Coniine synthesis by ammonia cyclization of a 1,5-dibromoalkane Steps:

• Coniine is 2-propylpiperidine, a 6-membered ring with N and a propyl substituent at position 2 (C₈H₁₇N).
• Ammonia reacts with a dibromide via double alkylation to form the ring, requiring 5 carbons between Br atoms for piperidine.
• The propyl group (3 carbons) attaches to one terminal carbon, making the total chain an 8-carbon straight chain.
• Thus, Br at positions 1 and 5 on octane places the propyl (carbons 6-8) correctly for the 2-position after cyclization.

Why D is correct:

• 1,5-Dibromooctane (Br-CH₂-(CH₂)₃-CHBr-(CH₂)₂-CH₃) has 5 carbons from C1 to C5; N links C1 and C5, forming piperidine with propyl on C5 (position 2).

Why the others are wrong:

• A: 1,2-Dibromopropane (C₃H₆Br₂) has only 2 carbons between Br, forming aziridine, not piperidine.
• B: 1,1-Dibromo-2-propylcyclopentane is a geminal dibromide on a 5-membered ring, preventing linear chain cyclization to piperidine.
• C: 1,4-Dibromooctane has 4 carbons between Br (C1 to C4), forming 5-membered pyrrolidine with butyl substituent, not 2-propylpiperidine.

Final answer: D