Alcohols Phenols and Ethers — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Alcohols Phenols and Ethers" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Alcohols Phenols and Ethers" — 8 important questions with detailed answers for CBSE…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Chemistry — Alcohols Phenols and Ethers. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Alcohols Phenols and Ethers — All 8 Questions Solved
Q1. What is the difference between an alcohol and a phenol? Give structural formulas and one example of each.
Alcohol: A compound with an -OH group bonded to an sp3 hybridized carbon atom (aliphatic carbon).
Example: CH3CH2OH (ethanol, alcohol)
Structural formula: R-OH where R is an alkyl group
Phenol: A compound with an -OH group bonded directly to an aromatic benzene ring (sp2 hybridized carbon).
Example: C6H5OH (phenol)
Structural formula: Ar-OH where Ar is an aromatic group
Key differences:
1. Bonding: In alcohols, OH is on sp3 carbon; in phenols, OH is on sp2 (aromatic) carbon
2. Acidity: Phenols are weakly acidic (Ka ~ 10^-10), alcohols are neutral or very weakly acidic (alcohols do not show acidic properties in aqueous solution)
3. Chemical reactions: Phenols undergo aromatic substitution; alcohols undergo nucleophilic substitution and elimination
4. Physical properties: Phenols are crystalline solids at room temperature; many alcohols are liquids
5. Esterification: Both form esters, but phenol esters are more readily hydrolyzed
6. Color with FeCl3: Phenols give colored complexes (usually violet or blue); alcohols do not
Q2. Explain why phenols are more acidic than alcohols. Write the equation for the ionization of phenol in water.
Ionization equation:
C6H5OH ⇌ C6H5O- + H+
Ka = 1.3 × 10^-10 (pKa ≈ 10)
Why phenols are more acidic than alcohols:
1. Resonance stabilization of phenoxide ion (C6H5O-):
The negative charge on the oxygen is delocalized into the aromatic ring through resonance. The electron-rich aromatic ring can accommodate the negative charge through resonance structures:
- O- can form π bonds with the ring carbons
- This distributes the negative charge across the aromatic ring
- Resonance stabilization lowers the energy of the phenoxide ion
2. Comparison with alcoholate ion (RO-):
In an alcohol, the alkoxide ion (R-O-) cannot be stabilized by resonance because the carbon is sp3 hybridized and cannot form π bonds with a non-existent aromatic system.
3. Inductive effect:
The sp2 carbon in phenol is more electronegative than the sp3 carbon in alcohols, making it more electron-withdrawing. This further stabilizes the negative charge on oxygen.
4. Stability of conjugate base:
The phenoxide ion is more stable than the alcoholate ion due to resonance stabilization. This favors ionization.
Quantitative comparison:
- Ethanol: Ka ~ 10^-16 (pKa ≈ 16), essentially neutral
- Phenol: Ka ~ 10^-10 (pKa ≈ 10), weakly acidic
Note: Phenols are still weak acids and do not show typical acidic properties in aqueous solution, but they are strong enough to react with strong bases like NaOH to form sodium phenoxide (C6H5ONa), which is soluble in water.
Q3. Write the equation for the reaction of phenol with bromine water. What is the product, and why is this reaction so vigorous?
Reaction equation:
C6H5OH + 3 Br2 → C(Br)3 C(Br)3... (Actually forms 2,4,6-tribromophenol)
Correct equation:
C6H5OH + 3 Br2 (in H2O) → 2,4,6-C6H2(OH)(Br)3 + 3 HBr
Product: 2,4,6-tribromophenol (white precipitate)
Why is this reaction so vigorous?
1. High reactivity of the benzene ring:
The -OH group is an ortho-para directing activating group. It strongly activates the benzene ring toward electrophilic aromatic substitution.
2. Resonance activation:
The lone pair on oxygen resonates into the ring, increasing electron density at ortho and para positions. This makes the ring extremely electron-rich and highly reactive toward electrophiles like Br2.
3. Mechanism:
The reaction proceeds through electrophilic aromatic substitution:
- Br2 (or Br+ from Br2 + H2O) acts as the electrophile
- The electron-rich ring attacks the electrophile
- A stable intermediate forms with the hydrogen/bromide
- The product is a brominated phenol
4. Multiple substitutions:
The -OH group is such a strong activating group that substitution occurs at all three positions (ortho, ortho, para) even with just bromine water. No catalyst is needed.
5. Comparison:
Benzene itself requires a Lewis acid catalyst (FeBr3) for bromination, but phenol reacts vigorously with Br2 water without any catalyst. This demonstrates the extreme activation effect of the -OH group.
Note: The white precipitate formed is sparingly soluble in water, making this a useful test for phenols.
Q4. What are ethers? Give the general structural formula and two examples with their IUPAC names.
Ethers: Compounds with two alkyl or aryl groups bonded to the same oxygen atom through single bonds. General formula: R-O-R' where R and R' are alkyl or aryl groups.
Structural features:
- The oxygen is sp3 hybridized
- C-O-C bond angle ~ 104°
- Two C-O single bonds
- No H on oxygen (differs from alcohols)
Examples with IUPAC names:
1. CH3-O-CH3 (dimethyl ether)
IUPAC name: methoxyethane (incorrect) → actually called dimethyl ether
IUPAC nomenclature: When naming ethers, use the group with more carbons as the main chain, and the other group as a substituent with 'oxy' prefix
CH3-O-C2H5: Methoxyethane (ethyl group is main chain, methoxy is substituent)
2. C2H5-O-C2H5 (diethyl ether)
IUPAC name: Ethoxyethane
Also called ethyl ether or ether
Structure: CH3CH2-O-CH2CH3
3. C6H5-O-CH3 (methyl phenyl ether or anisole)
IUPAC name: Methoxybenzene
Structure: C6H5-O-CH3
Naming convention:
- Choose the larger alkyl/aryl group as the base
- The smaller group becomes a substituent with 'oxy' prefix
- For symmetric ethers (same groups on both sides), use 'di' prefix
Q5. Explain the relative solubility of alcohols and ethers in water. Why is diethyl ether slightly soluble in water while diethyl disulfide is insoluble?
Solubility in water:
Alcohols (e.g., ethanol CH3CH2OH):
- Highly soluble in water due to strong hydrogen bonding
- The -OH group can form hydrogen bonds with water molecules
- Can act as both hydrogen bond donor (through H of OH) and acceptor (through O)
- Example: Ethanol is miscible with water in all proportions
Ethers (e.g., diethyl ether C2H5-O-C2H5):
- Slightly soluble in water (5.6 g/100 mL at 20°C)
- The ether oxygen can act as a hydrogen bond acceptor only (has no H on oxygen)
- Cannot act as a hydrogen bond donor
- The alkyl groups provide hydrophobic character
- Net result: Limited hydrogen bonding interactions
Comparison:
- Diethyl ether (C2H5-O-C2H5): Slightly soluble, forms limited H-bonds through C=O lone pairs
- Diethyl disulfide (C2H5-S-S-C2H5): Insoluble (or very slightly soluble)
Reasons:
1. Sulfur has lower electronegativity than oxygen
2. Sulfur lone pairs are less polarized and less available for hydrogen bonding
3. Sulfur is larger and less effective at hydrogen bonding
4. The hydrophobic ethyl groups dominate, making the molecule non-polar
General trend:
- Alcohols: Very soluble (hydrogen bonding donors and acceptors)
- Ethers: Moderately soluble (hydrogen bonding acceptors only)
- Thioethers/disulfides: Insoluble (poor hydrogen bonding capability)
The key difference is the ability of the oxygen in ethers to act as a hydrogen bond acceptor, whereas sulfur is a poor hydrogen bond acceptor compared to oxygen.
Q6. Describe the reaction of ethers with dilute mineral acids. Write an equation for the reaction of diethyl ether with HBr.
Reaction with dilute HBr:
Equation: C2H5-O-C2H5 + HBr → C2H5Br + C2H5OH
Full reaction: (C2H5)2O + HBr → C2H5Br + C2H5OH
(diethyl ether + hydrobromic acid → ethyl bromide + ethanol)
Mechanism (SN1):
1. Protonation of ether oxygen by HBr:
C2H5-O-C2H5 + H+ → [C2H5-O+-C2H5]
The lone pair on oxygen is protonated
2. Formation of carbocation:
[C2H5-O+-C2H5] → C2H5+ + O-C2H5
The O-C bond breaks, forming a carbocation intermediate
(In practice, this is concerted with the next step)
3. Nucleophilic attack by bromide:
C2H5+ + Br- → C2H5Br
Br- attacks the carbocation
4. Deprotonation of protonated alcohol:
[C2H5-O-H] + Base → C2H5OH + H+
Or water deprotonates the protonated ethoxy group
Overall: Breaking of one C-O bond in the ether
Reactivity order:
Symmetric ethers give equal amounts of alkyl halide and alcohol.
Asymmetric ethers (R-O-R'): The bond between oxygen and the larger alkyl group tends to break preferentially due to formation of a more stable (or less hindered) carbocation.
Example: C2H5-O-CH3 + HBr → C2H5Br + CH3OH (mainly)
The larger ethyl group forms the carbocation, leaving methoxy to accept the proton.
Conditions:
- Dilute mineral acids (HBr, HI, HCl in dilute form)
- Heat is usually applied
- The reaction is slow and requires forcing conditions
Note: Ethers are generally very stable compounds. Their resistance to bases and neutral conditions is a key feature making them useful as protecting groups in organic synthesis.
Q7. How would you distinguish between an alcohol and an ether using chemical tests?
Chemical tests to distinguish alcohols from ethers:
Test 1: Sodium metal test (most reliable)
- Alcohols: Vigorous reaction with sodium metal
R-OH + Na → R-ONa + H2↑
Hydrogen gas is evolved with vigorous bubbling and heat
- Ethers: No reaction with sodium metal
R-O-R' + Na → No reaction
Conclusion: If H2 gas is evolved, the compound is an alcohol. No gas = ether
Test 2: Ferric chloride test
- Alcohols (aliphatic): No color change
- Ethers: No color change
(Neither pure ethers nor aliphatic alcohols give color with FeCl3)
Note: This test is useful for phenols (which give color), not for distinguishing alcohols from ethers
Test 3: Iodoform test (for methyl alcohols/ketones)
- Alcohols with CH3CH(OH)- group (secondary methyl alcohols):
Positive test: yellow precipitate of iodoform (CHI3)
Example: CH3CH(OH)CH3 + I2 + NaOH → CHI3↓ (yellow) + other products
- Ethers: No reaction
- Limitation: Only works for specific types of alcohols
Test 4: Ester formation with acid chlorides or anhydrides
- Alcohols: React to form esters
R-OH + CH3COCl → CH3CO-O-R + HCl
The HCl gas evolved can be detected
- Ethers: No reaction with acid chlorides
Most reliable: Sodium metal test
When sodium metal is added:
- Alcohols: Vigorous reaction, H2 gas evolved
- Ethers: No reaction
- This clearly distinguishes the two compound classes
Q8. Explain the preparation of phenol by the cumene process. Write the overall equation and describe the steps involved.
Cumene process (isopropylbenzene process) for phenol preparation:
Overall equation:
2 C6H6 + C3H6 → C6H5CH(CH3)2 (cumene) → C6H5OH (phenol) + CH3COCH3 (acetone)
Step 1: Formation of cumene (isopropylbenzene)
C6H6 (benzene) + C3H6 (propene) → C6H5-CH(CH3)2 (cumene)
Conditions: Phosphoric acid catalyst, 200-300°C, 25-30 atm pressure
Mechanism: Friedel-Crafts alkylation with isopropyl carbocation (formed from propene + H+)
Step 2: Oxidation of cumene
C6H5CH(CH3)2 (cumene) + O2 → C6H5-C(CH3)2-OOH (cumene hydroperoxide)
Conditions: Oxidation with air/oxygen at 50°C, using acid catalyst (dilute H2SO4 or HCl)
Mechanism:
- Formation of hydroperoxide intermediate through free radical mechanism
- The O-O bond forms due to oxidative conditions
Step 3: Acid-catalyzed rearrangement (Pillar rearrangement)
2 C6H5-C(CH3)2-OOH + dilute H2SO4 → C6H5OH + CH3COCH3 + H2O
Conditions: Dilute sulfuric acid or hydrochloric acid, room temperature
Mechanism:
- Protonation of hydroperoxide oxygen
- Rearrangement to form phenol and acetone
- Phenyl group migrates from carbon to oxygen
Advantages of cumene process:
1. Selective preparation of phenol
2. Byproduct acetone is valuable and can be sold
3. Economic process (cumene is cheap, easily available)
4. Higher yields compared to older chlorobenzene process
5. Less production of environmental pollutants
Historical note: Earlier methods (chlorobenzene hydrolysis, benzenesulfonic acid hydrolysis) are now obsolete due to the superiority of the cumene process
Alcohols Phenols and Ethers — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the IUPAC name for CH3-CH(OH)-CH2-CH3?
- A. Butan-2-ol
- B. 2-Butanol
- C. sec-Butanol
- D. Butanol
Answer: A. Butan-2-ol — The longest carbon chain containing the hydroxyl group has four carbons, making it butane. The hydroxyl group is on the second carbon, hence Butan-2-ol.
MCQ 2. Which of the following reagents is used for the preparation of primary alcohols from aldehydes?
- A. LiAlH4
- B. NaBH4
- C. Both LiAlH4 and NaBH4
- D. Grignard reagent followed by hydrolysis
Answer: C. Both LiAlH4 and NaBH4 — Both lithium aluminium hydride (LiAlH4) and sodium borohydride (NaBH4) are common reducing agents used to convert aldehydes into primary alcohols.
MCQ 3. Phenol is more acidic than ethanol. This is because:
- A. Phenoxide ion is more stable due to resonance
- B. Ethanol has a larger alkyl group
- C. Phenol has a benzene ring
- D. Ethanol forms stronger hydrogen bonds
Answer: A. Phenoxide ion is more stable due to resonance — Phenol is more acidic than ethanol because the phenoxide ion formed after losing a proton is stabilized by resonance with the benzene ring, whereas ethoxide ion has no such stabilization.
MCQ 4. When diethyl ether is heated with concentrated HI, the products formed are:
- A. Ethanol and ethyl iodide
- B. Ethanol and water
- C. Ethyl iodide and water
- D. Two moles of ethyl iodide
Answer: D. Two moles of ethyl iodide — Ethers undergo cleavage with strong acids like HI. Since HI is in excess and heating is involved, the initially formed ethanol further reacts with HI to give ethyl iodide.
MCQ 5. Which of the following is the best method for the preparation of an unsymmetrical ether like methyl tert-butyl ether?
- A. Heating methanol with tert-butanol in the presence of acid
- B. Reaction of sodium methoxide with tert-butyl bromide
- C. Reaction of tert-butoxide with methyl bromide
- D. Dehydration of a mixture of methanol and tert-butanol
Answer: C. Reaction of tert-butoxide with methyl bromide — Williamson synthesis works best when a primary alkyl halide reacts with a secondary or tertiary alkoxide. Using a tertiary alkyl halide (like tert-butyl bromide) with an alkoxide will lead to elimination (E2) predominantly.
MCQ 6. Which of the following compounds has the highest boiling point?
- A. n-Butane
- B. Diethylether
- C. n-Propanol
- D. Ethanol
Answer: C. n-Propanol — Alcohols have higher boiling points than ethers and alkanes of comparable molecular mass due to the presence of extensive intermolecular hydrogen bonding. n-Propanol has a higher molecular mass than ethanol, leading to stronger van der Waals forces.
MCQ 7. What is the product when a primary alcohol is subjected to oxidation with PCC (Pyridinium Chlorochromate)?
- A. Aldehyde
- B. Carboxylic acid
- C. Ketone
- D. Alkene
Answer: A. Aldehyde — PCC is a mild oxidizing agent that oxidizes primary alcohols specifically to aldehydes and secondary alcohols to ketones, without further oxidizing aldehydes to carboxylic acids.
MCQ 8. The Reimer-Tiemann reaction involves the electrophilic substitution of phenol with:
- A. Chloroform in the presence of NaOH
- B. Carbon dioxide in the presence of NaOH
- C. Methyl chloride in the presence of AlCl3
- D. Nitric acid in the presence of H2SO4
Answer: A. Chloroform in the presence of NaOH — The Reimer-Tiemann reaction converts phenol to salicylaldehyde using chloroform (CHCl3) and aqueous NaOH, involving a dichlorocarbene intermediate as an electrophile.
MCQ 9. Which of the following is a key intermediate in the commercial preparation of phenol from cumene?
- A. Cumene hydroperoxide
- B. Benzene sulfonic acid
- C. Chlorobenzene
- D. Aniline
Answer: A. Cumene hydroperoxide — In the cumene process, cumene is oxidized by air to cumene hydroperoxide, which then reacts with dilute acid to produce phenol and acetone.
MCQ 10. Dehydration of ethanol with concentrated H2SO4 at 443 K yields:
- A. Ethene
- B. Diethyl ether
- C. Ethane
- D. Acetone
Answer: A. Ethene — At higher temperatures (443 K or 170°C), concentrated sulfuric acid dehydrates ethanol to form ethene through an E1 mechanism. At lower temperatures (413 K or 140°C), diethyl ether is formed.
MCQ 11. Which reagent is used in the Lucas test to distinguish between primary, secondary, and tertiary alcohols?
- A. Concentrated HCl and anhydrous ZnCl2
- B. Aqueous NaOH
- C. Tollens' reagent
- D. Fehling's solution
Answer: A. Concentrated HCl and anhydrous ZnCl2 — The Lucas test uses a mixture of concentrated HCl and anhydrous ZnCl2 (Lucas reagent). Tertiary alcohols react immediately, secondary alcohols react within 5-10 minutes, and primary alcohols react very slowly or not at all.
MCQ 12. Which of the following is the strongest acid?
- A. Ethanol
- B. Phenol
- C. Water
- D. o-Nitrophenol
Answer: D. o-Nitrophenol — o-Nitrophenol is a stronger acid than phenol because the electron-withdrawing nitro group stabilizes the phenoxide ion through resonance and inductive effects, increasing its acidity. Water and ethanol are weaker acids than phenol.
MCQ 13. Phenol reacts with bromine water to give:
- A. 2,4,6-Tribromophenol
- B. o-Bromophenol
- C. p-Bromophenol
- D. Bromobenzene
Answer: A. 2,4,6-Tribromophenol — Phenol is highly activated towards electrophilic aromatic substitution. With bromine water (an excess of bromine), substitution occurs at all available ortho and para positions (2, 4, and 6), leading to the formation of 2,4,6-Tribromophenol.
MCQ 14. Reaction of Grignard reagent (RMgX) with an aldehyde followed by hydrolysis produces:
- A. Primary alcohol
- B. Secondary alcohol
- C. Tertiary alcohol
- D. Ketone
Answer: B. Secondary alcohol — Grignard reagents react with formaldehyde to produce primary alcohols, with other aldehydes to produce secondary alcohols, and with ketones to produce tertiary alcohols.
MCQ 15. What is the IUPAC name for CH3-O-CH2-CH3?
- A. Methoxyethane
- B. Ethyl methyl ether
- C. Ethane methyl ether
- D. Methyl ethyl ether
Answer: A. Methoxyethane — In IUPAC nomenclature for ethers, the smaller alkyl group along with the oxygen atom is named as an alkoxy group. Here, -OCH3 is methoxy and the remaining part is ethane.
MCQ 16. Kolbe's reaction is the carboxylation of phenoxide ion with carbon dioxide to yield:
- A. Salicylic acid
- B. Salicylaldehyde
- C. Phenol
- D. Benzoic acid
Answer: A. Salicylic acid — Kolbe's reaction involves treating sodium phenoxide with carbon dioxide at high temperature and pressure, followed by acidification, to yield salicylic acid.
MCQ 17. When ethanol reacts with PCl5, the main organic product formed is:
- A. Chloroethane
- B. Ethanoyl chloride
- C. Ethyl chloride
- D. Both a and c
Answer: D. Both a and c — Ethanol reacts with PCl5 to replace the -OH group with -Cl, forming chloroethane (also known as ethyl chloride), along with POCl3 and HCl.
MCQ 18. The cleavage of ethers by concentrated HI generally follows which mechanism for unsymmetrical ethers involving a methyl group?
- A. SN2
- B. SN1
- C. E1
- D. E2
Answer: A. SN2 — For unsymmetrical ethers, especially those with a methyl group or primary alkyl groups, the cleavage with HI typically follows an SN2 mechanism, where the iodide attacks the less sterically hindered carbon.
MCQ 19. Anisole undergoes Friedel-Crafts alkylation with CH3Cl in the presence of anhydrous AlCl3 to give mainly:
- A. o-Methoxy toluene and p-methoxy toluene
- B. m-Methoxy toluene
- C. Toluene
- D. Phenol
Answer: A. o-Methoxy toluene and p-methoxy toluene — The methoxy group (-OCH3) is an activating and ortho-para directing group. Therefore, Friedel-Crafts alkylation of anisole will predominantly yield a mixture of o-methoxy toluene and p-methoxy toluene.
MCQ 20. A compound 'X' gives turbidity immediately with Lucas reagent. 'X' could be:
- A. Propan-1-ol
- B. Propan-2-ol
- C. 2-Methylpropan-2-ol
- D. Ethanol
Answer: C. 2-Methylpropan-2-ol — The Lucas test distinguishes between primary, secondary, and tertiary alcohols. Tertiary alcohols react immediately with Lucas reagent to give turbidity, as 2-methylpropan-2-ol is a tertiary alcohol.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.