The formulas of A and B for the following reaction sequence are

Fructose is a ketose sugar with molecular formula $$\mathrm{C_6H_{12}O_6}$$ containing a ketone group at the $$\mathrm{C2}$$ position.

Formation of product $$\mathrm{A}$$:

Fructose reacts with hydrogen cyanide $$\mathrm{(HCN)}$$ through nucleophilic addition at the carbonyl carbon to form a cyanohydrin.

Acidic hydrolysis using $$\mathrm{H_3O^+}$$ converts the cyano group $$\mathrm{(-C \equiv N)}$$ into a carboxylic acid group $$\mathrm{(-COOH)}$$.

This increases the carbon chain length from $$\mathrm{6}$$ to $$\mathrm{7}$$ carbons, producing a polyhydroxy carboxylic acid with formula:

$$\mathrm{C_7H_{14}O_8}$$

Formation of product $$\mathrm{B}$$:

Sodium borohydride $$\mathrm{(NaBH_4)}$$ reduces the ketone group of fructose into a secondary alcohol, forming a sugar alcohol with formula:

$$\mathrm{C_6H_{14}O_6}$$

Further reduction using $$\mathrm{HI/P}$$ removes all oxygen-containing groups by replacing them with hydrogen atoms.

The six-carbon chain remains intact, giving:

$$\mathrm{n\text{-}hexane,\ C_6H_{14}}$$

Correct Option: $$\mathrm{A = C_7H_{14}O_8,\ B = C_6H_{14}}$$