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Saturday, 31 October

JEE Main 2022 · ChemistryMultiple choiceDiagram basedMediumMulti-step

JEE Main 27 June 2022, Shift 1, Chemistry Q75

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

The major product of the following reaction is : [figure: reactant drawn as a benzene ring with a substituent F ortho to a side chain and NO2\mathrm{NO_2} para to it; the side chain is a zigzag carbon chain whose carbon attached to the ring bears I on a bold wedge bond] →DMFPhS⊖ Na⊕\xrightarrow[\text{DMF}]{\mathrm{PhS^{\ominus}\ Na^{\oplus}}}

The figure, in words

Skeletal structures. Reactant: benzene ring, F at the position next to the side chain, NO2 at the position opposite the side chain; side chain is a zigzag carbon chain, the carbon bonded to the ring carries I on a bold wedge. Reagent over the arrow: PhS(-) Na(+), below the arrow DMF. Options A-D are four skeletal structures differing in whether I/F are retained or replaced by SPh and in the wedge/hash drawing of the bond to the chain carbon.
  1. (1)[figure: same ring with F and NO2\mathrm{NO_2} unchanged; the chain carbon on the ring bears SPh on a hashed bond]Official answer
  2. (2)[figure: chain carbon on the ring still bears I on a bold wedge bond; SPh is on the ring where F was; NO2\mathrm{NO_2} unchanged]
  3. (3)[figure: same ring with F and NO2\mathrm{NO_2} unchanged; the chain carbon on the ring bears SPh on a bold wedge bond]
  4. (4)[figure: chain carbon on the ring bears SPh on a bold wedge bond; a second SPh is on the ring where F was; NO2\mathrm{NO_2} unchanged]

Official answer

Option 1

NTA final key.