Organic Compounds Containing Oxygen: Chemistry | JEE Main

Match List I with List II: is reacted with reagents in List I to form products in List II. Choose the correct answer from the options given below:

Benzene diazonium chloride reactant and various reactions: with aniline (forming azobenzene derivative), HBF4/heat (fluorobenzene), Cu/HCl (chlorobenzene), CuCN/KCN (benzonitrile).

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AA-IV, B-III, C-II, D-IBA-I, B-III, C-IV, D-IICA-III, B-I, C-II, D-IVDA-III, B-I, C-IV, D-II

Step-by-step solutionView

Correct answer

The correct matching is A-III, B-I, C-IV, D-II based on selective carbonyl reductions, double-bond reductions, and Clemmensen reduction.

Option analysis

Why each option works or fails

A · A-IV, B-III, C-II, D-I

Assuming that Clemmensen reduction (Zn(Hg)/HCl) reduces the alkene rather than fully reducing the carbonyl to a methylene group, or confusing Luche/DIBAL-H with complete reduction. Recognize that Zn(Hg)/conc. HCl reduces carbonyl groups to $-\text{CH}_2-$ , yielding cyclohexene from cyclohex-2-en-1-one (A-III).

B · A-I, B-III, C-IV, D-II

Confusing the reagents Zn(Hg)/conc. HCl and $\text{CeCl}_3 \cdot 7\text{H}_2\text{O} / \text{NaBH}_4$ , leading to swapped assignments between cyclohexene and cyclohex-2-en-1-ol. Associate the Luche reagent ( $\text{NaBH}_4 + \text{CeCl}_3$ ) with selective 1,2-reduction of $\alpha,\beta$ -unsaturated ketones to allylic alcohols (B-I).

C · A-III, B-I, C-II, D-IV

Swapping the outcomes of $\text{H}_2/\text{Pd}$ and $\text{LiAlH}_4$ reductions for $\alpha,\beta$ -unsaturated ketones. Recall that $\text{H}_2/\text{Pd}$ selectively reduces the conjugated alkene to give cyclohexanone (C-IV), whereas $\text{LiAlH}_4$ reduces both the double bond and the carbonyl group to give cyclohexanol (D-II).

D · A-III, B-I, C-IV, D-II

Correct option: reflects accurate knowledge of selective reductions for $\alpha,\beta$ -unsaturated ketones. Zn(Hg)/conc. HCl gives cyclohexene (A-III); Luche reduction gives cyclohex-2-en-1-ol (B-I); $\text{H}_2/\text{Pd}$ (1 eq.) gives cyclohexanone (C-IV); $\text{LiAlH}_4$ reduces both double bond and carbonyl to yield cyclohexanol (D-II).

Reviewed route

Solution

StepWorking

01identify

The substrate is 4-tert-butylcyclohexanone. List I contains various reagents: A. $\text{CH}_3\text{MgBr}$ followed by $\text{H}_3\text{O}^+$ , B. $\text{LiAlH}_4$ , C. $\text{CH}_3\text{MgBr}/\text{H}_3\text{O}^+$ followed by $\text{H}_2\text{SO}_4/\Delta$ , and D. $\text{LiAlH}_4$ followed by $\text{H}_2\text{SO}_4/\Delta$ .

02mechanism

For A: Nucleophilic addition of $\text{CH}_3\text{MgBr}$ to the carbonyl carbon gives a tertiary alcohol, 4-tert-butyl-1-methylcyclohexan-1-ol (Structure III). Thus, A matches III.

03mechanism

For B: Hydride reduction of the ketone with $\text{LiAlH}_4$ gives a secondary alcohol, 4-tert-butylcyclohexan-1-ol (Structure I). Thus, B matches I.

04mechanism

For C: Grignard addition forms the tertiary alcohol (III), which upon acid-catalyzed dehydration ( $\text{H}_2\text{SO}_4/\Delta$ ) eliminates water to yield the most stable, endocyclic trisubstituted alkene: 4-tert-butyl-1-methylcyclohex-1-ene (Structure IV). Thus, C matches IV.

05mechanism

For D: Reduction forms the secondary alcohol (I), which upon acid-catalyzed dehydration ( $\text{H}_2\text{SO}_4/\Delta$ ) gives 4-tert-butylcyclohex-1-ene (Structure II). Thus, D matches II.

06product

Matching pairs are: A -> III, B -> I, C -> IV, D -> II.

✓verify

Matching A-III and B-I immediately rules out options 0, 1, and 2, confirming Option 3 as the unique correct answer.

✓ 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

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Students also ask

Why does dehydration of 4-tert-butyl-1-methylcyclohexanol give IV instead of an exocyclic double bond?

Zaitsev's rule dictates formation of the more stable, more highly substituted endocyclic alkene (trisubstituted).

Answer

The correct matching is A-III, B-I, C-IV, D-II based on selective carbonyl reductions, double-bond reductions, and Clemmensen reduction.

Why each option works or fails

A: A-IV, B-III, C-II, D-I - Assuming that Clemmensen reduction (Zn(Hg)/HCl) reduces the alkene rather than fully reducing the carbonyl to a methylene group, or confusing Luche/DIBAL-H with complete reduction. Recognize that Zn(Hg)/conc. HCl reduces carbonyl groups to $-\text{CH}_2-$ , yielding cyclohexene from cyclohex-2-en-1-one (A-III).
B: A-I, B-III, C-IV, D-II - Confusing the reagents Zn(Hg)/conc. HCl and $\text{CeCl}_3 \cdot 7\text{H}_2\text{O} / \text{NaBH}_4$ , leading to swapped assignments between cyclohexene and cyclohex-2-en-1-ol. Associate the Luche reagent ( $\text{NaBH}_4 + \text{CeCl}_3$ ) with selective 1,2-reduction of $\alpha,\beta$ -unsaturated ketones to allylic alcohols (B-I).
C: A-III, B-I, C-II, D-IV - Swapping the outcomes of $\text{H}_2/\text{Pd}$ and $\text{LiAlH}_4$ reductions for $\alpha,\beta$ -unsaturated ketones. Recall that $\text{H}_2/\text{Pd}$ selectively reduces the conjugated alkene to give cyclohexanone (C-IV), whereas $\text{LiAlH}_4$ reduces both the double bond and the carbonyl group to give cyclohexanol (D-II).
D · correct: A-III, B-I, C-IV, D-II - Correct option: reflects accurate knowledge of selective reductions for $\alpha,\beta$ -unsaturated ketones. Zn(Hg)/conc. HCl gives cyclohexene (A-III); Luche reduction gives cyclohex-2-en-1-ol (B-I); $\text{H}_2/\text{Pd}$ (1 eq.) gives cyclohexanone (C-IV); $\text{LiAlH}_4$ reduces both double bond and carbonyl to yield cyclohexanol (D-II).

Step-by-step solution

identify: The substrate is 4-tert-butylcyclohexanone. List I contains various reagents: A. $\text{CH}_3\text{MgBr}$ followed by $\text{H}_3\text{O}^+$ , B. $\text{LiAlH}_4$ , C. $\text{CH}_3\text{MgBr}/\text{H}_3\text{O}^+$ followed by $\text{H}_2\text{SO}_4/\Delta$ , and D. $\text{LiAlH}_4$ followed by $\text{H}_2\text{SO}_4/\Delta$ .
mechanism: For A: Nucleophilic addition of $\text{CH}_3\text{MgBr}$ to the carbonyl carbon gives a tertiary alcohol, 4-tert-butyl-1-methylcyclohexan-1-ol (Structure III). Thus, A matches III.
mechanism: For B: Hydride reduction of the ketone with $\text{LiAlH}_4$ gives a secondary alcohol, 4-tert-butylcyclohexan-1-ol (Structure I). Thus, B matches I.
mechanism: For C: Grignard addition forms the tertiary alcohol (III), which upon acid-catalyzed dehydration ( $\text{H}_2\text{SO}_4/\Delta$ ) eliminates water to yield the most stable, endocyclic trisubstituted alkene: 4-tert-butyl-1-methylcyclohex-1-ene (Structure IV). Thus, C matches IV.
mechanism: For D: Reduction forms the secondary alcohol (I), which upon acid-catalyzed dehydration ( $\text{H}_2\text{SO}_4/\Delta$ ) gives 4-tert-butylcyclohex-1-ene (Structure II). Thus, D matches II.
product: Matching pairs are: A -> III, B -> I, C -> IV, D -> II.
verify: Matching A-III and B-I immediately rules out options 0, 1, and 2, confirming Option 3 as the unique correct answer.