A and B in the following reactions are:

The Reaction Name: The transformation of Product A into Product B using

$$SnCl_2$$, $$HCl$$, and $$H_3O^+$$ is a specific named reaction known as the Stephen reduction (or Stephen aldehyde synthesis).

• The Starting Material: Product A, formed from the first step (diazotization and reaction with $$KCN$$), is benzonitrile.
• Step 1 - Partial Reduction: Stannous chloride ($$SnCl_2$$) and hydrochloric acid ($$HCl$$) act as a reducing agent. They supply hydrogen to the carbon-nitrogen triple bond of benzonitrile. However, they only reduce it partially to a double bond, forming a stannic chloride byproduct and an intermediate salt called an imine hydrochloride.
• Equation for Reduction: $$C_6H_5-C\equiv N + 2[H] + HCl \xrightarrow{SnCl_2} C_6H_5-CH=NH \cdot HCl$$
• Step 2 - Acidic Hydrolysis: The imine hydrochloride is highly susceptible to hydrolysis. When the acidic water ($$H_3O^+$$) is added, it cleaves the carbon-nitrogen double bond. The nitrogen atom is removed (forming ammonium chloride) and is replaced by an oxygen atom.
• Equation for Hydrolysis: $$C_6H_5-CH=NH \cdot HCl + H_2O \xrightarrow{H_3O^+} C_6H_5-CHO + NH_4Cl$$
• The Final Product: Because the nitrile group ($$-C\equiv N$$) has been formally replaced by an aldehyde group ($$-CHO$$), your final Product B is benzaldehyde.