What is the product obtained in the reaction?
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CHO} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}_4^{+}]{\text {(i) }} \text { product }$
The advantage of $\mathrm{LiAlH}_4$ is that it does not reduce the isolated olefinic bond and hence, it can reduce unsaturated aldehydes and ketones to unsaturated alcohols.
