JEE Main 2023 (10 Apr, Shift 1) Chemistry Q74: Hydrocarbons

Question

The major product ‘P’ formed in the given reaction is
[structure: naphthalene ring bearing $\mathrm{CH_2CH_3}$ at one position of one ring, $\mathrm{CH{=}CH_2}$ at the para (opposite) position of the same ring, and $\mathrm{CH_2COOCH_3}$ on the other ring] $\xrightarrow[	ext{(ii) } \mathrm{H_3O^+}]{	ext{(i) alk.}\mathrm{KMnO_4},\ \Delta}$ P (Major Product)

Accepted Answer

Option 4: [figure: naphthalene bearing $\mathrm{COOH}$ at the two para positions of one ring and $\mathrm{COOH}$ (directly attached) on the other ring]

Answer

[figure: naphthalene bearing $\mathrm{CH_2COOH}$ at one position, $\mathrm{CH_2COOH}$ on the other ring, and a diol side chain drawn as a Fischer projection (HO–H, HO–H, H) at the position para to the first]

Answer

[figure: naphthalene bearing $\mathrm{COOH}$ at the two para positions of one ring and $\mathrm{CH_2COOH}$ on the other ring]

Answer

[figure: naphthalene bearing $\mathrm{CH_2COOH}$ at one position, $\mathrm{COCH_3}$ at the para position of the same ring, and $\mathrm{CH_2COOH}$ on the other ring]

Answer

[figure: naphthalene bearing $\mathrm{COOH}$ at the two para positions of one ring and $\mathrm{COOH}$ (directly attached) on the other ring]

JEE Main 10 April 2023, Shift 1, Chemistry Q74

Same idea in other shifts