See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The molecule shown is a branched alkene. Reading the structure: there is a carbon bearing two methyl groups connected by a C=C double bond to a CH group, followed by a CH group, then a CH group, then a CH2 or CH3 terminus. - He: on the carbon that is sp3, bears two methyl groups, and is directly attached to the double bond carbon. This carbon has three carbon substituents and is adjacent to the double bond, making He a 3° allylic hydrogen. - Hd: on the sp2 carbon of the double bond (the =CH- carbon). This is a vinyl hydrogen. - Hc: on the sp3 carbon directly adjacent to the double bond on the right side. This carbon is secondary (attached to two carbons) and allylic. So Hc is a 2° allylic hydrogen. - Hb: on an sp3 carbon that is secondary (attached to two carbons) but NOT adjacent to the double bond. So Hb is a 2° alkyl hydrogen. - Ha: on the terminal sp3 carbon (primary, one carbon substituent, not adjacent to double bond). So Ha is a 1° alkyl hydrogen. Step 2: Rank ease of abstraction (radical or ionic abstraction, generally by stability of resulting radical/carbocation). Stability order of resulting radicals: 3° allylic > 2° allylic > 2° alkyl > 1° alkyl > vinyl - He gives a 3° allylic radical (most stable) → easiest to abstract. - Hc gives a 2° allylic radical → second easiest. - Hb gives a 2° alkyl radical → third. - Ha gives a 1° alkyl radical → fourth. - Hd gives a vinyl radical (least stable, highest BDE) → hardest to abstract. Step 3: Confirm assignments match given answer. - Ha = 1° alkyl ✓ - Hb = 2° alkyl ✓ - Hc = 2° allyl ✓ - Hd = vinyl ✓ - He = 3° allyl ✓ Decreasing ease of abstraction: He > Hc > Hb > Ha > Hd ✓ Therefore, the correct answer is {"A": {"Ha": "1° alkyl", "Hb": "2° alkyl", "Hc": "2° allyl", "Hd": "vinyl", "He": "3° allyl"}, "B": "He>Hc>Hb>Ha>Hd"}.