Consider the following reactions giving major product. Identify the correct reaction.

All four alternatives try to place a $$-CN$$ group on the aromatic ring that originally bears an $$-NH_2$$ group (aniline). The only reliable laboratory route is to pass through a diazonium salt and then carry out a Sandmeyer cyanation. Option A is the only alternative that follows this exact sequence, so it alone furnishes the nitrile as the major product.

Step 1 (Diazotisation)
Aniline is treated at $$0-5^{\circ}{\rm C}$$ with $$NaNO_2/HCl$$. This converts the basic amino group into the corresponding diazonium chloride: $$C_6H_5NH_2 \xrightarrow{NaNO_2/HCl,\;0-5^{\circ}{\rm C}} C_6H_5N_2^+Cl^-$$.

Step 2 (Sandmeyer cyanation)
The freshly prepared diazonium salt is then reacted with $$CuCN$$ (generally in aqueous $$HCl$$). Copper(I) cyanide replaces the diazonium group by $$-CN$$, expelling nitrogen gas: $$C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN + N_2 \uparrow$$.

Thus Option A produces benzonitrile ($$C_6H_5CN$$) in good yield. The overall transformation is

$$C_6H_5NH_2 \;\xrightarrow[\;CuCN\;]{\;NaNO_2/HCl,\;0-5^{\circ}{\rm C}\;} C_6H_5CN + N_2 \uparrow$$.

Why the other options fail

Case B: Direct treatment of aniline with $$NaCN$$ (or $$KCN$$) cannot displace the strongly basic $$-NH_2$$ group from an aromatic ring, so no nitrile is formed.

Case C: If the diazonium salt is first obtained but then warmed before adding $$CuCN$$, it decomposes (mostly to phenol) and gives very poor or negligible nitrile yield.

Case D: Adding copper(II) salts and cyanide ions directly to aniline, without the diazonium intermediate, again fails because $$-NH_2$$ is not a leaving group; polymeric tars predominate.

Therefore the only reaction that cleanly affords the required nitrile is that listed in Option A.

Final answer: Option A which is: aniline → diazotisation → CuCN (Sandmeyer) → benzonitrile.