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JEE MAIN Chemistry QUESTION #1015
Question 1
Toluene (excess) is treated with \(\text{CrO}_2\text{Cl}_2\), \(\text{CS}_2\), then \(\text{H}_3\text{O}^+\), then \(\text{NaHSO}_3\) to give Residue (A). Residue (A) + HCl (dil) \(\to\) Compound (B). Structure of Residue (A) and Compound (B) respectively is:
  • [A] = benzaldehyde (CHO), [B] = sodium benzoate (COONa)
  • [A] = benzaldehyde bisulphite (PhCH(OH)SO\(_3\)Na), [B] = benzaldehyde
  • [A] = \(\text{CH(OCrOHCl}_2)_2\) (chromate acetal), [B] = benzaldehyde
  • [A] = bisulphite addition product (\(\text{PhCH(OH)SO}_3\text{Na}\)), [B] = benzaldehyde (PhCHO)✔️
Correct Answer Explanation
Etard reaction: toluene \(+\text{CrO}_2\text{Cl}_2\to\) chromium complex \(\xrightarrow{\text{H}_3\text{O}^+}\) benzaldehyde. Benzaldehyde \(+\text{NaHSO}_3\to\) bisulphite addition product (Residue A). Bisulphite adduct \(+\text{HCl}\to\) benzaldehyde regenerated (Compound B). So A = bisulphite adduct, B = benzaldehyde.