See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The molecule is a cyclohexene derivative. Looking at the structure: there is an endocyclic double bond. One sp2 carbon (C1) bears a methyl group and is also connected to a carbon chain bearing Hd and He. The other sp2 carbon (C2) bears Hc directly on the double bond. Ha is on a ring carbon that is allylic (two carbons away from the double bond on one side), Hb is on a ring carbon further from the double bond. Step 2: Classify each hydrogen: - Hc: directly on a C=C carbon (sp2 carbon of the ring double bond) → vinyl H - He: on the terminal carbon of the side chain (CH2 or CH3 attached to the carbon bearing Hd), with no adjacent double bond influence making it primary → 1° alkyl H - Ha: on a ring CH2 or CH carbon that is NOT directly adjacent to the double bond → 2° alkyl H - Hb: on a ring CH carbon adjacent to Ha's carbon, also 2° but let's check allylic status. Hb is on a carbon that is allylic (next to the allylic carbon or directly allylic to the double bond) → 2° allyl H - Hd: on a carbon directly adjacent to the sp2 carbon of the double bond (allylic position), and that carbon is also bonded to three carbons (the ring carbon on each side and the side chain carbon bearing He), making it tertiary allylic → 3° allyl H Step 3: Arrange in decreasing order of C-H bond energy (bond dissociation energy). General trend for C-H BDE: - Vinyl C-H (sp2) > 1° alkyl C-H > 2° alkyl C-H > 2° allylic C-H > 3° allylic C-H This is because: - Vinyl C-H bonds are the strongest due to higher s-character of sp2 carbon - Primary alkyl C-H > secondary alkyl C-H (more substituted = weaker bond, easier radical/anion formation) - Allylic C-H bonds are weaker than simple alkyl C-H bonds because the resulting radical/carbanion is resonance-stabilized - 3° allylic is weakest among these because both tertiary substitution and allylic stabilization lower the BDE Therefore: Hc (vinyl) > He (1° alkyl) > Ha (2° alkyl) > Hb (2° allyl) > Hd (3° allyl) This matches the given answer exactly. Therefore, the correct answer is {"A": {"Ha": "2° alkyl", "Hb": "2° allyl", "Hc": "vinyl", "Hd": "3° allyl", "He": "1° alkyl"}, "B": "Hc>He>Ha>Hb>Hd"}.