Question

N-methylpiperidine, when reacted with excess methyl iodide and then moist silver oxide, produces an alkene on heating (Hofmann elimination). Name the alkene formed.

• A. Penta-1,4-diene
• B. 1,3-Butadiene
• C. Penta-1,3-diene
• D. Ethene

Hint:

A cyclic amine needs two rounds of exhaustive methylation and Hofmann elimination to open the ring and remove nitrogen completely.

Answer 1

The Correct Option is A

Step 1: N-methylpiperidine is a cyclic tertiary amine, a six-membered ring containing one nitrogen already bearing a methyl group.

Step 2: Treating it with excess methyl iodide converts the ring nitrogen into a quaternary ammonium iodide.

Step 3: Moist silver oxide exchanges iodide for hydroxide, and heating causes Hofmann elimination.

Step 4: Because nitrogen is part of a ring, this first elimination opens the ring into an open-chain amino-alkene, \((CH_3)_2N-CH_2-CH_2-CH_2-CH=CH_2\).

Step 5: This open-chain amine is again methylated, treated with \(Ag_2O\), and heated a second time, removing trimethylamine and forming a second double bond.

Step 6: The double bonds end up at carbons 1 and 4 of the five-carbon chain, giving \(CH_2=CH-CH_2-CH=CH_2\), penta-1,4-diene.

\[\boxed{\text{Penta-1,4-diene}}\]