An alkyl bromide $X\left(\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Br}\right)$ undergoes hydrolysis in a two step mechanism $X$ is converted to Grignard reagent and then reacted with $\mathrm{CO}_2$ in dry ether followed by acidification gave $Y$. What is $Y$ ?
Consider the following sequence of reactions
$$ \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa} \xrightarrow[\Delta]{\mathrm{NaOH} / \mathrm{CaO}} X \xrightarrow[\text { Anhy } \cdot \mathrm{AlCl}_3]{\text { CO+ } \mathrm{HCl}} Y \xrightarrow[\text { NaOH }]{\text { Conc. }} A+B $$
If $A$ is the reduction product of $Y$, what is $B$ ?
$$ \text { What is } A \text { in the following reaction? } $$

The correct statement regarding $X$ and $Y$ formed in the following reaction is
$$ \left(\mathrm{CH}_3\right)_3 \mathrm{COC}_2 \mathrm{H}_5 \xrightarrow[\Delta]{\mathrm{HI}} \text { halide }(X)+\text { alcohol }(Y) $$
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