Draw the product formed when the $\mathbf{\alpha }\mathbf{,}\mathbf{\beta }$-unsaturated ketone A is treated with each reagent.

$$\begin{gathered} \left(a\right){\mathbf{NaBH}}_{\mathbf{4}}\mathbf{,}{\mathbf{CH}}_{\mathbf{3}}\mathbf{OH} \\ \left(b\right)\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{,}\left(1\mathrm{equiv}\right)\mathbf{,}\mathbf{}\mathbf{Pd}\mathbf{-}\mathbf{C} \\ \left(c\right)\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}\left(\mathrm{excess}\right)\mathbf{,}\mathbf{}\mathbf{Pd}\mathbf{-}\mathbf{C} \\ \left(d\right)\mathbf{}\left[1\right]\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{Li}\mathbf{;}\mathbf{}\left[2\right]\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \left(e\right)\mathbf{}\left[1\right]\mathbf{}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{MgBr}\mathbf{;}\mathbf{}\left[2\right]\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \left(f\right)\mathbf{}\left[1\right]\mathbf{}{\left({\mathrm{CH}}_2={\mathrm{CH}}_2\right)}_{\mathbf{2}}\mathbf{}\mathbf{CuLi}\mathbf{}\mathbf{;}\mathbf{}\left[2\right]\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

Metal hydrides such as ${\mathbf{NaBH}}_{\mathbf{4}}$are strong reducing agents which can reduce an aldehyde into primary alcohol and a ketone into secondary alcohol.

Representation of the product formed from reagent

Hydrogens can selectively reduce alkenes into alkanes. But if they are used in excess, when both alkene and a carbonyl group are present, they can reduce both the functional group i.e. alkenes into alkanes and carbonyl groups into alcohols.

When lithium is used, the halogen and metal exchange to form the organolithium reagent and function as a nucleophilic agent and also as a base.

When Mg is used, it gets inserted into the carbon-halogen bond to form the Grignard reagent and functions as a nucleophilic agent and also as a base.

When Cu is used, it gets inserted into the carbon-carbon bond to form the organocuprate reagent and functions as a nucleophilic agent and also as a base.