CUET UG Chemistry Booster Test -3 Chemical Reactions and Mechanisms -

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CUET UG Chemistry Booster Test -3 Chemical Reactions and Mechanisms

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test -3 Chemical Reactions and Mechanisms

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QUESTION 1 OF 20

Match the substituted functional group with the reagent used in the substitution reaction of an alkyl halide:

List-I	List-II
1. Alkyl nitrite	a. NaOR'
2. Nitroalkane	b. AgNO₂
3. Isonitrile	c. AgCN
4. Ether	d. KNO₂

Answer Review

1 Match the substituted functional group with the reagent used in the substitution reaction of an alkyl halide:

List-I	List-II
1. Alkyl nitrite	a. NaOR'
2. Nitroalkane	b. AgNO₂
3. Isonitrile	c. AgCN
4. Ether	d. KNO₂

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

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

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

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

💡 Short Explanation

�� KNO₂ forms alkyl nitrites. �� AgNO₂ forms nitroalkanes. �� AgCN forms isonitriles.

✅ Correct Answer Explanation

Alkyl halides react with ambident nucleophiles to give different products depending on the reagent. KNO₂ gives alkyl nitrites, AgNO₂ gives nitroalkanes, AgCN gives isonitriles, and sodium alkoxide NaOR' gives ethers.

❓ Why Other Options Are Incorrect

�� Option B → Interchanges alkyl nitrite and nitroalkane reagents.
�� Option C → Incorrectly matches isonitrile with NaOR'.
�� Option D → Incorrectly matches alkyl nitrite with NaOR'.

🎯 Strategy

Used

Option Grouping

Application:

�� Match each reagent with its known substitution product.

Final Logic:

�� Correct matching is 1-d, 2-b, 3-c, 4-a.

🧠 Memory Trick

KNO₂ = Nitrite, AgNO₂ = Nitro.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Nucleophilic Substitution with Ambident Nucleophiles

2 In the competition between substitution and elimination for a secondary alkyl halide, which combination of factors dictates the dominant pathway?

A Exclusively the molecular mass of the halide.

B The strength, size of the base/nucleophile, and reaction conditions.

C The atmospheric pressure and the number of chiral centers.

D The dipole moment of the C-X bond only.

💡 Short Explanation

�� Secondary halides can undergo substitution or elimination. �� Bulky bases favour elimination. �� Small nucleophiles can favour substitution.

✅ Correct Answer Explanation

For secondary alkyl halides, both substitution and elimination are possible. The dominant pathway depends on the strength and size of the nucleophile/base, solvent, temperature, and reaction conditions. Bulky strong bases usually favour elimination, while suitable small nucleophiles may favour substitution.

❓ Why Other Options Are Incorrect

�� Option A → Molecular mass alone cannot decide the pathway.
�� Option C → Atmospheric pressure is not the key deciding factor.
�� Option D → Dipole moment alone is insufficient.

🎯 Strategy

Used

Elimination

Application:

�� Remove options using extreme words like "exclusively" and "only."

Final Logic:

�� Multiple reaction factors decide substitution vs elimination.

🧠 Memory Trick

Pathway depends on base, size, solvent, heat.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Substitution vs Elimination

3 Why does the reaction of haloalkanes with KCN form predominantly alkyl cyanides, while AgCN forms mostly isocyanides?

A KCN is covalent, providing free nitrogen.

B AgCN is ionic, completely dissociating into cyanide ions.

C KCN is ionic and the attack occurs mainly through the carbon atom as the C-C bond is more stable; AgCN is covalent, leaving nitrogen free to donate electron pair.

D Silver selectively binds to carbon, blocking it from reaction.

💡 Short Explanation

�� KCN gives free CN⁻ ions. �� CN⁻ attacks mainly through carbon. �� AgCN is covalent and attacks through nitrogen.

✅ Correct Answer Explanation

KCN is ionic and provides cyanide ions. In cyanide ion, carbon attack is favoured because it forms a more stable C-C bond, producing alkyl cyanides. AgCN is more covalent, so carbon is less available and nitrogen donates the electron pair, forming isocyanides.

❓ Why Other Options Are Incorrect

�� Option A → KCN is ionic, not covalent.
�� Option B → AgCN is covalent, not completely ionic.
�� Option D → This is not the standard explanation.

🎯 Strategy

Used

Conceptual/Tonal Matching

Application:

�� Use ionic/covalent nature of KCN and AgCN.

Final Logic:

�� KCN gives cyanide; AgCN gives isocyanide.

🧠 Memory Trick

KCN = Cyanide, AgCN = Isonitrile.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Ambident Nucleophiles

4 Which value corresponds to the bond length of the C-Br bond in bromomethane, establishing its relative leaving group ability?

A 139 pm

B 178 pm

C 193 pm

D 214 pm

💡 Short Explanation

�� C-Br bond length is 193 pm. �� It is longer than C-Cl. �� It is shorter than C-I.

✅ Correct Answer Explanation

Carbon-halogen bond length increases down the halogen group due to increasing halogen size. The C-Br bond length in bromomethane is 193 pm. This longer bond compared to C-Cl makes bromide a better leaving group than chloride.

❓ Why Other Options Are Incorrect

�� Option A → C-F bond length.
�� Option B → C-Cl bond length.
�� Option D → C-I bond length.

🎯 Strategy

Used

Data Recall

Application:

�� Recall standard C-X bond length values.

Final Logic:

�� C-Br bond length is 193 pm.

🧠 Memory Trick

Br bond length = 193 pm.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Carbon-Halogen Bond Length

5 Arrange the following compounds in decreasing order of reactivity towards SN2 displacement:
(A) 1-Bromopentane
(B) 2-Bromo-2-methylbutane
(C) 2-Bromopentane
(D) Bromomethane

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

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

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

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

💡 Short Explanation

�� SN2 favours least steric hindrance. �� Methyl halides are fastest. �� Tertiary halides are slowest.

✅ Correct Answer Explanation

SN2 reactions occur through backside attack. Therefore, steric hindrance strongly controls the reaction rate. The order of SN2 reactivity is: Methyl > Primary > Secondary > Tertiary So the decreasing order is: Bromomethane > 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane Therefore: (D), (A), (C), (B)

❓ Why Other Options Are Incorrect

�� Option A → Places tertiary halide too high.
�� Option B → Places primary halide above methyl halide.
�� Option D → Gives almost reverse order.

🎯 Strategy

Used

Steric Hindrance Analysis

Application:

�� Compare crowding around the carbon bearing bromine.

Final Logic:

�� Less crowded substrate reacts faster in SN2.

🧠 Memory Trick

SN2: Methyl > 1° > 2° > 3°.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: SN2 Reactivity

6 When (-)-2-bromooctane is reacted with sodium hydroxide, (+)-octan-2-ol is formed. What does the change in the optical rotation sign from (-) to (+) signify in this context?

A A complete racemisation has occurred.

B The reaction underwent SN1 mechanism.

C The nucleophile attached itself on the side opposite to the halogen, resulting in inversion.

D Retention of configuration has taken place.

💡 Short Explanation

�� SN2 proceeds through backside attack. �� Configuration gets inverted. �� Optical rotation changes sign.

✅ Correct Answer Explanation

When hydroxide ion attacks (-)-2-bromooctane through an SN2 mechanism, it attacks from the side opposite to the leaving bromide ion. This causes inversion of configuration (Walden inversion). The observed change from (-) to (+) indicates formation of the opposite enantiomer.

❓ Why Other Options Are Incorrect

�� Option A → Racemisation produces a mixture of enantiomers.
�� Option B → SN1 generally leads to racemisation.
�� Option D → Retention would not indicate backside attack.

🎯 Strategy

Used

Mechanism-Based Analysis

Application:

�� Connect SN2 with Walden inversion.

Final Logic:

�� Backside attack causes inversion.

🧠 Memory Trick

SN2 = Backside Attack = Inversion.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: SN2 Stereochemistry

7 Why do allylic and benzylic halides show exceptionally high reactivity towards SN1 reactions?

A The halogens are extremely large and polarizable.

B They experience high steric hindrance preventing SN2.

C The carbocations formed are strongly stabilised through resonance.

D They exclusively form carbanion intermediates.

💡 Short Explanation

�� SN1 involves carbocation formation. �� Allylic and benzylic carbocations are resonance stabilized. �� Greater stability increases reaction rate.

✅ Correct Answer Explanation

The rate-determining step of an SN1 reaction is carbocation formation. Allylic and benzylic carbocations are highly stable because the positive charge is delocalized through resonance. This stabilization greatly increases the rate of SN1 reactions.

❓ Why Other Options Are Incorrect

�� Option A → Halogen size is not the main reason.
�� Option B → Steric hindrance is not the key factor here.
�� Option D → SN1 forms carbocations, not carbanions.

🎯 Strategy

Used

Carbocation Stability Analysis

Application:

�� Identify the factor increasing SN1 rate.

Final Logic:

�� Resonance stabilization promotes SN1.

🧠 Memory Trick

Resonance-stabilized carbocation = Fast SN1.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: SN1 Reactivity

8 Which statements correctly define the outcomes of an SN1 reaction on an optically active alkyl halide?
It proceeds with complete inversion of configuration.
It results in a mixture of products with the same and opposite configurations.
The intermediate formed is a planar, achiral carbocation.
The process is accompanied by racemisation.

A 1, 2

B 3, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� SN1 forms a planar carbocation. �� Attack occurs from both sides. �� Racemisation results.

✅ Correct Answer Explanation

In an SN1 reaction, the leaving group departs first, producing a planar carbocation. Since the carbocation is planar, the nucleophile can attack from either side. This produces both retention and inversion products, leading to racemisation.

❓ Why Other Options Are Incorrect

�� Statement 1 is incorrect because complete inversion is characteristic of SN2.
�� Option A includes statement 1.
�� Option B omits statement 2.
�� Option D includes statement 1.

🎯 Strategy

Used

Elimination

Application:

�� Remove all options containing statement 1.

Final Logic:

�� SN1 gives racemisation through planar carbocation.

🧠 Memory Trick

SN1 = Planar Carbocation = Racemisation.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: SN1 Stereochemistry

9 Identify the dominant reaction type when a primary alkyl halide is treated with a small, unhindered strong nucleophile like an OH⁻ ion.

A SN1 Substitution

B SN2 Substitution

C Electrophilic Addition

D Zaitsev Elimination

💡 Short Explanation

�� Primary halides have low steric hindrance. �� OH⁻ is a strong nucleophile. �� SN2 pathway is favoured.

✅ Correct Answer Explanation

A primary alkyl halide provides easy access to the carbon atom for backside attack. Since hydroxide ion is a strong, unhindered nucleophile, the reaction proceeds predominantly through the SN2 mechanism.

❓ Why Other Options Are Incorrect

�� Option A → Primary carbocations are unstable, so SN1 is unlikely.
�� Option C → Electrophilic addition occurs in alkenes.
�� Option D → Elimination is less favoured with small nucleophiles on primary halides.

🎯 Strategy

Used

Mechanism Selection

Application:

�� Evaluate substrate type and nucleophile.

Final Logic:

�� Primary halide + OH⁻ = SN2.

🧠 Memory Trick

Primary + Strong Nucleophile = SN2.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: SN2 Mechanism

10 Arrange the following isomeric alkanes of molecular formula C₅H₁₂ in decreasing order of the number of monochloro structural isomers they yield on photochemical chlorination:
(A) Isopentane (2-methylbutane)
(B) n-Pentane (pentane)
(C) Neopentane (2,2-dimethylpropane)

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

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

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

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

💡 Short Explanation

�� More distinct hydrogen environments give more products. �� Isopentane has the highest number. �� Neopentane has the lowest.

✅ Correct Answer Explanation

Photochemical chlorination can replace hydrogen atoms located in different chemical environments. Isopentane (2-methylbutane) gives 4 monochloro structural isomers. n-Pentane gives 3 monochloro structural isomers. Neopentane gives 1 monochloro structural isomer because all twelve hydrogens are equivalent. Therefore: (A) > (B) > (C)

❓ Why Other Options Are Incorrect

�� Option B → Reverse order.
�� Option C → Places neopentane too high.
�� Option D → Places n-pentane above isopentane.

🎯 Strategy

Used

Symmetry Analysis

Application:

�� Count unique hydrogen environments.

Final Logic:

�� More unique hydrogens give more monochloro products.

🧠 Memory Trick

More symmetry = Fewer chlorination products.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Free Radical Halogenation of Alkanes

11 If during a transformation, no bond to the stereocentre is broken, yet the sign of optical rotation changes from (+) to (-), this process strictly represents:

A Inversion of configuration

B Retention of configuration

C Racemisation

D Elimination

💡 Short Explanation

�� No bond at the stereocentre is broken. �� Configuration remains unchanged. �� Optical rotation sign alone does not determine configuration.

✅ Correct Answer Explanation

Retention of configuration means the three-dimensional arrangement around the stereocentre remains unchanged throughout the reaction. The sign of optical rotation (+ or −) is an experimentally observed property and is not directly related to R/S configuration. Therefore, if no bond to the stereocentre is broken, the process represents retention of configuration.

❓ Why Other Options Are Incorrect

�� Option A → Inversion requires backside attack and change in configuration.
�� Option C → Racemisation produces a mixture of enantiomers.
�� Option D → Elimination does not describe stereochemical retention.

🎯 Strategy

Used

Conceptual Analysis

Application:

�� Distinguish optical rotation from configuration.

Final Logic:

�� Unchanged stereocentre means retention.

🧠 Memory Trick

Same arrangement = Retention.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Stereochemistry

12 Give the IUPAC name of the chiral molecule formed among the following examples.

A 2-Bromopropanoic acid

B Dichloromethane

C Propan-2-ol

D Tetrachloromethane

💡 Short Explanation

�� Chiral carbon must have four different groups. �� 2-Bromopropanoic acid satisfies this condition. �� Other compounds are achiral.

✅ Correct Answer Explanation

In 2-bromopropanoic acid, the carbon attached to bromine is bonded to four different groups: H Br CH₃ COOH Therefore, it is a chiral molecule. The other compounds possess symmetry or identical substituents and are achiral.

❓ Why Other Options Are Incorrect

�� Option B → Carbon has two identical Cl atoms.
�� Option C → Carbon has two identical CH₃ groups.
�� Option D → All four substituents are identical.

🎯 Strategy

Used

Chiral Centre Identification

Application:

�� Check for four different groups around carbon.

Final Logic:

�� Only 2-bromopropanoic acid is chiral.

🧠 Memory Trick

Four different groups = Chiral carbon.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Chirality

13 What fundamental structural property dictates that two enantiomers differ with respect to the rotation of plane polarised light?

A The presence of a double bond.

B The lack of molecular symmetry along with non-superimposability of mirror images.

C Differences in van der Waals forces.

D The difference in molecular weight between the two isomers.

💡 Short Explanation

�� Enantiomers are mirror images. �� They are non-superimposable. �� Lack of symmetry causes optical activity.

✅ Correct Answer Explanation

Enantiomers are non-superimposable mirror images that lack symmetry. This structural feature allows them to interact differently with plane-polarised light, causing rotation in opposite directions.

❓ Why Other Options Are Incorrect

�� Option A → Double bonds do not necessarily produce chirality.
�� Option C → Van der Waals forces are not responsible.
�� Option D → Enantiomers have identical molecular weights.

🎯 Strategy

Used

Definition Matching

Application:

�� Recall the structural basis of optical activity.

Final Logic:

�� Chirality arises from non-superimposable mirror images.

🧠 Memory Trick

No symmetry = Possible optical activity.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Enantiomers

14 Why is a racemic modification optically inactive?

A The chiral centers undergo continuous inversion.

B The molecules become achiral when mixed.

C The rotation due to one isomer cancels out the rotation due to the other isomer.

D The carbon-halogen bonds break spontaneously.

💡 Short Explanation

�� Racemic mixture contains equal enantiomers. �� Rotations are equal and opposite. �� Net rotation becomes zero.

✅ Correct Answer Explanation

A racemic mixture contains equal amounts of dextrorotatory and laevorotatory enantiomers. Since they rotate plane-polarised light by equal amounts in opposite directions, the overall rotation cancels out, making the mixture optically inactive.

❓ Why Other Options Are Incorrect

�� Option A → Continuous inversion does not occur.
�� Option B → Molecules remain chiral individually.
�� Option D → Bond cleavage is unrelated.

🎯 Strategy

Used

Conceptual Recall

Application:

�� Apply the definition of racemic mixture.

Final Logic:

�� Equal and opposite rotations cancel.

🧠 Memory Trick

Racemic = (+) + (−) = 0.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Racemic Mixtures

15 What is the IUPAC name of the major alkene formed from the dehydrohalogenation of 2,2,3-Trimethyl-3-bromopentane?

A 2,3,4-Trimethylpent-2-ene

B 3,4,4-Trimethylpent-2-ene

C 2,3,3-Trimethylpent-1-ene

D 2-Ethyl-3,3-dimethylbut-1-ene

💡 Short Explanation

�� Elimination follows Zaitsev's rule. �� More substituted alkene is favoured. �� Option B represents the major product.

✅ Correct Answer Explanation

During dehydrohalogenation, bromine leaves from carbon-3 and hydrogen is removed from the adjacent β-carbon in accordance with Zaitsev's rule. The most substituted and therefore most stable alkene is formed as the major product. The correct IUPAC name is: 3,4,4-Trimethylpent-2-ene

❓ Why Other Options Are Incorrect

�� Option A → Incorrect numbering after alkene formation.
�� Option C → Less substituted alkene.
�� Option D → Not the preferred IUPAC representation.

🎯 Strategy

Used

Reaction Product Prediction

Application:

�� Apply Zaitsev's rule and IUPAC nomenclature.

Final Logic:

�� More substituted alkene is the major product.

🧠 Memory Trick

Zaitsev → Most substituted alkene.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Dehydrohalogenation and Zaitsev Rule

16 The formulation of Zaitsev's rule directly addresses which chemical phenomenon?

A Regioselectivity in beta-elimination forming the most substituted alkene.

B Stereoselectivity in SN2 inversion.

C Racemisation in SN1.

D Stability of carbocations in Friedel-Crafts alkylation.

💡 Short Explanation

�� Zaitsev's rule applies to elimination reactions. �� More substituted alkene is favored. �� It predicts the major elimination product.

✅ Correct Answer Explanation

Zaitsev's rule states that during β-elimination, the major product is generally the more substituted alkene because it is thermodynamically more stable. Thus, the rule explains the regioselectivity observed in elimination reactions.

❓ Why Other Options Are Incorrect

�� Option B → Describes SN2 stereochemistry.
�� Option C → Describes SN1 stereochemistry.
�� Option D → Related to electrophilic substitution, not elimination.

🎯 Strategy

Used

Concept Identification

Application:

�� Recall the purpose of Zaitsev's rule.

Final Logic:

�� Zaitsev predicts the major alkene formed.

🧠 Memory Trick

Zaitsev = More substituted alkene.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Elimination Reactions

17 In the preparation of Grignard reagents, which metal serves as the central electropositive atom bonding with the carbon of the alkyl group?

A Sodium

B Copper

C Magnesium

D Potassium

💡 Short Explanation

�� Grignard reagents contain C–Mg bond. �� Magnesium is electropositive. �� General formula is R–Mg–X.

✅ Correct Answer Explanation

Grignard reagents are organomagnesium compounds prepared by reacting alkyl or aryl halides with magnesium metal in dry ether. The electropositive magnesium atom forms a polar bond with carbon. General formula: R–Mg–X

❓ Why Other Options Are Incorrect

�� Option A → Sodium is used in Wurtz reactions.
�� Option B → Copper is not used in Grignard reagent formation.
�� Option D → Potassium does not form standard Grignard reagents.

🎯 Strategy

Used

Formula Recall

Application:

�� Recall the composition of Grignard reagents.

Final Logic:

�� Grignard reagents contain magnesium.

🧠 Memory Trick

Grignard = Mg in the middle.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Grignard Reagents

18 In the Wurtz reaction, what dictates that the produced hydrocarbon contains double the number of carbon atoms?

A The solvent ether supplies the extra carbons.

B Two alkyl halide molecules couple together via reaction with sodium metal.

C The reaction breaks down a larger polymer.

D Alkyl halides rearrange into longer chains spontaneously.

💡 Short Explanation

�� Two alkyl groups join together. �� Sodium promotes coupling. �� Carbon chain length doubles.

✅ Correct Answer Explanation

The Wurtz reaction involves coupling of two molecules of an alkyl halide in the presence of sodium metal and dry ether. General reaction: 2R–X + 2Na → R–R + 2NaX Since two alkyl groups combine, the product contains approximately double the number of carbon atoms.

❓ Why Other Options Are Incorrect

�� Option A → Ether acts as solvent only.
�� Option C → No polymer breakdown occurs.
�� Option D → Chain extension occurs through coupling, not spontaneous rearrangement.

🎯 Strategy

Used

Reaction Mechanism Recall

Application:

�� Identify the source of additional carbon atoms.

Final Logic:

�� Two alkyl halides combine into one larger hydrocarbon.

🧠 Memory Trick

Wurtz = Double the carbon chain.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Wurtz Reaction

📄 Passage

Haloarenes exhibit nucleophilic resistance due to resonance, differences in hybridization, and phenyl cation instability. However, under extreme conditions or with activating groups, substitution can occur. For instance, chlorobenzene converts to phenol at 623 K and 300 atmospheres. The presence of an electron-withdrawing group like -NO₂ greatly increases this reactivity when located at the ortho- and para-positions because the carbanion intermediate is stabilized by resonance. Conversely, if the -NO₂ group is at the meta-position, no resonating structure places the negative charge on the carbon bearing the nitro group, thus offering no stabilization to the intermediate.

According to the passage, why does a nitro group at the meta-position fail to increase the reactivity of a haloarene toward nucleophiles?

A It acts as an electron-donating group at the meta position.

B The negative charge in the carbanion never localizes on the meta-carbon, so the nitro group cannot stabilize it.

C It causes immense steric hindrance.

D The meta-carbon immediately ejects the nitro group upon attack.

💡 Short Explanation

�� Nucleophilic substitution proceeds through a carbanion intermediate. �� Meta nitro group cannot participate in resonance stabilization. �� Reactivity enhancement is absent.

✅ Correct Answer Explanation

The nitro group enhances nucleophilic substitution only when it can stabilize the negatively charged intermediate through resonance. At the meta position, no resonance structure places the negative charge on the carbon attached to the nitro group. Therefore, stabilization does not occur and reactivity is not significantly increased.

❓ Why Other Options Are Incorrect

�� Option A → Nitro group is electron-withdrawing, not electron-donating.
�� Option C → Steric hindrance is not the main reason.
�� Option D → No such mechanism occurs.

🎯 Strategy

Used

Passage-Based Analysis

Application:

�� Apply resonance stabilization concepts.

Final Logic:

�� Meta nitro group cannot stabilize the intermediate.

🧠 Memory Trick

Nitro activates only at ortho and para.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Nucleophilic Substitution in Haloarenes

📄 Passage

What specific temperature and pressure are required to convert chlorobenzene to phenol without the presence of electron-withdrawing groups?

A 298 K and 1 atm

B 373 K and 100 atm

C 623 K and 300 atm

D 1000 K and 50 atm

💡 Short Explanation

�� Chlorobenzene is resistant to nucleophilic substitution. �� Harsh conditions are required. �� NCERT specifies 623 K and 300 atm.

✅ Correct Answer Explanation

Due to resonance stabilization and partial double bond character of the C–Cl bond, chlorobenzene is much less reactive than haloalkanes. Therefore, conversion of chlorobenzene to phenol requires severe reaction conditions of approximately 623 K temperature and 300 atmospheres pressure.

❓ Why Other Options Are Incorrect

�� Option A → Conditions are too mild.
�� Option B → Insufficient temperature and pressure.
�� Option D → Not the NCERT conditions.

🎯 Strategy

Used

Data Recall

Application:

�� Recall standard NCERT reaction conditions.

Final Logic:

�� Chlorobenzene → Phenol requires 623 K and 300 atm.

🧠 Memory Trick

Chlorobenzene → Phenol = 623 K, 300 atm.

📚 Class XII – Chemistry – Chapter: Haloalkanes and Haloarenes – Topic: Nucleophilic Substitution of Haloarenes

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