CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Name the reagents used in the following reactions:
Friedel–Crafts alkylation of anisole [CBSE (F) 2014]
Answer:
Anhydrous aluminium chloride (anhyd. AlCl₃)

Teacher's Note:
1. The full reagent set is an alkyl halide (e.g. CH₃Cl) with anhydrous AlCl₃; the products are the o- and p-alkyl anisoles.

Teacher's Note:
1. Neutral FeCl₃ is the standard test for phenols: any compound with an OH on an aromatic ring gives a violet/purple colour, alcohols do not.

Question 3. Write the IUPAC name of the given compound: [CBSE Delhi 2015]

Answer:
2,5-Dinitrophenol

Teacher's Note:
1. Number the ring from the carbon bearing –OH as C-1 and go in the direction that gives the nitro groups the lowest locants (2,5 rather than 3,6).

Question 4. Write the IUPAC name of the given compound:
C₆H₅–CH₂–CH₂–OH [CBSE OD Central 2016]
Answer:
2-Phenylethanol.

Teacher's Note:
1. Students often go wrong with the ‘-ol’ suffix: the OH carbon is C-1 of the ethane chain and the phenyl group sits on C-2.

Question 5. Write the IUPAC name of the following:
CH₃–C(CH₃)(Br)–CH₂–CH₂OH [CBSE Comptt. OD 2012]
Answer:
3-Bromo-3-methylbutan-1-ol.

Teacher's Note:
1. Number the chain from the carbon carrying –OH (the principal group) and list the prefixes bromo and methyl in alphabetical order.

Question 6. Write the IUPAC name of the following:
(CH₃)₃C–OH, i.e. CH₃–C(CH₃)(OH)–CH₃ [CBSE Comptt. OD 2012]
Answer:
2-Methylpropan-2-ol.

Teacher's Note:
1. The longest chain containing the OH carbon has only three carbons; the fourth carbon is a methyl branch – this is tert-butyl alcohol.

Question 7. Write the IUPAC name of the following compound:
CH₂=CH–C(CH₃)₂–OH [CBSE Foreign Set-2 2017]
Answer:
2-Methylbut-3-en-2-ol

Teacher's Note:
1. The OH carbon must get the lowest locant, even though this puts the double bond at position 3.

Question 8. Draw the structural formula of 2-methyl propan-2-ol molecule. [CBSE Delhi 2012]
Answer:
2-Methylpropan-2-ol:
CH₃–C(OH)(CH₃)–CH₃, i.e. (CH₃)₃C–OH

Teacher's Note:
1. Propan-2-ol is CH₃CH(OH)CH₃; adding a methyl on C-2 replaces its last H, giving a tertiary alcohol.

Question 9. Write the IUPAC name of the following:
CH₃–C(CH₃)(C₂H₅)–CH(OH)–CH₃ [CBSE Delhi/OD 2018]
Answer:
3,3-Dimethylpentan-2-ol

Teacher's Note:
1. The ethyl group is part of the longest chain: count CH₃–CH(OH)–C–CH₂–CH₃ as a five-carbon pentane chain with two methyls on C-3.

Question 10. Draw the structure of hex-1-en-3-ol compound. [CBSE Delhi 2012]
Answer:
CH₃–CH₂–CH₂–CH(OH)–CH=CH₂

Teacher's Note:
1. Read the name from the end: six carbons (hex), double bond from C-1, OH on C-3.

Question 11. Which of the following isomers is more volatile: o-nitrophenol or p-nitrophenol? [CBSE Delhi 2014]
Answer:
o-Nitrophenol.

Teacher's Note:
1. Intramolecular H-bonding in the ortho isomer prevents association of molecules, so it is more volatile (and steam volatile).

Question 12. Write the chemical reaction to explain Kolbe’s reaction. [CBSE Comptt. Delhi 2013]
Answer:
C₆H₅OH (Phenol) \( \xrightarrow{NaOH} \) C₆H₅ONa \( \xrightarrow[(ii)\ H^+]{(i)\ CO_2} \) o-HO–C₆H₄–COOH (2-Hydroxybenzoic acid / Salicylic acid)

Teacher's Note:
1. Do not confuse it with Kolbe’s electrolytic reaction; in Kolbe’s reaction of phenol, the weak electrophile CO₂ attacks the strongly activated phenoxide ion at the ortho position.

Question 13. Give a chemical test to distinguish between 2-Pentanol and 3-Pentanol. [CBSE Comptt. Delhi 2013]
Answer:
Iodoform test: 2-Pentanol, CH₃–CH(OH)–CH₂CH₂CH₃, on warming with I₂ and NaOH gives a yellow precipitate of iodoform (CHI₃) as it contains the CH₃–CH(OH)– group, but 3-Pentanol, CH₃CH₂–CH(OH)–CH₂CH₃, does not give a yellow precipitate.

Teacher's Note:
1. Both are secondary alcohols, so the Lucas test cannot separate them; always write the observation (yellow precipitate of CHI₃), not just the name of the test.

Question 14. How would you obtain ethane-1,2-diol from ethanol? [CBSE Comptt. OD 2013]
Answer:
CH₃CH₂OH (Ethanol) \( \xrightarrow[443\,K]{\text{Conc. }H_2SO_4} \) CH₂=CH₂ (Ethylene) \( \xrightarrow{\text{cold dil. }KMnO_4} \) HOCH₂–CH₂OH (Ethane-1,2-diol)

Teacher's Note:
1. Cold dilute alkaline KMnO₄ (Baeyer’s reagent) adds two OH groups across the double bond; hot or acidic KMnO₄ would break the C=C bond instead.

Question 15. How would you obtain phenol from benzene? [CBSE Comptt. OD 2013]
Answer:
C₆H₆ (Benzene) \( \xrightarrow{\text{Oleum}} \) C₆H₅SO₃H (Benzene sulphonic acid) \( \xrightarrow{NaOH} \) C₆H₅SO₃Na (Sodium benzene sulphonate) \( \xrightarrow[625\,K]{NaOH/\Delta} \) C₆H₅OH (Phenol) + Na₂SO₃

Teacher's Note:
1. Always show the conditions: sulphonation with oleum, then fusion with NaOH at high temperature; the phenoxide formed is finally acidified to phenol.

Short Answer Questions (2 Marks)

Question 16. Write the equations involved in the following reactions:
(i) Reimer-Tiemann reaction
(ii) Williamson synthesis [CBSE Delhi 2014]
Answer:
(i) Reimer-Tiemann reaction:
C₆H₅OH (Phenol) \( \xrightarrow[340\,K]{CHCl_3 + NaOH(aq)} \) [o-NaO–C₆H₄–CHCl₂] (intermediate) \( \xrightarrow{NaOH} \) o-NaO–C₆H₄–CHO \( \xrightarrow{H_3O^+} \) o-HO–C₆H₄–CHO (Salicylaldehyde / 2-hydroxybenzaldehyde)
(ii) Williamson synthesis:
R–X (Alkyl halide) + R′–O⁻Na⁺ (Sodium alkoxide) → R–O–R′ + NaX
CH₃–CH₂–Br (Ethyl bromide) + Na^{+ −}O–CH₂–CH₃ (sodium ethoxide) → CH₃–CH₂–O–CH₂–CH₃ (Diethyl ether) + NaBr

Teacher's Note:
1. While writing reactions, do not forget the reagents and conditions, and make sure the equations are balanced.

Question 17. Write the mechanism of acid dehydration of ethanol to yield ethene. [CBSE SQP 2018-2019]
Answer:
Step 1: Formation of protonated alcohol.
CH₃–CH₂–O–H + H⁺ \( \xrightarrow{\text{Fast}} \) CH₃–CH₂–O⁺H₂ (Protonated ethanol / Ethyl oxonium ion)
Step 2: Formation of carbocation.
CH₃–CH₂–O⁺H₂ \( \xrightarrow{\text{Slow}} \) CH₃–CH₂⁺ + H₂O
Step 3: Formation of ethene by elimination of a proton.
H–CH₂–CH₂⁺ ⇄ CH₂=CH₂ (Ethene) + H⁺

Teacher's Note:
1. Mention each step: protonation (fast), loss of water to give the carbocation (slow, rate determining) and loss of H⁺ to give the alkene.

Question 18. (i) Write the mechanism of the following reaction:
CH₃CH₂OH \( \xrightarrow{HBr} \) CH₃CH₂Br + H₂O
(ii) Write the equation involved in Reimer-Tiemann reaction. [CBSE Delhi 2014]
Answer:
(i) H–Br ⇄ H⁺ + Br⁻
CH₃–CH₂–OH + H⁺ ⇄ CH₃CH₂O⁺H₂ (Ethyl oxonium ion / Protonated 1° alcohol)
CH₃–CH₂–O⁺H₂ → CH₃–CH₂⁺ + H₂O
CH₃–CH₂⁺ + Br⁻ → CH₃CH₂Br
Overall reaction: CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O
(ii) Reimer-Tiemann reaction:
C₆H₅OH (Phenol) \( \xrightarrow[340\,K]{CHCl_3 + NaOH(aq)} \) [o-NaO–C₆H₄–CHCl₂] (intermediate) \( \xrightarrow{NaOH} \) o-NaO–C₆H₄–CHO \( \xrightarrow{H_3O^+} \) o-HO–C₆H₄–CHO (Salicylaldehyde / 2-hydroxybenzaldehyde)

Teacher's Note:
1. Protonation is the key first step: it turns the poor leaving group OH⁻ into H₂O, a good leaving group.

Question 19. (i) Give chemical tests to distinguish between the following pairs of compounds: Methanol and Phenol
(ii) o-nitro phenol is more acidic than o-methoxy phenol. Explain why. [CBSE Comptt. OD 2013]
Answer:
(i) Phenol gives a violet colouration with FeCl₃ solution while methanol does not.
(ii) The electron withdrawing groups like –NO₂ increase the stability of phenoxide ion by dispersal of negative charge. The electron releasing groups like –OCH₃ decrease the acidic strength of phenol by destabilising the phenoxide ion by concentrating negative charge. Thus, o-nitrophenol is more acidic than o-methoxy phenol.

Teacher's Note:
1. In a distinguishing test, always write the observation for both compounds; underlining the observation helps the examiner.

Question 20. Explain the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol. [CBSE OD 2012]
Answer:
>C=C< (Alkene) + H₂O \( \xrightarrow{H^+} \) >CH–C(OH)<
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺:
H₂O + H⁺ → H₃O⁺
>C=C< + H₃O⁺ ⇄ >CH–C⁺< + H₂O
Step 2: Nucleophilic attack of water on carbocation:
>CH–C⁺< + H₂O ⇄ >CH–C(O⁺H₂)<
Step 3: Deprotonation to form an alcohol:
>CH–C(O⁺H₂)< + H₂O ⇄ >CH–C(OH)< + H₃O⁺

Teacher's Note:
1. Write the mechanism step by step and show the charges and lone pairs; the regenerated H₃O⁺ in the last step shows that the acid is a catalyst.

Question 21. (i) Describe the mechanism of acid dehydration of ethanol to yield ethene.
(ii) Describe a chemical test to distinguish between ethanol and phenol. [CBSE Comptt. Delhi/OD 2015; Comptt. Delhi 2012]
Answer:
(i) Acid dehydration of ethanol to ethene: When ethanol is heated with concentrated sulphuric acid at 453 K, it undergoes dehydration and ethene is formed.
CH₃CH₂OH \( \xrightarrow{H_2SO_4,\ 453\,K} \) CH₂=CH₂ + H₂O
Mechanism:
Step 1: Formation of protonated alcohol.
CH₃–CH₂–O–H + H⁺ \( \xrightarrow{\text{Fast}} \) CH₃–CH₂–O⁺H₂ (Protonated ethanol / Ethyl oxonium ion)
Step 2: Formation of carbocation: It is the slowest step or rate determining step.
CH₃–CH₂–O⁺H₂ \( \xrightarrow{\text{Slow}} \) CH₃–CH₂⁺ + H₂O
Step 3: Formation of ethene by elimination of a proton.
H–CH₂–CH₂⁺ ⇄ CH₂=CH₂ (Ethene) + H⁺
(ii) On reaction with FeCl₃ phenol gives violet, blue and red colour, while ethanol gives no reaction.

Teacher's Note:
1. Step 2 is the slow, rate determining step – mention it, as examiners often award a mark for identifying it.

Question 22. Write the mechanism of the following reaction:
2CH₃CH₂OH \( \xrightarrow[413\,K]{\text{conc. }H_2SO_4} \) CH₃CH₂–O–CH₂CH₃ [CBSE Delhi 2016]
Answer:
CH₃–CH₂–OH + H⁺ → CH₃–CH₂–O⁺H₂ (Ethyl oxonium ion)
CH₃–CH₂–O(H) + CH₃–CH₂–O⁺H₂ → CH₃–CH₂–O⁺(H)–CH₂–CH₃ + H₂O
CH₃–CH₂–O⁺(H)–CH₂–CH₃ → CH₃–CH₂–O–CH₂–CH₃ + H⁺

Teacher's Note:
1. Temperature decides the product: at 413 K an alcohol molecule attacks the protonated alcohol (S_N2) to give ether, at 443 K elimination gives ethene.

Question 23. Explain the following behaviours:
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.
(ii) ortho-nitrophenol is more acidic than ortho-methoxyphenol. [CBSE OD 2012]
Answer:
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses because alcohols form hydrogen bonds with water.
(ii) The electron withdrawing groups like –NO₂ increase the stability of phenoxide ion by dispersal of negative charge. The electron releasing groups like –OCH₃ decrease the acidic strength of phenol by destabilising the phenoxide ion by concentrating negative charge. Thus, o-nitrophenol is more acidic than o-methoxyphenol.

Teacher's Note:
1. Solubility falls as the alkyl part grows, because the larger hydrophobic chain outweighs the H-bonding of the single OH group.

Question 24. (i) Arrange the following compounds in the increasing order of their acid strength: p-cresol, p-nitrophenol, phenol
(ii) Write the mechanism (using curved arrow notation) of the following reaction:
CH₂=CH₂ + H₃O⁺ → CH₃–CH₂⁺ + H₂O [CBSE OD Set-1, 2, 3 2017]
Answer:
(i) p-cresol < Phenol < p-nitrophenol
(ii) The π electrons of the C=C bond attack a hydrogen of H₃O⁺ (curved arrow from the double bond to H), and the O–H bond electrons move onto oxygen (curved arrow from the O–H bond to O):
CH₂=CH₂ + H–O⁺(H)–H ⇄ H–CH₂–CH₂⁺ + H₂O

Teacher's Note:
1. Use arrows to show the transfer of electrons: each curved arrow starts at an electron pair (a bond or a lone pair) and ends where that pair goes.

Question 25. Write the structures of the products when Butan-2-ol reacts with the following:
(i) CrO₃
(ii) SOCl₂ [CBSE OD Set-1, 2, 3 2017]
Answer:
(i) CH₃–CO–CH₂–CH₃ (Butan-2-one)
(ii) CH₃–CHCl–CH₂–CH₃ (2-Chlorobutane)

Teacher's Note:
1. A secondary alcohol is oxidised by CrO₃ only up to the ketone stage; SOCl₂ replaces OH by Cl and the by-products SO₂ and HCl escape as gases.

Question 26. How do you convert:
(i) Phenol to toluene
(ii) Formaldehyde to Ethanol [CBSE OD Set-2 2016]
Answer:
(i) C₆H₅OH (Phenol) + Zn \( \xrightarrow{\Delta} \) C₆H₆ (Benzene) + ZnO
C₆H₆ + CH₃Cl \( \xrightarrow{\text{anhyd. }AlCl_3} \) C₆H₅CH₃ (Toluene) + HCl
(ii) HCHO (Formaldehyde) + CH₃MgBr \( \xrightarrow{\text{dry ether}} \) CH₃CH₂OMgBr \( \xrightarrow{H_2O/H^+} \) CH₃CH₂OH (Ethanol) + Mg(OH)Br

Teacher's Note:
1. Grignard reagents with HCHO give 1° alcohols with one carbon more than the alkyl group; with other aldehydes they give 2° and with ketones 3° alcohols.

Short Answer Questions (3 Marks)

Question 27. The following is not an appropriate reaction for the preparation of tert.-butyl ethyl ether:
C₂H₅O⁻Na⁺ + (CH₃)₃C–Cl → (CH₃)₃C–OC₂H₅
(i) What would be the major product of the given reaction?
(ii) Write a suitable reaction for the preparation of tert-butyl ethyl ether, specifying the names of reagents used.
Justify your answer in both cases. [CBSE Sample Paper 2016]
Answer:
(i) The major product of the given reaction is 2-methylprop-1-ene. It is because sodium ethoxide (CH₃–CH₂O⁻Na⁺) is a strong nucleophile as well as a strong base. Thus, elimination reaction predominates over substitution reaction.
(CH₃)₃C–Cl (tert-butyl chloride) + CH₃–CH₂O⁻Na⁺ (Sodium ethoxide) → CH₃–C(CH₃)=CH₂ (Isobutylene, 2-methyl prop-1-ene) + NaCl + CH₃CH₂OH
(ii) To prepare tertiary butylethyl ether, the alkyl halide should be primary, i.e., ethyl chloride and the nucleophile should be tertiary, i.e., tertiary butoxide ion. It is because the reaction occurs by S_N2 mechanism and primary alkyl halides are most reactive in S_N2 reactions.
(CH₃)₃C–O⁻Na⁺ (Sodium-tert-butoxide) + CH₃CH₂Cl (Ethyl chloride) → (CH₃)₃C–OC₂H₅ (tert-Butyl ethyl ether) + NaCl

Teacher's Note:
1. In Williamson synthesis the bulky group goes on the alkoxide and the halide must be primary – reversing them gives an alkene.

Question 28. Answer the following questions:
(i) Why are ethers insoluble in water?
(ii) Complete the reaction equation:
? → CH₃CH₂CH₂–C(OH)(CH₃)–CH₂CH₂CH₃
(iii) How will you know whether a given OH group is alcoholic or phenolic in nature?
Answer:
(i) Ethers are insoluble in water because due to the bigger size of the alkyl groups, the oxygen atom in ethers fails to form intermolecular H-bonds with water.
(ii) CH₃CH₂CH=C(CH₃)–CH₂CH₂CH₃ (4-Methylhept-3-ene) →^H₂O/H+ CH₃CH₂CH₂–C(OH)(CH₃)–CH₂CH₂CH₃ (4-Methylheptan-4-ol)
(iii) Phenolic OH group gives blue or violet colouration with neutral FeCl₃, while alcoholic OH group does not.

Teacher's Note:
1. In (ii), Markovnikov hydration puts –OH on the carbon that gives the 3° carbocation, which is C-4 bearing the methyl group.

Question 29. A compound ‘A’ having molecular formula C₄H₁₀O is found to be soluble in concentrated sulphuric acid. It does not react with sodium metal or potassium permanganate. On heating with excess of HI, it gives a single alkyl halide. Deduce the structure of compound A and explain all the reactions.
Answer:
(i) As compound A does not react with sodium metal or potassium permanganate, it cannot be an alcohol.
(ii) As compound A dissolves in conc. H₂SO₄, it may be an ether.
(iii) As compound A on heating with excess of HI gives a single alkyl halide, therefore, compound A must be a symmetrical ether.
(iv) The only symmetrical ether having molecular formula C₄H₁₀O is diethyl ether. Thus compound ‘A’ is diethylether, CH₃–CH₂–O–CH₂–CH₃.
CH₃CH₂–O–CH₂CH₃ (Diethyl ether, A) →^{conc. H₂SO₄} [CH₃–CH₂–O⁺(H)–CH₂–CH₃]HSO₄⁻ (Soluble oxonium salt)
CH₃CH₂OCH₂CH₃ (Diethyl ether, A) →^{HI, Δ} 2CH₃CH₂–I (Ethyl iodide)

Teacher's Note:
1. With excess HI, the alcohol first formed is also converted to the alkyl iodide, so a symmetrical ether gives just one product.

Question 30. Write the structures of the main products in the following reactions:
(i) Methyl (2-oxocyclohexyl)ethanoate (cyclohexanone ring carrying –CH₂–CO–OCH₃ at C-2) →^NaBH₄
(ii) C₆H₅–CH=CH₂ + H₂O →^H+
(iii) C₆H₅–OC₂H₅ + HI → [CBSE 2018]
Answer:
(i) Methyl (2-oxocyclohexyl)ethanoate →^NaBH₄ Methyl (2-hydroxycyclohexyl)ethanoate (the ring C=O becomes CH–OH; the –CH₂–CO–OCH₃ ester group is unchanged)
(ii) C₆H₅–CH=CH₂ (Ethenyl benzene) + H₂O →^H+ C₆H₅–CH(OH)–CH₃ (1-Phenylethanol)
(iii) C₆H₅–OC₂H₅ (Ethoxy benzene) + HI → C₆H₅–OH (Phenol) + C₂H₅–I (Iodoethane)

Teacher's Note:
1. NaBH₄ reduces aldehydes and ketones but not esters, so only the ring ketone is reduced in (i).

Question 31. Write the products of the following reactions:
(i) CH₃–CH₂–O–CH₃ + HI →
(ii) C₆H₅OH + Br₂ →^CS₂
(iii) C₆H₅OCH₃ + CH₃COCl →^{anhyd. AlCl₃}
Answer:
(i) CH₃–CH₂–O–CH₃ (Ethyl methyl ether) + HI → CH₃–I (Methyl iodide) + CH₃–CH₂–OH (Ethyl alcohol)
(ii) C₆H₅OH (Phenol) + Br₂ →^{CS₂, 273 K} 2-Bromophenol (Minor) + 4-Bromophenol (Major)
(iii) C₆H₅OCH₃ (Anisole) + CH₃–CO–Cl →^{Anhyd. AlCl₃} 2-Methoxyacetophenone (Minor) + 4-Methoxyacetophenone (Major)

Teacher's Note:
1. In a non-polar solvent like CS₂ at low temperature, phenol gives only monobromophenols; with bromine water it gives 2,4,6-tribromophenol.

Question 32. What happens when
(i) (CH₃)₃C–OH is treated with Cu at 573 K,
(ii) Anisole is treated with CH₃Cl/anhydrous AlCl₃,
(iii) Phenol is treated with Zn dust?
Write chemical equations in support of your answer. [CBSE (F) 2017]
Answer:
(i) H₃C–C(CH₃)(OH)–CH₃ (tert-butyl alcohol) →^{Cu, 573 K} H₃C–C(CH₃)=CH₂ (Isobutylene) + H₂O
(ii) C₆H₅–O–CH₃ (Anisole) →^{+ CH₃–Cl, Anhyd. AlCl₃} 2-methoxytoluene (Minor) + 4-methoxytoluene (Major)
(iii) C₆H₅OH (Phenol) + Zn → C₆H₆ (Benzene) + ZnO

Teacher's Note:
1. With hot copper, 1° and 2° alcohols are dehydrogenated to aldehydes and ketones, but a 3° alcohol has no α-H and is dehydrated to an alkene instead.

Question 33. What happens when
(i) Salicylic acid is treated with (CH₃CO)₂O/H⁺?
(ii) Phenol is oxidised with Na₂Cr₂O₇/H⁺?
(iii) Anisole is treated with CH₃Cl/anhydrous AlCl₃?
Write chemical equation in support of your answer. [CBSE 2019 (56/3/1)]
Answer:
(i) o-HO–C₆H₄–COOH (Salicylic acid) + (CH₃CO)₂O (Acetic anhydride) →^H+ o-CH₃COO–C₆H₄–COOH (Aspirin) + CH₃COOH
(ii) C₆H₅OH (Phenol) →^{Na₂Cr₂O₇, H₂SO₄} Benzoquinone (O=C₆H₄=O, p-quinone)
(iii) C₆H₅OCH₃ (Anisole) →^{+ CH₃Cl, Anhyd. AlCl₃/CS₂} 2-methoxytoluene (Minor) + 4-methoxytoluene (Major)

Teacher's Note:
1. Aspirin formation is acetylation of the phenolic –OH; the –COOH group of salicylic acid is untouched.

Question 34. Write the main product(s) in each of the following reactions:
(i) (CH₃)₃C–O–CH₃ + HI → ?
(ii) CH₃–CH=CH₂ \( \xrightarrow[(ii)\ 3H_2O_2/OH^-]{(i)\ B_2H_6} \) ?
(iii) C₆H₅–OH \( \xrightarrow[(ii)\ CO_2,\ H^+]{(i)\ aq.\,NaOH} \) ? [CBSE Delhi 2016]
Answer:
(i) (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH (Methanol)
(ii) CH₃–CH=CH₂ \( \xrightarrow[(ii)\ 3H_2O_2/OH^-]{(i)\ B_2H_6} \) CH₃CH₂CH₂OH (Propanol)
(iii) C₆H₅OH \( \xrightarrow[(ii)\ CO_2,\ H^+]{(i)\ aq.\,NaOH} \) o-HO–C₆H₄–COOH, 2-Hydroxybenzoic acid (Salicylic acid)

Teacher's Note:
1. With a tertiary alkyl group the ether cleaves by S_N1, so the halide goes to the tertiary carbon; with two primary/methyl groups (S_N2) the iodide goes to the smaller group.

Question 35. Write the final product(s) in each of the following reactions:
(i) (CH₃)₃C–O–CH₃ + HI →
(ii) CH₃–CH₂–CH(OH)–CH₃ \( \xrightarrow{Cu/573\,K} \)
(iii) C₆H₅–OH \( \xrightarrow[(ii)\ H^+]{(i)\ CHCl_3 + aq.\,NaOH} \) [CBSE OD Central 2016]
Answer:
(i) (CH₃)₃C–O–CH₃ + 2HI → (CH₃)₃C–I + CH₃I + H₂O
2-Iodo-2-methylpropane
(ii) CH₃–CH₂–CH(OH)–CH₃ \( \xrightarrow{Cu/573\,K} \) CH₃–CH₂–CO–CH₃ + H₂
Butan-2-one
(iii) C₆H₅–OH \( \xrightarrow[(ii)\ H^+]{(i)\ CHCl_3 + aq.\,NaOH} \) o-HO–C₆H₄–CHO (Salicylaldehyde)

Teacher's Note:
1. With one mole of HI the products of (i) are (CH₃)₃CI and CH₃OH; excess HI converts the CH₃OH further to CH₃I.

Question 36. How are the following conversions carried out?
(i) Propene to propane-2-ol
(ii) Benzyl chloride to Benzyl alcohol
(iii) Anisole to p-Bromoanisole [CBSE Comptt. Delhi 2015]
Answer:
(i) CH₃CH=CH₂ + H₂O \( \xrightarrow{H^+} \) CH₃–CH(OH)–CH₃
(ii) C₆H₅CH₂Cl + aq. NaOH → C₆H₅CH₂OH + NaCl
(iii) C₆H₅OCH₃ (Anisole) \( \xrightarrow{Br_2\ \text{in ethanoic acid}} \) p-Br–C₆H₄–OCH₃ (p-bromoanisole, major) + o-Br–C₆H₄–OCH₃

Teacher's Note:
1. Acid catalysed hydration follows Markovnikov’s rule, so propene gives the secondary alcohol propan-2-ol, not propan-1-ol.

Question 37. How are the following conversions carried out?
(i) Benzyl chloride to Benzyl alcohol
(ii) Ethyl magnesium chloride to Propan-1-ol
(iii) Propene to Propan-2-ol [CBSE Comptt. OD 2015]
Answer:
(i) C₆H₅CH₂Cl (Benzyl chloride) \( \xrightarrow{aq.\,NaOH} \) C₆H₅CH₂OH (Benzyl alcohol)
(ii) CH₃–CH₂MgCl (Ethyl magnesium chloride) + H–CO–H \( \xrightarrow{\text{Dry ether}} \) CH₃–CH₂–CH₂–OMgCl \( \xrightarrow{H_3O^+} \) CH₃–CH₂–CH₂–OH (Propan-1-ol)
(iii) CH₃CH=CH₂ (Propene) \( \xrightarrow[H_2SO_4]{H_2O} \) CH₃–CH(OH)–CH₃ (Propan-2-ol)

Teacher's Note:
1. To lengthen the chain by one carbon and get a primary alcohol, react the Grignard reagent with formaldehyde (HCHO).

Question 38. Write the major products in the following equations:
(i) CH₃–CH₂OH \( \xrightarrow{PCl_5} \) ?
(ii) C₆H₅OH + CH₃Cl \( \xrightarrow{\text{anhyd. }AlCl_3} \) ?
(iii) CH₃–Cl + CH₃CH₂–ONa → ? [CBSE Comptt. OD 2015]
Answer:
(i) CH₃–CH₂OH \( \xrightarrow{PCl_5} \) CH₃CH₂Cl
(ii) C₆H₅OH + CH₃Cl \( \xrightarrow{\text{Anhyd. }AlCl_3} \) o-CH₃–C₆H₄–OH + p-CH₃–C₆H₄–OH
(iii) CH₃Cl + CH₃CH₂–ONa → CH₃CH₂–O–CH₃

Teacher's Note:
1. The –OH group is ortho/para directing, so Friedel-Crafts methylation of phenol gives a mixture of o- and p-cresol.

Long Answer Questions (5 Marks)

Question 39. An aromatic compound ‘A’ on treatment with CHCl₃/KOH gives two compounds ‘B’ and ‘C’. Both B and C give the same product ‘D’ when distilled with zinc dust. Oxidation of D gives E having molecular formula C₇H₆O₂. The sodium salt of E on heating with sodalime gives F which may also be obtained by distilling A with zinc dust. Identify A to F.
Answer:
A = C₆H₅OH (Phenol);   B = o-HO–C₆H₄–CHO (o-Hydroxy benzaldehyde);   C = p-HO–C₆H₄–CHO (p-Hydroxy benzaldehyde)
D = C₆H₅CHO (Benzaldehyde);   E = C₆H₅COOH (Benzoic acid);   F = C₆H₆ (Benzene)
Reactions involved:
C₆H₅OH (A) →^CHCl₃/KOH o-HO–C₆H₄–CHO (B) + p-HO–C₆H₄–CHO (C)
B or C →^{Zn (dust) distillation} C₆H₅CHO (D) →^(O) C₆H₅COOH (E) →^{(i) NaOH; (ii) NaOH/CaO, 630 K} C₆H₆ (F)
C₆H₅OH (A) →^{Zn (dust) distillation} C₆H₆ (F)

Teacher's Note:
1. C₇H₆O₂ for E points straight to benzoic acid; decarboxylation with sodalime removes CO₂ to give benzene.

Question 40. (i) Give equations of the following reactions:
(a) Phenol is treated with conc. HNO₃.
(b) Propene is treated with B₂H₆ followed by H₂O₂/OH⁻.
(c) Sodium t-butoxide is treated with CH₃Cl.
(ii) How will you distinguish between butan-1-ol and butan-2-ol?
(iii) Arrange the following in increasing order of acidity:
Phenol, ethanol, water [CBSE 2019 (56/2/1)]
Answer:
(i) (a) C₆H₅OH (Phenol) →^{Conc. HNO₃} 2,4,6-Trinitrophenol
(b) CH₃–CH=CH₂ (Propene) →^{(i) B₂H₆; (ii) H₂O₂/OH−} CH₃CH₂CH₂OH (Propanol)
(c) (CH₃)₃C–O⁻Na⁺ (Sodium tert.butoxide) + CH₃–Cl → (CH₃)₃C–O–CH₃ (Methyl tert. butyl ether) + NaCl
(ii) Butan-2-ol when warmed with I₂ in NaOH (NaOI) gives yellow precipitate of iodoform while butan-1-ol does not respond to this test.
CH₃–CH₂–CH(OH)–CH₃ (Butan-2-ol) + 4I₂ + 6NaOH →^Δ CHI₃↓ (Iodoform, yellow ppt) + CH₃CH₂COONa + 5NaI + 5H₂O
(iii) Ethanol < water < phenol

Teacher's Note:
1. The iodoform test is positive only for alcohols with the CH₃–CH(OH)– unit (and for ethanol), so it separates butan-2-ol from butan-1-ol.