MCQEasyJEE Main 2024 · 1 February, Shift 1Aldehydes & Ketones

Chemistry Question from JEE Main 2024 · 1 February, Shift 1

Identify the reagents AA, BB and CC used for the following conversion:

The starting material is a cyclohexane ring carrying a CH2COOCH3-\text{CH}_2\text{COOCH}_3 group at C-11, a CH2CHO-\text{CH}_2\text{CHO} group at the adjacent C-22, and an OH-\text{OH} group at C-44 — that is, methyl 22-[44-hydroxy-22-(22-oxoethyl)cyclohexyl]acetate. It is taken through three steps:

Starting material  A  P1  B  P2  C  P3\text{Starting material} \xrightarrow{\;A\;} P_1 \xrightarrow{\;B\;} P_2 \xrightarrow{\;C\;} P_3
  • P1P_1: the same ring, now carrying CH2CHO-\text{CH}_2\text{CHO} at both C-11 and C-22, with the C-44 OH-\text{OH} untouched.
  • P2P_2: a bicyclic hydrindane — the two side chains have joined to close a five-membered ring fused to the cyclohexane — in which the new five-membered ring carries a C=C\text{C}=\text{C} bearing a CHO-\text{CHO} group; the ring OH-\text{OH} is still present.
  • P3P_3: the same bicyclic skeleton, with that CHO-\text{CHO} group replaced by CH3-\text{CH}_3.
  • A

    A=LiAlH4A = \text{LiAlH}_4, B=NaOH(aq)B = \text{NaOH(aq)}, C=NH2NH2/KOHC = \text{NH}_2-\text{NH}_2/\text{KOH}, ethylene glycol

  • B

    A=LiAlH4A = \text{LiAlH}_4, B=NaOH(alc)B = \text{NaOH(alc)}, C=Zn/HClC = \text{Zn/HCl}

  • C

    A=DIBAL-HA = \text{DIBAL-H}, B=NaOH(aq)B = \text{NaOH(aq)}, C=NH2NH2/KOHC = \text{NH}_2-\text{NH}_2/\text{KOH}, ethylene glycol

  • D

    A=DIBAL-HA = \text{DIBAL-H}, B=NaOH(alc)B = \text{NaOH(alc)}, C=Zn/HClC = \text{Zn/HCl}

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