Consider the following reactions sequence ![JEE Main 2026 Chemistry, Amines — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Amines/69d8073081239f6567c4f2d0/question_1__q_69d8073081239f6567c4f2d0__cdn-question-pool.getmarks.app__02APRS2__q75_d1__25b338ebb6_final_ppt_sync.png) When the product (P) is subjected to Carius analysis using $AgNO_3$, $1.0$ g of the product (P) will produce _______ g of the precipitate of AgBr. (Nearest Integer) (Given: molar mass in g mol$^{-1}$ C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)

JEE Main 2026 — Chemistry Organic Chemistry

2026integerhard

Consider the following reactions sequence

JEE Main 2026 Chemistry, Amines — question figure

When the product (P) is subjected to Carius analysis using $AgNO_3$ , $1.0$ g of the product (P) will produce _______ g of the precipitate of AgBr. (Nearest Integer)

(Given: molar mass in g mol $^{-1}$ C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The given reactant is p-nitrotoluene.

Step (i): Reduction with Sn/HCl followed by $OH^-$ converts the $-NO_2$ group to an $-NH_2$ group, yielding p-toluidine.

Step (ii): Reaction with acetic anhydride, $(CH_3CO)_2O$ , protects the amino group by forming N-(4-methylphenyl)acetamide.

Step (iii): Bromination with $Br_2/AlBr_3$ directs the incoming bromine to the ortho position relative to the strongly activating $-NHCOCH_3$ group (the para position is blocked by the methyl group). This yields 2-bromo-4-methylacetanilide.

Step (iv): Acid hydrolysis with $H_3O^+$ removes the acetyl group, yielding 2-bromo-4-methylaniline as the major product (P).

The molecular formula of product P (2-bromo-4-methylaniline) is $C_7H_8NBr$ .

Molar mass of P = $(7 \times 12) + (8 \times 1) + 14 + 80 = 186 \text{ g mol}^{-1}$ .

In Carius analysis, the bromine in the organic compound is quantitatively converted to silver bromide (AgBr).

Molar mass of AgBr = $108 + 80 = 188 \text{ g mol}^{-1}$ .

Since one molecule of P contains one bromine atom, 1 mole of P produces 1 mole of AgBr.

Moles of P in $1.0 \text{ g} = \dfrac{1.0}{186} \text{ mol}$ .

Moles of AgBr produced = $\dfrac{1.0}{186} \text{ mol}$ .

Mass of AgBr produced = $\dfrac{1.0}{186} \times 188 \text{ g} \approx 1.01 \text{ g}$ .

Rounding to the nearest integer, the mass of the AgBr precipitate is 1 g.

Answer: $1$

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JEE Main Organic Chemistry in other years

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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

2026integerhard

Consider the following reactions sequence

JEE Main 2026 Chemistry, Amines — question figure

When the product (P) is subjected to Carius analysis using $AgNO_3$ , $1.0$ g of the product (P) will produce _______ g of the precipitate of AgBr. (Nearest Integer)

(Given: molar mass in g mol $^{-1}$ C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The given reactant is p-nitrotoluene.

Step (i): Reduction with Sn/HCl followed by $OH^-$ converts the $-NO_2$ group to an $-NH_2$ group, yielding p-toluidine.

Step (ii): Reaction with acetic anhydride, $(CH_3CO)_2O$ , protects the amino group by forming N-(4-methylphenyl)acetamide.

Step (iv): Acid hydrolysis with $H_3O^+$ removes the acetyl group, yielding 2-bromo-4-methylaniline as the major product (P).

The molecular formula of product P (2-bromo-4-methylaniline) is $C_7H_8NBr$ .

Molar mass of P = $(7 \times 12) + (8 \times 1) + 14 + 80 = 186 \text{ g mol}^{-1}$ .

In Carius analysis, the bromine in the organic compound is quantitatively converted to silver bromide (AgBr).

Molar mass of AgBr = $108 + 80 = 188 \text{ g mol}^{-1}$ .

Since one molecule of P contains one bromine atom, 1 mole of P produces 1 mole of AgBr.

Moles of P in $1.0 \text{ g} = \dfrac{1.0}{186} \text{ mol}$ .

Moles of AgBr produced = $\dfrac{1.0}{186} \text{ mol}$ .

Mass of AgBr produced = $\dfrac{1.0}{186} \times 188 \text{ g} \approx 1.01 \text{ g}$ .

Rounding to the nearest integer, the mass of the AgBr precipitate is 1 g.

Answer: $1$

Your note

JEE Main Organic Chemistry in other years

Keep practicing Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

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Solution

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JEE Main Organic Chemistry in other years

More JEE Main Organic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Inorganic Chemistry

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