JEE Main202427 Jan 2024Evening ShiftChemistryAlcohols Phenols and EthersActual
Match List-I with List-II. List I (Reaction) List II (Reagent(s)) (A) (I) Na 2 Cr 2 O 7 / H 2 SO 4 (B) (II) (i) NaOH (ii) CH 3 Cl (C) (III) (i) NaOH , CHCl 3 (ii) NaOH (iii) HCl (D) (IV) (i) NaOH (ii) CO 2 (iii) HCl Choose the correct answer from the options given below:
Options
- A(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- B(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
- C(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
- D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Correct answer
D. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Step-by-step solution
(A) → Kolbe Schmidt Reaction-It is a type of addition reaction, which starts with phenol that reacts with carbon dioxide and sodium hydroxide to give salicylic acid as the end product. (B) → Reimer Tiemann Reaction-When phenol, i.e. C 6 H 5 OH , is treated with CHCl 3 (chloroform) in the presence of NaOH (sodium hydroxide), an aldehyde group - CHO is introduced at the ortho position of the benzene ring, leading to the formation of o-hydroxybenzaldehyde. The reaction is popularly known as the Reimer Tiemann reaction