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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3co260

Which statements are correct for the compound shown?

  1. A in methanol reduces only the ketone, leaving the ester
  2. BExcess followed by aqueous work-up gives a diol with a secondary and a primary alcohol group
  3. CHydrazine and KOH at high temperature (Wolff-Kishner) convert the ketone to ethyl and leave the methyl ester intact
  4. DIt gives a positive iodoform test

Answer: A, B, D

The drawn compound is methyl 4-acetylbenzoate. NaBH4 is selective for the ketone. LiAlH4 reduces both: CH3CH(OH)- and CH2OH. KOH/glycol at high temperature hydrolyses the ester, so statement 3 is wrong. The COCH3 unit gives CHI3 with I2/NaOH, so the iodoform test is positive.
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