Q.58 Chemistry ⭐ 360° Option Elimination Active

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NEET 2025 Chemistry Question 58 Solution

Topic: Hydrocarbons & Organic Chemistry · Official NTA Booklet Code 45

Question

Which one of the following compounds can exist as cis-trans isomers?

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Verified Against NTA Final Answer Key

Option (4) : 1,2-Dimethylcyclohexane

📚 Hydrocarbons & Organic Chemistry Easy

📋 Given Parameters & Problem Setup

Option 1 Compound Pent-1-ene (terminal alkene)

Option 2 Compound 2-Methylhex-2-ene (alkene with gem-dimethyl)

Option 3 Compound 1,1-Dimethylcyclopropane (gem-disubstituted cycloalkane)

Option 4 Compound 1,2-Dimethylcyclohexane (1,2-disubstituted cycloalkane)

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The 30-Second Exam Hall Shortcut: ⚡ Two Different Groups Geometrical Test

⏱️ Save ~30 seconds

To exhibit cis-trans isomerism, each carbon with restricted rotation must be bonded to TWO DIFFERENT GROUPS! Pent-1-ene has =CH₂ (identical H). 2-Methylhex-2-ene has =C(CH₃)₂ (identical CH₃). 1,1-Dimethylcyclopropane has gem-dimethyl. Only 1,2-Dimethylcyclohexane has different groups (-H and -CH₃) on both C1 and C2!

1 Rule: Geometrical isomerism requires: (1) Restricted rotation, and (2) Both stereocenters must each have two DIFFERENT substituent groups.

2 Test Pent-1-ene: CH₂=CH-CH₂-CH₂-CH₃. C1 has two identical -H atoms ⟹ NO cis-trans isomerism.

3 Test 2-Methylhex-2-ene: (CH₃)₂C=CH-CH₂-CH₂-CH₃. C2 has two identical -CH₃ groups ⟹ NO cis-trans isomerism.

4 Test 1,1-Dimethylcyclopropane: C1 has two identical -CH₃ groups ⟹ NO cis-trans isomerism.

5 Test 1,2-Dimethylcyclohexane: The cyclohexane ring provides restricted rotation. C1 is bonded to -H and -CH₃; C2 is bonded to -H and -CH₃. Both carbons have two different groups!

6 Cis-isomer: Both -CH₃ groups on the same face (e.g. cis-1,2). Trans-isomer: -CH₃ groups on opposite faces (trans-1,2).

7 Select Option (4) in under 15 seconds!

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Why You Can 100% Trust This Shortcut: Geometrical Isomerism in Alkenes and Cyclic Compounds (NCERT Chemistry · Organic Chemistry: Some Basic Principles and Techniques).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Fundamental Criteria for Geometrical (Cis-Trans) Isomerism

Cis-trans isomerism arises when rotation is restricted (either by a double bond or a ring structure) and each of the two bridgehead/olefinic carbons carries two distinct groups:

\text{For C}_a \text{ and C}_b: \quad \text{Substituents on C}_a \text{ must be } X \neq Y, \; \text{Substituents on C}_b \text{ must be } P \neq Q

• If either carbon bears two identical groups, interchange of groups produces the identical molecule, precluding isomerism.

Step 2

Analyze Options (1), (2), and (3)

Examine each candidate compound against the distinct-substituent criterion:

\text{Pent-1-ene: } \text{H}_2\text{C}=\text{CH-Pr} \implies \text{Two identical H atoms on C1}

• 2-Methylhex-2-ene: (CH₃)₂C=CH-Pr ⟹ Two identical methyl groups on C2.
• 1,1-Dimethylcyclopropane: C1 is bonded to two identical methyl groups, while C2 and C3 are identical -CH₂- groups. Rotation restriction does not produce diastereomers.

Step 3

Verify Cis-Trans Isomerism in 1,2-Dimethylcyclohexane

In 1,2-dimethylcyclohexane, the closed ring restricts free rotation:

\text{C1 has: } -\text{CH}_3, -\text{H}; \quad \text{C2 has: } -\text{CH}_3, -\text{H}

• When both methyl groups lie on the same side of the ring plane (e.g. 1-axial, 2-equatorial or both up), it forms the cis-isomer.
• When the methyl groups lie on opposite sides (one up, one down), it forms the trans-isomer (e.g. diequatorial 1e, 2e).
• These are distinct, non-superimposable geometrical diastereomers.
• Matches Option (4) perfectly.

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Terminal alkene with two identical hydrogen atoms on C1.

⚠️

Student Fallacy (Why candidates pick this trap): Thinking all alkenes show cis-trans isomerism.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. C2 has two identical methyl groups.

⚠️

Student Fallacy (Why candidates pick this trap): Failing to check that C2 has two identical methyl groups.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Both methyl groups are on the same carbon (geminal).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming any substituted ring shows cis-trans isomerism.

4 D

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. 1,2-disubstitution on a saturated ring exhibits well-known cis and trans diastereomers.

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Why this is airtight: None; textbook stereochemical principles.

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NTA Trap Alert: Looking Only at Alkenes for Cis-Trans Isomerism

Many candidates only associate cis-trans isomerism with C=C double bonds, ignoring disubstituted cycloalkanes! Because both alkenes in the options (Pent-1-ene and 2-Methylhex-2-ene) have identical groups, the answer is the cycloalkane!

🛡️ How to avoid the -1 mark: Rings have restricted rotation just like double bonds: 1,2-disubstituted rings show cis-trans isomerism!

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Hydrocarbons Standard Benchmark

Core Scientific Principle Tested:

"Ring structures restrict C-C rotation; 1,2-disubstituted cycloalkanes exhibit cis-trans isomerism."

Syllabus Focus: Geometrical Isomerism in Alkenes and Cycloalkanes

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Check the carbons: Two identical groups = NO cis-trans; 1,2 on a ring = YES cis-trans!

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NEET 2025 Chemistry Question 58 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Two Different Groups Geometrical Test

Step-by-Step Rigorous Workflow & Mathematical Proof

Fundamental Criteria for Geometrical (Cis-Trans) Isomerism

Analyze Options (1), (2), and (3)

Verify Cis-Trans Isomerism in 1,2-Dimethylcyclohexane

360° Option Elimination Engine Key Feature

NTA Trap Alert: Looking Only at Alkenes for Cis-Trans Isomerism

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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