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Saturday, 31 October

JEE Main 2022 · ChemistryMultiple choiceDiagram basedHardApplication

JEE Main 25 June 2022, Shift 2, Chemistry Q75

Question 75 of 90 in this shift, Chemistry question 15 of 30, Section A.

Amongst the following, the major product of the given chemical reaction is [drawn: six-membered ring containing one O, with a double bond drawn on the ring edge on the side next to the O] →CH3OHBr2\xrightarrow[\mathrm{CH_3OH}]{\mathrm{Br_2}} Major Product

The figure, in words

Reactant drawn as a six-membered ring with one O at the bottom and a double bond on the ring edge on the right side (between the carbon next to O and its neighbour). Reagents over the arrow: Br2 above, CH3OH below. Options are four six-membered O-containing rings with Br and OCH3 on adjacent ring carbons, differing in wedge/hash bonds.
  1. (1)[figure: six-membered ring with one O; two ring carbons on the right side carry substituents; upper carbon: Br on a hashed bond, lower carbon (next to ring O): OCH3\mathrm{OCH_3} on a bold wedge]Official answer
  2. (2)[figure: same ring; upper carbon: Br on a bold wedge, lower carbon (next to ring O): OCH3\mathrm{OCH_3} on a bold wedge]
  3. (3)[figure: same ring; upper carbon: OCH3\mathrm{OCH_3} on a hashed bond, lower carbon (next to ring O): Br on a bold wedge]
  4. (4)[figure: same ring; upper carbon: OCH3\mathrm{OCH_3} on a bold wedge, lower carbon (next to ring O): Br on a hashed bond]

Official answer

Option 1

NTA final key.