The end product (B) formed in the reaction sequence. Glucose $\underset{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\overset{\mathrm{HCN}}{\to }}\mathrm{A}\underset{∆}{\overset{\mathrm{HI},\mathrm{P}}{\to }}\mathrm{B}.$
 

When glucose reacts with HCN, a cyanohydrin is formed. On hydrolysis, this converts into a higher carboxylic acid. When this compound is treated with ∆ and HI, the –COOH group is reduced and the whole molecule gets converted into the corresponding alkane with one extra carbon atom.

Since glucose (C₆ unit) finally gives a 7-carbon straight chain alkane, the end product (B) is heptane.

This is a classic reaction where addition of HCN increases the carbon chain length by one and reduction with HI removes oxygen functionalities, giving a pure alkane.

Correct Answer: (a) Heptane

By using cyanohydrin formation + hydrolysis + reduction, we can perform chain lengthening of glucose, ending up with a higher alkane