See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: Morpholine (P) is a six-membered heterocyclic ring containing one oxygen atom and one NH group. Its structure is O-CH2-CH2-NH-CH2-CH2 (cyclic). Step 2 - First Hoffmann Exhaustive Methylation: The NH of morpholine is treated with excess methyl iodide to form the N,N,N-trimethyl quaternary ammonium salt (the nitrogen gets three methyl groups). Then treatment with Ag2O/H2O followed by heating causes elimination. The Hoffmann elimination removes a beta-hydrogen from one of the ethylene bridges adjacent to nitrogen, breaking the ring and forming a terminal alkene. After the first exhaustive methylation, the ring opens to give: O-CH2-CH2-N(CH3)2-CH=CH2, i.e., 2-(dimethylamino)ethyl vinyl ether... Let me reconsider: Morpholine ring opening via Hofmann elimination. The nitrogen has two -CH2CH2- groups, one connected to O and one free. First elimination: quaternary ammonium salt forms, Hofmann elimination occurs preferentially at the carbon beta to N that gives the less hindered alkene. The ring opens. The nitrogen is connected to -CH2CH2O- on one side and -CH2CH2- on the other side. Hofmann elimination from the -CH2CH2- side (not adjacent to O, since O withdraws electrons making those beta-H less acidic and also the C-O bond can break differently) opens the ring to give: CH2=CH-N(CH3)2 + HOCH2CH2... but ring must open fully. Actually, for morpholine: First Hofmann exhaustive methylation converts N-H to N+(CH3)3, then elimination opens the ring. The two possible beta-carbons are: (1) -CH2- adjacent to O, and (2) -CH2- on the all-carbon side. Hofmann elimination preferentially gives less substituted alkene. Elimination from either position opens the ring. If elimination occurs on the carbon NOT adjacent to O, we get: (CH3)2N-CH2CH2-O-CH2-CH=CH2. If from carbon adjacent to O: (CH3)2N-CH=CH2 + HOCH2CH2- but that cleaves differently. Step 3 - After ring opening with first Hofmann: The open-chain product is N,N-dimethyl-2-(2-vinyloxyethyl)amine or similar, containing a dimethylamino group and a vinyl ether moiety. The actual product after first Hofmann on morpholine ring opening is: CH2=CH-O-CH2CH2-N(CH3)2 (dimethylaminoethyl vinyl ether). Step 4 - Second Hofmann Exhaustive Methylation: The dimethylamino group -N(CH3)2 is methylated again to -N+(CH3)3, then Hofmann elimination of the adjacent -CH2CH2- (beta to N) gives another elimination. The -N+(CH3)3-CH2-CH2-O-CH=CH2 undergoes elimination to give CH2=CH-O-CH2-CH=CH2 (divinyl ether) + (CH3)3N. Step 5 - Final product: CH2=CH-O-CH=CH2, divinyl ether, which matches option (a). Why other options fail: (b) would require N to remain in the product, but both Hofmann eliminations expel N as trimethylamine. (c) and (d) would require C-O bond cleavage which does not occur under Hofmann conditions; the oxygen remains in the product. Therefore, the correct answer is A.