Hydrocarbons: JEE Main Chemistry Question with Solution
The correct sequence of reagents for the preparation of Q and R is :
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Hint 1 of 3
What reaction converts n-heptane into the aromatic precursor needed for this synthesis?
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Step-by-step solutionView
Correct answer
n-Heptane undergoes aromatization with Cr2O3 at 770 K,20 atm to yield toluene, which is oxidized by the Étard reaction (CrO2Cl2,H3O+) to benzaldehyde, and subsequently disproportionates via the Cannizzaro reaction (NaOH followed by acidification) into benzyl alcohol (Q) and benzoic acid (R).
Option analysis
Why each option works or fails
A · (i) CrO2Cl2,H3O+; (ii) Cr2O3,770 K,20 atm; (iii) NaOH; (iv) H3O+
Belief that the Étard reagent (CrO2Cl2) can oxidize an open-chain alkane directly before aromatization occurs. CrO2Cl2 selectively oxidizes benzylic methyl groups attached to an aromatic ring; aromatization of n-heptane to toluene must occur first.
B · (i) KMnO4,OH−; (ii) Mo2O3,Δ; (iii) NaOH; (iv) H3O+
Belief that KMnO4 selectively converts an alkane to toluene, or that Mo2O3 oxidizes toluene to benzaldehyde. Alkanes do not react with KMnO4 under normal conditions; catalytic reforming/aromatization requires transition metal oxides like Cr2O3 or Mo2O3 at high temperature and pressure, and the Étard reaction uses CrO2Cl2 to stop at the aldehyde.
C · (i) Cr2O3,770 K,20 atm; (ii) CrO2Cl2,H3O+; (iii) NaOH; (iv) H3O+
None. This sequence correctly orders aromatization, partial benzylic oxidation to benzaldehyde, and disproportionation to benzyl alcohol and benzoic acid. This is the correct reaction sequence.
D · (i) Mo2O3,Δ; (ii) CrO2Cl2,H3O+; (iii) NaOH; (iv) H3O+
Assuming that heating with Mo2O3 alone without specifying pressure or that Cr2O3 at 770 K,20 atm is not the standard NCERT condition for aromatization of n-heptane. The standard high-pressure aromatization reagent explicitly taught in NCERT is Cr2O3,770 K,20 atm (or supported on alumina).
Reviewed route
Solution
StepWorking
01identify
The reaction transforms n-heptane into an aromatic compound P (toluene). Next, toluene is oxidized specifically to benzaldehyde (Q). Finally, a base-catalyzed Cannizzaro reaction yields benzoic acid (R) and benzyl alcohol.
02mechanism
Step 1: Aromatization of n-heptane using Cr2O3 (or V2O5/Mo2O3) supported on alumina at 770 K,10−20 atm gives toluene (P).
Step 2: Selective oxidation of the methyl group in toluene to an aldehyde group using chromyl chloride (Etard reaction: CrO2Cl2 followed by H3O+) yields benzaldehyde (Q).
Step 3 & 4: Cannizzaro reaction of benzaldehyde (lacking α-H) upon treatment with concentrated NaOH followed by acidification (H3O+) gives benzoic acid (R) along with benzyl alcohol.
03product
The reagent sequence is: (i) Cr2O3,770 K,20 atm, (ii) CrO2Cl2,H3O+, (iii) NaOH, (iv) H3O+. This matches Option (2).
✓verify
Option (0) attempts Etard oxidation directly on n-heptane, which fails. Option (1) oxidizes with KMnO4 directly on alkane, which fails. Option (3) uses Mo2O3 without specifying the exact high-pressure/temperature conditions typically cited with Cr2O3 for n-heptane aromatization in NCERT. Hence, Option (2) is correct.
Hints that build this answer step by step
What reaction converts n-heptane into the aromatic precursor needed for this synthesis?
Aromatization using Cr2O3 at 770 K and 20 atm to form toluene
Which reagent converts toluene selectively into benzaldehyde (Étard reaction)?
CrO2Cl2 followed by H3O+
What transformation occurs when benzaldehyde is treated with concentrated NaOH followed by acid workup?
Cannizzaro reaction yielding benzyl alcohol and benzoic acid
Can CrO2Cl2 oxidize aliphatic alkanes directly to aldehydes?
No, chromyl chloride specifically oxidizes benzylic methyl groups via a chromium complex intermediate (Etard reaction); it does not act on linear unactivated alkanes.
