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Saturday, 31 October

JEE Main 2020 · ChemistryMultiple choiceSingle correctMediumConceptual

JEE Main 2 September 2020, Shift 2, Chemistry Q42

Question 42 of 75 in this shift, Chemistry question 17 of 25.

Consider the reaction sequence given below : (1) [drawn structure]−Br-\mathrm{Br} →H2OOH⊖\xrightarrow[\mathrm{H_2O}]{\mathrm{OH^{\ominus}}} [drawn structure]−OH+Br⊖-\mathrm{OH} + \mathrm{Br^{\ominus}} ......... (1) rate =k[t-BuBr]= k[\text{t-BuBr}] (2) [arrow drawn down from the first line, then right] →C2H5OHOH⊖\xrightarrow[\mathrm{C_2H_5OH}]{\mathrm{OH^{\ominus}}} H2C\mathrm{H_2C} [drawn double bond with CH3\mathrm{CH_3} and CH3\mathrm{CH_3}] +HOH+Br⊖+ \mathrm{HOH} + \mathrm{Br^{\ominus}} .... (2) rate =k[t-BuBr][OH⊖]= k[\text{t-BuBr}][\mathrm{OH^{\ominus}}] Which of the following statements is true :

The figure, in words

Reaction scheme drawn in skeletal (line) form: a branched skeletal structure with a Br attached, arrow labelled OH(-) over H2O giving the same skeleton with OH, plus Br(-), dashed leader to label (1), 'rate=k[t-BuBr]' below. A second arrow drops from the left of the first line and goes right, labelled OH(-) over C2H5OH, giving a drawn structure with printed labels H2C (double bond to a carbon) and two CH3 (one drawn above, one lower right), plus HOH + Br(-), dashed leader to label (2); below: 'rate=k[t-BuBr][OH(-)]'.
  1. (1)Doubling the concentration of base will double the rate of both the reactions.
  2. (2)Changing the concentration of base will have no effect on reaction (2).
  3. (3)Changing the concentration of base will have no effect on reaction (1).Official answer
  4. (4)Changing the base from OH⊖\mathrm{OH^{\ominus}} to ⊖OR^{\ominus}\mathrm{OR} will have no effect on reaction (2).

Official answer

Option 3

NTA final key (Sep 2020).

Chapter
Haloalkanes and Haloarenes
Topic
SN1 and SN2 Mechanisms
Idea tested
Rate Law and Concentration Dependence

Same topic in other shifts

All SN1 and SN2 Mechanisms questions
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  6. 4 Apr 2024, Shift 2 · Q76Find out the major product formed from the following reaction. [Me : −CH3-\mathrm{CH_3}] →Me2NH (2 equiv)\xrightarrow{\mathrm{Me_2NH\ (2\ equiv)}}HardDiagram basedHas 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.