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JEE Advanced 2026 · Paper 1, Chemistry question 16

Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option.

List-I

  • P
    2-Bromo-5-nitrobenzaldehyde oxime with aqueous NaOH.
  • Q
    2-Bromo-5-nitrobenzaldehyde oxime with (1) (CH₃CO)₂O, (2) Na₂CO₃.
  • R
    2-Bromo-5-nitroacetophenone oxime with aqueous NaOH.
  • S
    The O-acetyl oxime of 2-bromo-5-nitroacetophenone, ArC(CH₃)=N-O-COCH₃, with aqueous Na₂CO₃.

List-II

  • 1
    2-Hydroxy-5-nitrobenzonitrile.
  • 2
    2-Bromo-5-nitrobenzonitrile.
  • 3
    N-(2-Bromo-5-nitrophenyl)acetamide.
  • 4
    3-Methyl-5-nitro-1,2-benzisoxazole.
  • 5
    2-Bromo-5-nitroacetophenone oxime (unchanged starting oxime).
Choose one option
Answer and worked solution

Official answer: B

In each substrate the Br is ortho to the oxime carbon and para to the nitro group, so it is activated for nucleophilic aromatic substitution.

(P) Aqueous NaOH deprotonates the aldoxime, and the oxime O⁻ displaces Br to form 5-nitro-1,2-benzisoxazole. A 3-unsubstituted benzisoxazole has a C3-H, so base opens it (Kemp elimination) to 2-hydroxy-5-nitrobenzonitrile → 1.

(Q) Acetic anhydride gives the O-acetyl aldoxime, and NaX2COX3\ce{Na2CO3} eliminates acetic acid across C(H)=N to give the nitrile. Br is untouched: 2-bromo-5-nitrobenzonitrile → 2.

(R) The ketoxime anion displaces Br in the same way, giving 3-methyl-5-nitro-1,2-benzisoxazole. There is no C3-H, so the ring stays closed → 4.

(S) The O-acetyl ketoxime has no C-H on the oxime carbon to eliminate. Mild aqueous NaX2COX3\ce{Na2CO3} only hydrolyses the acetate, returning 2-bromo-5-nitroacetophenone oxime → 5.

P→1, Q→2, R→4, S→5.

Topics:HaloarenesCyanides and isocyanidesNucleophilic addition
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