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Saturday, 31 October

JEE Main 2020 · ChemistryMultiple choiceSingle correctMediumMulti-step

JEE Main 9 January 2020, Shift 2, Chemistry Q45

Question 45 of 75 in this shift, Chemistry question 20 of 25, Section A.

Consider the following reactions, [P] →(i) NaNO2/HCl, 0-5 °C; (ii) β-naphthol/NaOH\xrightarrow{\text{(i) NaNO}_2\text{/HCl, 0-5 °C; (ii) }\beta\text{-naphthol/NaOH}} Colored Solid [P] →Br2/H2O\xrightarrow{Br_2/H_2O} C7H6NBr3\mathrm{C_7H_6NBr_3} The compound [P] is :

The figure, in words

Scheme: compound [P] gives a coloured solid with NaNO2/HCl then beta-naphthol/NaOH, and gives C7H6NBr3 with Br2/H2O. Options are substituted benzenes with NH2/NHCH3 and CH3 groups.
  1. (1)[figure: o-toluidine (2-methylaniline)]
  2. (2)[figure: m-toluidine (3-methylaniline)]Official answer
  3. (3)[figure: p-toluidine (4-methylaniline)]
  4. (4)[figure: N-methyl-3-methylaniline (NHCH3 and CH3 meta on benzene ring)]

Official answer

Option 2

NTA final key (Jan 2020).

Chapter
Amines
Topic
Diazonium Salts
Idea tested
Coupling with Beta-Naphthol

Same idea in other shifts

Asked 2× in all
  1. 26 Jun 2022, Shift 2 · Q78The final product 'C' in the following series of reactions [drawn benzene ring with NO2] --Sn/HCl--> A --NaNO2\mathrm{NaNO_2}, HCl--> B…MediumSingle correctHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.