Match List I with List II:

List-I (Compound)	List-II (pKₐ$pK_a$ value)
A. Ethanol	I. 10.0
B. Phenol	II. 15.9
C. m-Nitrophenol	III. 7.1
D. p-Nitrophenol	IV. 8.3

Choose the correct answer from the options given below :

Core Logic

Aliphatic alcohols (Bu-OH$\text{Bu-OH}$) are least acidic due to +I$+I$ effects. Among phenols, electron-donating groups like -OMe$-\text{OMe}$ (+M$+M$ effect) reduce acidity relative to plain phenol, while electron-withdrawing groups (-NO₂$-\text{NO}_2$, -M$-M$ and -I$-I$ effects) significantly enhance stability of the conjugate phenoxide base. Two -NO₂$-\text{NO}_2$ groups heighten acidity maximally.

Step 1: Order verification

Pattern Recognition

+M$+M$ groups lower acidity; -M$-M$ groups raise it. Alcohols are always less acidic than resonant phenols.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

Statement I is correct: At the para-position, the -NO₂$-\text{NO}_2$ group exerts both powerful -I$-I$ and -M$-M$ effects, maximizing electron withdrawal from the phenoxide ion. Intramolecular hydrogen bonding reduces the acidity of o-nitrophenol.

Statement II is incorrect: Lucas reagent (conc. HCl + anhydrous ZnCl₂$\text{conc. HCl} + \text{anhydrous ZnCl}_2$) reacts instantly with tertiary alcohols to give immediate turbidity. Primary alcohols like ethanol do not show turbidity at room temperature without prolonged heating.

Pattern Recognition

Para-nitrophenol acidity maximization vs Lucas test thresholds (3° > 2° > 1°$3^{\circ} > 2^{\circ} > 1^{\circ}$). Primary alcohols react very slowly.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

The substrate has two reactive functional groups towards HBr$HBr$ (in excess):

• An aromatic ether (anisole-type) linkage: -O-CH₂-CH₃$-O-CH_2-CH_3$
• An isolated alkene (vinyl) group attached to the aromatic ring: -CH=CH₂$-CH=CH_2$

Reaction 1: Ether Cleavage The ether linkage reacts with HBr$HBr$. Protonation of the ether oxygen occurs first, forming an oxonium ion. The Br^-$Br^-$ ion then attacks the less hindered (and sp³$sp^3$ hybridized) alkyl group (SN2$S_N2$ mechanism), specifically the ethyl group. The C(aryl)-O$C(\text{aryl})-O$ bond is much stronger due to partial double bond character from resonance, so it does NOT break. This yields a phenol group on the ring and ethyl bromide (CH₃-CH₂-Br$CH_3-CH_2-Br$).

Reaction 2: Electrophilic Addition to Alkene The vinyl group (-CH=CH₂$-CH=CH_2$) undergoes electrophilic addition with HBr$HBr$. Protonation yields the more stable secondary benzylic carbocation (Markovnikov's rule). The Br^-$Br^-$ then attacks this carbocation to form a 1$1$-bromoethyl group attached to the ring.

Step 1: Detailed Mechanism

Final product: The ring retains an -OH$-OH$ group (phenol) at the original ether position, and the vinyl group is converted into a -CH(Br)-CH₃$-CH(Br)-CH_3$ group.

Pattern Recognition

Excess HBr$HBr$ with an aryl-alkyl ether always cleaves the alkyl C-O$C-O$ bond to give phenol + alkyl bromide. Never break the aryl C-O$C-O$ bond.

Chapter Mix

Class 12 Chemistry: Alcohols Phenols and Ethers Class 11 Chemistry: Hydrocarbons

Core Logic

Salicylaldehyde is synthesized from phenol via the Reimer-Tiemann reaction.

In this reaction, phenol is treated with chloroform (CHCl₃$CHCl_3$ or HCCl₃$HCCl_3$) and aqueous sodium hydroxide (NaOH$NaOH$) to introduce an aldehyde group (-CHO$-CHO$) at the ortho position of the benzene ring.

Pattern Recognition

Reimer-Tiemann = Phenol + CHCl₃$CHCl_3$ + NaOH$NaOH$ arrow$\rightarrow$ Salicylaldehyde. Kolbe's = Phenol + CO₂$CO_2$ + NaOH$NaOH$ arrow$\rightarrow$ Salicylic acid.

Chapter Mix

Class 12 Chemistry: Alcohols Phenols and Ethers