Consider the following sequence of reactions. ![JEE Main 2026 Chemistry, Amines — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Amines/69d8075081239f6567c4f444/question_1__q_69d8075081239f6567c4f444__cdn-question-pool.getmarks.app__06APRS1__q72_d1__8b61bb6b65_final_ppt_sync.png) The percentage of nitrogen in the yellow product (X) formed is _______%. (Nearest Integer) (Given Molar mass in g mol$^{-1}$ H:1, C:12, N:14)

The given starting material is diazoaminobenzene ($C_6H_5-N=N-NH-C_6H_5$). When diazoaminobenzene is treated with an acid catalyst (HCl) and excess aniline, and warmed to $40-45^\circ \text{C}$, it undergoes a rearrangement reaction (or dissociation followed by coupling) to form $p$-aminoazobenzene. The reaction sequence is as follows: 1. Diazoaminobenzene reacts with HCl to form benzenediazonium chloride and aniline. 2. The benzenediazonium ion then couples with the excess aniline at the para position to yield $p$-aminoazobenzene. 3. Cooling and adding acetic acid helps in precipitating the free base, $p$-aminoazobenzene, which is a yellow dye (often called Aniline Yellow). The structure of the yellow product (X), $p$-aminoazobenzene, is $C_6H_5-N=N-C_6H_4-NH_2$. The molecular formula of $p$-aminoazobenzene is $C_{12}H_{11}N_3$. Let's calculate its molar mass: Molar mass $= (12 \times 12) + (11 \times 1) + (3 \times 14)$ Molar mass $= 144 + 11 + 42 = 197 \text{ g mol}^{-1}$ The mass of nitrogen in one mole of the compound is $3 \times 14 = 42 \text{ g}$. The percentage of nitrogen in the product is: $\% \text{ N} = \dfrac{42}{197} \times 100 \approx 21.32\%$ Rounding to the nearest integer, we get $21$. Answer: $21$

JEE Main 2026 — Chemistry Organic Chemistry

2026integermedium

Consider the following sequence of reactions.

JEE Main 2026 Chemistry, Amines — question figure

The percentage of nitrogen in the yellow product (X) formed is _______%. (Nearest Integer) (Given Molar mass in g mol $^{-1}$ H:1, C:12, N:14)

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The given starting material is diazoaminobenzene ( $C_6H_5-N=N-NH-C_6H_5$ ).

When diazoaminobenzene is treated with an acid catalyst (HCl) and excess aniline, and warmed to $40-45^\circ \text{C}$ , it undergoes a rearrangement reaction (or dissociation followed by coupling) to form $p$ -aminoazobenzene.

The reaction sequence is as follows:

Diazoaminobenzene reacts with HCl to form benzenediazonium chloride and aniline.
The benzenediazonium ion then couples with the excess aniline at the para position to yield $p$ -aminoazobenzene.
Cooling and adding acetic acid helps in precipitating the free base, $p$ -aminoazobenzene, which is a yellow dye (often called Aniline Yellow).

The structure of the yellow product (X), $p$ -aminoazobenzene, is $C_6H_5-N=N-C_6H_4-NH_2$ .

The molecular formula of $p$ -aminoazobenzene is $C_{12}H_{11}N_3$ .

Let's calculate its molar mass:

Molar mass $= (12 \times 12) + (11 \times 1) + (3 \times 14)$

Molar mass $= 144 + 11 + 42 = 197 \text{ g mol}^{-1}$

The mass of nitrogen in one mole of the compound is $3 \times 14 = 42 \text{ g}$ .

The percentage of nitrogen in the product is:

$\% \text{ N} = \dfrac{42}{197} \times 100 \approx 21.32\%$

Rounding to the nearest integer, we get $21$ .

Answer: $21$

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2026medium

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2026medium

Given below are two statements: Statement I: $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{C}}$ is more stable than $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{3}$ as nine hyperconjugation interactions are possible in $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{C}}$. Statement II: $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{3}$ is less stable than $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{C}}$ as only three hyperconjugation interactions are possible in $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{3}$. In the light of the above statements, choose the correct answer from the options given below

2026medium

Which of the following reactions is an example of Wurtz reaction?

2026easy

A hydroxy compound $(\mathrm{X})$ with molar mass $122 \mathrm{~g} \mathrm{~mol}^{-1}$ is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of $290 \mathrm{~g} \mathrm{~mol}^{-1}$. The number of hydroxyl groups present in compound (X) is:

2026medium

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JEE Main 2026 — Chemistry Organic Chemistry

2026integermedium

Consider the following sequence of reactions.

JEE Main 2026 Chemistry, Amines — question figure

The percentage of nitrogen in the yellow product (X) formed is _______%. (Nearest Integer) (Given Molar mass in g mol $^{-1}$ H:1, C:12, N:14)

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The given starting material is diazoaminobenzene ( $C_6H_5-N=N-NH-C_6H_5$ ).

The reaction sequence is as follows:

Diazoaminobenzene reacts with HCl to form benzenediazonium chloride and aniline.
The benzenediazonium ion then couples with the excess aniline at the para position to yield $p$ -aminoazobenzene.
Cooling and adding acetic acid helps in precipitating the free base, $p$ -aminoazobenzene, which is a yellow dye (often called Aniline Yellow).

The structure of the yellow product (X), $p$ -aminoazobenzene, is $C_6H_5-N=N-C_6H_4-NH_2$ .

The molecular formula of $p$ -aminoazobenzene is $C_{12}H_{11}N_3$ .

Let's calculate its molar mass:

Molar mass $= (12 \times 12) + (11 \times 1) + (3 \times 14)$

Molar mass $= 144 + 11 + 42 = 197 \text{ g mol}^{-1}$

The mass of nitrogen in one mole of the compound is $3 \times 14 = 42 \text{ g}$ .

The percentage of nitrogen in the product is:

$\% \text{ N} = \dfrac{42}{197} \times 100 \approx 21.32\%$

Rounding to the nearest integer, we get $21$ .

Answer: $21$

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JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main Organic Chemistry in other years

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