\mathrm{Y})$ is obtained as :
$3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(X)+2 \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+8 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow$ $3 \mathrm{CH}_3 \mathrm{COOH}(Y)+2 \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{~K}_2 \mathrm{SO}_4$ $+11 \mathrm{H}_2 \mathrm{O}$
Carboxylic acid undergo reduction with $\mathrm{LiAlH}_4$ to give primary alcohol as
So, $X$ is $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ and $Y$ is $\mathrm{CH}_3 \mathrm{COOH}$
