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 | Subject: Chemistry
Chapter: Alcohols, Phenols and Ethers (Part 1)
Maximum Marks: 30 | Time Allowed: 1 Hour
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 (7 MARKS)
Q1. The correct IUPAC name of the compound
CH₃–CH(OH)–CH₂–CH(CH₃)–CH₃ is: [1]
A. 4-Methylpentan-2-ol
B. 2-Methylpentan-4-ol
C. Isohexanol
D. 1,1-Dimethylbutan-3-ol
Q2. Which of the following compounds will give phenol when treated with aqueous NaOH at high temperature and pressure (623 K, 300 atm) followed by acidification? [1]
A. Chlorobenzene
B. Benzene sulfonic acid
C. Aniline
D. Cumene
Q3. Arranging the following molecules in the increasing order of their acidic strength, the correct sequence is: [1]
I. Ethanol
II. Phenol
III. p-Nitrophenol
IV. p-Methylphenol
A. I < IV < II < III
B. I < II < IV < III
C. III < II < IV < I
D. IV < I < II < III
Q4. Hydroboration–Oxidation of propene (CH₃–CH=CH₂) yields which of the following principal organic products? [1]
A. Propan-1-ol
B. Propan-2-ol
C. Propane-1,2-diol
D. Propanal
Q5. When an alcohol reacts with an active metal like sodium (Na), the bond that undergoes complete cleavage is the: [1]
A. C–O bond
B. O–H bond
C. C–H bond
D. C–C bond
Direction for Q6–Q7:
For the following questions, two statements are given — one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer from the options below:
A. Both A and R are true, R is the correct explanation of A
B. Both A and R are true, R is not the correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q6. [1]
Assertion (A): Phenol is significantly more acidic than ethanol.
Reason (R): Phenoxide ion obtained after losing a proton is stabilized by resonance delocalization of the negative charge.
Q7. [1]
Assertion (A): Acid-catalyzed hydration of but-1-ene gives butan-1-ol as the primary product.
Reason (R): The addition of a water molecule across an unsymmetrical alkene follows Markovnikov's rule via a stable carbocation intermediate.
SECTION B (8 MARKS)
Q8. Write the IUPAC names of the following compounds: [2]
(i) CH₃–CH(CH₃)–CH(OH)–CH₂–CH₂–OH
(ii) 2-Bromophenol
Q9. Give chemical equations for the industrial preparation of phenol from cumene. Why is this commercial pathway highly preferred? [2]
Q10. Predict the major organic product and write the complete chemical equation for the following reactions: [2]
(i) Phenol with acetyl chloride (CH₃COCl) in the presence of pyridine.
(ii) tert-Butyl alcohol ((CH₃)₃COH) with metallic aluminum (Al).
Q11. Explain why primary alcohols react faster than tertiary alcohols when reacting with active metals, whereas the trend reverses during reactions involving C–O bond cleavage. [2]
SECTION C (6 MARKS)
Q12. Outline the detailed step-by-step reaction mechanism for the acid-catalyzed hydration of ethene (CH₂=CH₂) to produce ethanol (CH₃CH₂OH). [3]
Q13. How will you synthesize phenol using the following starting raw materials? Give chemical equations for each step: [3]
(i) Aniline (via benzene diazonium chloride formation)
(ii) Benzene (via sulfonation pathway)
SECTION D (5 MARKS)
Q14. Account for the following observations with appropriate chemical reasoning and structures: [5]
(i) The presence of an electron-withdrawing group (EWG) like –NO₂ increases the acidity of phenol, while an electron-donating group (EDG) like –OCH₃ decreases it.
(ii) Alcohols act as weak Bronsted bases as well as weak Bronsted acids. Explain both characters with suitable structural examples or equations.
(iii) Phenol reacts with metallic sodium to evolve hydrogen gas, but does it dissolve or react with aqueous NaHCO₃? State Yes/No with justification.
SECTION E (4 MARKS)
Q15. Case Study Based Question [4]
Organic reactions involving alcohols and phenols can fundamentally be classified based on whether cleavage occurs at the oxygen–hydrogen (O–H) bond or the carbon–oxygen (C–O) bond. Reactions that retain the basic skeleton and show cleavage at the O–H bond demonstrate the acidic behavior of these families.
Esterification is a classic example of O–H bond cleavage, where an alcohol or phenol reacts with carboxylic acids, acid chlorides, or acid anhydrides in an equilibrium pathway to generate fragrant esters. Pyridine is frequently introduced when acid chlorides are used to continuously neutralize hydrochloric acid waste and shift the equilibrium toward the product side.
(i) Name the organic compound formed when salicylic acid (2-hydroxybenzoic acid) undergoes acetylation with acetic anhydride in an acidic medium. State its common medicine name. [1]
(ii) Write the balanced chemical equation for the reaction between ethanol and acetic acid in the presence of concentrated H₂SO₄. [1]
(iii) Arrange Methanol, Water, and Phenol in decreasing order of acidic strength. Provide a concise rationale based on inductive effects or charge stability. [2]
— End of Paper —
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026–27
Class: XII | Subject: Chemistry
Chapter: Alcohols, Phenols and Ethers (Part 1)
ANSWER KEY WITH MARKING SCHEME
SECTION A (7 MARKS)
Q1. Correct Answer: A. 4-Methylpentan-2-ol [1]
Longest chain containing –OH group = Pentane
Numbering is done from the end nearest to –OH.
Structure:
CH₃–CH(OH)–CH₂–CH(CH₃)–CH₃
OH at carbon-2 and methyl group at carbon-4.
Marking Scheme:
Correct option selected = 1 mark
Q2. Correct Answer: A. Chlorobenzene [1]
Chlorobenzene on heating with aqueous NaOH at 623 K and 300 atm forms sodium phenoxide, which on acidification gives phenol.
Reaction:
C₆H₅Cl + NaOH → C₆H₅ONa → C₆H₅OH
Marking Scheme:
Correct option selected = 1 mark
Q3. Correct Answer: A. I < IV < II < III [1]
Acidity order:
Ethanol < p-Methylphenol < Phenol < p-Nitrophenol
Reason:
Alcohols are least acidic.
–CH₃ group donates electrons and decreases acidity.
–NO₂ group withdraws electrons and increases acidity.
Marking Scheme:
Correct option selected = 1 mark
Q4. Correct Answer: A. Propan-1-ol [1]
Hydroboration–oxidation follows anti-Markovnikov addition.
CH₃–CH=CH₂ → CH₃–CH₂–CH₂OH
Product = Propan-1-ol
Marking Scheme:
Correct option selected = 1 mark
Q5. Correct Answer: B. O–H bond [1]
Alcohol reacts with sodium by cleavage of the O–H bond.
2ROH + 2Na → 2RONa + H₂↑
Marking Scheme:
Correct option selected = 1 mark
Q6. Correct Answer: A [1]
Assertion (A): True
Phenol is more acidic than ethanol.
Reason (R): True
Phenoxide ion is stabilized by resonance, while ethoxide ion is not.
Therefore, R correctly explains A.
Marking Scheme:
Correct option selected = 1 mark
Q7. Correct Answer: D [1]
Assertion (A): False
Hydration of but-1-ene follows Markovnikov’s rule, producing butan-2-ol as the major product, not butan-1-ol.
Reason (R): True
Unsymmetrical alkene hydration proceeds through a more stable carbocation intermediate.
Marking Scheme:
Correct option selected = 1 mark
SECTION B (8 MARKS)
Q8. Write IUPAC names. [2]
(i) CH₃–CH(CH₃)–CH(OH)–CH₂–CH₂–OH
Longest chain = Pentane
OH groups get priority numbering.
Name = 4-Methylpentane-1,3-diol
(ii) 2-Bromophenol
Correct IUPAC name = 2-Bromophenol (already IUPAC)
Marking Scheme:
Part (i) correct = 1 mark
Part (ii) correct = 1 mark
Q9. Industrial preparation of phenol from cumene. [2]
Step 1: Oxidation of cumene
C₆H₅CH(CH₃)₂ + O₂ → Cumene hydroperoxide
Step 2: Acidic cleavage
Cumene hydroperoxide + H⁺ → C₆H₅OH + CH₃COCH₃
Products:
Phenol
Acetone
Reason for preference:
This method is preferred because valuable acetone is obtained as a useful by-product, making the process economical.
Marking Scheme:
Correct equations = 1 mark
Correct reason = 1 mark
Q10. Predict product and write equation. [2]
(i) Phenol + Acetyl chloride
C₆H₅OH + CH₃COCl → C₆H₅OCOCH₃ + HCl
Product = Phenyl ethanoate (Phenyl acetate)
Pyridine absorbs HCl.
(ii) tert-Butyl alcohol + Aluminum
2Al + 6(CH₃)₃COH → 2Al[OC(CH₃)₃]₃ + 3H₂↑
Product = Aluminum tert-butoxide
Marking Scheme:
Part (i) equation/product = 1 mark
Part (ii) equation/product = 1 mark
Q11. Explain reaction trend. [2]
With active metals:
Primary alcohols react faster because steric hindrance near oxygen is minimum, making O–H bond cleavage easier.
Trend:
1° > 2° > 3°
During C–O bond cleavage:
Tertiary alcohols react faster because tertiary carbocations are more stable.
Trend:
3° > 2° > 1°
Marking Scheme:
Correct explanation of O–H bond cleavage = 1 mark
Correct explanation of C–O bond cleavage = 1 mark
SECTION C (6 MARKS)
Q12. Acid-catalyzed hydration of ethene mechanism. [3]
Step 1: Protonation of alkene
CH₂=CH₂ + H⁺ → CH₃–CH₂⁺
(Ethyl carbocation formed)
Step 2: Nucleophilic attack by water
CH₃–CH₂⁺ + H₂O → CH₃CH₂OH₂⁺
Step 3: Deprotonation
CH₃CH₂OH₂⁺ → CH₃CH₂OH + H⁺
Product = Ethanol
Marking Scheme:
Step 1 = 1 mark
Step 2 = 1 mark
Step 3/final product = 1 mark
Q13. Synthesis of phenol. [3]
(i) From aniline
Step 1: Diazotization
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Step 2: Hydrolysis
C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl
(ii) From benzene (Sulfonation route)
Step 1: Sulfonation
C₆H₆ + H₂SO₄ → C₆H₅SO₃H
Step 2:
C₆H₅SO₃H + NaOH → C₆H₅ONa
Step 3: Acidification
C₆H₅ONa + HCl → C₆H₅OH + NaCl
Marking Scheme:
Aniline route = 1.5 marks
Benzene route = 1.5 marks
SECTION D (5 MARKS)
Q14. Explain observations. [5]
(i) Effect of EWG and EDG on acidity [2]
–NO₂ group withdraws electrons by −I effect, stabilizing phenoxide ion.
Hence acidity increases.
–OCH₃ donates electrons by +M effect, destabilizing phenoxide ion.
Hence acidity decreases.
(ii) Alcohols as Bronsted acid and base [2]
As Bronsted acid:
CH₃OH + Na → CH₃ONa + H₂
(Donates H⁺)
As Bronsted base:
CH₃OH + H⁺ → CH₃OH₂⁺
(Accepts H⁺)
(iii) Reaction with NaHCO₃ [1]
No
Phenol does not react with aqueous NaHCO₃ because it is a weaker acid than carbonic acid and cannot liberate CO₂.
Marking Scheme:
Part (i) explanation = 2 marks
Part (ii) explanation/equations = 2 marks
Part (iii) correct justification = 1 mark
SECTION E (4 MARKS)
Q15. Case Study Based Question [4]
(i) Product of salicylic acid acetylation [1]
Compound formed = 2-Acetoxybenzoic acid
Common medicine name = Aspirin
(ii) Ethanol + Acetic acid reaction [1]
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
(conc. H₂SO₄)
Product = Ethyl ethanoate
(iii) Decreasing order of acidity [2]
Phenol > Water > Methanol
Reason:
Phenol forms resonance-stabilized phenoxide ion.
Methanol shows +I effect, reducing acidity.
Water is more acidic than methanol because it lacks electron-donating alkyl group.
Marking Scheme:
Correct order = 1 mark
Correct rationale = 1 mark
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