CBSE Class 12 Chemistry Alcohols Phenols and Ethers Case Study Questions Set 01

Case Study 1

Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more at o- and p-positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of electrophilic substitution reaction. Reimer-Tiemann reaction introduces an aldehyde group on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method for the synthesis of substituted salicylaldehydes. In this reaction, treatment of phenol with chloroform in presence of aqueous sodium or potassium hydroxide at 340 K followed by hydrolysis of the reacting product gives 2-hydroxybenzaldehyde as the major product.
Phenol →^{CHCl₃ + (aq) NaOH, 340 K} [o-Cl₂HC–C₆H₄–O⁻Na⁺] →^NaOH o-OHC–C₆H₄–O⁻Na⁺ →^H₃O+ Salicylaldehyde (2-Hydroxybenzaldehyde)

Question 1. What is the electrophile used in Reimer-Tiemann reaction?
Answer:
:CCl₂

Teacher's Note:
1. Dichlorocarbene is formed in situ: CHCl₃ + OH⁻ → ⁻CCl₃ + H₂O, then ⁻CCl₃ → :CCl₂ + Cl⁻.

Question 2. Draw the structure of the intermediate in the following reaction:

Answer:

Sodium salt of 2-(dichloromethyl)-4-methylphenol, i.e. the –CHCl₂ group ortho to –O⁻Na⁺ and CH₃ para to it.

Teacher's Note:
1. The CHCl₂ group is later hydrolysed by NaOH to –CHO; the para position is already blocked by CH₃, so substitution goes ortho.

Question 3. Write the IUPAC name of the product formed in the equation given in above question.
Answer:
2-hydroxy-5-methylbenzaldehyde

Teacher's Note:
1. In the name, the –CHO carbon is C-1 because benzaldehyde is the parent, so –OH is at 2 and CH₃ at 5.

Question 4. Which is more reactive, phenol or benzene towards electrophilic substitution?
Answer:
Phenol is more reactive than benzene towards electrophilic substitution reaction because in case of phenol, the intermediate carbocation is more resonance stabilized.

Teacher's Note:
1. The +R effect of –OH raises electron density at the ortho and para positions, which is why phenol reacts even without a catalyst.

Question 5. What happens when carbon tetrachloride is used in the reaction instead of chloroform?
Answer:
Salicylic acid is formed as the major product.

Teacher's Note:
1. With CCl₄ the ortho group introduced is –CCl₃, which hydrolyses to –COOH instead of –CHO.

Case Study 2

Williamson synthesis is one of the best methods for the preparation of ethers. It involves the treatment of an alkyl halide with a suitable sodium alkoxide. Williamson’s synthesis involves nucleophilic substitution of the halogen atom of an alkyl halide by an alkoxide group as shown below:
R–O⁻Na⁺ + R′X → R–O–R′ + NaX
When alkyl halide used in the reaction is primary, Williamson’s synthesis proceeds via S_N2 mechanism leading to the formation of an ether. This method is a versatile method for the synthesis of both symmetrical and unsymmetrical ethers.

Question 6. Why is Williamson’s synthesis not applicable when the alkyl halide used is tertiary?
Answer:
This is because 3° alkyl halides are highly susceptible to dehydrohalogenation in the presence of sodium alkoxides and elimination reaction occurs.

Teacher's Note:
1. Alkoxides are strong bases as well as nucleophiles; with a crowded 3° halide they remove a β-hydrogen and give an alkene.

Question 7. How would you obtain allyl phenyl ether?
Answer:
Allyl phenyl ether can be obtained by heating CH₂=CHCH₂Br and C₆H₅–O⁻Na⁺.
CH₂=CHCH₂Br + C₆H₅ONa → CH₂=CHCH₂–O–C₆H₅ + NaBr

Teacher's Note:
1. The phenyl group must come from the phenoxide, never from bromobenzene, because aryl halides do not undergo S_N2 reactions.

Question 8. What happens when benzyl ethyl ether reacts with HI?
Answer:
Benzyl iodide and ethyl alcohol will be formed after the reaction of benzyl ethyl ether with HI.
C₆H₅CH₂–O–CH₂CH₃ + HI → C₆H₅CH₂I + CH₃CH₂OH

Teacher's Note:
1. Cleavage occurs on the benzyl side because the benzyl carbocation is resonance stabilised (S_N1).

Question 9. Name the alkyl halide and sodium alkoxide used to synthesise tert-butyl ethyl ether.
Answer:
Ethyl bromide and sodium tert-butoxide.

Teacher's Note:
1. Always choose the 1° alkyl halide and put the bulky 3° group on the alkoxide.

Question 10. (CH₃)₃C–O–CH₃ on reaction with HI gives (CH₃)₃C–I and CH₃–OH as the main products and not (CH₃)₃C–OH and CH₃–I. Give reason.
Answer:
The reaction between (CH₃)₃COCH₃ and HI follows S_N1 mechanism. For an S_N1 reaction, the formation of product is controlled by stability of the carbocation formed in the slowest step.
Since tert.butyl carbonium ion (CH₃)₃C⁺ formed after the cleavage of C–O bond in the slowest step is more stable than methyl carbonium ion (CH₃⁺) therefore (CH₃)₃C–I and CH₃OH are the main products.

Teacher's Note:
1. Compare with methyl ethyl ether + HI, which is S_N2: there iodide attacks the smaller methyl group, giving CH₃I.

Case Study 3

Alcohols are weak acids even weaker than water. Their acidic character, however, decreases in the order : 1° alcohols > 2° alcohols > 3° alcohols and the basic character of their alkoxides in the reverse order. The presence of electron-withdrawing groups further increases the acidity of alcohols. Alcohols react with halogen acids to form the corresponding haloalkanes. Chloroalkanes can, however, be prepared by action of SOCl₂, PCl₃ and PCl₅.

Question 11. Arrange the following alcohols in order of their increasing acid strength ; CH₃CH₂OH, CF₃CH₂OH, CCl₃CH₂OH.
Answer:
Since F has stronger –I-effect than Cl, therefore, acid strength increases in the order :
CH₃CH₂OH < CCl₃CH₂OH < CF₃CH₂OH.

Teacher's Note:
1. Electron-withdrawing groups stabilise the alkoxide ion by dispersing its negative charge; the stronger the –I effect, the stronger the acid.

Question 12. Arrange the following alcohols in order of their increasing reactivity towards sodium.
2-methyl-2-propanol, 1-butanol and 2-butanol.
Answer:
Since acid strength of alcohols increases in the order : 3° < 2° < 1°, therefore, reactivity with Na metal increases in the same order.
2-methyl-2-propanol (3°) (least reactive) < 2-butanol (2°) (more reactive) < 1-butanol (1°) (most reactive)

Teacher's Note:
1. Reaction with sodium is an acid-base reaction (H₂ is released from the O–H), so it follows the acidity order.

Question 13. Why is thionyl chloride method considered better than phosphorus halides method for the preparation of chloroalkanes ?
Answer:
The chloroalkanes obtained by thionyl chloride method is almost pure because bye products SO₂ and HCl are gases.
R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑

Teacher's Note:
1. With PCl₅ or PCl₃ the by-products (POCl₃, H₃PO₃) are liquids or solids that must be separated from the product.

Question 14. t-Butoxide ion is a stronger base than methoxide ion but methoxide ion is a stronger nucleophile than t-butoxide ion. Explain.
Answer:
Due to +I-effect of CH₃ groups, (CH₃)₃CO⁻ is a much stronger base than CH₃O⁻. Nucleophilicity, on the other hand, depends upon the size of the alkoxide ion. Since CH₃O⁻ ion is much smaller in size than (CH₃)₃CO⁻ ion, therefore, CH₃O⁻ is a stronger nucleophile than (CH₃)₃CO⁻ ion.

Teacher's Note:
1. Basicity is attack on a small proton, so bulk hardly matters; nucleophilicity is attack on a carbon, where bulk (steric hindrance) matters a lot.

Case Study 4

Phenols are acidic in nature. They turn blue litmus red and give violet colouration with FeCl₃. The presence of electron-donating groups decreases while electron-withdrawing groups increases the acidity of phenols. A characteristic feature of phenols is that they undergo electrophilic substitution reactions such as halogenation, nitration, sulphonation, etc. Since OH group is a strong electron-donating group, substitution occurs predominantly at o- and p-positions.

Question 15. Out of o-nitrophenol and o-methoxyphenol which is more acidic and why ?
Answer:
–NO₂ group is electron-withdrawing while –OCH₃ group is electron-donating, therefore, o-nitrophenol is a stronger acid than o-methoxyphenol.

Teacher's Note:
1. An electron-withdrawing group stabilises the phenoxide ion; an electron-donating group destabilises it.

Question 16. With Br₂ water, phenol gives 2, 4, 6-tribromophenol but with Br₂ in CS₂, it predominantly gives p-bromophenol.
Answer:
In aqueous solution, phenol ionizes to form phenoxide ion which activates the benzene ring to a larger extent. As a result, ring gets highly activated and hence trisubstitution occurs.
On the other hand, in non-polar solvents like CS₂, ionization is greatly suppressed. Consequently, oxygen of the OH group donates electrons to the benzene ring to a small extent. As a result, ring gets activated to a small extent and hence monosubstitution occurs. Further, due to steric hindrance at the o-position, p-substitution product usually predominates.

Teacher's Note:
1. The solvent controls the outcome: a polar solvent also polarises the Br₂ molecule, adding to the effect.

Question 17. p-Hydroxybenzenesulphonic acid on treatment with Br₂-water gives 2, 4, 6-tribromophenol rather than 2, 6-dibromophenol-4-sulphonic acid. Explain why so ?
Answer:
Since SO₃H is a good leaving group, therefore, during bromination, it is knocked out and hence 2, 4, 6-tribromophenol is formed.

Teacher's Note:
1. Replacement of a group other than H by an electrophile (here –SO₃H by Br) is called ipso substitution.

Question 18. Give reasons for the following :
(i) Anisole is less reactive than phenol towards electrophilic substitution reactions.
(ii) Unlike phenol, picric acid is soluble in aqueous NaHCO₃ solution.
Answer:
(i) The resonance structures of phenol (with a positive charge on oxygen) become stable by releasing a highly stable proton, but the corresponding resonance structures of anisole cannot stabilize themselves by releasing a highly unstable CH₃⁺ carbocation. Therefore, anisole is less reactive than phenol towards electrophilic substitution reactions.
(ii) Presence of three –NO₂ groups at o- and p-positions makes picric acid a very strong acid as compared to phenol, therefore, picric acid decomposes NaHCO₃ to release CO₂ but phenol does not.

Teacher's Note:
1. Picric acid (pK_a about 0.4) is stronger than carbonic acid, while phenol (pK_a 10) is weaker; that decides the reaction with NaHCO₃.

Case Study 5

The most versatile method for the synthesis of both symmetrical and unsymmetrical ethers is Williamson’s synthesis which involves the reaction between an alkyl/aralkyl halide (but not aryl halide) with an alkoxide ion derived from a primary, secondary or a tertiary alcohol. Since ethers are Lewis bases, they do not react with organic and inorganic bases but form oxonium salts with protonic acids or complexes with Lewis acids. With halogen acids, preferably HI or HBr having good nucleophilic anions, on heating at 373 K undergo cleavage to form the corresponding alcohol and the alkyl halide. This cleavage may occur either by S_N1 or S_N2 mechanism depending upon the structure of the ether and polarity of the solvent.

Question 19. Why are Grignard reagents prepared in ethers and not in benzene ? Explain.
Answer:
Due to presence of lone pairs of electrons on the oxygen atom, ethers form coordinate complexes with Grignard reagents but benzene does not. As a result, Grignard reagents dissolve in ether.
2 (CH₃CH₂)₂O (Diethyl ether) + RMgX → RMgX·2O(CH₂CH₃)₂ (Mg coordinated to the two ether oxygens)

Teacher's Note:
1. Benzene has no lone pairs to donate, so it cannot solvate (stabilise) the magnesium atom.

Question 20. Using Williamson’s synthesis how will you prepare anisole ?
Answer:
Anisole is prepared by treating sodium phenoxide with methyl iodide
C₆H₅–O⁻Na⁺ (Sod. phenoxide) + CH₃–I \( \xrightarrow[-NaI]{S_N2} \) C₆H₅–OCH₃ (Anisole)

Teacher's Note:
1. Iodobenzene with sodium methoxide would not work, because aryl halides do not undergo S_N2 reactions.

Question 21. Identify Z in the following sequence :
CH₃CH₂CH=CH₂ \( \xrightarrow{HBr/H_2O_2} \) Y \( \xrightarrow{C_2H_5ONa} \) Z
Answer:
CH₃CH₂–CH=CH₂ \( \xrightarrow[(\text{Anti-Mark addition})]{HBr/H_2O_2} \) CH₃CH₂CH₂–CH₂Br (Y) \( \xrightarrow[-NaBr]{CH_3CH_2\text{-}ONa} \) CH₃CH₂CH₂CH₂–O–CH₂CH₃ (Z)
Z is 1-ethoxybutane (butyl ethyl ether).

Teacher's Note:
1. The peroxide effect (anti-Markovnikov addition) works only with HBr, giving the 1° bromide that is ideal for Williamson synthesis.

Question 22. Phenyl methyl ether reacts with HI to give phenol and methyl iodide and not iodobenzene and methyl alcohol. Why ?
Answer:
Protonation of phenyl methyl ether (anisole) gives oxonium ion. In this ion, O–CH₃ bond is weaker than O–C₆H₅ bond which has some double bond character due to resonance between the lone pair of electrons on the oxygen atom and the sp²-hybridized carbon atom of the phenyl ring. Therefore, attack by I⁻ ion exclusively breaks the weaker O–CH₃ bond forming methyl iodide and phenol.
C₆H₅–O–CH₃ \( \xrightarrow{H^+} \) C₆H₅–O⁺(H)–CH₃ \( \xrightarrow{I^-} \) C₆H₅–OH (Phenol) + CH₃–I (Methyl iodide)

Teacher's Note:
1. Remember the rule for aryl alkyl ethers: the halogen always ends up on the alkyl group.

Case Study 6

The most versatile method used for the synthesis of all the three classes of alcohols (1°, 2° and 3°) is the action of a suitable Grignard reagent on aldehydes, ketones or esters followed by acid hydrolysis. The other useful method for the preparation of alcohols is from alkenes either by acid-catalysed hydration or by hydroboration-oxidation and oxymercuration-demercuration.

Case Study 7

Alcohols and phenols are acidic in nature. Presence of electron-donating groups decrease while that of electron-withdrawing groups increase the acidity of phenols. Alcohols undergo nucleophilic substitution reactions with hydrogen halides to give alkyl halides. In contrast, presence of OH groups in phenols activates the ring towards electrophilic substitution reactions such as halogenation, nitration, sulphonation and Friedel-Crafts reaction.