CBSE Class 12 Chemistry Alcohol Phenol Ether Exam Notes

CBSE Class 12 Chemistry Alcohol Phenol Ether Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Alcohols, phenols and ethers are an important part of the Organic Chemistry syllabus for CBSE Class 12 Chemistry. These exam notes on Alcohol, Phenol and Ether cover the reasoning questions on boiling point, acidity, reactivity and solubility, the mechanisms of dehydration, hydration and ether cleavage, the name reactions (hydroboration-oxidation, Kolbe's reaction, Reimer-Tiemann reaction and Williamson synthesis), the test to distinguish alcohols, and the conversions between these compounds.

ALCOHOL, PHENOL AND ETHER

This chapter is divided into the following parts from the exam point of view:
1. Reasoning
2. IUPAC naming
3. Mechanisms
4. Name reactions
5. Distinguish between two compounds
6. Conversions

REASONING

Q1. Arrange the following compounds in increasing order of boiling point: pentanol, butanol, butan 2-ol, propan-1-ol, methanol.

Answer: Methanol < ethanol < propan-1-ol < butan-2-ol (branched) < butan-1-ol < pentan-1-ol. Boiling point is directly related to molecular mass and inversely related to branching.

Q2. Arrange the following in the increasing order of the property given:
(a) CH₃CH₂OH, CF₃CH₂OH, CCl₃CH₂OH: acid strength.
(b) 2-methyl-2-propanol, 1-butanol and 2-butanol: reactivity towards Na.

Answer:
(a) CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH. The halogen atoms withdraw electron density by the −I effect, which weakens the O - H bond and makes it easier to release a proton. Fluorine is more electronegative than chlorine, so the −I effect of CF₃ is stronger than that of CCl₃.
(b) 2-methyl-2-propanol < 2-butanol < 1-butanol. The reactivity with Na depends on the acidity of the alcohol. The +I effect of alkyl groups makes the alcohol less acidic, and the tertiary alcohol has the most alkyl groups.

Q3. Unlike phenols, alcohols are easily protonated.

Answer: In phenol, the lone pair of electrons on the oxygen atom is delocalised over the benzene ring by resonance, so it is not easily available for protonation. In alcohols, the lone pair on oxygen is localised and easily available for protonation.

Q4. How do you account for the fact that unlike phenol, 2,4-dinitrophenol and 2,4,6-trinitrophenol are soluble in Na₂CO₃ (aq) solution?

Answer: The nitro groups (NO₂) are strong electron-withdrawing groups. Two and three NO₂ groups make these phenols much more acidic, so they react with Na₂CO₃ to form soluble sodium salts. Phenol is not acidic enough for this reaction.

Q5. Alcohols react with halogen acids or phosphorus halides to form haloalkanes, but phenol does not form halobenzene.

Answer: In alcohols, the carbon attached to the OH group is sp³ hybridised, so nucleophilic substitution takes place easily: R - OH + H⁺ → R - X + H₂O. In phenol, the C - OH carbon is sp² hybridised, and the C - O bond gets partial double-bond character by resonance. So the C - O bond is not easily broken, and phenol cannot give a halobenzene by nucleophilic substitution.

Q6. Explain why dehydration of alcohols to form alkenes is always carried out with conc. H₂SO₄ and not with conc. HCl or HNO₃.

Answer: Conc. H₂SO₄ protonates the alcohol and the leaving water molecule forms a carbocation, which then loses a proton to give the alkene. The HSO₄⁻ ion is a poor nucleophile, so it does not substitute the protonated alcohol and the elimination occurs. Conc. HCl gives Cl⁻, which is a good nucleophile, and it forms the alkyl chloride by substitution. With conc. HNO₃, the NO₃⁻ ion is a nucleophile and gives a substitution reaction. HCl is also described as an oxidising agent, which oxidises the alcohol to an aldehyde and then to an acid.

Q7. Di-tert-butyl ether cannot be made by Williamson's synthesis.

Answer: Preparation of di-tert-butyl ether requires tert-butyl bromide, which prefers to undergo elimination (giving 2-methylpropene) rather than substitution with the alkoxide ion.

Q8. Explain why ethers are cleaved only by acids and not by bases.

Answer: Acids protonate the ether oxygen, which converts the leaving group into a neutral alcohol, and a strong nucleophile can then attack and cleave the C - O bond. Bases cannot do this, because an alkoxide ion is a very poor leaving group.

Q9. Explain why anisole is less reactive than phenol towards electrophilic substitution.

Answer: Both - OH and - OCH₃ groups activate the ring by their +R (electron-releasing by resonance) effect. In phenol, the intermediate is further stabilised because the - OH group can release a proton (H⁺), which anisole cannot do, so phenol is more reactive than anisole.

Q10. Ortho and para nitrophenol are more acidic than phenol.

Answer: The NO₂ group is an electron-withdrawing group. In the ortho and para positions it stabilises the phenoxide ion by the −R (resonance) effect, which delocalises the negative charge onto the nitro group. So the conjugate base is more stable and the acid is stronger.

Q11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene, and why?

Answer: This is a Try Yourself question, and the appropriate set of reactants is given below. The appropriate set is sodium 4-nitrophenoxide with methyl iodide (or dimethyl sulphate). This is a Williamson synthesis in which the methyl halide is primary and undergoes SN2 attack. The other route, 4-nitrochlorobenzene with sodium methoxide, is not suitable because an aryl halide does not undergo SN2 substitution.

Q12. Why do alcohols have higher boiling points than ethers or alkanes?

Answer: Alcohols form strong intermolecular hydrogen bonds. Ethers have only weak dipole-dipole interactions, and alkanes have only weak van der Waals forces.

Q13. Why are alcohols more soluble in water than alkanes?

Answer: Alcohols can form intermolecular hydrogen bonds with water molecules, whereas alkanes cannot.

Q14. While separating a mixture of ortho and para nitrophenol by steam distillation, name the isomer which is steam volatile.

Answer: Ortho-nitrophenol is steam volatile. It forms intramolecular hydrogen bonding, so its molecules do not associate with each other and it escapes with steam. Para-nitrophenol forms strong intermolecular hydrogen bonds and is not steam volatile.

Q15. Why is phenol more acidic than alcohols?

Answer: In phenol, the phenoxide ion formed after losing H⁺ is stabilised by resonance, which spreads the negative charge over the benzene ring. In alcohols, the alkoxide ion has no such resonance stabilisation.

Q17. Preparation of ethers by acid catalysed dehydration of 2° and 3° alcohols is not a suitable method. Why?

Answer: Secondary and tertiary alcohols mainly undergo elimination to give alkenes. The carbocations formed from them are prone to elimination, and steric hindrance prevents the SN2 attack that is needed for ether formation.

Q18. Define:
(a) Rectified spirit
(b) Absolute alcohol
(c) Denaturation of alcohol
(d) Fermentation

Answer:
(a) Rectified spirit is ethanol containing about 4.4% water by mass (about 95.6% ethanol), obtained by fractional distillation of fermented wash.
(b) Absolute alcohol is ethanol that is free from water, containing 100% ethanol.
(c) Denaturation of alcohol is the addition of poisonous substances such as methanol or pyridine to ethanol, to make it unfit for drinking while keeping it useful for industrial purposes.
(d) Fermentation is the conversion of sugars such as glucose into ethanol and carbon dioxide by the action of enzymes (zymase) present in yeast.

Q19. Phenol has a smaller dipole moment than methanol.

Answer: In phenol, the lone pair on oxygen is delocalised into the benzene ring by resonance. This gives the C - O bond partial double-bond character, and it reduces the net polarity of the molecule. Methanol has no such resonance, so its C - O and O - H dipoles add up to a larger net dipole moment (about 1.7 D, compared with about 1.4 D for phenol).

Q20. Diethyl ether does not react with sodium. Explain.

Answer: Diethyl ether has no acidic hydrogen atom (no O - H group), so it cannot release hydrogen gas with sodium. Alcohols react with sodium because the O - H hydrogen is acidic.

Q21. Dimethyl ether is completely miscible with H₂O, but diethyl ether is soluble in water only to a small extent.

Answer: The oxygen atom of an ether forms hydrogen bonds with water molecules. Dimethyl ether has small methyl groups, so it can form these hydrogen bonds freely and is completely miscible with water. Diethyl ether has larger ethyl groups, which hinder hydrogen bonding with water and increase the hydrocarbon character, so only a small amount dissolves.

Q22. Explain why cleavage of alkyl phenyl ethers with HBr always produces phenol and alkyl bromide.

Answer: The ether oxygen is protonated first. The bromide ion then attacks the alkyl carbon by SN2 substitution, which gives the alkyl bromide. The bond between oxygen and the phenyl carbon is not broken, because the sp² carbon of the benzene ring does not undergo nucleophilic substitution. So phenol is formed.

MECHANISMS

1. Acidic dehydration of alcohol (ethanol to ethene)
Step 1: Formation of protonated alcohol (fast): CH₃CH₂ - O - H + H⁺ ⇌ CH₃CH₂ - O⁺H₂ (ethyl oxonium ion).
Step 2: Formation of carbocation (slow, rate determining step): CH₃CH₂ - O⁺H₂ → CH₃CH₂⁺ + H₂O.
Step 3: Formation of ethene by elimination of a proton: CH₃CH₂⁺ → CH₂=CH₂ + H⁺.

2. Acid catalysed hydration of alkene
Step 1: Protonation of the alkene by H₃O⁺ to form a carbocation (electrophilic attack).
Step 2: Nucleophilic attack of water on the carbocation, giving a protonated alcohol.
Step 3: Deprotonation of the protonated alcohol to form the alcohol and regenerate H⁺.

3. Dehydration of alcohol to form ether (ethanol to diethyl ether)
(i) CH₃CH₂ - O - H + H⁺ → CH₃CH₂ - O⁺H₂
(ii) CH₃CH₂ - O - H + CH₃CH₂ - O⁺H₂ → CH₃CH₂ - O⁺(H) - CH₂CH₃ + H₂O (SN2 attack of ethanol on the protonated ethanol)
(iii) Loss of a proton gives diethyl ether: CH₃CH₂ - O - CH₂CH₃ + H⁺

4. Reaction of ether with HI
Step 1: Protonation of the ether oxygen: CH₃ - O - CH₂CH₃ + H - I ⇌ CH₃ - O⁺(H) - CH₂CH₃ + I⁻.
Step 2: Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion by SN2 and displaces an alcohol molecule. In the cleavage of mixed ethers with two different alkyl groups, the alcohol and alkyl iodide formed depend on the nature of the alkyl groups. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms the alkyl iodide (SN2 reaction). Example: I⁻ + CH₃ - O⁺(H) - CH₂CH₃ → CH₃ - I + CH₃CH₂ - OH.
Step 3: The alcohol formed is protonated and then replaced by iodide: CH₃CH₂ - OH + H⁺ + I⁻ → CH₃CH₂I + H₂O.
If HI is in excess and the temperature is high, all products formed are haloalkanes.

NAME REACTIONS

1. Hydroboration-oxidation reaction
Alkene + BH₃ in THF, followed by H₂O₂ and OH⁻, gives an alcohol with anti-Markovnikov addition of water. Example: 2-methylprop-1-ene (H₃C - C(=CH₂) - CH₃) gives 2-methylpropan-1-ol (H₃C - CH(CH₃) - CH₂OH).

2. Kolbe's reaction
Phenol + NaOH → sodium phenoxide. Sodium phenoxide + CO₂ (under pressure, at about 125 °C), then H⁺ → 2-hydroxybenzoic acid (salicylic acid).

3. Reimer-Tiemann reaction
Phenol + CHCl₃ + aqueous NaOH → sodium salicylaldehyde intermediate (sodium salt of 2-hydroxybenzal chloride, CHCl₂ group), which on treatment with NaOH and then H⁺ gives salicylaldehyde (2-hydroxybenzaldehyde).

4. Williamson synthesis
R - X + R′ - O⁻Na⁺ → R - O - R′ + NaX
The reaction involves SN2 attack of an alkoxide ion on a primary alkyl halide. Better results are obtained if the alkyl halide is primary. With secondary and tertiary alkyl halides, elimination competes with substitution.
Example: (CH₃)₃C - O⁻Na⁺ + CH₃ - Br → (CH₃)₃C - O - CH₃ + NaBr.

DISTINGUISH BETWEEN TWO COMPOUNDS

Lucas test: This test is used to distinguish between primary, secondary and tertiary alcohols. Lucas reagent is a solution of zinc chloride in concentrated hydrochloric acid. A positive test is indicated by a change from colourless to turbid, signalling the formation of a chloroalkane. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides: ROH + HCl → RCl + H₂O.
Tertiary alcohols react immediately with Lucas reagent, as shown by turbidity owing to the low solubility of the organic chloride in the aqueous mixture. Secondary alcohols react within five or so minutes (depending on their solubility). Primary alcohols do not react appreciably with Lucas reagent at room temperature.

CONVERSIONS

Q1. Show how the following alcohols are prepared by the reaction of a suitable Grignard reagent on methanal.
(i) CH₃ - CH(CH₃) - CH₂OH
(ii) Cyclohexylmethanol

Answer:
(i) HCHO + (CH₃)₂CH - MgBr → (CH₃)₂CH - CH₂ - OMgBr; then H₂O gives 2-methylpropan-1-ol (CH₃ - CH(CH₃) - CH₂OH) and Mg(OH)Br.
(ii) HCHO + cyclohexylmagnesium bromide → cyclohexyl-CH₂ - OMgBr; then H₂O gives cyclohexylmethanol and Mg(OH)Br.

Q2. Write the structures of the products of the following reactions.
(i) CH₃ - CH=CH₂ with H₂O/H⁺
(ii) Methyl 2-oxocyclohexyl ethanoate with NaBH₄
(iii) 2-methylbutanal (CH₃ - CH₂ - CH(CH₃) - CHO) with NaBH₄

Answer:
(i) CH₃ - CH(OH) - CH₃: propan-2-ol (Markovnikov addition of water).
(ii) Methyl (2-hydroxycyclohexyl)ethanoate: NaBH₄ reduces the ketone group to a secondary alcohol.
(iii) 2-methylbutan-1-ol (CH₃ - CH₂ - CH(CH₃) - CH₂OH).

Q3. Write the structures of the major products expected from the following reactions.
(a) Mononitration of 3-methylphenol
(b) Dinitration of 3-methylphenol
(c) Mononitration of phenyl methanoate

Answer:
(a) The - OH group and the - CH₃ group together decide the position of the incoming nitro group. The major products are 3-methyl-4-nitrophenol and 5-methyl-2-nitrophenol.
(b) The major product is 5-methyl-2,4-dinitrophenol (the same as 3-methyl-4,6-dinitrophenol).
(c) The product is 4-nitrophenyl ester (OCOCH₃ group with NO₂ at the para position), as the ester group directs the nitro group to the para position.

Q4. Give the structures of the products you would expect when each of the following alcohols reacts with (a) HCl - ZnCl₂, (b) HBr and (c) SOCl₂.
(i) Butan-1-ol
(ii) 2-Methylbutan-2-ol

Answer:
(a) (i) Butan-1-ol with HCl - ZnCl₂: no reaction at room temperature, since it is a primary alcohol.
(a) (ii) 2-Methylbutan-2-ol with HCl - ZnCl₂: 2-chloro-2-methylbutane, with white turbidity (tertiary alcohol).
(b) (i) Butan-1-ol with HBr: 1-bromobutane (CH₃CH₂CH₂CH₂Br).
(b) (ii) 2-Methylbutan-2-ol (3°) with HBr: 2-bromo-2-methylbutane (CH₃CH₂C(CH₃)₂Br) + H₂O (white turbidity).
(c) (i) Butan-1-ol with SOCl₂: 1-chlorobutane, with SO₂ and HCl as by-products.
(c) (ii) 2-Methylbutan-2-ol with SOCl₂: mainly alkenes by elimination, since the tertiary alcohol forms a stable carbocation.

Q4 (b). Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol.

Answer:
(i) 1-Methylcyclohexene (major product), with water as the by-product.
(ii) But-2-ene (major product), with water as the by-product, following Saytzeff's rule.

Q5. The following is not an appropriate reaction for the preparation of tert-butyl ethyl ether: (CH₃)₃C - Cl + C₂H₅ONa → (CH₃)₃C - O - C₂H₅.
(i) What would be the major product of this reaction?
(ii) Write a suitable reaction for the preparation of tert-butyl ethyl ether.

Answer:
(i) The major product is 2-methylprop-1-ene. Sodium ethoxide is a strong nucleophile as well as a strong base, and the tertiary halide undergoes elimination rather than substitution.
(ii) (CH₃)₃C - O⁻Na⁺ + CH₃CH₂ - Br → (CH₃)₃C - O - C₂H₅ + NaBr. The primary halide ethyl bromide undergoes SN2 substitution.

Q6. Give the major products that are formed by heating each of the following ethers with HI.
(i) CH₃ - CH₂ - CH(CH₃) - CH₂ - O - CH₂ - CH₃
(ii) CH₃ - CH₂ - CH₂ - O - C(CH₃)₂ - CH₂ - CH₃
(iii) C₆H₅ - CH₂ - O - C₆H₅

Answer:
(i) CH₃CH₂CH(CH₃)CH₂OH (2-methylbutan-1-ol) + CH₃CH₂I. Iodide attacks the less hindered ethyl carbon by SN2.
(ii) CH₃CH₂CH₂OH (propan-1-ol) + CH₃CH₂C(CH₃)₂I (2-iodo-2-methylbutane). The tertiary alkyl group forms a stable carbocation, so the C - O bond on that side is cleaved.
(iii) C₆H₅CH₂I (benzyl iodide) + C₆H₅OH (phenol).

Parts (iv), (v) and (vi) of Q6 are reactant and product pairs:
(iv) CH₃ - CH₂ - CH(CH₃) - CH₂OH + CH₃CH₂I
(v) CH₃CH₂CH₂OH + CH₃CH₂ - C(CH₃)₂ - I
(vi) C₆H₅ - CH₂I + C₆H₅ - OH

Q7. Predict the products of the following reactions.
(i) CH₃CH₂CH₂ - O - CH₃ + HBr
(ii) Ethoxybenzene + HBr
(iii) Ethoxybenzene with conc. H₂SO₄ and conc. HNO₃
(iv) (CH₃)₃C - O - C₂H₅ + HI

Answer:
(i) CH₃CH₂CH₂OH (propan-1-ol) + CH₃Br (bromomethane).
(ii) Phenol + C₂H₅Br (bromoethane).
(iii) 4-Ethoxynitrobenzene (major) and 2-ethoxynitrobenzene (minor).
(iv) (CH₃)₃C - I (tert-butyl iodide) + C₂H₅OH (ethanol).

Q8. You are given benzene, conc. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.

Answer:
Benzene + conc. H₂SO₄ (heat) → benzenesulphonic acid.
Benzenesulphonic acid + NaOH (fusion, heat) → sodium phenoxide.
Sodium phenoxide + H₂SO₄ (dilute acid) → phenol.

Q9. Show how you will synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) Pentan-1-ol using a suitable alkyl halide.

Answer:
(i) Acid catalysed hydration of phenylethene (styrene) gives 1-phenylethanol: C₆H₅CH=CH₂ + H₂O (H⁺) → C₆H₅CH(OH)CH₃.
(ii) Chloromethylcyclohexane + NaOH → cyclohexylmethanol + NaCl.
(iii) 1-Chloropentane + NaOH → pentan-1-ol + NaCl.

Q10. Give equations of the following reactions.
(i) Oxidation of propan-1-ol with alkaline KMnO₄ solution.
(ii) Bromine in CS₂ with phenol.
(iii) Dilute HNO₃ with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.

Answer:
(i) CH₃CH₂CH₂OH (alk. KMnO₄) → CH₃CH₂COOH (propanoic acid).
(ii) Phenol + dilute HNO₃ → o-nitrophenol + p-nitrophenol.
(iii) Phenol + dilute HNO₃ → 2-nitrophenol (ortho-nitrophenol) and 4-nitrophenol (para-nitrophenol).
(iv) Phenol + CHCl₃ + aqueous NaOH (Reimer-Tiemann reaction) → salicylaldehyde (2-hydroxybenzaldehyde).

Q11. How are the following conversions carried out?
(i) Propene → propan-2-ol.
(ii) Benzyl chloride → benzyl alcohol.
(iii) Ethyl magnesium chloride → propan-1-ol.
(iv) Methyl magnesium bromide → 2-methylpropan-2-ol.

Answer:
(i) Propene + H₂O in the presence of an acid catalyst (H⁺) gives propan-2-ol: CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃.
(ii) Benzyl chloride + NaOH (aqueous), followed by acidification, gives benzyl alcohol.
(iii) Ethyl magnesium chloride + methanal gives an adduct, CH₃CH₂CH₂OMgCl, which on hydrolysis gives propan-1-ol.
(iv) Methyl magnesium bromide is treated with propanone, and an adduct is formed which gives 2-methylpropan-2-ol on hydrolysis.

Q12. Name the reagents used in the following reactions.
(i) Oxidation of a primary alcohol to a carboxylic acid.
(ii) Oxidation of a primary alcohol to an aldehyde.
(iii) Bromination of phenol to 2,4,6-tribromophenol.
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propan-2-ol to propene.
(vi) Butan-2-one to butan-2-ol.

Answer:
(i) Acidified potassium permanganate (or acidified potassium dichromate).
(ii) Pyridinium chlorochromate (PCC).
(iii) Bromine water.
(iv) Acidified potassium permanganate (or acidified potassium dichromate).
(v) 85% phosphoric acid.
(vi) NaBH₄ or LiAlH₄, followed by water.

Q13. Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesis.
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane

Answer:
(i) Sodium propoxide + 1-bromopropane → 1-propoxypropane + NaBr.
(ii) Sodium phenoxide + bromoethane → ethoxybenzene + NaBr.
(iii) Sodium 2-methylpropan-2-olate (sodium tert-butoxide) + bromomethane → 2-methoxy-2-methylpropane + NaBr. The tertiary halide is avoided because it would undergo elimination.
(iv) Sodium ethoxide + bromomethane → 1-methoxyethane + NaBr.

Q14. Write the equation of the reaction of hydrogen iodide with:
(i) 1-Propoxypropane
(ii) Methoxybenzene
(iii) Benzyl ethyl ether

Answer:
(i) CH₃CH₂CH₂ - O - CH₂CH₂CH₃ + HI (373 K) → CH₃CH₂CH₂OH (propan-1-ol) + CH₃CH₂CH₂I (1-iodopropane).
(ii) Methoxybenzene + HI → phenol + CH₃I (iodomethane).
(iii) Benzyl ethyl ether + HI → C₆H₅CH₂I (benzyl iodide) + C₂H₅OH (ethanol).

Q15. Show how you would synthesise the following alcohols from appropriate alkenes.
(i) 1-Methylcyclohexan-1-ol
(ii) 4-Methylheptan-4-ol
(iii) Pentan-2-ol
(iv) 2-Cyclohexylbutan-2-ol

Answer:
(i) 1-Methylcyclohexene + H₂O (H⁺) → 1-methylcyclohexan-1-ol (Markovnikov addition). Methylenecyclohexane (cyclohexane with =CH₂) + H₂O (H⁺) also gives 1-methylcyclohexan-1-ol.
(ii) 4-Methylhept-3-ene + H₂O (H⁺) → 4-methylheptan-4-ol.
(iii) Pent-1-ene + H₂O (H⁺) → pentan-2-ol (Markovnikov addition).
(iv) 2-Cyclohexylbut-2-ene + H₂O (H⁺) → 2-cyclohexylbutan-2-ol.

Q16. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: CH₃ - CH(OH) - CH(CH₃) - CH₃ + HBr → CH₃ - CH₂ - C(CH₃)₂ - Br. Give a mechanism for this reaction.

Answer:
Step 1: The OH group is protonated by HBr to form an oxonium ion, CH₃ - CH(OH₂⁺) - CH(CH₃) - CH₃.
Step 2: Loss of water gives a secondary carbocation at C-2 (less stable).
Step 3: A 1,2-hydride shift from C-3 to C-2 gives a tertiary carbocation, CH₃ - CH₂ - C⁺(CH₃)₂ (more stable).
Step 4: Bromide ion attacks the tertiary carbocation to give 2-bromo-2-methylbutane.

Q6.Explain why dehydration of alcohols to form alkenes is always carried out with conc. H2SO4 and not with conc. HCl or HNO3.
Ans.Reaction with acid will give carbocation. In case HSO4 – being non-nucleophile cannot substitute the Halogen group but in case Cl– will give substition rx. in presence of HNO3 as NO3- is a nucleophile. HCl being oxidising agent, and oxidise alcohol to aldehyde then to acid.

Explore more exam notes and revision material for this chapter on the site. CBSE Class 12 Chemistry Alcohol Phenol Ether Exam