Alcohols Phenols and Ethers Previous Year Questions — JEE Main Chemistry
20 past-year Alcohols Phenols and Ethers questions from JEE Main (Chemistry) — page 2.
Q1392 (2024)
Which of the following compound will most easily be attacked by an electrophile?
- Chlorobenzene
- Toluene
- Benzoic acid
- Phenol
### 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
- $Cl$ (in chlorobenzene): shows weak +M effect but strong -I effect, causing net deactivation.
- $CH_3$ (in toluene): shows +I effect and hyperconjugation, slightly activating the ring.
- $COOH$ (in benzoic acid): shows strong -M and -I effect, highly deactivating.
- $OH$ (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
Q1572 (2024)
The ascending order of acidity of -OH group in the following compounds is:
Choose the correct answer from the options given below:
- $(A) < (D) < (C) < (B) < (E)$
- $(C) < (A) < (D) < (B) < (E)$
- $(C) < (D) < (B) < (A) < (E)$
- $(A) < (C) < (D) < (B) < (E)$
### Core Logic
Aliphatic alcohols ($\text{Bu-OH}$) are least acidic due to $+I$ effects. Among phenols, electron-donating groups like $-\text{OMe}$ ($+M$ effect) reduce acidity relative to plain phenol, while electron-withdrawing groups ($-\text{NO}_2$, $-M$ and $-I$ effects) significantly enhance stability of the conjugate phenoxide base. Two $-\text{NO}_2$ groups heighten acidity maximally. {{SOL_IMG1}}
### Step 1: Order verification
$$\text{Bu-OH (A)} < \text{p-methoxyphenol (C)} < \text{phenol (D)} < \text{p-nitrophenol (B)} < \text{2,4-dinitrophenol (E)}$$
### Pattern Recognition
$+M$ groups lower acidity; $-M$ groups raise it. Alcohols are always less acidic than resonant phenols.
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q71 (2024)
Given below are two statements:
Statement (I): p-nitrophenol is more acidic than m-nitrophenol and o-nitrophenol.
Statement (II) : Ethanol will give immediate turbidity with Lucas reagent.
In the light of the above statements, choose the correct answer from the options given below :
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
- Statement I is false but Statement II is true
### Core Logic
Statement I is correct: At the para-position, the $-\text{NO}_2$ group exerts both powerful $-I$ and $-M$ effects, maximizing electron withdrawal from the phenoxide ion. Intramolecular hydrogen bonding reduces the acidity of o-nitrophenol.
Statement II is incorrect: Lucas reagent ($\text{conc. HCl} + \text{anhydrous ZnCl}_2$) 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 ($3^{\circ} > 2^{\circ} > 1^{\circ}$). Primary alcohols react very slowly.
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q1657 (2024)
The major product(P) in the following reaction is
{{IMG1}}
- ""
- ""
- ""
- ""
### Core Logic
The substrate has two reactive functional groups towards $HBr$ (in excess):
1. An aromatic ether (anisole-type) linkage: $-O-CH_2-CH_3$
2. An isolated alkene (vinyl) group attached to the aromatic ring: $-CH=CH_2$
**Reaction 1: Ether Cleavage**
The ether linkage reacts with $HBr$. Protonation of the ether oxygen occurs first, forming an oxonium ion. The $Br^-$ ion then attacks the less hindered (and $sp^3$ hybridized) alkyl group ($S_N2$ mechanism), specifically the ethyl group. The $C(\text{aryl})-O$ 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 ($CH_3-CH_2-Br$).
**Reaction 2: Electrophilic Addition to Alkene**
The vinyl group ($-CH=CH_2$) undergoes electrophilic addition with $HBr$. Protonation yields the more stable secondary benzylic carbocation (Markovnikov's rule). The $Br^-$ then attacks this carbocation to form a $1$-bromoethyl group attached to the ring.
### Step 1: Detailed Mechanism
{{SOL_IMG1}}
{{SOL_IMG2}}
Final product: The ring retains an $-OH$ group (phenol) at the original ether position, and the vinyl group is converted into a $-CH(Br)-CH_3$ group.
### Pattern Recognition
Excess $HBr$ with an aryl-alkyl ether always cleaves the alkyl $C-O$ bond to give phenol + alkyl bromide. Never break the aryl $C-O$ bond.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols Phenols and Ethers
Class 11 Chemistry: Hydrocarbons
Q1742 (2024)
Salicylaldehyde is synthesized from phenol, when reacted with
- $\text{CO}_2 \text{, NaOH}$
- $\text{CCl}_4 \text{, NaOH}$
- $\text{HCCl}_3 \text{, NaOH}$
### Core Logic
Salicylaldehyde is synthesized from phenol via the Reimer-Tiemann reaction.
In this reaction, phenol is treated with chloroform ($CHCl_3$ or $HCCl_3$) and aqueous sodium hydroxide ($NaOH$) to introduce an aldehyde group ($-CHO$) at the ortho position of the benzene ring.
{{SOL_IMG1}}
### Pattern Recognition
Reimer-Tiemann = Phenol + $CHCl_3$ + $NaOH$ $\rightarrow$ Salicylaldehyde.
Kolbe's = Phenol + $CO_2$ + $NaOH$ $\rightarrow$ Salicylic acid.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols Phenols and Ethers
