JEE Main 2020 · ChemistryMultiple choiceDiagram basedHardMulti-step

JEE Main 3 September 2020, Shift 2, Chemistry Q44

Question 44 of 75 in this shift, Chemistry question 19 of 25.

The compound A in the following reactions is : A

\xrightarrow[\text{(ii) Conc. }\mathrm{H_2SO_4}/\Delta]{\text{(i) }\mathrm{CH_3MgBr/H_2O}}

\xrightarrow[\text{(ii) Zn/}\mathrm{H_2O}]{\text{(i) }\mathrm{O_3}}

C + D C

\xrightarrow[\text{(ii) }\Delta]{\text{(i) Conc.KOH}}

[figure: benzene ring]

-\mathrm{COO^{\ominus}K^+}

+ [figure: benzene ring]

-\mathrm{CH_2OH}

\xrightarrow[\Delta]{\mathrm{Ba(OH)_2}}

\mathrm{H_3C-C(CH_3)=CH-C(=O)-CH_3}

The figure, in words

Reaction scheme as printed: the first arrow from A (reagents (i) CH3MgBr/H2O, (ii) Conc. H2SO4/Delta) has no product written at its head; B starts the next line. In the C step the products are drawn as a benzene ring (hexagon with circle) bonded to COO with a circled minus and K with a plus, plus a benzene ring bonded to CH2OH. D gives H3C-C(CH3)=CH-C(=O)-CH3 drawn with CH3 on the second carbon via a vertical bond and O double-bonded (drawn above) to the fourth carbon. In option 1 the CH carbon is drawn with two CH3 groups on branches.

(1) $\mathrm{C_6H_5-C(=O)-CH(CH_3)_2}$
(2) $\mathrm{C_6H_5-C(=O)-CH_3}$
(3) $\mathrm{C_6H_5-CH_2-C(=O)-CH_3}$ Official answer
(4) $\mathrm{C_6H_5-C(=O)-CH_2CH_3}$

Official answer

Option 3

NTA final key (Sep 2020).

Chapter: Aldehydes, Ketones and Carboxylic Acids
Topic: Aldol and Cannizzaro
Idea tested: Cannizzaro Reaction

JEE Main 3 September 2020, Shift 2, Chemistry Q44

Same idea in other shifts