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Saturday, 31 October

JEE Main 2021 · ChemistryMultiple choiceOtherMediumMulti-step

JEE Main 4 August 2021, Shift 2, Chemistry Q43

Question 43 of 90 in this shift, Chemistry question 13 of 30, Section A.

Neopentyl bromide on reaction with ethyl alcohol gives, as the major product,

The figure, in words

Each option is a drawn structure printed with condensed labels: (1) CH3 and CH3 bonded to a C that is double-bonded (C = C) to a C bearing H and CH3; (2) CH3 - C - CH2CH3 with the central C double-bonded (||) to CH2 below; (3) CH3-C-CH2-CH2-C-CH3 with CH3 above and below each of the two C atoms (vertical bonds); (4) CH2 = CH - C = CH2 with CH3 below the third carbon (vertical bond).
  1. (1)[structure drawn: CH3\mathrm{CH_3} and CH3\mathrm{CH_3} on the left carbon, double bond to a right carbon bearing H and CH3\mathrm{CH_3}]Official answer
  2. (2)[structure printed: CH3−C−CH2CH3\mathrm{CH_3-C-CH_2CH_3} with the central C double-bonded to CH2\mathrm{CH_2} drawn below it]
  3. (3)[structure printed: CH3−C(CH3)(CH3)−CH2−CH2−C(CH3)(CH3)−CH3\mathrm{CH_3-C(CH_3)(CH_3)-CH_2-CH_2-C(CH_3)(CH_3)-CH_3} drawn with one CH3\mathrm{CH_3} above and one below each of the two C atoms]
  4. (4)[structure printed: CH2=CH−C=CH2\mathrm{CH_2=CH-C=CH_2} with CH3\mathrm{CH_3} drawn below the third C]

Official answer

Option 1

NTA final key.