The major product of the following reaction is: C H 3 C H = C H C O 2 C H 3 → L i A l H 4

1. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{C}{\mathrm{H}}_3$
2. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{H}=\mathrm{C}\mathrm{H}\mathrm{C}{\mathrm{H}}_2\mathrm{O}\mathrm{H}$
3. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{O}\mathrm{H}$
4. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}\mathrm{H}\mathrm{O}$

Solution

$\mathrm{L}\mathrm{i}\mathrm{A}\mathrm{l}{\mathrm{H}}_4$ convert ester into alcohol by reduction, it does not reduce $\mathrm{\pi }$ bond between carbon atoms.

Asked in: JEE Main 2019 (09 Apr Shift 1)

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