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MCQMediumJEE Main 2023 · 25 January, Shift 1Phenols

Chemistry Question from JEE Main 2023 · 25 January, Shift 1

Identify the product formed (AA and EE) in the following reaction sequence:

Reaction sequence starting from p-nitrotoluene treated with bromine, then Sn/HCl, then NaNO2/HCl at 273 to 278 K, then H3PO2/H2O, and finally KMnO4/KOH followed by H3O+ to form product E.
  • A
    Option A shows A as dibromo nitrotoluene with bromine substituents on both ortho positions relative to methyl, and E as dibromobenzoic acid retaining two bromine substituents.
  • B
    Option B shows A as monobromo p-nitrotoluene and E as monobromobenzoic acid, both with bromine ortho to methyl and nitro converted to hydrogen in the final product.
  • C
    Option C shows A as monobromo p-nitrotoluene with bromine ortho to methyl, and E as bromotoluene-like final structure lacking oxidation to carboxylic acid.
  • D
    Option D shows A as monobromo nitrotoluene and E as bromohydroxy benzoic acid containing both carboxylic acid and hydroxyl substituents.

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