The above reduction can be best carried out by

Schiff's base is prepared from:

Match the reactants in Column-$I$ with the major products in Column-$II$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A)$ $HCHO + \text{Cyclohexyl-MgBr} \rightarrow X \xrightarrow{H_2O} Y$	$(i)$ $\text{Benzenesulfonic acid}$
$(B)$ $\text{2-oxocyclohexylacetate} \xrightarrow{NaBH_4}$	$(ii)$ $\text{Cyclohexylmethanol}$
$(C)$ $HCHO + CH_3-CH_2-CH(CH_3)-CH_2-MgBr \rightarrow X \xrightarrow{H_2O} Y$	$(iii)$ $\text{2-(2-hydroxycyclohexyl)acetate}$
$(D)$ $\text{Benzene} + \text{Oleum}$	$(iv)$ $Y = \text{3-methylpentan-1-ol}$

Silver Mirror test is given by which one of the following compounds?

Acetyl bromide reacts with excess of $CH_3MgI$ followed by treatment with a saturated solution of $NH_4Cl$ gives:

Which of the following reagents would carry out the following transformation?
$C_6H_5COCH_3 \rightarrow C_6H_5CH(OD)CH_3$ (Note: The product shows a deuterium atom attached to the chiral carbon,not the oxygen).