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) $\text{(i)}NaOH\text{(ii)}{CH}_{3}Cl$

(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:

Question

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)	$\text{(i)}NaOH\text{(ii)}{CH}_{3}Cl$
(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:

Quantrex Academy · Accepted Answer

Correct answer: D. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

(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. (C) → Oxidation of phenol to p-benzoquinone the first step, there will be oxidation of phenol with an acidified solution of sodium dichromate. The acidified solution means that there is presence of an acid like sulphuric acid. Now, in the reaction we can see the product formed is benzoquinone, that is a conjugated diketone. (D) Williamson ether synthesis-The Williamson ether synthesis is an S _ N 2 reaction in which an alkoxide ion is a nucleophile that displaces a halide ion from an alkyl halide to give an ether. → PhOH + NaOH → H _ 2 O + PhO ^ -

Match List-I with List-II. List I (Reaction) List II (Reagent(s))...

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