Learn

Saturday, 31 October

JEE Main 2020 · ChemistryMultiple choiceDiagram basedMediumMulti-step

JEE Main 8 January 2020, Shift 2, Chemistry Q45

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

The major product [B] in the following sequence of reactions is : CH3−CCH(CH3)2=CH−CH2CH3 →(ii) H2O2, O⊖H(i) B2H6 [A] →Δdil. H2SO4 [B]\mathrm{CH_3-\underset{CH(CH_3)_2}{C}=CH-CH_2CH_3}\ \xrightarrow[\mathrm{(ii)\ H_2O_2,\ \overset{\ominus}{O}H}]{\mathrm{(i)\ B_2H_6}}\ \mathrm{[A]}\ \xrightarrow[\Delta]{\mathrm{dil.\ H_2SO_4}}\ \mathrm{[B]}

The figure, in words

In the stem and options, CH(CH3)2 is printed directly below the carbon it is attached to, joined by a vertical line; in option 4 the double bond is drawn vertically down from the central C to a C bearing H3C (left) and CH3 (right).
  1. (1)CH3−CHCH(CH3)2−CH=CH−CH3\mathrm{CH_3-\underset{CH(CH_3)_2}{CH}-CH=CH-CH_3}
  2. (2)CH2=CCH(CH3)2−CH2CH2CH3\mathrm{CH_2=\underset{CH(CH_3)_2}{C}-CH_2CH_2CH_3}
  3. (3)CH3−CCH(CH3)2=CH−CH2CH3\mathrm{CH_3-\underset{CH(CH_3)_2}{C}=CH-CH_2CH_3}
  4. (4)CH3−C−CH2CH2CH3\mathrm{CH_3-C-CH_2CH_2CH_3} with the central C double-bonded (drawn downwards) to a C that carries H3C\mathrm{H_3C} and CH3\mathrm{CH_3}Official answer

Official answer

Option 4

NTA final key (Jan 2020).