Class 12 Chemistry Test Paper Chapter 1 AND 6 | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026-27
CLASS: XII (CBSE) | TIME: 1 Hour | MAX. MARKS: 30 SUBJECT: CHEMISTRY (Unit 1: Solutions & Unit 6: Haloalkanes and Haloarenes)
GENERAL INSTRUCTIONS:
Read all questions carefully before answering. — Misreading the question and then blaming the paper will not increase your marks.
Show proper steps wherever required. — “Sir, answer toh yahi aana tha” is not an accepted mathematical method.
Write neatly and clearly. — If your handwriting requires a decoder machine, checking may become an adventure.
Manage your time wisely. — Spending 45 minutes on one question and calling the rest “optional” is not a strategy.
If you do not know the answer, you may cry silently. — Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.
SECTION A (Objective Type Questions)
[7 × 1 = 7 Marks]
Q1. An equimolar binary liquid solution of substance A and substance B exhibits a negative deviation from Raoult's law. Which of the following conditions is perfectly true regarding this mixture?
A. Delta_mix H > 0 and Delta_mix V > 0
B. Intermolecular A-B interactions are weaker than A-A and B-B molecular interactions
C. Delta_mix H < 0 and Delta_mix V < 0
D. The total vapour pressure of the actual solution becomes higher than predicted by Raoult's law
Q2. Which of the following aqueous solution mixtures will possess the highest boiling point elevation magnitude at 1 atm external atmospheric pressure?
A. 0.1 M Glucose solution B. 0.1 M BaCl2 solution C. 0.1 M NaCl solution D. 0.1 M Urea solution
Q3. When 2-Bromobutane undergoes an SN1 nucleophilic substitution reaction sequence using a strong aqueous KOH solution, the net optical nature of the isolated collection of product molecules is:
A. Dextrorotatory only due to 100% molecular inversion
B. Laevorotatory only due to 100% molecular retention
C. Optically inactive due to partial or complete racemisation
D. Highly unstable meso-isomer complex configuration
Q4. Arrange the following organic molecular compounds in the exact order of their increasing rate of reactivity towards an SN2 bimolecular substitution pathway:
A. 2-Bromobutane < 1-Bromobutane < 2-Bromo-2-methylpropane
B. 2-Bromo-2-methylpropane < 2-Bromobutane < 1-Bromobutane
C. 1-Bromobutane < 2-Bromobutane < 2-Bromo-2-methylpropane
D. 2-Bromo-2-methylpropane < 1-Bromobutane < 2-Bromobutane
Q5. An organic compound 'X' with molecular formula C3H7Cl on treatment with metallic sodium in dry ether medium yields 2,3-dimethylbutane. Identify the starting alkyl halide structural isomer 'X':
A. 1-Chloropropane B. 2-Chloropropane C. 1-Chlorobutane D. 2-Chlorobutane
For Question 6 and Question 7, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following:
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q6. Assertion (A): Elevation in boiling point is a colligative property that is directly dependent on the total number of solute particles present in the solution volume.
Reason (R): The addition of a non-volatile solute to a volatile liquid solvent always increases its net vapour pressure at a given temperature.
Q7. Assertion (A): Chlorobenzene undergoes nucleophilic aromatic substitution reactions under much more drastic reaction conditions compared to chloroethane.
Reason (R): In chlorobenzene, the C-Cl bond acquires a partial double bond character due to the resonance delocalisation of lone pairs of electrons of chlorine with the benzene ring.
SECTION B (Very Short Answer Questions)
[4 × 2 = 8 Marks]
Q8. State Henry's law regarding the solubility of a gas in a liquid medium. Why do aquatic animals feel much more comfortable in cold water environments rather than in warm water environments? Explain clearly using the law. [2]
Q9. Complete and balance the following organic chemical reaction transformations: [2] (i) CH3-CH2-CH=CH2 + HBr + (Benzoyl Peroxide) -> [A] (ii) CH3-CH2-Cl + NaI + (Dry Acetone / Delta) -> [B]
Q10. An antifreeze solution is prepared inside a laboratory by dissolving 31 g of pure ethylene glycol (C2H6O2) completely in 500 g of water. Calculate the total molality (m) of this solution mixture. (Molar mass of C2H6O2 = 62 g/mol) [2]
Q11. Explain why haloarenes are structurally and electronically less reactive towards nucleophilic substitution pathways compared to standard haloalkanes. State any two distinct chemical or structural reasons. [2]
SECTION C (Short Answer Questions)
[2 × 3 = 6 Marks]
Q12. Write the structural formulas and systematic IUPAC nomenclature for the primary organic product generated during the following chemical synthesis routes: [3] (i) Benzene diazonium chloride heated with warm aqueous KI solution. (ii) 2-Chlorobutane reacted with hot ethanolic KOH solution (Identify the major alkene product using Zaitsev's rule). (iii) Methyl chloride treated with dry silver fluoride (AgF).
Q13. A solution containing 1.5 g of a completely non-volatile, non-electrolyte organic solute dissolved in 50 g of pure benzene (C6H6) lowered the freezing point of benzene by 0.40 K. Calculate the precise molar mass of this unknown solute compound. (The cryoscopic freezing point depression constant Kf for benzene is given as 5.12 K kg/mol). [3]
SECTION D (Long Answer Question)
[1 × 5 = 5 Marks]
Q14. Carry out the following conceptual evaluations: (a) What is meant by an abnormal molecular mass of a solute? Explain how the van 't Hoff factor (i) can be dynamically used to determine the exact degree of association or dissociation of a chemical solute in a volatile solvent. [2] (b) Account for the fact that Chloroform (CHCl3) is stored in completely filled dark-amber coloured glass bottles away from light sources. Write the chemical equation involved. [1.5] (c) A solution of 1.00 g of non-volatile Aluminium chloride (AlCl3) in 50.0 g of water exhibits an elevation in boiling point equal to 0.224 K. Assuming 100% ionic dissociation, evaluate the true van 't Hoff factor 'i' for this salt in water. (Kb for water = 0.512 K kg/mol, Molar mass of AlCl3 = 133.5 g/mol) [1.5]
SECTION E (Case Study Based Question)
[1 × 4 = 4 Marks]
Q15. Read the following text passage carefully and answer the questions that follow:
Nucleophilic Substitution Dynamics and Chiral Stereochemistry: The mechanism of nucleophilic substitution reactions of alkyl halides depends heavily on the structural layout of the carbon center bound to the leaving group. Tertiary alkyl halides preferentially follow the SN1 (Substitution Nucleophilic Unimolecular) pathway, which proceeds in two discrete stages via a flat, stable carbocation intermediate. Nucleophilic attack can happen with equal probability from either side of this intermediate, resulting in a 50:50 mixture of enantiomers if the starting material was optically active. Conversely, primary alkyl halides react via the SN2 (Substitution Nucleophilic Bimolecular) pathway. This mechanism takes place in a single concerted step, involving a pentacoordinate transition state where the nucleophile attacks from the backside, exactly 180 degrees away from the departing halide ion. This results in complete spatial inversion of configuration, commonly referred to as Walden Inversion. Secondary alkyl halides occupy an intermediate position and can follow either route depending on solvent polarity, nucleophile strength, and steric factors.
(a) Out of the pair of enantiomers 1-bromobutane and 2-bromobutane, which specific compound will react faster via an SN2 mechanism, and why? [1] (
b) Why does the SN1 reaction pathway of an optically active chiral alkyl halide invariably lead to partial or complete racemisation of the final reaction mix? [1]
(c) Hydrolysis of an optically active haloalkane 'Y' produces a racemic alcohol mixture. If 'Y' is an isomer of C4H9Br, draw the structural formula of 'Y' and explain the stereochemical mechanism showing its transition state or intermediate formation. [2]
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026–27
CLASS: XII (CBSE) | SUBJECT: CHEMISTRY
Unit 1: Solutions & Unit 6: Haloalkanes and Haloarenes
SOLVED ANSWER KEY WITH MARKING SCHEME
Max. Marks: 30
SECTION A – OBJECTIVE TYPE QUESTIONS
[7 × 1 = 7 Marks]
Q1.
Negative deviation from Raoult’s law occurs when:
A–B intermolecular forces are stronger than A–A and B–B
Heat is released
Therefore:
Δmix H < 0
Δmix V < 0
Correct Answer: C. Δmix H < 0 and Δmix V < 0
Marking Scheme: 1 Mark
Q2.
Boiling point elevation:
ΔTb = iKb m
Higher van’t Hoff factor (i) gives higher boiling point elevation.
Glucose → i = 1
Urea → i = 1
NaCl → i = 2
BaCl2 → i = 3
Highest value:
BaCl2
Correct Answer: B. 0.1 M BaCl2 solution
Marking Scheme: 1 Mark
Q3.
2-Bromobutane undergoes SN1 reaction.
SN1 proceeds via planar carbocation.
Nucleophile attacks from both sides.
Hence racemisation occurs.
Product becomes optically inactive.
Correct Answer: C. Optically inactive due to partial or complete racemisation
Marking Scheme: 1 Mark
Q4.
SN2 reactivity:
Tertiary < Secondary < Primary
Hence:
2-Bromo-2-methylpropane < 2-Bromobutane < 1-Bromobutane
Correct Answer: B
Marking Scheme: 1 Mark
Q5.
Wurtz reaction:
2R–Cl + 2Na → R–R + 2NaCl
Product = 2,3-dimethylbutane
Starting alkyl group:
(CH3)2CH–
Hence alkyl halide:
CH3–CH(Cl)–CH3
= 2-Chloropropane
Correct Answer: B. 2-Chloropropane
Marking Scheme: 1 Mark
Q6.
Assertion: True
Elevation in boiling point depends on number of solute particles.
Reason: False
Non-volatile solute decreases vapour pressure.
Correct Answer: C
(A is true, R is false)
Marking Scheme: 1 Mark
Q7.
Assertion: True
Haloarenes require drastic conditions.
Reason: True
C–Cl bond has partial double bond character due to resonance.
This explains lower reactivity.
Correct Answer: A
(Both true and R is correct explanation)
Marking Scheme: 1 Mark
SECTION B – VERY SHORT ANSWER QUESTIONS
[4 × 2 = 8 Marks]
Q8. Henry’s Law [2]
Statement:
At constant temperature, solubility of a gas in liquid is directly proportional to partial pressure of gas above liquid.
Formula:
p = KH × x
Where:
p = partial pressure
KH = Henry’s constant
x = mole fraction of gas
Why aquatic animals prefer cold water?
Cold water dissolves more oxygen.
Hence aquatic animals get more dissolved oxygen for respiration.
Marking Scheme:
Henry’s law statement/formula = 1 Mark
Explanation = 1 Mark
Q9. Complete the Reactions [2]
(i)
CH3–CH2–CH=CH2 + HBr + Benzoyl Peroxide
Peroxide effect → Anti-Markovnikov addition
Product [A]:
CH3–CH2–CH2–CH2Br
= 1-Bromobutane
(ii)
CH3–CH2–Cl + NaI (Dry acetone)
Finkelstein reaction:
Product [B]:
CH3–CH2–I + NaCl
= Ethyl iodide
Marking Scheme:
Part (i) = 1 Mark
Part (ii) = 1 Mark
Q10. Molality of Antifreeze Solution [2]
Given:
Mass of ethylene glycol = 31 g
Molar mass = 62 g/mol
Mass of water = 500 g = 0.5 kg
Moles of solute:
n = 31/62
n = 0.5 mol
Molality:
m = n/kg solvent
m = 0.5/0.5
m = 1 mol kg^-1
Final Answer:
Molality = 1 m
Marking Scheme:
Mole calculation = 1 Mark
Final answer = 1 Mark
Q11. Haloarenes are Less Reactive [2]
Reason 1: Partial double bond character
Due to resonance, C–Cl bond becomes stronger and difficult to break.
Reason 2: sp² hybridised carbon
Carbon attached to Cl is sp² hybridised, making bond shorter and stronger.
Marking Scheme:
First reason = 1 Mark
Second reason = 1 Mark
SECTION C – SHORT ANSWER QUESTIONS
[2 × 3 = 6 Marks]
Q12. Product and IUPAC Name [3]
(i)
Benzene diazonium chloride + KI
Reaction:
C6H5N2+Cl^- + KI → C6H5I + N2
Product:
Iodobenzene
(ii)
2-Chlorobutane + alcoholic KOH
Elimination reaction
Major product by Saytzeff rule:
CH3–CH=CH–CH3
= But-2-ene
(iii)
CH3Cl + AgF
Swarts reaction
Product:
CH3F
= Fluoromethane
Marking Scheme:
Each correct product = 1 Mark
Q13. Molar Mass Calculation [3]
Given:
w₂ = 1.5 g
w₁ = 50 g = 0.05 kg
ΔTf = 0.40 K
Kf = 5.12 K kg/mol
Formula:
ΔTf = Kf × m
m = 0.40/5.12
m = 0.0781 mol/kg
Now:
m = n/w₁
0.0781 = n/0.05
n = 0.003905 mol
Molar mass:
M = mass/moles
M = 1.5/0.003905
M = 384.1 g/mol
Final Answer:
Molar mass = 384 g/mol (approx.)
Marking Scheme:
Formula setup = 1 Mark
Molality calculation = 1 Mark
Final molar mass = 1 Mark
SECTION D – LONG ANSWER QUESTION
[1 × 5 = 5 Marks]
Q14.
(a) Abnormal Molecular Mass & van’t Hoff Factor [2]
Abnormal molecular mass:
When experimentally determined molecular mass differs from actual molecular mass due to association or dissociation.
van’t Hoff factor:
i = Normal molar mass / Observed molar mass
For dissociation:
i > 1
For association:
i < 1
Degree of dissociation:
α = (i − 1)/(n − 1)
Degree of association:
α = (1 − i)/(n − 1)
Marking Scheme:
Definition = 1 Mark
van’t Hoff factor explanation = 1 Mark
(b) Storage of Chloroform [1.5]
Chloroform gets oxidised in presence of air and sunlight to poisonous gas phosgene (COCl2).
Reaction:
2CHCl3 + O2 → 2COCl2 + 2HCl
Therefore stored:
In dark coloured bottles
Completely filled
Away from sunlight
Marking Scheme:
Reason = 1 Mark
Equation = 0.5 Mark
(c) van’t Hoff Factor of AlCl3 [1.5]
Given:
w₂ = 1 g
M = 133.5 g/mol
w₁ = 50 g = 0.05 kg
ΔTb = 0.224 K
Kb = 0.512
Formula:
ΔTb = iKb m
Molality:
m = (1/133.5)/0.05
m = 0.1498
Now:
0.224 = i × 0.512 × 0.1498
i = 0.224/0.0767
i = 2.92
Final Answer:
van’t Hoff factor:
i = 2.92
Marking Scheme:
Molality = 0.5 Mark
Formula application = 0.5 Mark
Final answer = 0.5 Mark
SECTION E – CASE STUDY QUESTION
[1 × 4 = 4 Marks]
Q15.
(a) Faster SN2 Reaction [1]
1-Bromobutane reacts faster.
Reason:
Primary alkyl halide has less steric hindrance, favouring backside attack.
Marking Scheme:
Correct compound + reason = 1 Mark
(b) Why SN1 causes Racemisation? [1]
SN1 reaction forms a planar carbocation intermediate.
Nucleophile attacks from both front and back sides.
Hence racemic mixture forms.
Marking Scheme:
Correct explanation = 1 Mark
(c) Haloalkane ‘Y’ [2]
Since racemic alcohol formed:
Haloalkane must be optically active secondary alkyl bromide.
Compound:
CH3–CH(Br)–CH2–CH3
= 2-Bromobutane
Mechanism:
Step 1:
Formation of planar carbocation
CH3–CH⁺–CH2–CH3
Step 2:
OH⁻ attacks from both sides
Forms equal mixture of:
(R)-2-butanol and (S)-2-butanol
(racemic mixture)
Marking Scheme:
Correct structure = 1 Mark
Mechanism explanation = 1 Mark
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