Class 12 Chemistry Test Paper Chapter 7 Alcohol Phenol and Ether | 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 7: Alcohols, Phenols and Ethers)
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 unknown alcohol is treated with Lucas reagent (conc. HCl + ZnCl2) at room temperature. The solution turns turbid immediately. When the same alcohol is oxidized with copper at 573 K, it forms a volatile alkene instead of a carbonyl compound. The alcohol is: A. Butan-1-ol B. Butan-2-ol C. 2-Methylpropan-2-ol D. Propan-1-ol
Q2. Which of the following structural arrangements represents the correct decreasing order of acid strength among the substituted phenols? A. p-Nitrophenol > p-Cresol > Phenol > o-Nitrophenol B. p-Nitrophenol > o-Nitrophenol > Phenol > p-Cresol C. p-Cresol > Phenol > o-Nitrophenol > p-Nitrophenol D. o-Nitrophenol > p-Nitrophenol > Phenol > p-Cresol
Q3. During the synthesis of an ether via Williamson synthesis, an alkyl halide is reacted with a sodium alkoxide. To synthesize tert-butyl methyl ether with the highest efficiency, the ideal combination of reactants should be: A. Sodium tert-butoxide and Methyl bromide B. Sodium methoxide and tert-Butyl bromide C. tert-Butyl alcohol and Methanol in conc. H2SO4 at 443 K D. Sodium methoxide and tert-Butyl alcohol
Q4. When Phenol is systematically treated with excess bromine water (Br2 / H2O), the primary product obtained is a white precipitate of: A. 2-Bromophenol B. 4-Bromophenol C. 2,4,6-Tribromophenol D. 2,4-Dibromophenol
Q5. What is the systematic IUPAC name for the organic compound given below?
CH3 - CH(OH) - CH2 - CH(CH3) - CH2 - CH3
A. 4-Methylhexan-2-ol
B. 3-Methylhexan-5-ol
C. 2-Methylhexan-4-ol
D. Isoheptanol
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): The boiling points of isomeric alcohols follow the sequence: primary > secondary > tertiary. Reason (R): With branching, the surface area of the molecule decreases, which significantly reduces the magnitude of van der Waals forces.
Q7. Assertion (A): The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle (109.5 degrees). Reason (R): This decrease in bond angle is due to the severe steric repulsion between the bulky unshared lone pairs of electrons on the oxygen atom.
SECTION B (Very Short Answer Questions)
[4 × 2 = 8 Marks]
Q8. Predict the structures and write down the systematic IUPAC names of the organic products generated in the following reaction processes: [2] (i) Propene reacts with water in the presence of dilute H2SO4 catalyst. (ii) Hydroboration-Oxidation of Propene using B2H6 followed by H2O2 / NaOH.
Q9. Give a clear chemical test along with balanced equations to distinguish between the following pairs of chemical compounds: [2] (i) Phenol and Ethanol (ii) Propan-1-ol and Propan-2-ol
Q10. Explain why Ethers possess significantly lower boiling points compared to their isomeric alcohols, despite having comparable molecular masses. Elaborate based on intermolecular forces. [2]
Q11. Account for the following observations: [2] (i) Phenol has a smaller dipole moment compared to Methanol. (ii) The carbon-oxygen bond length in Phenol (136 pm) is significantly shorter than that in Methanol (142 pm).
SECTION C (Short Answer Questions)
[2 × 3 = 6 Marks]
Q12. Write the complete, step-by-step acid-catalyzed mechanism for the dehydration of Ethanol to yield Ethene gas at 443 K. Show the formation of all intermediate species clearly. [3]
Q13. Write the chemical equations and conditions involved in the industrial or laboratory synthesis of the following transformations: [3] (i) Reimer-Tiemann Reaction starting from Phenol. (ii) Kolbe's Reaction to synthesize Salicylic acid from Phenol. (iii) Industrial preparation of Phenol from Cumene (isopropylbenzene).
SECTION D (Long Answer Question)
[1 × 5 = 5 Marks]
Q14. Carry out the following conceptual evaluations: An organic compound 'A' with molecular formula C7H8O is insoluble in aqueous NaHCO3 but dissolves readily in dilute NaOH solution. When 'A' is treated with a mixture of conc. HNO3 and conc. H2SO4, it forms a mixture of two structural isomers 'B' and 'C'. However, when 'A' is treated with catalytic hydrogen in the presence of nickel under extreme pressure, it forms an optically inactive cyclic alcohol 'D' (C6H12O) with elimination of a gaseous byproduct.
(a) Identify the structural formulas of organic compounds 'A', 'B', 'C', and 'D'. [2] (b) Write all the chemical equations for the reactions involving 'A' transforming into 'B', 'C', and 'D'. [2] (c) Compare the relative ease of nitration of compound 'A' with that of pure Benzene. Give one structural reason. [1]
SECTION E (Case Study Based Question)
[1 × 4 = 4 Marks]
Q15. Read the following text passage carefully and answer the questions that follow:
Nucleophilic Cleavage of Ethers and Electronic Shifts: The ether linkage (C-O-C) is quite stable under basic and mild acidic conditions, but can be cleaved using halogen acids (HX), preferably hydrogen iodide (HI) or hydrogen bromide (HBr) at high temperatures. The mechanism of cleavage of asymmetric ethers depends fundamentally on the nature of the alkyl groups attached to the oxygen atom. If both alkyl groups are primary or secondary, the reaction follows an SN2 pathway. The halide ion attacks the less sterically hindered alkyl group, yielding an alcohol from the larger group and an alkyl halide from the smaller fragment. However, if even one of the alkyl groups is tertiary, the reaction mechanism shifts entirely to an SN1 pathway. The protonated ether undergoes slow heterolysis to generate a highly stable tertiary carbocation intermediate, which is then rapidly attacked by the halide ion. In this case, the tertiary alkyl halide is the major product. When anisole (aryl alkyl ether) undergoes cleavage with HI, the reaction always forms phenol and methyl iodide due to the high stability of the sp2-hybridized aryl-oxygen bond which exhibits partial double bond character due to resonance.
(a) Predict the primary organic products formed when Ethyl methyl ether is heated with 1 equivalent of concentrated HI solution. [1]
(b) Why does the reaction of tert-Butyl methyl ether with HI exclusively yield tert-Butyl iodide and Methanol instead of Methyl iodide and tert-Butyl alcohol? [1]
(c) Write the complete, step-by-step chemical mechanism for the reaction of Anisole (Methoxybenzene) with concentrated HI at 373 K. Explain why Iodobenzene is never formed as a product in this cleavage. [2]
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026–27
CLASS: XII (CBSE) | SUBJECT: CHEMISTRY
Unit 7: Alcohols, Phenols and Ethers
SOLVED ANSWER KEY WITH MARKING SCHEME
Max. Marks: 30
SECTION A – OBJECTIVE TYPE QUESTIONS
[7 × 1 = 7 Marks]
Q1.
Lucas reagent gives immediate turbidity → tertiary alcohol.
On oxidation with Cu at 573 K, tertiary alcohol forms alkene by dehydration, not carbonyl compound.
Hence alcohol is:
2-Methylpropan-2-ol
Correct Answer: C. 2-Methylpropan-2-ol
Marking Scheme: 1 Mark
Q2.
Electron withdrawing group (–NO2) increases acidity.
Electron donating group (–CH3) decreases acidity.
p-Nitrophenol > o-Nitrophenol > Phenol > p-Cresol
(o-Nitrophenol slightly less acidic than para due to intramolecular H-bonding)
Correct Answer: B
Marking Scheme: 1 Mark
Q3.
Williamson synthesis works best with primary alkyl halide via SN2.
tert-Butyl halide undergoes elimination.
Hence best combination:
Sodium tert-butoxide + Methyl bromide
(CH3Br is primary)
Correct Answer: A
Marking Scheme: 1 Mark
Q4.
Phenol with excess bromine water forms white ppt. of:
2,4,6-Tribromophenol
Correct Answer: C
Marking Scheme: 1 Mark
Q5.
Longest chain = Hexane
OH gets lowest number.
CH3–CH(OH)–CH2–CH(CH3)–CH2–CH3
OH at carbon 2
Methyl at carbon 4
Name:
4-Methylhexan-2-ol
Correct Answer: A
Marking Scheme: 1 Mark
Q6.
Assertion: True
Boiling point:
Primary > Secondary > Tertiary
Reason: True
Branching reduces surface area and van der Waals force.
This explains lower boiling point.
Correct Answer: A
Marking Scheme: 1 Mark
Q7.
Assertion: True
Bond angle slightly less than 109.5°.
Reason: True
Repulsion due to lone pairs compresses bond angle.
Reason correctly explains assertion.
Correct Answer: A
Marking Scheme: 1 Mark
SECTION B – VERY SHORT ANSWER QUESTIONS
[4 × 2 = 8 Marks]
Q8. Predict Products [2]
(i) Hydration of Propene
CH3–CH=CH2 + H2O
(dil. H2SO4)
Markovnikov addition:
Product:
CH3–CHOH–CH3
IUPAC Name: Propan-2-ol
(ii) Hydroboration-Oxidation
CH3–CH=CH2
(B2H6 / H2O2, NaOH)
Anti-Markovnikov addition:
Product:
CH3–CH2–CH2OH
IUPAC Name: Propan-1-ol
Marking Scheme:
- First product + name = 1 Mark
- Second product + name = 1 Mark
Q9. Distinguish Between [2]
(i) Phenol and Ethanol
Neutral FeCl3 test
Phenol gives violet colour.
Ethanol shows no colour.
Reaction:
Phenol + FeCl3 → Violet complex
(ii) Propan-1-ol and Propan-2-ol
Lucas Test
Propan-2-ol (secondary alcohol)
→ Turbidity in 5 minutes.
Propan-1-ol (primary alcohol)
→ No turbidity at room temperature.
Marking Scheme:
- First distinction = 1 Mark
- Second distinction = 1 Mark
Q10. Why ethers have lower boiling point? [2]
Alcohols form intermolecular hydrogen bonding because of O–H bond.
Ethers cannot form intermolecular hydrogen bonding since they lack hydrogen attached to oxygen.
Hence weaker intermolecular force exists in ethers.
Therefore ethers have lower boiling point than isomeric alcohols.
Marking Scheme:
- Hydrogen bonding concept = 1 Mark
- Proper explanation = 1 Mark
Q11. Account for Observations [2]
(i) Phenol has smaller dipole moment than methanol
Due to resonance in phenol, lone pair on oxygen delocalises into benzene ring reducing polarity.
(ii) C–O bond in phenol shorter
In phenol, resonance gives partial double bond character to C–O bond.
Hence bond becomes shorter (136 pm).
Marking Scheme:
- Part (i) = 1 Mark
- Part (ii) = 1 Mark
SECTION C – SHORT ANSWER QUESTIONS
[2 × 3 = 6 Marks]
Q12. Mechanism of Dehydration of Ethanol [3]
At 443 K in presence of conc. H2SO4:
Step 1: Protonation of alcohol
CH3CH2OH + H+ → CH3CH2OH2+
(Ethyloxonium ion)
Step 2: Removal of water
CH3CH2OH2+ → CH3CH2+ + H2O
(Carbocation formation)
Step 3: Elimination of proton
CH3CH2+ → CH2=CH2 + H+
Final product:
Ethene gas
Overall reaction:
CH3CH2OH → CH2=CH2 + H2O
(Conc. H2SO4, 443 K)
Marking Scheme:
- Step 1 = 1 Mark
- Step 2 = 1 Mark
- Step 3 + final equation = 1 Mark
Q13. Important Reactions [3]
(i) Reimer-Tiemann Reaction
Phenol + CHCl3 + 3NaOH
→ o-Hydroxybenzaldehyde (Salicylaldehyde)
(ii) Kolbe’s Reaction
Sodium phenoxide + CO2
(under pressure)
→ Salicylic acid after acidification
(iii) Cumene Process
Cumene + O2
→ Cumene hydroperoxide
Acid hydrolysis:
→ Phenol + Acetone
Marking Scheme:
- Each reaction = 1 Mark
SECTION D – LONG ANSWER QUESTION
[1 × 5 = 5 Marks]
Q14.
(a) Identify A, B, C, D [2]
Given:
C7H8O
Dissolves in NaOH but not NaHCO3
→ Compound is Cresol (Methylphenol)
A = Phenol derivative: Cresol (CH3–C6H4–OH)
Nitration gives two isomers:
B = 2-Nitrocresol
C = 4-Nitrocresol
Hydrogenation:
Forms cyclohexanol derivative:
D = Methylcyclohexanol
(optically inactive)
Marking Scheme:
- A identification = 0.5 Mark
- B and C = 1 Mark
- D = 0.5 Mark
(b) Chemical Equations [2]
Nitration
Cresol + HNO3/H2SO4
→ o-Nitrocresol + p-Nitrocresol
Hydrogenation
Cresol + 3H2
(Ni, pressure)
→ Methylcyclohexanol
Marking Scheme:
- Nitration equation = 1 Mark
- Hydrogenation equation = 1 Mark
(c) Ease of nitration [1]
Compound A nitrates more easily than benzene.
Reason:
–OH group activates benzene ring by +R effect and increases electron density.
Marking Scheme:
- Correct comparison and reason = 1 Mark
SECTION E – CASE STUDY QUESTION
[1 × 4 = 4 Marks]
Q15.
(a) Ethyl methyl ether + HI [1]
CH3–O–CH2CH3 + HI
Less hindered methyl group attacked.
Products:
CH3I + CH3CH2OH
Marking Scheme:
Correct products = 1 Mark
(b) tert-Butyl methyl ether with HI [1]
tert-Butyl carbocation is highly stable.
Hence reaction follows SN1 pathway.
Products:
(CH3)3C–I + CH3OH
tert-Butyl iodide forms instead of methyl iodide.
Marking Scheme:
Correct reason = 1 Mark
(c) Anisole with HI at 373 K [2]
Step 1: Protonation
C6H5–O–CH3 + H+
→ Protonated ether
Step 2: Nucleophilic attack
I^- attacks methyl carbon by SN2.
C6H5–O–CH3 + HI
→ C6H5OH + CH3I
Products:
Phenol + Methyl iodide
Why iodobenzene not formed?
Aryl C–O bond has partial double bond character due to resonance and aryl carbon is sp² hybridised.
Hence cleavage of aryl–O bond does not occur.
Marking Scheme:
- Mechanism = 1 Mark
- Reason for no iodobenzene = 1 Mark
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