CUET UG Chemistry Booster Test -3 Nomenclature and Bond Nature -

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CUET UG Chemistry Booster Test -3 Nomenclature and Bond Nature

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test -3 Nomenclature and Bond Nature

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QUESTION 1 OF 20

What is the common name for the compound whose IUPAC name is Tetrachloromethane?

Answer Review

1 What is the common name for the compound whose IUPAC name is Tetrachloromethane?

A Chloroform

B Methylene chloride

C Carbon tetrachloride

D Bromoform

💡 Short Explanation

Tetrachloromethane has formula CCl₄. It contains four chlorine atoms. Common name is carbon tetrachloride.

✅ Correct Answer Explanation

Tetrachloromethane is a methane derivative in which all four hydrogen atoms are replaced by chlorine atoms. Formula: CCl₄ Its common name is carbon tetrachloride.

❓ Why Other Options Are Incorrect

A. Chloroform → Trichloromethane, CHCl₃.
B. Methylene chloride → Dichloromethane, CH₂Cl₂.
D. Bromoform → Tribromomethane, CHBr₃.

🎯 Strategy

Used

Nomenclature Recall

🧠 Memory Trick

Tetra chloro = Carbon tetrachloride

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Polyhalogen Compounds

2 The IUPAC name for the structural formulation (CCl₃)₃CCl is:

A 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

B Decachlorobutane

C 1,1,1,2,2,3,3-Heptachloro-2-methylpropane

D 2,2-Bis(trichloromethyl)tetrachloroethane

💡 Short Explanation

The structure contains a heavily chlorinated propane chain. One trichloromethyl group is attached at carbon-2. Correct IUPAC name follows substituted propane nomenclature.

✅ Correct Answer Explanation

The compound ((CCl₃)₃CCl) contains a central carbon attached to chlorine and three trichloromethyl groups. The correct systematic naming selects the suitable chlorinated propane parent chain and identifies the remaining trichloromethyl substituent. Thus, the correct name is: 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

❓ Why Other Options Are Incorrect

B. Decachlorobutane → Oversimplified and not the accepted IUPAC name.
C. Incorrect parent chain and substituent placement.
D. Incorrect structural description for the given formula.

🎯 Strategy

Used

IUPAC Structural Mapping

🧠 Memory Trick

Find parent chain first, then name substituents.

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Nomenclature of Polyhalogen Compounds

3 In understanding the stereochemistry of alkyl halides, identifying stereocenters is critical. Which carbon is the asymmetric carbon (stereocenter) in 2-chlorobutane?

A C-1

B C-2

C C-3

D C-4

💡 Short Explanation

C-2 is attached to four different groups. It is bonded to Cl, H, CH₃ and C₂H₅. Therefore, C-2 is the stereocenter.

✅ Correct Answer Explanation

2-Chlorobutane has the structure: CH₃–CH(Cl)–CH₂–CH₃ Carbon-2 is attached to: H Cl CH₃ CH₂CH₃ Since all four groups are different, C-2 is an asymmetric carbon or stereocenter.

❓ Why Other Options Are Incorrect

A. C-1 → Has three hydrogen atoms, not chiral.
C. C-3 → Attached to two hydrogens, not chiral.
D. C-4 → Terminal CH₃ carbon, not chiral.

🎯 Strategy

Used

Stereocenter Identification

🧠 Memory Trick

Chiral carbon = four different groups

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Stereochemistry

4 Consider the stereochemical aspects of nucleophilic substitution on chiral alkyl halides:
(I) Sₙ2 reactions are accompanied by an inversion of configuration.
(II) Sₙ1 reactions are accompanied by racemization.
(III) Retention of configuration implies the spatial arrangement of bonds to an asymmetric center is completely preserved.
Which statements are correct?

A I and II

B II and III

C I and III

D I, II, and III

💡 Short Explanation

SN2 gives inversion. SN1 gives racemisation. Retention means same configuration is preserved.

✅ Correct Answer Explanation

SN2 reactions occur by backside attack, causing inversion of configuration. SN1 reactions form a planar carbocation intermediate, so nucleophile can attack from either side, leading to racemisation. Retention of configuration means the spatial arrangement around the chiral center remains unchanged. Therefore, all three statements are correct.

❓ Why Other Options Are Incorrect

A, B and C omit one correct statement.

🎯 Strategy

Used

Statement Verification

🧠 Memory Trick

SN2 inverts, SN1 racemises

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Stereochemistry of Substitution Reactions

5 Match List-I (Structure) with List-II (Name):

List-I	List-II
1. p-ClC₆H₄CH₂CH(CH₃)₂	a. 1-Chloro-4-(2-methylpropyl)benzene
2. m-ClCH₂C₆H₄CH₂C(CH₃)₃	b. 1-Chloromethyl-3-(2,2-dimethylpropyl)benzene
3. o-Br-C₆H₄CH(CH₃)CH₂CH₃	c. 1-Bromo-1-(2-bromophenyl)butane
4. CH₃C(p-ClC₆H₄)₂CH(Br)CH₃	d. 2-Bromo-3,3-bis(4-chlorophenyl)butane

A 1-a, 2-b, 3-c, 4-d

B 1-c, 2-d, 3-a, 4-b

C 1-d, 2-a, 3-b, 4-c

D 1-b, 2-c, 3-d, 4-a

💡 Short Explanation

Para chloro is matched with 1,4-substitution. Meta chloromethyl gives 1,3-substitution. Ortho bromo derivative gives 2-bromophenyl name. Bis chlorophenyl butane matches the complex structure.

✅ Correct Answer Explanation

The correct structural matches are: 1 → a 2 → b 3 → c 4 → d The para structure corresponds to 1-chloro-4-substitution. The meta structure corresponds to 1,3-substitution. The ortho bromo compound contains a 2-bromophenyl group. The last structure contains two p-chlorophenyl groups and bromobutane skeleton.

❓ Why Other Options Are Incorrect

B, C and D mismatch aryl substitution positions and side-chain structures.

🎯 Strategy

Used

Substituent Position Mapping

🧠 Memory Trick

o = 1,2; m = 1,3; p = 1,4

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Nomenclature of Haloarenes

6 Why does electrophilic substitution in haloarenes occur predominantly at ortho- and para-positions despite the halogen's electron-withdrawing inductive effect?

A Inductive effect completely dominates at the meta position.

B Halogen releases electrons through resonance, which stabilizes the intermediate carbocation predominantly at ortho- and para-positions.

C Halogen's large atomic size sterically blocks the meta position.

D The sp² carbon forces meta-deactivation exclusively.

💡 Short Explanation

Halogens withdraw by –I effect. They donate by resonance. Resonance stabilizes ortho and para intermediates.

✅ Correct Answer Explanation

Haloarenes are deactivated due to the strong –I effect of halogen. However, halogen also donates lone pair electrons to the benzene ring by resonance. This resonance donation increases electron density at ortho and para positions and stabilizes the corresponding sigma complexes. Therefore, halogens are deactivating but ortho/para directing.

❓ Why Other Options Are Incorrect

A. Does not explain ortho/para direction.
C. Steric blocking is not the main reason.
D. Haloarenes do not direct exclusively to meta.

🎯 Strategy

Used

Electronic Effect Analysis

🧠 Memory Trick

Halogen deactivates but directs o/p

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Electrophilic Substitution in Haloarenes

7 In the strictly standardized IUPAC system, both geminal (alkylidene) and vicinal (alkylene) dihalides are identically classified and named under the general category of:

A Dihaloalkenes

B Dihaloalkanes

C Haloarenes

D Alkyl halides

💡 Short Explanation

Gem and vic are common classifications. IUPAC names them as dihaloalkanes. Both contain two halogens in alkane chain.

✅ Correct Answer Explanation

Geminal dihalides have both halogens on the same carbon, while vicinal dihalides have halogens on adjacent carbons. In IUPAC nomenclature, both are named as dihaloalkanes with proper locants.

❓ Why Other Options Are Incorrect

A. Dihaloalkenes contain double bonds.
C. Haloarenes are aromatic halogen compounds.
D. Alkyl halides usually refer to monohaloalkanes.

🎯 Strategy

Used

Nomenclature Classification

🧠 Memory Trick

Two halogens on alkane = dihaloalkane

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Dihaloalkanes

8 Identify reaction type: The laboratory addition of bromine in CCl₄ to an unknown molecule discharges the reddish-brown color, forming a colourless vic-dibromide. This reaction specifically detects the presence of:

A A single bond

B An aromatic ring

C A double bond

D A primary alcohol

💡 Short Explanation

Bromine in CCl₄ detects unsaturation. Alkene decolourises bromine. Product is vic-dibromide.

✅ Correct Answer Explanation

Bromine in CCl₄ is used to test for carbon-carbon double bonds. Alkenes add bromine across the double bond: C=C + Br₂ → Br–C–C–Br The reddish-brown colour disappears and a colourless vic-dibromide is formed.

❓ Why Other Options Are Incorrect

A. Single bonds do not decolourise bromine in CCl₄.
B. Aromatic rings do not undergo simple bromine addition.
D. Primary alcohols do not give this test.

🎯 Strategy

Used

Reaction Test Identification

🧠 Memory Trick

Br₂/CCl₄ colour loss = C=C present

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Dihalo Compounds

📄 Passage

The observation of Louis Pasteur that crystals of certain compounds exist in the form of mirror images laid the foundation of modern stereochemistry. Van't Hoff and Le Bel independently argued that if all substituents attached to a tetrahedral carbon are different, the mirror image is not superimposable; such a carbon is a stereocentre. Objects or molecules non-superimposable on their mirror images are said to be chiral. The stereoisomers related to each other as non-superimposable mirror images are called enantiomers. A mixture containing two enantiomers in equal proportions will have zero optical rotation, and such a mixture is known as a racemic mixture.

Based on the passage, molecules that are completely non-superimposable on their mirror images are termed:

A Achiral

B Chiral

C Racemic

D Symmetrical

💡 Short Explanation

Chiral objects have non-superimposable mirror images. Chiral molecules usually contain stereocentres. They show optical activity.

✅ Correct Answer Explanation

According to the passage, molecules that are non-superimposable on their mirror images are called chiral. This property is common in molecules containing an asymmetric carbon atom attached to four different groups.

❓ Why Other Options Are Incorrect

A. Achiral → Superimposable on mirror image.
C. Racemic → Equal mixture of two enantiomers.
D. Symmetrical → Usually achiral.

🎯 Strategy

Used

Passage Evidence

🧠 Memory Trick

Chiral = mirror image not superimposable

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Stereochemistry

📄 Passage

Based on the passage, the optical rotation of a strictly 50:50 mixture of two enantiomeric position/structural stereoisomers (a racemic mixture) is:

A Always positive

B Always negative

C Zero

D Dependent on the solvent

💡 Short Explanation

A racemic mixture has equal enantiomers. Their rotations cancel each other. Net optical rotation is zero.

✅ Correct Answer Explanation

A racemic mixture contains equal amounts of two enantiomers: one rotates plane-polarized light clockwise and the other anticlockwise. Since their rotations are equal and opposite, the total optical rotation becomes zero.

❓ Why Other Options Are Incorrect

A. Always positive → Incorrect because rotations cancel.
B. Always negative → Incorrect because rotations cancel.
D. Dependent on solvent → Not the deciding factor here.

🎯 Strategy

Used

Passage Evidence

🧠 Memory Trick

Racemic = right + left cancel

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Optical Isomerism

11 Consider the mechanism of nucleophilic substitution reactions on polarized C-X bonds:
(I) The nucleophile attacks the electron-deficient part of the substrate.
(II) The halogen atom departs as a leaving group (halide ion).
(III) The incoming nucleophile must always possess a positive charge.
Which are analytically accurate?

A I and II

B II and III

C I and III

D I, II, and III

💡 Short Explanation

Nucleophiles attack electron-deficient carbon. Halogen leaves as halide ion. Nucleophiles are electron-rich, not necessarily positive.

✅ Correct Answer Explanation

In alkyl halides, the C-X bond is polar: Cδ⁺—Xδ⁻ The carbon attached to halogen becomes electron-deficient, so the nucleophile attacks this carbon. During substitution, the halogen leaves as a halide ion. Statement III is incorrect because nucleophiles are electron-rich species. They may be negatively charged or neutral, but they are not always positively charged.

❓ Why Other Options Are Incorrect

B. II and III → Statement III is incorrect.
C. I and III → Statement III is incorrect.
D. I, II, and III → Includes incorrect Statement III.

🎯 Strategy

Used

Statement Elimination

🧠 Memory Trick

Nucleophile = Nucleus-loving, electron-rich

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Nucleophilic Substitution Reactions

12 Analytically, the C—Cl bond length in haloarenes (169 pm) is shorter than in haloalkanes (177 pm). This bond contraction is fundamentally caused by:

A The smaller size of the sp² hybridized carbon and its greater s-character.

B The lack of any resonance in the benzene ring.

C The massive steric hindrance of the phenyl ring.

D The sp³ hybridization of the halogen atom.

💡 Short Explanation

sp² carbon has more s-character than sp³ carbon. It holds electrons more tightly. Haloarene C—Cl bond becomes shorter.

✅ Correct Answer Explanation

In haloarenes, the halogen is bonded to an sp² hybridized carbon of the benzene ring. An sp² carbon has greater s-character than an sp³ carbon. Greater s-character makes the carbon more electronegative and allows it to hold the C—Cl bond pair more tightly. Also, resonance gives partial double bond character to the C—Cl bond, further shortening it.

❓ Why Other Options Are Incorrect

B. Lack of resonance → Incorrect; resonance is present.
C. Steric hindrance is not the fundamental cause of shorter bond length.
D. Halogen is not described by sp³ hybridization in this context.

🎯 Strategy

Used

Hybridisation Analysis

🧠 Memory Trick

More s-character = shorter, stronger bond

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Low Reactivity of Haloarenes

13 Based on NCERT values, what is the specific value of the carbon-fluorine (C—F) bond length in pm?

A 139

B 178

C 193

D 214

💡 Short Explanation

Fluorine is the smallest halogen. C—F bond is the shortest C—X bond. Its bond length is 139 pm.

✅ Correct Answer Explanation

Among carbon-halogen bonds, C—F is the shortest because fluorine has the smallest atomic size. NCERT value: C—F bond length = 139 pm

❓ Why Other Options Are Incorrect

• B. 178 → Approximate C	Cl bond length.
• C. 193 → Approximate C	Br bond length.
• D. 214 → Approximate C	I bond length.

🎯 Strategy

Used

Data Recall

🧠 Memory Trick

F is smallest, C—F = 139 pm

📚 Class XII Chemistry – Haloalkanes and Haloarenes – C—X Bond Length

14 Arrange the following substrate types in decreasing order of their general bond-cleavage reactivity in an Sₙ1 pathway (based on bond enthalpy and leaving group ability):

(A) R	Cl
(B) R	F
(C) R	I
(D) R	Br

A (C), (D), (A), (B)

B (A), (B), (C), (D)

C (B), (A), (D), (C)

D (D), (C), (A), (B)

💡 Short Explanation

SN1 involves C—X bond cleavage. Iodide is the best leaving group. Fluoride is the poorest leaving group.

✅ Correct Answer Explanation

In SN1 reactions, the first slow step is cleavage of the C—X bond. Leaving group ability follows: I⁻ > Br⁻ > Cl⁻ > F⁻ Bond enthalpy decreases in the order: C—F > C—Cl > C—Br > C—I So C—I cleaves most easily and C—F cleaves least easily. Therefore, decreasing SN1 reactivity is: R—I > R—Br > R—Cl > R—F Hence: (C), (D), (A), (B)

❓ Why Other Options Are Incorrect

B, C and D do not follow leaving-group ability or bond enthalpy trends.

🎯 Strategy

Used

Leaving Group Analysis

🧠 Memory Trick

I leaves first, F leaves last

📚 Class XII Chemistry – Haloalkanes and Haloarenes – SN1 Reactivity and Leaving Group Ability

15 Analytically, despite fluorine being highly electronegative, CH₃Cl possesses a slightly higher dipole moment (1.860 D) than CH₃F (1.847 D). According to general principles, this anomaly is because dipole moment depends on:

A Only the charge separation

B Only the bond length

C The product of charge separation and internuclear distance (bond length)

D The atomic mass of the central carbon

💡 Short Explanation

Dipole moment depends on charge and distance. C—Cl bond is longer than C—F. This makes CH₃Cl slightly more polar in dipole value.

✅ Correct Answer Explanation

Dipole moment is calculated as: μ = q × r where: q = magnitude of charge separation r = internuclear distance Although C—F has greater charge separation due to fluorine's high electronegativity, the C—F bond is very short. The C—Cl bond has slightly less charge separation but a longer bond length, giving CH₃Cl a slightly higher dipole moment than CH₃F.

❓ Why Other Options Are Incorrect

A. Only charge separation → Incomplete; distance also matters.
B. Only bond length → Incomplete; charge separation also matters.
D. Atomic mass of carbon → Not related to dipole moment.

🎯 Strategy

Used

Dimensional/Concept Analysis

🧠 Memory Trick

Dipole moment = charge × distance

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Dipole Moment of Alkyl Halides

16 Arrange the following halomethanes in decreasing order of their density (g/mL):
(A) CCl₄
(B) CHCl₃
(C) CH₂Cl₂

A (A), (B), (C)

B (C), (B), (A)

C (B), (A), (C)

D (A), (C), (B)

💡 Short Explanation

Density increases with the number of chlorine atoms. CCl₄ is the heaviest among the given compounds. CH₂Cl₂ has the lowest density among these.

✅ Correct Answer Explanation

Among halomethanes, increasing chlorine content generally increases molecular mass and density. CCl₄ has four chlorine atoms, CHCl₃ has three chlorine atoms, and CH₂Cl₂ has two chlorine atoms. Therefore, decreasing order of density is: CCl₄ > CHCl₃ > CH₂Cl₂ So the correct order is: (A), (B), (C)

❓ Why Other Options Are Incorrect

B → Gives reverse order.
C → Places CHCl₃ above CCl₄ incorrectly.
D → Places CH₂Cl₂ above CHCl₃ incorrectly.

🎯 Strategy

Used

Molecular Mass Trend Analysis

🧠 Memory Trick

More Cl atoms = Higher density

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Physical Properties of Polyhalogen Compounds

17 The severe inhibiting effect on the Sₙ2 mechanism caused by the presence of large, bulky alkyl groups near the carbon atom is known as:

A Inductive deactivation

B Resonance stabilization

C Steric hindrance

D Racemization

💡 Short Explanation

Bulky groups block nucleophile approach. SN2 requires backside attack. This blocking effect is steric hindrance.

✅ Correct Answer Explanation

In Sₙ2 reactions, the nucleophile must approach the carbon bearing the leaving group from the backside. Large bulky alkyl groups around or near this carbon obstruct the nucleophile's approach. This spatial blocking effect is called steric hindrance. This is why tertiary alkyl halides are very slow or unreactive in Sₙ2 reactions.

❓ Why Other Options Are Incorrect

A → Inductive effect deals with electron donation or withdrawal, not physical blocking.
B → Resonance stabilization affects intermediate stability, mainly in Sₙ1.
D → Racemization is an optical effect commonly linked with Sₙ1 reactions.

🎯 Strategy

Used

Mechanism-Based Concept Identification

🧠 Memory Trick

Steric = Space blocking

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Sₙ2 Mechanism

18 Match List-I (Substrate) with List-II (Preferred Mechanism Pathway):

List-I	List-II
1. Primary alkyl halide	a. Preferred Sₙ2 pathway due to low steric hindrance
2. Tertiary alkyl halide	b. Preferred Sₙ1 pathway due to high carbocation stability
3. Allylic halide	c. Highly reactive in Sₙ1 due to resonance stabilization (aliphatic C=C adjacent)
4. Benzylic halide	d. Highly reactive in Sₙ1 due to resonance stabilization (aromatic ring adjacent)

A 1-a, 2-b, 3-c, 4-d

B 1-c, 2-a, 3-b, 4-d

C 1-b, 2-c, 3-d, 4-a

D 1-a, 2-d, 3-c, 4-b

💡 Short Explanation

Primary halides favor Sₙ2. Tertiary halides favor Sₙ1. Allylic and benzylic halides form resonance-stabilized carbocations.

✅ Correct Answer Explanation

Primary alkyl halides have low steric hindrance, so they favor Sₙ2 reactions. Tertiary alkyl halides form stable tertiary carbocations, so they favor Sₙ1 reactions. Allylic halides form allylic carbocations stabilized by resonance with a C=C bond. Benzylic halides form benzylic carbocations stabilized by resonance with an aromatic ring. Correct matching: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

B, C and D incorrectly match substrate types with unsuitable reaction pathways.

🎯 Strategy

Used

Concept Mapping

🧠 Memory Trick

Primary = Sₙ2; Tertiary = Sₙ1; Allylic/Benzylic = Resonance

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Nucleophilic Substitution Mechanisms

19 Arrange the following benzylic bromides in decreasing order of reactivity towards Sₙ1 mechanism:
(A) C₆H₅CH₂Br
(B) C₆H₅CH(C₆H₅)Br
(C) C₆H₅CH(CH₃)Br
(D) C₆H₅C(CH₃)(C₆H₅)Br

A (D), (B), (C), (A)

B (A), (C), (B), (D)

C (D), (C), (B), (A)

D (B), (D), (C), (A)

💡 Short Explanation

Sₙ1 rate depends on carbocation stability. More resonance and alkyl stabilization increases reactivity. Compound D forms the most stable carbocation.

✅ Correct Answer Explanation

In Sₙ1 reactions, the rate-determining step is carbocation formation. Carbocation stability order: C₆H₅C(CH₃)(C₆H₅)⁺ > C₆H₅CH(C₆H₅)⁺ > C₆H₅CH(CH₃)⁺ > C₆H₅CH₂⁺ Reason: D is stabilized by two phenyl groups and one methyl group. B is stabilized by two phenyl groups. C is stabilized by one phenyl and one methyl group. A is stabilized by only one phenyl group. Therefore, decreasing Sₙ1 reactivity: (D), (B), (C), (A)

❓ Why Other Options Are Incorrect

B → Gives nearly reverse order.
C → Places C above B incorrectly.
D → Places B above D incorrectly.

🎯 Strategy

Used

Carbocation Stability Comparison

🧠 Memory Trick

More phenyl groups = more resonance = faster Sₙ1

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Benzylic Halides and Sₙ1 Reactivity

20 Haloarenes do not generally undergo Sₙ1 reactions under normal conditions because:

A The sp³ carbon strongly attracts nucleophiles.

B The phenyl cation formed by self-ionization is not stabilized by resonance.

C The bulky phenyl ring promotes nucleophilic attack.

D Halogens perfectly stabilize the empty p-orbital.

💡 Short Explanation

Sₙ1 requires carbocation formation. Phenyl cation is extremely unstable. Haloarenes resist Sₙ1 reactions.

✅ Correct Answer Explanation

For an Sₙ1 reaction, the substrate must ionize to form a carbocation. In haloarenes, cleavage of the C—X bond would generate a phenyl cation. This phenyl cation is highly unstable and is not stabilized by resonance in the usual way. Also, the aryl C—X bond has partial double bond character, making cleavage difficult. Thus haloarenes do not generally undergo Sₙ1 reactions under normal conditions.

❓ Why Other Options Are Incorrect

A → Haloarenes involve sp² carbon, not sp³ carbon.
C → Bulky phenyl ring does not promote nucleophilic attack.
D → Halogens do not perfectly stabilize the phenyl cation.

🎯 Strategy

Used

Intermediate Stability Analysis

🧠 Memory Trick

Haloarene Sₙ1 fails because phenyl cation is unstable

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Low Reactivity of Haloarenes

CUET UG Chemistry Booster Test -3 Nomenclature and Bond Nature

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