Consider the molecule 2-methylpentane. Let four of its carbon atoms be labelled as follows: P is the carbon atom of the methyl substituent, Q is the carbon atom at position 2 of the principal chain, R is the carbon atom at position 3, and S is the carbon atom at position 5. Heterolytic cleavage of a C-H bond to release a hydride ion ( H^- ) from which of the labelled carbons will generate the most stable carbocation?

Step-by-step solution

The stability of carbocations follows the order: tertiary ( 3^ ) > secondary ( 2^ ) > primary ( 1^ ) > methyl. In 2-methylpentane, the labelled carbons have the following degrees: P (methyl substituent): Primary ( 1^ ) carbon. Q (C-2): Tertiary ( 3^ ) carbon, as it is bonded to three other carbons. R (C-3): Secondary ( 2^ ) carbon, as it is bonded to two other carbons. S (C-5): Primary ( 1^ ) carbon. Loss of a hydride ion ( H^- ) from carbon Q generates a tertiary carbocation, which is the most stable among the pos