Consider the following sequence of reactions to give the major product $(X)$ ![JEE Main 2026 Chemistry, Hydrocarbons — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Hydrocarbons/69d8075681239f6567c4f48f/question_1__q_69d8075681239f6567c4f48f__cdn-question-pool.getmarks.app__04APRS1__q72_d1__58e5696e69_final_ppt_sync.png) $P$ g of the major product $(X)$ formed is reacted with $NaHCO_3$ solution to liberate a gas which occupied $11.2$ dm$^3$ at STP. $P = $ _____ g. (Given molar mass in g mol$^{-1}$ H:$1$, C:$12$, O:$16$, Cl:$35.5$)

JEE Main 2026 — Chemistry Organic Chemistry

2026numerical valuemedium

Consider the following sequence of reactions to give the major product $(X)$

JEE Main 2026 Chemistry, Hydrocarbons — question figure

$P$ g of the major product $(X)$ formed is reacted with $NaHCO_3$ solution to liberate a gas which occupied $11.2$ dm $^3$ at STP. $P =$ _____ g. (Given molar mass in g mol $^{-1}$ H: $1$ , C: $12$ , O: $16$ , Cl: $35.5$ )

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

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Solution

Step 1: The first reaction is a Friedel-Crafts alkylation. Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene.

Step 2: The second reaction is electrophilic aromatic substitution. Toluene reacts with $Cl_2$ in the presence of $FeCl_3$ to form a mixture of o-chlorotoluene and p-chlorotoluene. Due to steric hindrance at the ortho position, the major product is p-chlorotoluene.

Step 3: The third reaction is the oxidation of the alkyl side chain. p-chlorotoluene is oxidized by $K_2Cr_2O_7 / H_2SO_4$ to form p-chlorobenzoic acid, which is the major product $(X)$ .

The molecular formula of p-chlorobenzoic acid is $C_7H_5ClO_2$ .

Molar mass of $(X) = (7 \times 12) + (5 \times 1) + 35.5 + (2 \times 16) = 84 + 5 + 35.5 + 32 = 156.5 \text{ g mol}^{-1}$ .

Step 4: Reaction of p-chlorobenzoic acid with $NaHCO_3$ liberates $CO_2$ gas:

$C_6H_4ClCOOH + NaHCO_3 \rightarrow C_6H_4ClCOONa + H_2O + CO_2 \uparrow$

From the stoichiometry, $1$ mole of p-chlorobenzoic acid produces $1$ mole of $CO_2$ gas.

Volume of $CO_2$ liberated at STP = $11.2 \text{ dm}^3$ .

Moles of $CO_2$ liberated = $\dfrac{11.2}{22.4} = 0.5 \text{ mol}$ .

Therefore, moles of p-chlorobenzoic acid reacted = $0.5 \text{ mol}$ .

Mass of $(X)$ reacted, $P = 0.5 \text{ mol} \times 156.5 \text{ g mol}^{-1} = 78.25 \text{ g}$ .

Answer: $78.25$

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JEE Main Organic Chemistry in other years

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JEE Main 2026 — Chemistry Organic Chemistry

2026numerical valuemedium

Consider the following sequence of reactions to give the major product $(X)$

JEE Main 2026 Chemistry, Hydrocarbons — question figure

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

Step 1: The first reaction is a Friedel-Crafts alkylation. Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene.

Step 3: The third reaction is the oxidation of the alkyl side chain. p-chlorotoluene is oxidized by $K_2Cr_2O_7 / H_2SO_4$ to form p-chlorobenzoic acid, which is the major product $(X)$ .

The molecular formula of p-chlorobenzoic acid is $C_7H_5ClO_2$ .

Molar mass of $(X) = (7 \times 12) + (5 \times 1) + 35.5 + (2 \times 16) = 84 + 5 + 35.5 + 32 = 156.5 \text{ g mol}^{-1}$ .

Step 4: Reaction of p-chlorobenzoic acid with $NaHCO_3$ liberates $CO_2$ gas:

$C_6H_4ClCOOH + NaHCO_3 \rightarrow C_6H_4ClCOONa + H_2O + CO_2 \uparrow$

From the stoichiometry, $1$ mole of p-chlorobenzoic acid produces $1$ mole of $CO_2$ gas.

Volume of $CO_2$ liberated at STP = $11.2 \text{ dm}^3$ .

Moles of $CO_2$ liberated = $\dfrac{11.2}{22.4} = 0.5 \text{ mol}$ .

Therefore, moles of p-chlorobenzoic acid reacted = $0.5 \text{ mol}$ .

Mass of $(X)$ reacted, $P = 0.5 \text{ mol} \times 156.5 \text{ g mol}^{-1} = 78.25 \text{ g}$ .

Answer: $78.25$

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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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