JEE Main 2025 Jan 23 Shift 2, Chemistry Q22: Preparation of Ether

JEE Main2025Jan 23, Shift 2Chemistry

Consider the following sequence of reactions.

Your answer

Solution

The starting amine is 4-ethoxyaniline. Cold NaNO $_{2}$ /HCl converts the $- N H_{2}$ to a diazonium salt (A): $4 - E tO - C_{6} H_{4} - N_{2}^{+} C l^{-}$ .

The diazonium salt couples with phenoxide (phenol in NaOH) at the position para to $- O H$ , giving the azo dye B: $4 - E tO - C_{6} H_{4} - N = N - C_{6} H_{4} - O H$ (formula C $_{14}$ H $_{14}$ N $_{2}$ O $_{2}$ ).

B is then alkylated by NaOH/ethyl bromide (Williamson ether synthesis) on the phenolic $- O H$ , capping it as an ethyl ether to give C: $4 - E tO - C_{6} H_{4} - N = N - C_{6} H_{4} - O E t$ (formula C $_{16}$ H $_{18}$ N $_{2}$ O $_{2}$ ).

Counting $s p^{3}$ carbons in C: the two aromatic rings and the $N = N$ unit involve only $s p^{2}$ (and $s p$ ) centres. The only $s p^{3}$ carbons are the four carbons in the two $- O - C H_{2} - C H_{3}$ ether arms.

$∴$ Number of $s p^{3}$ carbons $= 2 + 2 = 4$ .

Concept behind this question

Preparation of EtherNotes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 2, Chemistry Q22: Preparation of Ether

Concept behind this question

More previous year questions on Preparation of Ether

Practice more Chemistry