Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Alcohols react both as nucleophiles and electrophiles. Reason R: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen. In the light of the above statements, choose the correct answer from the options given below:

Related Formula

Core Logic

Comparing the acidic strength among given compounds:

• Ethanol is least acidic due to lack of resonance stabilization in ethoxide ion pKₐ = 15.9$\implies pK_a = 15.9$ (II).
• Phenol forms resonance-stabilized phenoxide pKₐ = 10.0$\implies pK_a = 10.0$ (I).
• m$m$-Nitrophenol experiences -I$-I$ effect of -NO₂$-NO_2$ group pKₐ = 8.3$\implies pK_a = 8.3$ (IV).
• p$p$-Nitrophenol experiences both strong -R$-R$ and -I$-I$ effects, maximizing structural stability pKₐ = 7.1$\implies pK_a = 7.1$ (III).

Step 1: Assembly

This yields the complete matching pattern: A-II, B-I, C-IV, D-III.

Pattern Recognition

Para nitro substitued positions create maximal charge delocalization through resonance, giving it the lowest pKₐ$pK_a$ value out of the options.

Chapter Mix

Class 12 Chemistry: Organic Compounds Containing Oxygen

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