Organic Compounds Containing Oxygen: Chemistry | JEE Main

The product (P) formed from the following multistep reaction is :-

Multistep reaction starting from 4-nitrotoluene treated with (i) Br2, (ii) H2/Pd, (iii) NaNO2, HCl, 0°C, (iv) H3PO2 to give product (P), which is 2-bromotoluene (or 3-bromotoluene).

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Step-by-step solutionView

Correct answer

The sequence undergoes addition and acid-catalyzed elimination to yield the more stable, conjugated alkene, followed by oxidative ozonolysis to form the dicarboxylic acid derivative shown in the correct option.

Option analysis

Why each option works or fails

A ·

Believing that elimination yields the less substituted (Hofmann/kinetic) alkene before ozonolysis cleavage. Under thermodynamic acid-catalyzed conditions, the elimination proceeds toward the more stable, Zaitsev/conjugated alkene.

B ·

Incorrectly assigning the cleavage products of the ozonolysis step or retaining the wrong carbonyl oxidation state. Follow the cleavage of the specific carbon-carbon double bond systematically and ensure the terminal/internal positions oxidize to the correct carbonyl/carboxylic acid functionalities.

C ·

Selecting the product derived from an endocyclic rather than exocyclic or alternatively conjugated double bond. Identify the most substituted, fully conjugated alkene intermediate before performing the double bond cleavage.

D ·

None. This is the correct product resulting from stable intermediate alkene formation followed by oxidative cleavage. Correctly determined the most stable alkene via Zaitsev elimination and cleaved it to yield the corresponding dicarbonyl/acid product.

Reviewed route

Solution

StepWorking

01identify

The reaction sequence starts with a 6-membered cyclic bromohydrin (trans-2-bromocyclohexan-1-ol), which undergoes: (i) intramolecular epoxide formation with strong base $\text{K}^{+}\text{t-BuO}^-$ , (ii) regioselective epoxide ring-opening with Grignard reagent $\text{CH}_3\text{MgBr}$ followed by $\text{H}_3\text{O}^+$ , and (iii) acid-catalyzed dehydration with $\text{H}_2\text{SO}_4/\Delta$ .

02mechanism

Step 1: $\text{t-BuO}^-$ deprotonates the hydroxyl group, forming an alkoxide that attacks the adjacent carbon bearing bromine via an intramolecular $\text{S}_\text{N}2$ mechanism to yield cyclohexene oxide (epoxide).

03mechanism

Step 2: Methylmagnesium bromide ( $\text{CH}_3\text{MgBr}$ ) attacks the symmetric cyclohexene oxide ring via nucleophilic ring-opening, and upon protonation with $\text{H}_3\text{O}^+$ , yields trans-2-methylcyclohexan-1-ol.

04mechanism

Step 3: Heating with $\text{H}_2\text{SO}_4$ protonates the alcohol group followed by loss of $\text{H}_2\text{O}$ to generate a secondary carbocation at C1. A 1,2-hydride shift occurs to form the more stable tertiary carbocation at C2 (bearing the methyl group). Subsequent loss of a proton generates the most substituted and thermodynamically stable alkene, 1-methylcyclohexene (Saytzeff product).

05product

The major product formed is 1-methylcyclohexene, which corresponds to option (3).

✓verify

The product 1-methylcyclohexene is a trisubstituted alkene, which is thermodynamically favored over 3-methylcyclohexene or methylenecyclohexane.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does the double bond form at the methyl-substituted carbon rather than the adjacent CH2?

Carbocation rearrangement occurs via a 1,2-hydride shift to form a stable tertiary carbocation, leading to the more substituted (Saytzeff) alkene, 1-methylcyclohexene.

Answer

The sequence undergoes addition and acid-catalyzed elimination to yield the more stable, conjugated alkene, followed by oxidative ozonolysis to form the dicarboxylic acid derivative shown in the correct option.

Why each option works or fails

A: - Believing that elimination yields the less substituted (Hofmann/kinetic) alkene before ozonolysis cleavage. Under thermodynamic acid-catalyzed conditions, the elimination proceeds toward the more stable, Zaitsev/conjugated alkene.
B: - Incorrectly assigning the cleavage products of the ozonolysis step or retaining the wrong carbonyl oxidation state. Follow the cleavage of the specific carbon-carbon double bond systematically and ensure the terminal/internal positions oxidize to the correct carbonyl/carboxylic acid functionalities.
C: - Selecting the product derived from an endocyclic rather than exocyclic or alternatively conjugated double bond. Identify the most substituted, fully conjugated alkene intermediate before performing the double bond cleavage.
D · correct: - None. This is the correct product resulting from stable intermediate alkene formation followed by oxidative cleavage. Correctly determined the most stable alkene via Zaitsev elimination and cleaved it to yield the corresponding dicarbonyl/acid product.

Step-by-step solution

identify: The reaction sequence starts with a 6-membered cyclic bromohydrin (trans-2-bromocyclohexan-1-ol), which undergoes: (i) intramolecular epoxide formation with strong base $\text{K}^{+}\text{t-BuO}^-$ , (ii) regioselective epoxide ring-opening with Grignard reagent $\text{CH}_3\text{MgBr}$ followed by $\text{H}_3\text{O}^+$ , and (iii) acid-catalyzed dehydration with $\text{H}_2\text{SO}_4/\Delta$ .
mechanism: Step 1: $\text{t-BuO}^-$ deprotonates the hydroxyl group, forming an alkoxide that attacks the adjacent carbon bearing bromine via an intramolecular $\text{S}_\text{N}2$ mechanism to yield cyclohexene oxide (epoxide).
mechanism: Step 2: Methylmagnesium bromide ( $\text{CH}_3\text{MgBr}$ ) attacks the symmetric cyclohexene oxide ring via nucleophilic ring-opening, and upon protonation with $\text{H}_3\text{O}^+$ , yields trans-2-methylcyclohexan-1-ol.
mechanism: Step 3: Heating with $\text{H}_2\text{SO}_4$ protonates the alcohol group followed by loss of $\text{H}_2\text{O}$ to generate a secondary carbocation at C1. A 1,2-hydride shift occurs to form the more stable tertiary carbocation at C2 (bearing the methyl group). Subsequent loss of a proton generates the most substituted and thermodynamically stable alkene, 1-methylcyclohexene (Saytzeff product).
product: The major product formed is 1-methylcyclohexene, which corresponds to option (3).
verify: The product 1-methylcyclohexene is a trisubstituted alkene, which is thermodynamically favored over 3-methylcyclohexene or methylenecyclohexane.