CUET UG Chemistry Booster Test -3 Introduction and Basic Concepts -

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CUET UG Chemistry Booster Test -3 Introduction and Basic Concepts

⏱️ 30 Minutes

❓ 20 Questions

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CUET UG Chemistry Booster Test -3 Introduction and Basic Concepts

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QUESTION 1 OF 20

Identify reaction type where alkyl iodides are prepared by the reaction of alkyl chlorides/bromides with NaI in dry acetone (halogen exchange):

Answer Review

1 Identify reaction type where alkyl iodides are prepared by the reaction of alkyl chlorides/bromides with NaI in dry acetone (halogen exchange):

A Sandmeyer

B Swarts

C Finkelstein

D Wurtz

💡 Short Explanation

Halogen exchange reaction. NaI in dry acetone is used. Alkyl chlorides/bromides are converted to alkyl iodides.

✅ Correct Answer Explanation

The Finkelstein reaction is a nucleophilic substitution reaction used to prepare alkyl iodides from alkyl chlorides or alkyl bromides. R-Cl + NaI dry acetoneR-I + NaCl R-Br + NaI dry acetone R-I + NaBr The reaction proceeds because NaCl and NaBr precipitate out in dry acetone, driving the equilibrium forward.

❓ Why Other Options Are Incorrect

A. Sandmeyer
Used for replacement of diazonium groups.
B. Swarts
Used for preparation of alkyl fluorides.
D. Wurtz
Couples alkyl halides using sodium metal.

🎯 Strategy

Used

Named Reaction Identification

🧠 Memory Trick

Finkelstein → Iodine Introduction

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Finkelstein Reaction

2 IUPAC name of the sp² hybridised haloarene commonly known as o-Chlorotoluene is:

A Benzyl chloride

B 1-Chloro-2-methylbenzene

C Chlorophenylmethane

D 2-Chloromethylbenzene

💡 Short Explanation

o- indicates ortho (1,2-position). Chloro and methyl groups are attached to adjacent carbons. IUPAC name uses numerical locants.

✅ Correct Answer Explanation

o-Chlorotoluene contains: One methyl group on benzene. One chlorine atom adjacent to it. The substituents occupy positions 1 and 2. Hence the IUPAC name is: 1-Chloro-2-methylbenzene

❓ Why Other Options Are Incorrect

A. Benzyl chloride
Structure is C₆H₅CH₂Cl.
C. Chlorophenylmethane
Not the correct IUPAC name.
D. 2-Chloromethylbenzene
Represents a different structure.

🎯 Strategy

Used

IUPAC Nomenclature

🧠 Memory Trick

Ortho = Adjacent = 1,2

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Nomenclature

3 Which of the following natural organic compounds relies strictly on a halogen for its essential biological function, and its deficiency leads to a thyroid-related disease?

A Chloramphenicol

B Thyroxine

C Chloroquine

D Dichloromethane

💡 Short Explanation

Thyroxine contains iodine. Iodine deficiency causes goiter. It is a natural thyroid hormone.

✅ Correct Answer Explanation

Thyroxine is an iodine-containing hormone produced by the thyroid gland. Functions: Regulates metabolism. Supports growth and development. Deficiency of iodine reduces thyroxine production and causes: Goiter

❓ Why Other Options Are Incorrect

A. Chloramphenicol
Antibiotic.
C. Chloroquine
Antimalarial drug.
D. Dichloromethane
Industrial solvent.

🎯 Strategy

Used

Biological Application

🧠 Memory Trick

Thyroxine = Iodine = Thyroid

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Occurrence

4 Certain synthetic fluorinated compounds are critically investigated for use as blood substitutes in surgery primarily because they:

A Act as highly reactive nucleophiles in blood plasma

B Easily oxidize to provide oxygen

C Possess extreme chemical stability, inertness, and non-toxicity

D Irreversibly bind to red blood cells

💡 Short Explanation

Fully fluorinated compounds are chemically inert. They are non-toxic. They can transport dissolved oxygen.

✅ Correct Answer Explanation

Perfluorinated compounds possess: High chemical stability Non-toxicity Biological inertness Because of these properties, they are investigated as potential blood substitutes capable of carrying oxygen.

❓ Why Other Options Are Incorrect

A. They are not reactive nucleophiles.
B. They do not generate oxygen through oxidation.
D. They do not irreversibly bind RBCs.

🎯 Strategy

Used

Application-Based Recall

🧠 Memory Trick

Fluorinated = Stable + Inert

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Applications

5 Analytical properties of industrial solvents:
(A) Methylene chloride directly harms the human central nervous system.
(B) Chloroform is stored in completely filled dark bottles to prevent oxidation.
(C) Depletion of the ozone layer by CCl₄ increases human exposure to UV rays.
(D) Freons are easily liquefiable, toxic, and highly corrosive gases.

A A, B, C

B B, C, D

C A, C, D

D A, B, D

💡 Short Explanation

Methylene chloride affects the CNS. Chloroform is protected from oxidation. CCl₄ contributes to ozone depletion. Freons are non-toxic and non-corrosive.

✅ Correct Answer Explanation

A ✔ Methylene chloride exposure can damage the CNS. B ✔ Chloroform is stored in dark bottles to prevent formation of toxic phosgene. C ✔ Ozone depletion increases UV exposure. D ✘ Freons are generally non-toxic and non-corrosive.

❓ Why Other Options Are Incorrect

Any option including statement D becomes incorrect.

🎯 Strategy

Used

Statement Analysis

🧠 Memory Trick

Freons = Stable, Non-Corrosive

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Polyhalogen Compounds

6 Match List-I (Medical/Clinical Compound) with List-II (Specific Property/Use)

List-I	List-II
1. Iodoform	a. Broad-spectrum antibiotic for typhoid fever
2. Chloroform	b. Currently used general anaesthetic
3. Halothane	c. Antiseptic properties due to liberation of free iodine
4. Chloramphenicol	d. Depresses the central nervous system but replaced due to phosgene toxicity

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

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

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

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

💡 Short Explanation

Iodoform → antiseptic. Chloroform → CNS depressant. Halothane → anaesthetic. Chloramphenicol → antibiotic.

✅ Correct Answer Explanation

Compound — Match Iodoform — Antiseptic due to free iodine Chloroform — CNS depressant, replaced because of toxicity Halothane — General anaesthetic Chloramphenicol — Antibiotic used against typhoid Hence: 1-c, 2-d, 3-b, 4-a

❓ Why Other Options Are Incorrect

They incorrectly assign therapeutic uses.

🎯 Strategy

Used

Match the Following

🧠 Memory Trick

Iodoform–Iodine, Halothane–Hospital

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Applications

7 The extreme persistence of DDT in the environment and food chains is governed conceptually by which analytical factor?

A Rapid breakdown by diverse soil bacteria

B High solubility in aqueous environmental systems

C High chemical stability and prominent fat solubility resisting metabolism

D Volatility that allows it to escape into the stratosphere easily

💡 Short Explanation

DDT is highly stable. It accumulates in fatty tissues. It undergoes biomagnification.

✅ Correct Answer Explanation

DDT: Is not rapidly metabolized. Is fat soluble. Persists in ecosystems. This causes bioaccumulation and biomagnification through food chains.

❓ Why Other Options Are Incorrect

A. Opposite of the actual behavior.
B. DDT is not highly water soluble.
D. Not the major cause of persistence.

🎯 Strategy

Used

Environmental Chemistry

🧠 Memory Trick

DDT = Difficult to Decompose

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Environmental Effects

8 The initiation of radical chain reactions in the stratosphere by Freon ultimately destroys the natural balance of which atmospheric molecule?

A Nitrogen

B Oxygen

C Ozone

D Carbon dioxide

💡 Short Explanation

Freons release chlorine radicals. Chlorine destroys ozone. Ozone depletion increases UV exposure.

✅ Correct Answer Explanation

In the stratosphere: CF2Cl2→Cl∙ Chlorine radicals catalytically destroy ozone: Cl∙+O3→ClO∙+O2 This reduces ozone concentration.

❓ Why Other Options Are Incorrect

Nitrogen, oxygen, and carbon dioxide are not the primary targets.

🎯 Strategy

Used

Environmental Impact

🧠 Memory Trick

Freon → Chlorine → Ozone Loss

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Ozone Depletion

9 Based on the passage, how does the sp² hybridised carbon in organic haloarenes influence the C—X bond compared to aliphatic haloalkanes?

A It makes the bond longer and substantially weaker

B It donates electrons to the halogen, making it reactive

C It is more electronegative, holding the electron pair more tightly

D It restricts the conjugation of π-electrons

💡 Short Explanation

sp² carbon has greater s-character. Greater s-character increases electronegativity. The C–X bond becomes stronger.

✅ Correct Answer Explanation

The carbon attached to halogen in haloarenes is sp² hybridised. Because sp² carbon contains greater s-character than sp³ carbon: It is more electronegative. It holds bonding electrons more tightly. The C–X bond becomes stronger.

❓ Why Other Options Are Incorrect

The passage directly supports option C.

🎯 Strategy

Used

Hybridisation Concept

🧠 Memory Trick

More s-character = More Electronegativity

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Haloarene Reactivity

10 According to the passage, the exceptional chemical stability of haloarenes towards nucleophilic substitution is primarily due to:

A The presence of a highly polar, purely ionic bond

B The partial double bond character of the C—X bond arising from resonance

C Extreme solubility of the aromatic ring in water

D Large atomic radius of the sp² hybridised carbon atom

💡 Short Explanation

Lone pair of halogen participates in resonance. Partial double bond character develops. Bond cleavage becomes difficult.

✅ Correct Answer Explanation

In haloarenes: Halogen lone pairs conjugate with the benzene π-system. Resonance gives the C–X bond partial double bond character. The bond becomes shorter and stronger. Therefore haloarenes are much less reactive towards nucleophilic substitution than haloalkanes.

❓ Why Other Options Are Incorrect

A.
C–X bond is covalent, not ionic.
C.
Solubility is unrelated.
D.
Not responsible for the observed stability.

🎯 Strategy

Used

Resonance Analysis

🧠 Memory Trick

Resonance Strengthens the C–X Bond

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Haloarenes vs Haloalkanes

11 The unit of bond enthalpy for the cleavage of the C—X bond is:

A V m⁻¹

B kJ mol⁻¹

C Debye

D S cm² mol⁻¹

💡 Short Explanation

Bond enthalpy measures the energy required to break a chemical bond. It is expressed per mole of bonds broken. The standard unit is kJ mol⁻¹.

✅ Correct Answer Explanation

Bond enthalpy is defined as the amount of energy required to break one mole of a particular bond in the gaseous state. For carbon-halogen bonds (C–F, C–Cl, C–Br, C–I), bond strengths are compared using bond enthalpy values expressed in: kJ mol⁻¹ (kilojoules per mole) This unit allows comparison of bond strengths and prediction of reactivity trends among haloalkanes and haloarenes.

❓ Why Other Options Are Incorrect

A. V m⁻¹
Unit of electric field strength.
C. Debye
Unit of dipole moment.
D. S cm² mol⁻¹
Related to molar conductivity.

🎯 Strategy

Used

Unit Identification

🧠 Memory Trick

Bond Energy = kJ per Mole

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Nature of C–X Bond

12 Arrange the following primary alkyl halides in decreasing order of reactivity towards SN2 nucleophilic displacement (consider steric hindrance):
(A) 1-Bromobutane
(B) 1-Bromo-2,2-dimethylpropane
(C) 1-Bromo-2-methylbutane
(D) 1-Bromo-3-methylbutane

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

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

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

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

💡 Short Explanation

SN2 reactions are highly sensitive to steric hindrance. Less branching near the reaction center increases reactivity. Neopentyl bromide is the least reactive.

✅ Correct Answer Explanation

SN2 reactions occur through backside attack of the nucleophile. Greater steric hindrance around the carbon attached to the leaving group reduces SN2 reactivity. Order of steric hindrance: 1-Bromobutane → least hindered 1-Bromo-3-methylbutane 1-Bromo-2-methylbutane 1-Bromo-2,2-dimethylpropane (Neopentyl bromide) → highly hindered Therefore: 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane or (A) > (D) > (C) > (B)

❓ Why Other Options Are Incorrect

They do not correctly account for increasing steric hindrance near the reaction center.

🎯 Strategy

Used

Steric Hindrance Analysis

🧠 Memory Trick

Less Branching = Faster SN2

📚 Class XII Chemistry – Haloalkanes and Haloarenes – SN2 Mechanism

13 During the SN1 mechanism of an sp³ hybridised alkyl halide, the slow cleavage of the C—X bond forms a carbocation intermediate which is:

A sp³ hybridised and tetrahedral

B sp² hybridised and planar

C sp hybridised and linear

D Unhybridised and asymmetric

💡 Short Explanation

SN1 proceeds through carbocation formation. Carbocations are electron-deficient species. They possess trigonal planar geometry.

✅ Correct Answer Explanation

In the SN1 mechanism: R-X R+ + X- The rate-determining step forms a carbocation. Characteristics of carbocations: sp² hybridised Trigonal planar geometry Empty p orbital perpendicular to the plane This planar structure allows nucleophilic attack from either side.

❓ Why Other Options Are Incorrect

A.
Represents alkane geometry.
C.
Carbocations are not linear.
D.
Carbocations possess definite hybridisation.

🎯 Strategy

Used

Mechanism-Based Concept

🧠 Memory Trick

Carbocation = sp² = Planar

📚 Class XII Chemistry – Haloalkanes and Haloarenes – SN1 Mechanism

14 Analytically, why do sp² bonded haloarenes undergo electrophilic substitution reactions more slowly and require more drastic conditions compared to benzene?

A Halogens strongly activate the ring via an intensive +I effect

B Halogen atoms withdraw electrons via a –I effect, causing net ring deactivation

C The sp² carbon causes extreme steric hindrance preventing attack

D Resonance perfectly negates any electrophilic approach at ortho/para positions

💡 Short Explanation

Halogens show a strong –I effect. Electron density of the ring decreases. Electrophilic substitution becomes slower than in benzene.

✅ Correct Answer Explanation

Halogens possess two opposing effects: –I effect (electron withdrawing) → strong +R effect (electron donation by resonance) → weaker The overall effect is ring deactivation. As a result: Haloarenes react more slowly than benzene. Stronger reaction conditions are required. However, resonance directs incoming electrophiles mainly to ortho and para positions.

❓ Why Other Options Are Incorrect

A.
Halogens exhibit –I, not +I effect.
C.
Steric hindrance is not the primary reason.
D.
Resonance does not completely prevent substitution.

🎯 Strategy

Used

Electronic Effects Analysis

🧠 Memory Trick

Haloarenes: o/p Directing but Deactivating

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

15 Grignard reagents are fundamental in organic synthesis. Why is it absolutely imperative that their synthesis via alkyl halides occurs in dry ether?

A They decompose into toxic halogen gases upon exposure to moisture

B They are highly reactive and instantly react with proton sources like water to form hydrocarbons

C Ether acts as the primary leaving group

D They form explosive inorganic peroxides when solvated by water

💡 Short Explanation

Grignard reagents are destroyed by moisture. Water acts as a proton source. Dry ether prevents decomposition.

✅ Correct Answer Explanation

Grignard reagents have the general formula: [ R-MgX ] They are extremely strong nucleophiles and bases. Even traces of water react immediately: [ R-MgX + H_2O \rightarrow RH + Mg(OH)X ] Thus: Grignard reagent is destroyed. Desired reaction cannot occur. Therefore, preparation and storage must be carried out in absolutely dry ether.

❓ Why Other Options Are Incorrect

A.
Toxic halogen gases are not produced.
C.
Ether is the solvent, not a leaving group.
D.
This is not the primary reason for dry conditions.

🎯 Strategy

Used

Reaction Mechanism Application

🧠 Memory Trick

Grignard + Water = Dead Grignard

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Organometallic Compounds (Grignard Reagents)

16 While chloroform is deeply important historically as a solvent and an anaesthetic, its medical use was abandoned fundamentally because:

A It exhibits complete insolubility in physiological fats

B It undergoes spontaneous combustion at room temperature

C Chronic exposure causes liver damage where it is metabolised to toxic phosgene

D It reacts with oxygen to form chemically inert, suffocating gases

💡 Short Explanation

Chloroform was earlier used as an anaesthetic. It can damage the liver. It may form toxic phosgene.

✅ Correct Answer Explanation

Chloroform was historically used as an anaesthetic, but its medical use was discontinued because of serious toxicity. On exposure to air and light, chloroform can form phosgene, a highly poisonous compound. In the body, chloroform can also cause liver damage due to toxic metabolic products. Therefore, its anaesthetic use was abandoned due to safety concerns.

❓ Why Other Options Are Incorrect

A.
Chloroform is not rejected due to complete insolubility in fats.
B.
It does not spontaneously combust at room temperature.
D.
The main toxic product is phosgene, not chemically inert gas.

🎯 Strategy

Used

Application-Based Elimination

🧠 Memory Trick

Chloroform → Phosgene Risk

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Polyhalogen Compounds

17 Chloramphenicol is an asymmetric organic molecule. A mixture containing equal proportions of its dextro (+) and laevo (-) enantiomers possesses zero optical rotation and is strictly termed a:

A Meso compound

B Racemic mixture

C Diastereomeric mixture

D Structural isomer mix

💡 Short Explanation

Equal amounts of (+) and (−) enantiomers are present. Their optical rotations cancel each other. The mixture is optically inactive.

✅ Correct Answer Explanation

A racemic mixture contains equal proportions of two enantiomers: Dextro-rotatory (+) Laevo-rotatory (−) Since the optical rotations are equal and opposite, the net optical rotation becomes zero. Therefore, such a mixture is called a racemic mixture or racemate.

❓ Why Other Options Are Incorrect

A. Meso compound
A single molecule with internal symmetry, not a mixture.
C. Diastereomeric mixture
Contains stereoisomers that are not mirror images.
D. Structural isomer mix
Involves different connectivity, not optical isomers.

🎯 Strategy

Used

Definition-Based Identification

🧠 Memory Trick

Racemic = Right + Left Equal

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Optical Activity and Stereochemistry

18 Chloroquine is a vital synthetic compound. Its development demonstrates how synthetic chemistry utilizing halogens produces:

A Clinically active artificial medicines for diseases like malaria

B Environmentally harmful ozone-depleting freons exclusively

C Inorganic complexes used heavily in metallurgy

D Reactive Grignard reagents acting as anaesthetics

💡 Short Explanation

Chloroquine is synthetic. It contains halogen. It is used as an antimalarial medicine.

✅ Correct Answer Explanation

Chloroquine is a synthetic halogen-containing drug used mainly in the treatment of malaria. Its development shows that organohalogen chemistry is not limited to industrial solvents or pollutants; it also contributes to important medicines.

❓ Why Other Options Are Incorrect

B.
Chloroquine is not a freon.
C.
It is not an inorganic complex used in metallurgy.
D.
Grignard reagents are not anaesthetics.

🎯 Strategy

Used

Application-Based Reasoning

🧠 Memory Trick

Chloroquine → Malaria Medicine

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Medicinal Applications of Organohalogen Compounds

19 When analyzing elimination reactions (β-elimination) within the scope of haloalkanes, Zaitsev's rule dictates that the preferred product is:

A The least sterically hindered alkyne

B The alkene possessing the greater number of alkyl groups attached to the doubly bonded carbons

C The substitution product instead of the elimination product

D Exclusively a vinylic halide

💡 Short Explanation

Zaitsev's rule predicts the major alkene. More substituted alkene is more stable. β-elimination gives the more substituted double bond.

✅ Correct Answer Explanation

In β-elimination of haloalkanes, hydrogen is removed from a β-carbon and halogen is removed from the α-carbon. According to Zaitsev's rule, the major alkene formed is the more substituted alkene, meaning the alkene in which the doubly bonded carbon atoms are attached to a greater number of alkyl groups. This product is favored because it is more stable due to hyperconjugation and alkyl group electron donation.

❓ Why Other Options Are Incorrect

A.
Zaitsev's rule applies to alkene formation, not alkynes.
C.
It predicts elimination product preference, not substitution.
D.
Vinylic halides are not the exclusive products.

🎯 Strategy

Used

Rule Application

🧠 Memory Trick

Zaitsev = More Substituted Alkene

📚 Class XII Chemistry – Haloalkanes and Haloarenes – Elimination Reactions

20 A core analytical objective is correlating molecular structure to kinetic mechanisms. Which substrate inherently demonstrates the highest reactivity rate via an SN1 mechanism due to intermediate stability?

A Methyl chloride

B 1-Chlorobutane

C 2-Chlorobutane

D tert-Butyl chloride

💡 Short Explanation

SN1 depends on carbocation stability. Tertiary carbocations are most stable. tert-Butyl chloride forms a tertiary carbocation.

✅ Correct Answer Explanation

SN1 reactions proceed through carbocation formation in the slow rate-determining step. Reactivity order: 3° > 2° > 1° > methyl tert-Butyl chloride forms a stable tertiary carbocation: (CH3)3C+ This high carbocation stability makes tert-butyl chloride the most reactive substrate in SN1 reactions.

❓ Why Other Options Are Incorrect

A. Methyl chloride
Cannot form a stable carbocation.
B. 1-Chlorobutane
Forms unstable primary carbocation.
C. 2-Chlorobutane
Forms secondary carbocation, less stable than tertiary.

🎯 Strategy

Used

Carbocation Stability Analysis

🧠 Memory Trick

SN1 Loves Stable Carbocations

📚 Class XII Chemistry – Haloalkanes and Haloarenes – SN1 Mechanism

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