CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Arrange the following in decreasing order of their acidic character:
(i) CH₃O–C₆H₄–OH (p-methoxyphenol)   (ii) C₆H₅OH   (iii) O₂N–C₆H₄–OH (p-nitrophenol)
Answer:
(iii) > (ii) > (i)

Teacher's Note:
1. –NO₂ (electron withdrawing) stabilises the phenoxide ion and –OCH₃ (electron releasing) destabilises it, with phenol in between.

Question 2. Among HI, HBr and HCl, HI is most reactive towards alcohols. Why?
Answer:
Due to lowest bond dissociation energy of HI.

Teacher's Note:
1. The H–I bond is the longest and weakest, and I⁻ is also the best nucleophile of the three: reactivity HI > HBr > HCl.

Question 3. Name a compound which is used as antiseptic as well as disinfectant.
Answer:
Solution of phenol: 0.2% antiseptic, 1% disinfectant.

Teacher's Note:
1. The same compound acts as an antiseptic or a disinfectant depending only on its concentration.

Question 4. What is nitrating mixture?
Answer:
Conc. (H₂SO₄ + HNO₃)

Teacher's Note:
1. Conc. H₂SO₄ protonates HNO₃ to produce the nitronium ion, NO₂⁺, the actual electrophile in nitration.

Question 5. Lower alcohols are soluble in water, higher alcohols are not. Why?
Answer:
Due to formation of hydrogen bonds with water molecules. In higher alcohols the large hydrophobic alkyl part prevents this hydrogen bonding, so they are not soluble.

Teacher's Note:
1. Solubility falls as the alkyl (hydrocarbon) part grows, because only the –OH end can hydrogen bond with water.

Question 6. What happens when CH₃CH₂OH is heated with red P and HI?
Answer:
C₂H₅OH + 2HI —(Red P)→ C₂H₆ + I₂ + H₂O
Ethanol is reduced to ethane.

Teacher's Note:
1. Red P removes the I₂ formed (as PI₃), which keeps the reduction going in the forward direction.

Question 7. Complete the following reaction:
C₆H₅OH + 3HNO₃ —(conc. H₂SO₄)→ [ ? ] + 3H₂O

Answer:
[ ? ] = 2,4,6-trinitrophenol (picric acid)

Teacher's Note:
1. With conc. HNO₃ all three activated positions (2, 4, 6) of phenol are nitrated; dilute HNO₃ gives only o- and p-nitrophenol.

Question 8. Ethanol has higher boiling point than methoxymethane. Give reason.
Answer:
Because of H-bonds: ethanol molecules are associated by intermolecular hydrogen bonding, which is absent in methoxymethane.

Teacher's Note:
1. Both are C₂H₆O isomers – the difference comes only from the O–H bond, which ethers lack.

Question 9. Explain Kolbe’s reaction with example.
Answer:
Sodium phenoxide reacts with CO₂ at 400 K and 4–7 atm to give sodium salicylate, which on acidification gives salicylic acid.
C₆H₅ONa + CO₂ —(400 K, 4–7 atm)→ o-HO–C₆H₄–COONa —(dil. HCl, −NaCl)→ o-HO–C₆H₄–COOH (salicylic acid)

Teacher's Note:
1. Phenoxide (not phenol) is used because the O⁻ makes the ring reactive enough to attack the weak electrophile CO₂.

Question 10. Fill in the blank: Glucose —(Zymase)→ ______ + CO₂
Answer:
C₂H₅OH

Teacher's Note:
1. Zymase (from yeast) ferments glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.

Question 11. Fill in the blank: C₆H₅OH + Br₂ —(H₂O)→ ______
Answer:
2,4,6-Tribromophenol (white precipitate)

Teacher's Note:
1. In water phenol ionises and the very reactive ring is brominated at all three o/p positions.

Question 12. Fill in the blank: Salicylic acid (o-HO–C₆H₄–COOH) + (CH₃CO)₂O —(H⁺)→ ______
Answer:
o-CH₃COO–C₆H₄–COOH (acetylsalicylic acid, aspirin)

Teacher's Note:
1. Only the phenolic –OH is acetylated; the –COOH group is unchanged. Acetic acid is the by-product.

Question 13. Fill in the blank: (CH₃)₂CHOCH₂CH₃ —(HI)→ ______
Answer:
(CH₃)₂CHOH and CH₃CH₂I

Teacher's Note:
1. With a 1° and a 2° group, I⁻ attacks the less hindered (ethyl) carbon by S_N2.

Question 14. Fill in the blank: Phenol + Zn (dust) → ______
Answer:
C₆H₆

Teacher's Note:
1. Zinc dust removes the oxygen of phenol (C₆H₅OH + Zn → C₆H₆ + ZnO).

Question 15. Fill in the blank: Phenol + Na → ______
Answer:
C₆H₅ONa (+ H₂)

Teacher's Note:
1. Evolution of H₂ with sodium shows the acidic O–H, but alcohols give it too – it does not distinguish phenol from alcohols.

Question 16. Fill in the blank: R₂CHOH + HCl + ZnCl₂ (anhydrous) → R₂CHCl + ______
Answer:
H₂O

Teacher's Note:
1. This is the Lucas test; a 2° alcohol gives turbidity of the insoluble alkyl chloride in about 5 minutes.

Short Answer Questions (2 Marks)

Question 17. Write mechanism of reaction of HI with methoxymethane.
Answer:
Step 1: Protonation of the ether: CH₃–Ö–CH₃ + H–I ⇄ CH₃–O⁺(H)–CH₃ + I⁻
Step 2: I⁻ attacks the methyl carbon (S_N2): I⁻ + CH₃–O⁺(H)–CH₃ → [I···CH₃···O(H)–CH₃]^δ+ → CH₃I + CH₃OH
With excess HI, CH₃OH is further converted into CH₃I.

Teacher's Note:
1. Protonation turns the poor leaving group CH₃O⁻ into the good leaving group CH₃OH – that is why the ether is cleaved only by a strong acid.

Question 18. Arrange in order of boiling points:
(i) C₂H₅–O–C₂H₅, C₄H₉COOH, C₄H₉OH
(ii) C₃H₇CHO, CH₃COC₂H₅, C₂H₅COOCH₃, (CH₃CO)₂O
Answer:
(i) C₄H₉COOH > C₄H₉OH > C₂H₅–O–C₂H₅
(ii) (CH₃CO)₂O > C₂H₅COOCH₃ > CH₃COC₂H₅ > C₃H₇CHO

Teacher's Note:
1. Carboxylic acids boil highest because they form H-bonded dimers; ethers have no O–H and cannot hydrogen bond among themselves.

Question 19. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?
(i) 1-bromo-4-nitrobenzene + CH₃ONa
(ii) sodium 4-nitrophenoxide + CH₃Br
Answer:
(ii) p-O₂N–C₆H₄–ONa + CH₃Br → p-O₂N–C₆H₄–OCH₃ + NaBr
It is because of double bond character of the C–Br bond in 1-bromo-4-nitrobenzene due to resonance; it is less reactive towards S_N reaction.

Teacher's Note:
1. In Williamson’s synthesis the halogen must be on an alkyl carbon (preferably CH₃ or 1°), never on the benzene ring.

Question 20. Ethers are relatively inert. Justify.
Answer:
Due to absence of any active site in their molecules: the divalent oxygen is linked to carbon atoms on both sides (C–Ö–C), and there is no reactive O–H bond.

Teacher's Note:
1. This low reactivity is exactly why diethyl ether is used as a solvent, e.g. for preparing Grignard reagents.

Question 21. How will you distinguish between CH₃OH and C₂H₅OH?
Answer:
By the iodoform test. Ethanol on warming with I₂ and Na₂CO₃ gives a yellow precipitate of iodoform:
C₂H₅OH + 4I₂ + 3Na₂CO₃ —(warm)→ CHI₃ (yellow) + HCOONa + 5NaI + 2H₂O + 3CO₂
CH₃OH does not give this test.

Teacher's Note:
1. In a distinguishing test always state the observation (yellow precipitate) for one compound and its absence for the other.

Question 22. Show how will you synthesise the following alcohol by the reaction of a suitable Grignard reagent and methanal ?
CH₃–CH(CH₃)–CH₂OH (CBSE 2019)
Answer:
Disect the given molecule as shown below :
CH₃–CH(CH₃)– | –CH₂OH : the part CH₃–CH(CH₃)– comes from the Grignard reagent and the part –CH₂OH comes from methanal.
Thus, the Grignard reagent needed for the given synthesis is isopropylmagnesium bromide.
CH₃–CH(CH₃)–MgBr (Isopropylmagnesium bromide) + HCHO (Methanal) → CH₃–CH(CH₃)–CH₂OMgBr \( \xrightarrow{H^+/H_2O} \) CH₃–CH(CH₃)–CH₂OH (2-Methylpropan-1-ol)

Teacher's Note:
1. Methanal with any Grignard reagent always gives a 1° alcohol RCH₂OH; other aldehydes give 2° and ketones give 3° alcohols.

Question 23. Arrange the following in increasing order of their boiling points.
Propanol, propane, propanal (CBSE 2017)
Answer:
Due to intermolecular H-bonding, the b.p. of propanol is much higher than that of propanal but due to weak van der Waals forces of attraction between the molecules of propane, its b.p. is much lower than those of propanal and propanol. In other words, the b.ps. of these three compounds increase in the order :
propane < propanal < propanol

Teacher's Note:
1. Propanal is in the middle because its molecules are held by dipole-dipole forces, stronger than van der Waals forces but weaker than H-bonds.

Question 24. Arrange the following in increasing order of basicity : H₂O, OH⁻, CH₃OH, CH₃O⁻
Answer:
Consider the reaction, CH₃O⁻ + H₂O → CH₃OH + OH⁻. Since a stronger base displaces a weaker base from its compounds, therefore, CH₃O⁻ is a stronger base than OH⁻ ion and CH₃OH is a stronger base than H₂O. Thus, the order of increasing basicity is :
H₂O, CH₃OH, OH⁻, CH₃O⁻

Teacher's Note:
1. The anion of the weaker acid is the stronger base: CH₃OH is a weaker acid than H₂O, so CH₃O⁻ is a stronger base than OH⁻.

Question 25. Why (CH₃)₃COH is less acidic than (CH₃)₃SiOH although carbon is more electronegative than Si ?
Answer:
Carbon does not have d-orbitals and hence does not form dπ – pπ bonds. In contrast, silicon has empty d-orbitals and hence it forms dπ – pπ bonds with oxygen. As a result, transference of electrons occurs from O to Si. In other words, electron-density around O decreases. It, therefore, withdraws electrons of the O–H bond towards itself.
Consequently, O–H bond in (CH₃)₃SiOH breaks more easily than in (CH₃)₃COH and hence (CH₃)₃SiOH is more acidic than (CH₃)₃COH.

Teacher's Note:
1. Electronegativity is not the only factor: back-donation of the O lone pair into the empty orbitals of Si pulls electron density away from the O–H bond.

Question 26. Acid-catalysed dehydration of t-butanol is faster than that of n-butanol. Give reasons in one or two sentences.
Answer:
Acid-catalysed dehydration of alcohols occurs through the intermediate formation of carbocations.
(CH₃)₃C–OH (t-Butanol) \( \xrightarrow[-H_2O]{+H^+} \) (CH₃)₃C⁺ (t-Butyl carbocation, more stable)
CH₃CH₂CH₂CH₂–OH (n-Butanol) \( \xrightarrow[-H_2O]{+H^+} \) CH₃CH₂CH₂C⁺H₂ (n-Butyl carbocation, less stable)
Since t-butyl carbocation is more stable than n-butyl carbocation, therefore, acid-catalysed dehydration of t-butanol occurs faster than n-butanol.

Teacher's Note:
1. The rate-determining step is carbocation formation, so the ease of dehydration follows carbocation stability: 3° > 2° > 1°.

Question 27. Although both allyl alcohol and 1-propanol are primary alcohols, they can still be distinguished by Lucas reagent. Explain how?
Answer:
The allyl carbocation is resonance stabilized and hence is as stable as a 3° carbocation.
CH₂=CH–CH₂–OH (Allyl alcohol) \( \xrightarrow[-H_2O]{+H^+} \) CH₂=CH–C⁺H₂ ↔ C⁺H₂–CH=CH₂ (Allyl cation)
Therefore, on treatment with Lucas reagent, allyl alcohol gives white turbidity in less than 1 min. 1-Propanol, on the other hand, does not react with Lucas reagent.
CH₂=CH–CH₂OH (Allyl alcohol) \( \xrightarrow[(\text{Lucas reagent})]{HCl + ZnCl_2} \) CH₂=CH–CH₂Cl (Turbidity in < 1 min.)
CH₃CH₂CH₂OH (1-Propanol) \( \xrightarrow[(\text{Lucas reagent})]{HCl + ZnCl_2} \) No action

Teacher's Note:
1. The Lucas test really measures carbocation stability, so allylic and benzylic 1° alcohols behave like 3° alcohols.

Question 28. How do you account for the fact that unlike phenol, 2,4-dinitrophenol and 2,4,6-trinitrophenol are soluble in aqueous sodium carbonate solution ?
Answer:
Due to the strong electron-withdrawing nature of –NO₂ groups, both 2,4-dinitrophenol and 2,4,6-trinitrophenol are more acidic than carbonic acid and hence dissolve in aq. Na₂CO₃ solution to form the corresponding sodium salts with the evolution of CO₂.

Teacher's Note:
1. Phenol (pK_a 10) is weaker than carbonic acid, so it does not react with Na₂CO₃/NaHCO₃; picric acid (pK_a 0.4) is strong enough to liberate CO₂.

Short Answer Questions (3 Marks)

Question 29. Write structures of the major products of the following:
(i) Mononitration of 3-methylphenol
(ii) Dinitration of 3-methylphenol
(iii) Mononitration of phenyl ethanoate
Answer:
–OH and –CH₃ are o- and p-directing groups. The products are:
(i) 3-methyl-4-nitrophenol and 5-methyl-2-nitrophenol (3-methyl-6-nitrophenol)
and
(ii) 3-methyl-4,6-dinitrophenol (5-methyl-2,4-dinitrophenol)

(iii) 4-nitrophenyl ethanoate

Teacher's Note:
1. –OH is a stronger activator than –CH₃, so the nitro groups enter ortho/para to –OH; the crowded position between the two groups is usually avoided.

Question 30. Complete the following reactions:
(i) CH₃CH₂CH₂CHO —(Pd/H₂ or Ni)→
(ii) CH₃CH=CHCH₂OH —(PCC)→
(iii) CH₃CH=CH₂ —((1) B₂H₆, (2) H₂O₂/OH⁻)→
(iv) C₆H₅OH —((1) aq. NaOH, (2) CO₂, H⁺)→
(v) CH₂Br–CH₂Br —(KOH (aq))→
(vi) C₂H₅NH₂ —(HNO₂)→
Answer:
(i) CH₃CH₂CH₂CH₂OH
(ii) CH₃CH=CHCHO
(iii) CH₃CH₂CH₂OH
(iv) o-HO–C₆H₄–COOH (salicylic acid)
(v) CH₂OH–CH₂OH
(vi) CH₃CH₂OH

Teacher's Note:
1. PCC is a mild oxidant: it converts the –CH₂OH to –CHO and leaves the C=C double bond untouched.

Question 31. Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO₄ solution.
(ii) Bromine in CS₂ with phenol.
(iii) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:
(i) CH₃–CH₂–CH₂OH + 2[O] —(alkaline KMnO₄)→ CH₃CH₂COO⁻K⁺ (on acidification gives propanoic acid)
(ii) C₆H₅OH + Br₂ —(CS₂, low temperature)→ o-bromophenol + p-bromophenol (major) + HBr
(iii) C₆H₅OH + CHCl₃ —(NaOH, 343 K)→ o-OHC–C₆H₄–ONa + p-OHC–C₆H₄–ONa (Reimer-Tiemann reaction); acidification gives salicylaldehyde (major).

Teacher's Note:
1. Solvent decides bromination of phenol: in non-polar CS₂ only monobromophenols form, while bromine water gives 2,4,6-tribromophenol.

Question 32. Describe the following reactions with examples:
(i) Reimer-Tiemann reaction
(ii) Kolbe’s reaction
(iii) Friedel-Crafts acylation of anisole
Answer:
(i) On treating phenol with chloroform in presence of NaOH, a –CHO group is introduced at the ortho position of the ring:
C₆H₅OH —(CHCl₃, NaOH)→ o-Na⁺O⁻–C₆H₄–CHCl₂ —(NaOH, −2NaCl)→ o-NaO–C₆H₄–CH(OH)₂ —(−H₂O, then H⁺)→ o-HO–C₆H₄–CHO (salicylaldehyde)
(ii) Sodium phenoxide reacts with CO₂ to give salicylic acid:
C₆H₅ONa + CO₂ —(400 K, 4–7 atm)→ o-HO–C₆H₄–COONa —(dil. HCl, −NaCl)→ o-HO–C₆H₄–COOH (salicylic acid)
(iii) Anisole reacts with acetyl chloride in presence of anhydrous AlCl₃ to give o- and p-methoxyacetophenone:
C₆H₅OCH₃ + CH₃COCl —(anhy. AlCl₃)→ 2-methoxyacetophenone [10%] + 4-methoxyacetophenone [90%]

Teacher's Note:
1. –OCH₃ is o/p-directing; the para product dominates because the ortho position is sterically hindered.

Question 33. Dehydration of alcohols to form an alkene is always carried out with conc. H₂SO₄ and not with conc. HCl or HNO₃. Explain.
Answer:
In acidic medium alcohols get protonated and then lose H₂O to form a carbocation.
With HCl, Cl⁻ is a strong nucleophile and causes nucleophilic substitution (alkyl chloride is formed instead of alkene); HNO₃ causes oxidation.

Teacher's Note:
1. HSO₄⁻ is a very weak nucleophile, so the carbocation loses H⁺ (elimination) instead of being attacked.

Question 34. How will you convert:
(i) Phenol to cyclohexanol
(ii) Benzyl chloride to benzyl alcohol
(iii) Anisole to phenol
Answer:
(i) C₆H₅OH + 3H₂ —(Ni, heat)→ C₆H₁₁OH (cyclohexanol)
(ii) C₆H₅CH₂Cl + KOH (aq) → C₆H₅CH₂OH + KCl
(iii) C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I

Teacher's Note:
1. Benzyl chloride hydrolyses easily with aqueous KOH because Cl is on an sp³ carbon, unlike chlorobenzene where Cl is on the ring.

Question 35. Arrange the following compounds in the increasing order of the property indicated against each. Give reasons for your answer.
(i) CH₃CH₂OH, CF₃CH₂OH, CCl₃CH₂OH — Acid strength
(ii) 2-methyl-2-propanol, 1-butanol and 2-butanol — Reactivity towards sodium.
Answer:
(i) Due to –I-effect of the halogen, the electron density in the O–H bond decreases. As a result of this electron-deficiency, the O–H bond weakens and thus facilitates the release of a proton as compared to CH₃CH₂OH. Further, since F has stronger –I-effect than Cl, therefore, CF₃CH₂OH is a stronger acid than CCl₃CH₂OH while CH₃CH₂OH is the weakest acid. Thus, acid strength increases in the order :
CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH
(ii) It is an acid-base type reaction since alcohols are acidic in nature and sodium, a highly reactive metal, displaces their acidic hydrogen. As such, the reactivity of these alcohols towards sodium increases as the acidic character of alcohols increases. Now since the acidic character of alcohols increases in the order : 3° < 2° < 1°, therefore, the reactivity of Na towards alcohols increases in the same order, i.e.,
2-methyl-2-propanol (3°) (least reactive) < 2-butanol (2°) (more reactive) < 1-butanol (1°) (most reactive)

Teacher's Note:
1. Electron-withdrawing groups increase the acidity of alcohols; electron-releasing alkyl groups decrease it.

Question 36. Explain why alcohols do not react with NaBr but when H₂SO₄ is added they form alkyl bromides.
Answer:
R–OH + Br⁻ (Strong nucleophile) ↛ R–Br + OH⁻ (Strong base, bad leaving group)
Although Br⁻ is a strong nucleophile yet OH⁻ being a strong base is a bad leaving group. Therefore, Br⁻ cannot displace OH⁻ from ROH to form R–Br. In other words, NaBr does not react with alcohols.
However, when H₂SO₄ is added, protonation of alcohol occurs, i.e., –OH is converted into –OH₂⁺. Since H₂O is a very weak base, therefore, it is a good leaving group. In other words, Br⁻ can displace H₂O from protonated alcohol to form RBr.
R–OH (Alcohol) + H⁺ ⇌ R–OH₂⁺ (Protonated alcohol)
Br⁻ (Strong nucleophile) + R–OH₂⁺ → R–Br (Alkyl bromide) + H₂O (Good leaving group)

Teacher's Note:
1. Weak bases are good leaving groups: acid converts the poor leaving group OH⁻ into the good leaving group H₂O.

Question 37. Give the structure of the compound, C (C₄H₈) which when treated with H₂O/H₂SO₄ gives C₄H₁₀O which cannot be resolved into optical isomers.
Answer:
Since alkene C (C₄H₈) on treatment with H₂O/H₂SO₄ gives an alcohol C₄H₁₀O, which cannot be resolved, therefore, the alkene must be 2-methylpropene, (CH₃)₂C=CH₂.
(CH₃)₂C=CH₂ (2-Methylpropene) (C) + H⁺ → (CH₃)₂C⁺–CH₃ \( \xrightarrow{H_2\ddot{O}:} \) (CH₃)₂C(O⁺H₂)–CH₃ \( \xrightarrow{-H^+} \) (CH₃)₃C–OH (Optically inactive alcohol, M.F. C₄H₁₀O)
The other two possible alkenes having M.F. C₄H₈ are 1-butene and 2-butene. These upon hydration will give 2-butanol which is optically active.
CH₃CH₂CH=CH₂ (1-Butene) \( \xrightarrow[H_2SO_4]{H_2O} \) CH₃CH₂–C*H(OH)–CH₃ (2-Butanol, optically active)
CH₃CH=CHCH₃ (2-Butene) \( \xrightarrow[H_2SO_4]{H_2O} \) CH₃CH₂–C*H(OH)–CH₃ (2-Butanol, optically active)

Teacher's Note:
1. An alcohol is optically inactive when its carbinol carbon carries two identical groups; (CH₃)₃COH has three identical CH₃ groups.

Question 38. 3,3-Dimethylbutan-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as the major product. Suggest a suitable mechanism.
Answer:
(CH₃)₃C–CH(OH)–CH₃ (3,3-Dimethylbutan-2-ol) \( \xrightarrow{+H^+} \) (CH₃)₃C–CH(O⁺H₂)–CH₃ \( \xrightarrow{-H_2O} \) (CH₃)₃C–C⁺H–CH₃ (2° Carbocation, less stable)
(CH₃)₃C–C⁺H–CH₃ \( \xrightarrow[\text{shift}]{1,2\text{-Methyl}} \) (CH₃)₂C⁺–CH(CH₃)–CH₃ (3° Carbocation, more stable)
(CH₃)₂C⁺–CH(CH₃)–CH₃ \( \xrightarrow{-H^+} \) (CH₃)₂C=C(CH₃)₂ (Tetramethylethylene or 2,3-Dimethyl-2-butene)

Teacher's Note:
1. Whenever a 2° carbocation is next to a carbon carrying CH₃ groups or H, look for a 1,2-shift that gives a more stable 3° carbocation.

Question 39. Predict the major product of the following reaction :
C₆H₆ + (CH₃)₂CHCH₂OH \( \xrightarrow{H_2SO_4} \)
Answer:
In presence of conc. H₂SO₄, isobutyl alcohol (I) first gives a 1° carbocation (II) which then rearranges to the more stable 3° carbocation (III) by 1,2-hydride shift. Carbocation (III), then reacts with benzene to form tert-butylbenzene.
(CH₃)₂CH–CH₂OH (Isobutyl alcohol) (I) \( \xrightarrow{H^+} \) (CH₃)₂CH–CH₂–O⁺H₂ \( \xrightarrow{-H_2O} \) (CH₃)₂CH–C⁺H₂ (1° Carbocation, less stable) (II)
(CH₃)₂CH–C⁺H₂ \( \xrightarrow[\text{shift}]{1,2\text{-Hydride}} \) (CH₃)₃C⁺ (3° Carbocation, more stable) (III)
(CH₃)₃C⁺ + C₆H₆ \( \xrightarrow[(\text{F.C. alkylation})]{-H^+} \) C₆H₅–C(CH₃)₃ (tert-Butylbenzene)

Teacher's Note:
1. In Friedel-Crafts alkylation the carbocation may rearrange before it attacks the ring, so the alkyl group in the product can differ from that in the reagent.

Question 40. Write down the structure of the product of the following reactions :
(i) The alcohol shown below (6,6-dimethylcyclohex-3-en-1-ol) \( \xrightarrow{H^+} \) ?
(ii) C₆H₅–CH=CH₂ + H₂O \( \xrightarrow{H^+} \) ? (CBSE 2018)

Answer:
(i) Protonation of alcohol (I) followed by loss of H₂O gives 2° carbocation (II) which undergoes 1,2-methyl shift to afford more stable 3° carbocation (III). Loss of a proton from (III) gives isolated diene (IV). Acid-catalysed isomerization of (IV) ultimately gives the more stable conjugated diene (V) as the final product.
6,6-Dimethylcyclohex-3-en-1-ol (I) \( \xrightarrow{+H^+} \) protonated alcohol (–OH₂⁺ on C-1) \( \xrightarrow{-H_2O} \) 2° carbocation at C-1 (II, less stable)
(II) \( \xrightarrow[\text{shift}]{1,2\text{-Methyl}} \) 3° carbocation at the ring carbon carrying the remaining CH₃ group (III, more stable)
(III) \( \xrightarrow{-H^+} \) 1,2-dimethylcyclohexa-1,4-diene (Diene IV, isolated) \( \xrightarrow{+H^+,\ -H^+} \) 1,2-dimethylcyclohexa-1,3-diene (Diene V, conjugated)
(ii) C₆H₅–CH=CH₂ (Styrene) + H⁺ → C₆H₅–C⁺H–CH₃ (Stabilized by resonance over the benzene ring) \( \xrightarrow{H_2O} \) C₆H₅–CH(O⁺H₂)–CH₃ \( \xrightarrow{-H^+} \) C₆H₅–CH(OH)–CH₃ (1-Phenylethanol)

Teacher's Note:
1. In acid-catalysed hydration of styrene, H⁺ adds to the terminal CH₂ so that the positive charge sits on the benzylic carbon, which the ring stabilises by resonance.

Long Answer Questions (5 Marks)

Question 41. Explain a process in which a biocatalyst is used in industrial preparation of a compound known to you. [NCERT Exemplar]
Answer:
Biocatalysts are complex organic compounds which act as catalysts in reaction taking place in living organism. These biocatalysts (enzymes) are used in the manufacture of ethanol.
Ethanol from sugar solution (molasses):
Molasses is a non-crystalline form of sugar obtained as the mother liquor after crystallization of sugar from sugar solution. This contains about 50% sugar. It is diluted to about 10% solution and yeast is added and kept for about 2-3 days. Yeast supplies the enzymes invertase and zymase. The enzyme invertase hydrolyses sucrose to glucose and fructose, which are converted by zymase to ethanol.
C₁₂H₂₂O₁₁ (Sucrose) + H₂O \( \xrightarrow{\text{Invertase}} \) C₆H₁₂O₆ (Glucose) + C₆H₁₂O₆ (Fructose)
C₆H₁₂O₆ (Glucose or fructose) \( \xrightarrow{\text{Zymase}} \) 2C₂H₅OH (Ethanol) + 2CO₂
In wine making, grapes are the source of sugar and yeast. As grapes ripen, the quantity of sugar increases and the yeast grows on the skin of the grapes. When the grapes are crushed, sugar and enzyme come in contact and fermentation starts. Fermentation takes place under anaerobic conditions (i.e. in the absence of air). During fermentation, CO₂ is released.
The action of enzyme is inhibited when the concentration of alcohol exceeds 14%. If air enters the fermentation mixture, the O₂ released from the air oxidizes ethanol to ethanoic acid which spoils the taste of alcoholic drinks and make it sour.

Teacher's Note:
1. Name both enzymes and their jobs: invertase (sucrose → glucose + fructose) and zymase (glucose/fructose → ethanol + CO₂).

Question 42. An alcohol A (C₄H₁₀O) on oxidation with acidified K₂Cr₂O₇ gives carboxylic acid ‘B’ (C₄H₈O₂). Compound ‘A’ when dehydrated with conc. H₂SO₄ at 443 K gives compound ‘C’. With aqueous H₂SO₄, ‘C’ gives compound ‘D’ (C₄H₁₀O) which is an isomer of ‘A’. Compound ‘D’ is resistant to oxidation but compound ‘A’ can be easily oxidised. Identify A, B, C and D and write their structures.
Answer:
A: (CH₃)₂CHCH₂OH (2-methylpropan-1-ol)
B: CH₃CH(CH₃)COOH (2-methylpropanoic acid)
C: (CH₃)₂C=CH₂ (2-methylpropene)
D: (CH₃)₃C–OH (2-methylpropan-2-ol)

Teacher's Note:
1. D resists oxidation, so it must be a 3° alcohol; the only 3° C₄ alcohol is tert-butyl alcohol, which fixes the carbon skeleton of A.