When $ CH_2 = CH -COOH$ is reduced with $LiAlH_4$ . the compound obtained will be

Question

$\begin{array}{1 1} CH_3-CH_2 -COOH \\ CH_2 = CH -CH-CH_2OH \\ CH_3 = CH_2 -CH_2-CH_2OH \\ CH_3-CH_2-CHO \end{array} $