JEE MAIN - Chemistry (2023 - 30th January Evening Shift - No. 8)
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In the above conversion of compound $(\mathrm{X})$ to product $(\mathrm{Y})$, the sequence of reagents to be used will be:
$\begin{array}{lll}\text { (i) } \mathrm{Br}_2, \mathrm{Fe} & \text { (ii) } \mathrm{Fe}, \mathrm{H}^{+} & \text {(iii) } \mathrm{LiAIH}_4\end{array}$
$\begin{array}{lll}\text { (i) } \mathrm{Br}_2(\mathrm{aq}) & \text { (ii) } \mathrm{LiAIH}_4 & \text { (iii) } \mathrm{H}_3 \mathrm{O}^{+}\end{array}$
(i) $\mathrm{Fe}, \mathrm{H}^{+}$ (ii) $\mathrm{Br}_2$ (aq) (iii) $\mathrm{HNO}_2$ (iv) $\mathrm{H}_3 \mathrm{PO}_2$
(i) $\mathrm{Fe}, \mathrm{H}^{+}$
(ii) $\mathrm{Br}_2(\mathrm{aq})$
(iii) $\mathrm{HNO}_2$
(iv) $\mathrm{CuBr}$
Explanation
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