Given below are two statements: Statement I: Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent. Statement II: Sodium metal can be used to dry diethyl ether and not ethyl alcohol. In the light of given statements, choose the correct answer from the options given below:

### Core Logic An electrophile seeks electrons. Higher electron density in the benzene ring makes it more susceptible (reactive) to electrophilic attack. The ring's electron density is governed by the inductive (I) and mesomeric/resonance (M) effects of the attached groups. ### Step 1: Evaluate Substituent Effects - Cl$Cl$ (in chlorobenzene): shows weak +M effect but strong -I effect, causing net deactivation. - CH_3$CH_3$ (in toluene): shows +I effect and hyperconjugation, slightly activating the ring. - COOH$COOH$ (in benzoic acid): shows strong -M and -I effect, highly deactivating. - OH$OH$ (in phenol): shows strong +M effect which dominates its weak -I effect, strongly activating the ring. ### Step 2: Conclusion Phenol has the highest electron density in the ring among the given options due to the strong +M effect of the -OH group. Thus, it is most easily attacked by an electrophile. ### Pattern Recognition Reactivity towards Electrophilic Aromatic Substitution (EAS): Strong +M (-OH, -NH2) > Weak +I/Hyperconjugation (-CH3) > Halogens (-Cl, net deactivating) > Strong -M (-COOH, -NO2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Alcohols, Phenols and Ethers

### Core Logic Statement I is correct: At the para-position, the -textNO_2$-\text{NO}_2$ 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 (textconc. HCl + textanhydrous ZnCl_2$\text{conc. HCl} + \text{anhydrous ZnCl}_2$) 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^circ > 2^circ > 1^circ$3^{\circ} > 2^{\circ} > 1^{\circ}$). Primary alcohols react very slowly. ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers

### Core Logic Aliphatic alcohols (textBu-OH$\text{Bu-OH}$) are least acidic due to +I$+I$ effects. Among phenols, electron-donating groups like -textOMe$-\text{OMe}$ (+M$+M$ effect) reduce acidity relative to plain phenol, while electron-withdrawing groups (-textNO_2$-\text{NO}_2$, -M$-M$ and -I$-I$ effects) significantly enhance stability of the conjugate phenoxide base. Two -textNO_2$-\text{NO}_2$ groups heighten acidity maximally.

### Step 1: Order verification textBu-OH (A) < textp-methoxyphenol (C) < textphenol (D) < textp-nitrophenol (B) < text2,4-dinitrophenol (E)$$\text{Bu-OH (A)} < \text{p-methoxyphenol (C)} < \text{phenol (D)} < \text{p-nitrophenol (B)} < \text{2,4-dinitrophenol (E)}$$ ### 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 The substrate has two reactive functional groups towards HBr$HBr$ (in excess): 1. An aromatic ether (anisole-type) linkage: -O-CH_2-CH_3$-O-CH_2-CH_3$ 2. An isolated alkene (vinyl) group attached to the aromatic ring: -CH=CH_2$-CH=CH_2$ **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 sp^3$sp^3$ hybridized) alkyl group (S_N2$S_N2$ mechanism), specifically the ethyl group. The C(textaryl)-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 (CH_3-CH_2-Br$CH_3-CH_2-Br$). **Reaction 2: Electrophilic Addition to Alkene** The vinyl group (-CH=CH_2$-CH=CH_2$) 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)-CH_3$-CH(Br)-CH_3$ 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. ### Evaluation Rubric / Model Answer null ### 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 (CHCl_3$CHCl_3$ or HCCl_3$HCCl_3$) 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 + CHCl_3$CHCl_3$ + NaOH$NaOH$ rightarrow$\rightarrow$ Salicylaldehyde. Kolbe's = Phenol + CO_2$CO_2$ + NaOH$NaOH$ rightarrow$\rightarrow$ Salicylic acid. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols Phenols and Ethers