The major product formed in the following reaction is :
This question is built on drawn structures, so the options cannot be reproduced as text. The chemistry it tests is set out below.
The two reagents together are the standard recipe for benzyne chemistry.
Step (i), NaNH2 in liquid ammonia — benzyne generation. Sodamide is a very strong base. It removes the proton ortho to the bromine, and the resulting aryl carbanion expels bromide to give benzyne, a highly strained intermediate carrying a formal triple bond in the ring. That extra bond is not a real alkyne π bond but a weak sideways overlap of two sp2 orbitals lying in the plane of the ring, which is why benzyne is so reactive and cannot be isolated.
Step (ii), furan — Diels-Alder trapping. Furan is an excellent diene, and benzyne is an outstanding dienophile precisely because of that strained, electron-poor in-plane bond. The two combine in a [4+2] cycloaddition.
The product is the bridged bicyclic adduct 1,4-dihydro-1,4-epoxynaphthalene — the oxygen of the furan becomes a one-atom oxygen bridge across the newly formed six-membered ring. Trapping benzyne with furan in this way is in fact the classic experiment that demonstrates benzyne's existence, since the intermediate itself is too short-lived to observe directly.
Options showing simple substitution of bromide by an amino group would correspond to a different pathway, and options without the oxygen bridge would not reflect a cycloaddition with furan.
Per the official final answer key the answer is option (A).
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