CBSE Class 12 Chemistry Alcohols Phenols and Ethers Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Fill in the blank: R–CHOH–R —([O])→ ______
Answer:
R–CO–R (a ketone)

Teacher's Note:
1. Secondary alcohols oxidise to ketones and stop there; primary alcohols go to aldehydes and then acids.

Question 2. Fill in the blank: CH₃–CH(OH)–CH₃ —(SOCl₂)→ CH₃–CH(Cl)–CH₃ + ______
Answer:
SO₂ + HCl

Teacher's Note:
1. SOCl₂ is preferred for making alkyl chlorides because both by-products are gases and escape, leaving the pure product.

Question 3. Fill in the blank: CH₃–CH₂–OH —(Bleaching powder)→ ______
Answer:
CHCl₃

Teacher's Note:
1. Bleaching powder first oxidises ethanol to acetaldehyde, chlorinates it to chloral, and its Ca(OH)₂ then hydrolyses chloral to chloroform.

Question 4. What are primary alcohols ? (Manipur Board 2011)
Answer:
Alcohols in which –OH group is attached to a primary carbon, i.e., alcohols which contain the grouping –CH₂OH are called primary (1°) alcohols. For example,
CH₃–CH₂OH (Ethanol), CH₃CH₂–CH₂OH (Propan-1-ol), CH₃CH₂CH₂–CH₂OH (Butan-1-ol)

Teacher's Note:
1. Methanol, CH₃OH, is also classed as a primary alcohol even though its carbon carries no alkyl group.

Question 5. What is the IUPAC name of the alcohol : HC≡C–CH₂OH ?
Answer:
Prop-2-yn-1-ol.

Teacher's Note:
1. Number from the end nearer the –OH group; the suffix “-ol” gets the lowest locant, so the triple bond is at C-2.

Question 6. Draw the structure of
(i) hex-1-en-3-ol (CBSE 2012)
(ii) 3-aminopentan-2-ol (UP Board 2012)
(iii) 4-tert-butyl-3-iodoheptane (CBSE Sample Paper 2016)
Answer:
(i) CH₃CH₂CH₂–CH(OH)–CH=CH₂ (Hex-1-en-3-ol)
(ii) CH₃CH₂–CH(NH₂)–CH(OH)–CH₃ (3-Aminopentan-2-ol)
(iii) CH₃CH₂–CH(I)–CH(C(CH₃)₃)–CH₂CH₂CH₃ (4-tert-Butyl-3-iodoheptane)

Teacher's Note:
1. Write the parent chain first, number it, and only then hang the substituents on the numbered carbons; this avoids misplacing groups.

Question 7. Give the IUPAC name of [(CH₃)₂CH]₃COH.
Answer:
Expand the given condensed formula into complete structural formula.
CH₃–CH(CH₃)–C(OH)(CH(CH₃)₂)–CH(CH₃)–CH₃
2,4-Dimethyl-3-(1-methylethyl)pentan-3-ol or 3-Isopropyl-2,4-dimethylpentan-3-ol

Teacher's Note:
1. All three isopropyl groups are identical; any two of them together with the carbinol carbon make up the five-carbon parent chain.

Question 8. Give the IUPAC name of the following compound :
(i) CH₃–O–C(CH₃)₂–CH₃ (CBSE 2017)
(ii) H₃C–C(CH₃)=C(Br)–CH₂–OH (CBSE 2017)
(iii) C₆H₅–CH=CH–CH₂–OH (CBSE 2017)
(iv) C₆H₅–CH₂–CH₂–OH (CBSE (Foreign) 2017)
(v) CH₂=CH–C(CH₃)₂–OH (CBSE (Foreign) 2017)
(vi) C₆H₅–C(CH₃)₂–OH (CBSE (Foreign) 2017)
Answer:
(i) 2-Methoxy-2-methylpropane
(ii) 2-Bromo-3-methylbut-2-en-1-ol
(iii) 3-Phenylprop-2-en-1-ol
(iv) 2-Phenylethanol
(v) 2-Methylbut-3-en-2-ol
(vi) 2-Phenylpropan-2-ol

Teacher's Note:
1. In an ether the larger alkyl group is the parent and the smaller one becomes the alkoxy prefix (methoxy here).

Question 9. Write the mechanism (using curved arrow notation) of the following reaction :
(i) CH₂=CH₂ \( \xrightarrow{H_3O^+} \) CH₃–CH₂⁺ + H₂O (CBSE 2017)
(ii) CH₃CH₂–O⁺H₂ \( \xrightarrow{CH_3CH_2OH} \) CH₃CH₂–O⁺(H)–CH₂CH₃ + H₂O (CBSE 2017)
Answer:
(i) The π-electrons of the C=C bond attack a proton of H₃O⁺ (curved arrow from the double bond to H, and from the H–O bond to O):
CH₂=CH₂ + H–O⁺H₂ ⇄ CH₃–CH₂⁺ + H₂O
(ii) The lone pair on the oxygen of a second ethanol molecule attacks the carbon of the protonated ethanol (S_N2), and the C–O bond breaks to release water:
CH₃CH₂–O(H) + CH₃CH₂–O⁺H₂ → CH₃CH₂–O⁺(H)–CH₂CH₃ + H₂O

Teacher's Note:
1. A curved arrow always starts at an electron pair (a lone pair or a bond) and points to the atom that receives it.

Question 10. Convert propene to propan-2-ol. (CBSE Sample Paper 2016)
Answer:
CH₃–CH=CH₂ (Propene) \( \xrightarrow{\text{Conc. }H_2SO_4} \) CH₃–CH(OSO₃H)–CH₃ (Isopropyl hydrogen sulphate) \( \xrightarrow[H_2O]{\text{Boiling}} \) CH₃–CH(OH)–CH₃ (Propan-2-ol)

Teacher's Note:
1. Addition follows Markovnikov’s rule: H goes to the terminal CH₂, so the –OH ends up on the middle carbon.

Question 11. Suggest one method for carrying out anti-Markovnikov’s addition of water indirectly to propylene. Give one example.
Answer:
Hydroboration-oxidation gives alcohols which correspond to anti-Markovnikov’s addition of H₂O to alkenes.
6 CH₃CH=CH₂ (Propene) \( \xrightarrow[(\text{Hydroboration})]{B_2H_6/THF} \) 2 (CH₃CH₂CH₂)₃B (Tri-n-propylborane) \( \xrightarrow[(\text{Oxidation})]{H_2O_2\text{-}NaOH} \) 6 CH₃CH₂CH₂OH (Propan-1-ol)

Teacher's Note:
1. Boron adds to the less substituted carbon of the double bond and is then replaced by –OH, so the product is the 1° alcohol.

Question 12. Convert ethanol to propan-2-ol. (CBSE 2019)
Answer:
First convert ethanol to ethanal. Addition of CH₃MgI to ethanal followed by acid hydrolysis gives propan-2-ol.
CH₃–CH₂–OH (Ethanol) \( \xrightarrow[\text{oxidation}]{PCC} \) CH₃–CHO (Ethanal) \( \xrightarrow[\text{Dry ether}]{CH_3MgI} \) (CH₃)₂CH–OMgI (Addition product) \( \xrightarrow[\text{Hydrolysis}]{H^+/H_2O} \) CH₃–CH(OH)–CH₃ (Propan-2-ol)

Teacher's Note:
1. PCC stops the oxidation at the aldehyde stage; a stronger oxidant would take ethanol to ethanoic acid.

Question 13. Convert propan-2-ol to 2-methylpropan-2-ol. (CBSE 2015, DSB 2015)
Answer:
CH₃–CH(OH)–CH₃ (Propan-2-ol) \( \xrightarrow[\text{or } C_5H_5NH^+CrO_3Cl^-]{Cu/573\text{ K}} \) CH₃–CO–CH₃ (Propanone) \( \xrightarrow[(ii)\ H_3O^+]{(i)\ CH_3MgBr} \) CH₃–C(CH₃)(OH)–CH₃ (2-Methylpropan-2-ol)

Teacher's Note:
1. A ketone with a Grignard reagent always gives a 3° alcohol; an aldehyde (other than HCHO) gives a 2° alcohol.

Question 14. How is ethyl methanoate converted into propan-2-ol? (Maharashtra Board 2012)
Answer:
Ethyl methanoate on reaction with excess of CH₃MgI followed by hydrolysis with dil. H₂SO₄ gives propan-2-ol.
H–CO–OC₂H₅ (Ethyl methanoate) \( \xrightarrow[-Mg(OC_2H_5)I]{CH_3MgI} \) [H–CO–CH₃] (Ethanal) \( \xrightarrow{CH_3MgI} \) CH₃–CH(OMgI)–CH₃ \( \xrightarrow[-Mg(OH)I]{H^+/H_2O} \) CH₃–CH(OH)–CH₃ (Propan-2-ol)

Teacher's Note:
1. An ester reacts with two molecules of Grignard reagent: the first converts it into a carbonyl compound, the second adds to that carbonyl.

Question 15. Name the reagent in the following conversion ?
CH₃–CO–CH₃ \( \xrightarrow{?} \) CH₃–CH(OH)–CH₃ (CBSE 2015, DSB 2015)
Answer:
Lithium aluminium hydride (LiAlH₄) or sodium borohydride (NaBH₄).

Teacher's Note:
1. Reduction of a ketone gives a 2° alcohol; catalytic hydrogenation (H₂/Ni or Pd) is also an acceptable answer.

Short Answer Questions (2 Marks)

Question 16. Unlike phenols, alcohols are easily protonated.
or Alcohols are easily protonated in comparison to phenols.
or Why do phenols not give protonation reactions readily ? (CBSE 2008)
Answer:
In phenols, the lone pairs of electrons on the oxygen atom are delocalized over the benzene ring due to resonance and hence are not easily available for protonation.
In contrast, in alcohols, the lone pairs of electrons on the oxygen atom are localized due to absence of resonance and hence are easily available for protonation.

Teacher's Note:
1. The same delocalisation that makes phenol a stronger acid makes it a weaker base than alcohols.

Question 17. What are alcohols? How are they classified? (J & K Board 2012)
Answer:
Alcohols are hydroxy derivatives of aliphatic hydrocarbons, i.e., compounds in which one or more hydrogen atoms of an alkane are replaced by –OH groups. Their general formula is R–OH (C_nH_2n+1OH for saturated monohydric alcohols).
Classification. (a) On the basis of number of –OH groups : monohydric (one –OH, e.g., CH₃CH₂OH), dihydric (two –OH, e.g., ethane-1,2-diol, HOCH₂CH₂OH), trihydric (three –OH, e.g., glycerol, HOCH₂CH(OH)CH₂OH) and polyhydric alcohols.
(b) Monohydric alcohols on the basis of hybridisation of the carbon carrying –OH :
(i) Compounds containing C(sp³)–OH bond : these are further classified as primary (1°, –CH₂OH, e.g., CH₃CH₂OH), secondary (2°, >CH–OH, e.g., (CH₃)₂CHOH) and tertiary (3°, R₃C–OH, e.g., (CH₃)₃COH) alcohols. Allylic (CH₂=CH–CH₂OH) and benzylic (C₆H₅CH₂OH) alcohols also belong to this class.
(ii) Compounds containing C(sp²)–OH bond : vinylic alcohols (CH₂=CH–OH) and phenols (C₆H₅OH).

Teacher's Note:
1. The 1°/2°/3° label depends on the number of carbon atoms attached to the carbinol carbon, not on the position of –OH in the chain.

Question 18. What are primary, secondary and tertiary alcohols ? Give one example of each.
Answer:
Primary alcohols (1°) : the –OH group is attached to a primary carbon (a carbon attached to one other carbon atom or none), i.e., they contain the group –CH₂OH. Example : ethanol, CH₃CH₂OH.
Secondary alcohols (2°) : the –OH group is attached to a secondary carbon (a carbon attached to two other carbon atoms), i.e., they contain the group >CHOH. Example : propan-2-ol, (CH₃)₂CHOH.
Tertiary alcohols (3°) : the –OH group is attached to a tertiary carbon (a carbon attached to three other carbon atoms). Example : 2-methylpropan-2-ol, (CH₃)₃COH.

Teacher's Note:
1. A quick check: count the H atoms on the carbinol carbon – two (or three) for 1°, one for 2° and none for 3°.

Question 19. How will you convert an alkyl halide into a primary alcohol containing two carbon atoms more than the alkyl halide ?
Answer:
The alkyl halide is first converted into a Grignard reagent which is then treated with ethylene oxide (oxirane), and the adduct is hydrolysed :
R–X + Mg \( \xrightarrow{\text{Dry ether}} \) R–MgX \( \xrightarrow{\text{Ethylene oxide}} \) R–CH₂CH₂–OMgX \( \xrightarrow{H^+/H_2O} \) R–CH₂CH₂–OH + Mg(OH)X
Example : CH₃Br \( \xrightarrow{Mg/\text{ether}} \) CH₃MgBr \( \xrightarrow[(ii)\ H^+/H_2O]{(i)\ \text{Ethylene oxide}} \) CH₃CH₂CH₂OH (Propan-1-ol)

Teacher's Note:
1. Formaldehyde adds one carbon (as –CH₂OH) while ethylene oxide adds two (as –CH₂CH₂OH); both give 1° alcohols.

Question 20. Convert acetone into tert-butyl alcohol.
Answer:
Acetone reacts with methylmagnesium bromide in dry ether and the adduct so formed on hydrolysis gives tert-butyl alcohol :
CH₃–CO–CH₃ (Acetone) + CH₃MgBr \( \xrightarrow{\text{Dry ether}} \) (CH₃)₃C–OMgBr (Adduct) \( \xrightarrow{H^+/H_2O} \) (CH₃)₃C–OH (tert-Butyl alcohol) + Mg(OH)Br

Teacher's Note:
1. All three methyl groups of the product: two come from acetone and one from the Grignard reagent.

Question 21. Give one example of Bouveault-Blanc reduction. (West Bengal Board 2013)
Answer:
Reduction of an ester with sodium and ethanol to a mixture of two primary alcohols is called Bouveault-Blanc reduction. For example,
CH₃COOC₂H₅ (Ethyl acetate) + 4[H] \( \xrightarrow{Na/C_2H_5OH} \) CH₃CH₂OH + C₂H₅OH (Ethanol)

Teacher's Note:
1. The acid part of the ester (RCOO–) becomes RCH₂OH, while the alkyl part is released as its own alcohol.

Question 22. What is hydroboration-oxidation reaction ? Explain with an example. (Hr. Board 2011)
or Write the main product of the following reaction
CH₃–CH=CH₂ \( \xrightarrow[(ii)\ 3H_2O_2/OH^-]{(i)\ B_2H_6} \) (CBSE 2016)
Answer:
Addition of diborane to an alkene to form a trialkylborane, followed by its oxidation with alkaline hydrogen peroxide to give an alcohol, is called hydroboration-oxidation. The overall result is the addition of water to the double bond against Markovnikov’s rule. For example,
6 CH₃–CH=CH₂ (Propene) + B₂H₆ \( \xrightarrow{THF} \) 2 (CH₃CH₂CH₂)₃B (Tri-n-propylborane) \( \xrightarrow{6H_2O_2/OH^-} \) 6 CH₃CH₂CH₂OH (Propan-1-ol) + 2 B(OH)₃
Thus, the main product is propan-1-ol.

Teacher's Note:
1. Hydroboration gives excellent yields and no rearrangement, because no carbocation is formed.

Question 23. Arrange the following in decreasing order of boiling points : 2-methyl-2-propanol, 1-butanol, 2-methyl-1-propanol and 2-butanol
Answer:
Boiling points decrease with increase in branching because the surface area (and hence van der Waals forces) decreases; among isomers, 1° > 2° > 3°, and a straight chain 1° alcohol boils higher than a branched 1° alcohol. Thus :
1-butanol (390 K) > 2-methyl-1-propanol (381 K) > 2-butanol (373 K) > 2-methyl-2-propanol (355 K)

Teacher's Note:
1. All four have the same H-bonding –OH group, so only the shape of the carbon skeleton decides the order here.

Question 24. Why are alcohols less acidic than water ?
Answer:
In alcohols, the alkyl group has +I-effect. It increases the electron density on the oxygen atom and hence in the O–H bond, so the O–H bond becomes stronger and releases H⁺ less readily than water.
Further, the alkoxide ion (RO⁻) formed is less stable than the hydroxide ion (OH⁻) because the alkyl group intensifies the negative charge on oxygen. Hence, alcohols are weaker acids than water. This is shown by the reaction : RO⁻Na⁺ + H₂O → ROH + NaOH (water, the stronger acid, protonates the alkoxide ion).

Teacher's Note:
1. Since water displaces alcohol from sodium alkoxide, alkoxide ions are stronger bases than hydroxide ions.

Question 25. The acidity of alcohols falls in the order : primary > secondary > tertiary. Explain.
Answer:
The 1°, 2° and 3° alcohols have one, two and three alkyl groups respectively on the carbon carrying the –OH group. Alkyl groups have +I-effect, so the electron density in the O–H bond increases in the order 1° < 2° < 3°. Therefore, the O–H bond becomes stronger and more difficult to break in the same order.
Also, the alkoxide ion formed becomes less stable as the number of electron-releasing alkyl groups increases (3° alkoxide is least stable and is also least solvated due to steric hindrance). Hence, acidity : primary > secondary > tertiary.

Teacher's Note:
1. Two words to use in the answer: +I effect (on the O–H bond) and stability/solvation of the alkoxide ion.

Question 26. Why 1° alcohols are more acidic than 2° alcohols? (Hr. Board 2012)
Answer:
A 1° alcohol has one alkyl group while a 2° alcohol has two alkyl groups on the carbon carrying the –OH group. Due to the +I-effect of two alkyl groups, the electron density on oxygen and in the O–H bond of a 2° alcohol is more than that in a 1° alcohol. So, the O–H bond of a 2° alcohol is stronger and releases a proton less readily. Also, the 2° alkoxide ion is less stable (and less solvated) than the 1° alkoxide ion. Hence, 1° alcohols are more acidic than 2° alcohols.

Teacher's Note:
1. The order of basicity of the alkoxide ions is just the reverse: 3° > 2° > 1°.

Short Answer Questions (3 Marks)

Question 27. Arrange the following compounds in the increasing order of their acidic character
p-Nitrophenol, ethanol, phenol
Answer:
Since ethoxide ion is less stable than ethanol (due to +I-effect of the CH₃CH₂ group)
CH₃CH₂–OH (Ethanol, more stable) \( \xrightarrow{-H^+} \) CH₃CH₂–O⁻ (Ethoxide ion, less stable) (destabilized by +I-effect of the CH₃CH₂– group)
C₆H₅OH (Phenol, less stable) \( \xrightarrow{-H^+} \) C₆H₅O⁻ (Phenoxide ion, more stable) (Due to greater stabilization over phenol due to resonance)
while phenoxide ion is more stable than phenol (due to greater resonance stabilization), therefore, phenol is more acidic than ethanol.
Further since electron-withdrawing groups (i.e., –NO₂) increase the acidity of the substituted phenols relative to phenol itself, p-nitrophenol is more acidic than phenol.
Combining both the trends, the overall acidity of the three compounds increases in the order : ethanol < phenol < p-nitrophenol.

Teacher's Note:
1. Acidity depends on how stable the anion is compared with the neutral molecule: resonance and –NO₂ stabilise the phenoxide ion, the ethyl group destabilises ethoxide.

Question 28. Haloalkanes can easily be prepared from alcohols while aryl halides cannot be prepared from phenols. Explain. (Hr. Board 2011)
Answer:
Alcohols being weak bases easily get protonated in presence of strong acids like HCl, HBr or HI to form the corresponding protonated alcohols. Due to the presence of +ve charge on the oxygen atom ; C–O bond in protonated alcohols becomes weak and hence can be easily cleaved by Cl⁻, Br⁻ or I⁻ ion to form the corresponding alkyl halide.
RCH₂–OH (Alcohol) \( \xrightarrow{H^+} \) RCH₂–O⁺H₂ (Protonated alcohol) \( \xrightarrow{X^-} \) RCH₂–X (Alkyl halide) + H₂O
In contrast, phenols are much weaker bases than alcohols due to delocalization of the lone pair of electrons of the oxygen atom over the benzene ring (due to resonance). As a result, phenols are not sufficiently protonated. Further, due to resonance, C–O bond in phenols has some double bond character and hence cannot be easily cleaved by the nucleophiles, such as Cl⁻, Br⁻ or I⁻ ion to form the corresponding aryl halide.
C₆H₅–OH (Phenol, resonance hybrid: δ+ on O, δ− on the ring) \( \xrightarrow{HX} \) × (no reaction) C₆H₅–X + H₂O

Teacher's Note:
1. Two reasons work together for phenol: poor protonation of O and partial double bond character of the C–O bond.

Question 29. PCl₅ reacts with ethanol to form chloroethane. However, with phenol, it does not give chlorobenzene but gives triphenylphosphate. Explain.
Answer:
To begin with both ethanol and phenol react with PCl₅ to give the corresponding chlorophosphate esters.
CH₃CH₂–OH (Ethanol) + Cl₄P–Cl → CH₃CH₂–O–PCl₄ (I) + H⁺ + Cl⁻
C₆H₅OH + Cl₄P–Cl → C₆H₅–O–PCl₄ (II) + H⁺ + Cl⁻
Since in ethanol, C–O bond has only single bond character, therefore, nucleophilic attack by Cl⁻ ion occurs easily on chlorophosphate ester (I) to give chloroethane.
Cl⁻ + CH₃CH₂–O–PCl₄ (I) → CH₃CH₂–Cl (Chloroethane) + POCl₃ + Cl⁻
However, in case of phenol, C–O bond has some double bond character due to resonance and hence difficult to break. Instead nucleophilic attack by C₆H₅OH occurs twice on the chlorophosphate ester (II) to form intermediate (III). Further, nucleophilic attack by C₆H₅OH on III does not occur due to steric hindrance. Instead (III) undergoes hydrolysis by water to form triphenylphosphate.
C₆H₅–O–PCl₄ (II) \( \xrightarrow[-2HCl]{2C_6H_5OH} \) (C₆H₅O)₃PCl₂ (III) \( \xrightarrow[-2HCl]{2H_2O} \) [(C₆H₅O)₃P(OH)₂] \( \xrightarrow{-H_2O} \) (C₆H₅O)₃P=O (Triphenylphosphate)

Teacher's Note:
1. The –OH of phenol cannot be replaced by halogen using PCl₅, HX or SOCl₂, because its C–O bond has partial double bond character.

Question 30. Which of the following is the correct method for synthesising methyl tert-butyl ether and why ?
(i) (CH₃)₃CBr + NaOMe →
(ii) CH₃Br + NaO-tert-Bu →
Answer:
(ii). Substitution always competes with elimination. Since 3° alkyl halides preferentially undergo elimination while 1° alkyl halides preferentially undergo substitution, therefore, reaction (i) will undergo elimination to give isobutylene while reaction (ii) will undergo substitution to give t-butyl methyl ether.
MeO⁻Na⁺ + H–CH₂–C(CH₃)₂–Br (3° Alkyl halide) \( \xrightarrow{\text{Elimination}} \) CH₂=C(CH₃)–CH₃ (Isobutylene) + NaBr + MeOH
(CH₃)₃C–O⁻Na⁺ + CH₃–Br (1° Alkyl halide) \( \xrightarrow{\text{Substitution}} \) (CH₃)₃C–OCH₃ (t-Butyl methyl ether) + NaBr

Teacher's Note:
1. Williamson rule: the halide must be 1° or methyl; any 2°/3° group of the ether should come from the alkoxide.

Question 31. Identify the product of the following reaction :
(CH₃)₃C–CH₂–Br \( \xrightarrow[\Delta]{EtOH} \)
Answer:
In presence of a weak nucleophile such as ethanol, neopentyl bromide ionizes to form first a 1° carbocation which rearranges to form the more stable 3° carbocation. This is then attacked by weak nucleophile ethanol followed by loss of a proton to yield ethyl tert-pentyl ether.
(CH₃)₃C–CH₂–Br (Neopentyl bromide or 2,2-Dimethyl-1-bromopropane) \( \xrightarrow[-Br^-]{\text{Ionization}} \) (CH₃)₃C–C⁺H₂ (1° Carbocation, less stable)
(CH₃)₃C–C⁺H₂ \( \xrightarrow[\text{shift}]{1,2\text{-Methyl}} \) (CH₃)₂C⁺–CH₂CH₃ (3° Carbocation, more stable) \( \xrightarrow{Et\ddot{O}H} \) (CH₃)₂C(O⁺(H)Et)–CH₂CH₃
(CH₃)₂C(O⁺(H)Et)–CH₂CH₃ \( \xrightarrow{-H^+} \) (CH₃)₂C(OEt)–CH₂CH₃ (Ethyl tert-pentyl ether or 2-Ethoxy-2-methylbutane)

Teacher's Note:
1. Weak nucleophile + heat favours S_N1 (solvolysis); with a neopentyl group always expect a 1,2-methyl shift before the nucleophile attacks.

Question 32. Ethers are cleaved only by acids and not by bases.
Answer:
The C–O–C bond in ethers like the C–OH bond in alcohols is quite strong. In order to weaken it, the oxygen atom must be protonated. A subsequent nucleophilic attack by a strong nucleophile such as Br⁻ or I⁻ ion on the less hindered carbon atom of the protonated ethers, C–O⁺(H)–C, brings about the cleavage of ethers to form an alcohol and an alkyl halide. For example,
CH₃CH₂–O–CH₃ (Ethyl methyl ether) \( \xrightarrow[-H^+]{+H^+} \) CH₃CH₂–O⁺(H)–CH₃ (Protonated ether) \( \xrightarrow{I^-} \) CH₃CH₂OH (Ethyl alcohol) + CH₃I (Methyl iodide)
On the other hand, bases cannot weaken the C–O–C bond because ethers, themselves being Lewis bases, are not attacked by strong bases such as OH⁻ or H⁻ (from LiAlH₄) ion. In other words, bases cannot cleave ethers.

Teacher's Note:
1. Ethers are so unreactive towards bases, oxidants and reductants that they are used as solvents, e.g. diethyl ether for Grignard reagents and LiAlH₄.

Question 33. Write down the structure of the product of the following reaction :
PhCH₂–O–CHPh₂ \( \xrightarrow{\text{aq. HI}} \) ? (AIPMT Mains 2009)
Answer:
Protonation of the given ether gives oxonium ion (I) which undergoes C–O cleavage to form benzyl alcohol and the more stable diphenylmethyl carbocation (II). Nucleophilic attack by I⁻ ion on (II) gives diphenylmethyl iodide.
PhCH₂–O–CHPh₂ \( \xrightarrow{H^+} \) PhCH₂–O⁺(H)–CHPh₂ (Oxonium ion I) \( \xrightarrow[\text{cleavage}]{C{-}O} \) PhCH₂OH (Benzyl alcohol) + Ph₂C⁺H (II)
Ph₂C⁺H (II) \( \xrightarrow[(\text{Nucleophilic attack})]{I^-} \) Ph₂CH–I (Diphenylmethyl iodide)
Thus, the products are benzyl alcohol and diphenylmethyl iodide.

Teacher's Note:
1. Between two groups that both form stabilised cations, cleavage gives the more stable carbocation – here Ph₂CH⁺ (two rings) beats PhCH₂⁺.

Question 34. 2, 2-Dimethyloxirane can be cleaved by acid (H⁺). Write its mechanism.
Answer:
In the acidic medium, the oxygen atom of the oxirane ring gets protonated. The presence of positive charge on the oxygen atom weakens the C–O bond more on the side of C₂ since the partial positive charge created on C₂ will be stabilized by the +I-effect of the two CH₃ groups in the transition state (T.S.). Consequently, the attack of the nucleophile, i.e., H₂O preferentially occurs on C₂ yielding 2-methylpropan-1, 2-diol.
(CH₃)₂C(–O–)CH₂ (2,2-Dimethyloxirane, C-2 = the carbon carrying two CH₃ groups, C-1 = CH₂) + H⁺ → protonated oxirane → [T.S.: C₂–O bond partly broken, δ+ on C₂, δ+ on O]
[T.S.] \( \xrightarrow{H_2\ddot{O}:} \) (CH₃)₂C(O⁺H₂)–CH₂–OH \( \xrightarrow{-H^+} \) (CH₃)₂C(OH)–CH₂OH (2-Methylpropane-1, 2-diol)

Teacher's Note:
1. Under acidic conditions the nucleophile attacks the more substituted ring carbon (S_N1-like); under basic conditions it attacks the less hindered carbon (S_N2).

Question 35. Anisole is less reactive than phenol towards electrophilic substitution reactions. Justify your answer with proper reasoning.
Answer:
Both –OH and –OCH₃ groups have +R-effect as shown below :
Phenol : I ↔ II ↔ III ↔ IV ↔ V, where I and V are the two Kekulé structures and in II, III and IV the oxygen carries a positive charge (C=O⁺–H) while the negative charge is on the two ortho carbons and the para carbon respectively.
Anisole : VI ↔ VII ↔ VIII ↔ IX ↔ X, where VI and X are the two Kekulé structures and in VII, VIII and IX the oxygen carries a positive charge (C=O⁺–CH₃) while the negative charge is on the two ortho carbons and the para carbon respectively.
Structures (II to IV) involve separation of +ve and –ve charges. Similarly, structures (VII–IX) involve separation of +ve and –ve charges. But structures (II to IV) can stabilize themselves by releasing a proton which is a very stable chemical species :
II (C=O⁺–H) → C=O (phenoxide-type structure with negative charge on the ring) + H⁺ (Stable species)
VII (C=O⁺–CH₃) → C=O + CH₃⁺ (Unstable species)
In contrast, structures (VII–IX) cannot stabilize themselves by releasing a methyl carbocation which is a very unstable chemical species.
It is because of this extra stability of structures II–IV through loss of a proton which is responsible for greater reactivity of phenol over anisole. For example, phenol reacts with Br₂ water to give 2,4,6-tribromophenol but anisole does not react with Br₂ water at all.

Teacher's Note:
1. Both –OH and –OCH₃ are o,p-directing activators; the difference lies only in how strongly they activate the ring.

Question 36. Give the IUPAC names of the following :
(i) CH₃–CH(OH)–CH₂–CH₂OCH₂CH₃
(ii) CH₃–C(CH₃)₂–CH₂–CH₂OH
(iii) C₆H₁₁–CH₂OH (cyclohexyl ring carrying –CH₂OH)
(iv) C₆H₅–CH₂CH₂OH (CBSE 2016)
(v) CH₃CH₂–C(CH₃)(Br)–CH(Br)–CH₂CH₂OH
(vi) CH₃–CH(OH)–CH(OH)–CH(Br)–CH₂OH
(vii) CH₃–C(CH₃)=C(Br)–CH₂OH
(viii) H₂C=CH–CH(OH)–CH₂–CH₂–CH₃ (CBSE 2009)
(ix) C₆H₅–CH₂–CH(OH)–CH₃ (CBSE 2010)
(x) ClCH₂CH₂–C(CH₃)₂–OH (Hr. Board 2011)
(xi) ClCH₂–C(CH₃)₂–OH (Hr. Board 2011)
(xii) CH₂=CH–CH₂OH (UP Board 2012)
(xiii) The phenol carrying –OH at C-1 and two –NO₂ groups at C-2 and C-5 (CBSE 2015)
(xiv) CH₃–C(CH₃)(C₂H₅)–CH(OH)–CH₃ (CBSE 2018)
Answer:
(i) 4-Ethoxybutan-2-ol
(ii) 3,3-Dimethylbutan-1-ol
(iii) Cyclohexylmethanol
(iv) 2-Phenylethanol
(v) 3,4-Dibromo-4-methylhexan-1-ol
(vi) 2-Bromopentane-1,3,4-triol
(vii) 2-Bromo-3-methylbut-2-en-1-ol
(viii) Hex-1-en-3-ol
(ix) 1-Phenylpropan-2-ol
(x) 4-Chloro-2-methylbutan-2-ol
(xi) 1-Chloro-2-methylpropan-2-ol
(xii) Prop-2-en-1-ol
(xiii) 2,5-Dinitrophenol
(xiv) 3,3-Dimethylpentan-2-ol

Teacher's Note:
1. In (xiv) the ethyl group is part of the longest chain; never name an ethyl branch on C-2 of a chain, it means the chain was chosen wrongly.

Question 37. Draw structures of all isomeric alcohols of molecular formula C₄H₁₀O and give their IUPAC names. (Hr. Board 2011 ; Maharashtra Board 2012)
Answer:
There are four isomeric alcohols of molecular formula C₄H₁₀O :
(i) CH₃CH₂CH₂CH₂OH : Butan-1-ol (1°)
(ii) CH₃CH₂–CH(OH)–CH₃ : Butan-2-ol (2°)
(iii) (CH₃)₂CH–CH₂OH : 2-Methylpropan-1-ol (1°)
(iv) (CH₃)₃C–OH : 2-Methylpropan-2-ol (3°)

Teacher's Note:
1. Butan-2-ol has a chiral carbon, so it exists as a pair of enantiomers; mention this if the question asks for stereoisomers too.

Question 38. Explain the mechanism of following reactions :
(i) Acid-catalysed hydration of an alkene forming an alcohol. (CBSE 2009, 2012)
(ii) Hydration of ethene to form ethanol. (CBSE 2010 S ; Hr. Board 2011)
or Write the mechanism of the following reaction : CH₂=CH₂ + H₂O \( \xrightarrow{H^+} \) CH₃CH₂OH (CBSE 2019)
(iii) Addition of Grignard reagents to carbonyl group of a compound forming an adduct followed by hydrolysis.
Answer:
(i) and (ii) The mechanism involves three steps (shown for ethene) :
Step 1. Protonation of the alkene to form a carbocation by electrophilic attack of H₃O⁺ : H₂O + H⁺ → H₃O⁺ ; CH₂=CH₂ + H–O⁺H₂ ⇄ CH₃–CH₂⁺ + H₂O
Step 2. Nucleophilic attack of water on the carbocation : CH₃–CH₂⁺ + H₂O ⇄ CH₃–CH₂–O⁺H₂
Step 3. Deprotonation to form the alcohol : CH₃–CH₂–O⁺H₂ + H₂O ⇄ CH₃CH₂OH + H₃O⁺
For an unsymmetrical alkene, the more stable carbocation is formed in Step 1, so the addition follows Markovnikov’s rule, e.g., propene gives propan-2-ol.
(iii) Step 1. Nucleophilic addition of the Grignard reagent to the carbonyl group : the C–Mg bond is polar (R^δ−–Mg^δ+X); the carbanion-like R attacks the carbonyl carbon and MgX attaches to oxygen forming an adduct : >C=O + R–MgX → >C(R)–OMgX (Adduct)
Step 2. Hydrolysis of the adduct with dilute acid gives the alcohol : >C(R)–OMgX + H₂O \( \xrightarrow{H^+} \) >C(R)–OH + Mg(OH)X
HCHO gives 1° alcohols, other aldehydes give 2° alcohols and ketones give 3° alcohols.

Teacher's Note:
1. Step 1 (formation of the carbocation) is the slow, rate-determining step of acid-catalysed hydration.

Long Answer Questions (5 Marks)

Question 39. An ether ‘A’ (C₅H₁₂O) when heated with excess of hot concentrated HI produced two alkyl halides which on hydrolysis form compounds B and C. Oxidation of B gives an acid D whereas oxidation of C gave a ketone E. Deduce the structures of A, B, C, D and E.
Answer:
A: CH₃CH₂OCH(CH₃)₂ (2-ethoxypropane)
B: CH₃CH₂OH
C: CH₃CHOHCH₃
D: CH₃COOH
E: CH₃COCH₃

Teacher's Note:
1. The acid tells you B is a 1° alcohol and the ketone tells you C is a 2° alcohol; their carbon counts (2 + 3) must add up to 5.

Question 40. Which of the following compounds gives fastest reaction with HBr and why?
(i) (CH₃)₃COH   (ii) CH₃CH₂CH₂OH   (iii) CH₃CH(OH)CH₃   (iv) CH₃CH(CH₃)CH₂OH
Answer:
(i) (CH₃)₃C–OH
Due to the three electron-donating (+I) methyl groups, it forms the most stable 3° carbocation, (CH₃)₃C⁺.

Teacher's Note:
1. Reaction of alcohols with HX goes through a carbocation, so the order of reactivity is 3° > 2° > 1°.