Consider the following reaction sequence Compound (x) [76.6%C, 6.38%H, vapour density 47] (i)CO2,NaOH,120,high pressure (ii)H3O+ Compound (y) (Major Product) Compound (y) develops characteristic colour with neutral FeCl3 solution. Identify the INCORRECT statement from the following for the above sequence.

Based on empirical data for (x): Molar Mass = 2 x Vapour Density = 2 x 47 = 94~g/mol. C = 76.6% (76.6/12) approx 6.38 H = 6.38% (6.38/1) approx 6.38 O = 17.02% (17.02/16) approx 1.06 Ratio C:H:O approx 6:6:1 C6H6O (Phenol). Compound (x) is Phenol. It reacts with CO2/NaOH via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y). (1) Both Phenol and Salicylic acid dissolve in NaOH. (True) (2) Salicylic acid (y) has a -COOH group and dissolves in NaHCO3, evolving CO2 gas. (True) (3) Salicylic acid (y) is much more acidic than Phenol (x) due to the presence of the carboxylic acid group and intramolecular hydrogen bonding in its conjugate base. Therefore, the statement "x is more acidic than y" is INCORRECT. (4) Being highly aromatic compounds, both burn with a sooty flame. (True) The reaction sequence Phenol + CO2/NaOH arrow Salicylic ; acid is a staple (Kolbe's Reaction). Always test functional group properties: Phenols dissolve in NaOH but not NaHCO3, while Carboxylic acids dissolve in both. Class 12 Chemistry: Alcohols Phenols and Ethers Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Consider the following reaction sequence Compound (x) [76.6%C, 6.38%H, vapour density 47] (i)CO2,NaOH,120,high pressure (ii)H3O+ Compound (y) (Major Product) Compound (y) develops characteristic colou

Based on empirical data for (x): Molar Mass = 2 x Vapour Density = 2 x 47 = 94~g/mol. C = 76.6% (76.6/12) approx 6.38 H = 6.38% (6.38/1) approx 6.38 O = 17.02% (17.02/16) approx 1.06 Ratio C:H:O approx 6:6:1 C6H6O (Phenol). Compound (x) is Phenol. It reacts with CO2/NaOH via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y). (1) Both Phenol and Salicylic acid dissolve in NaOH. (True) (2) Salicylic acid (y) has a -COOH group and dissolves in NaHCO3, evolving CO2 gas.

Consider the following reaction sequence Compound (x) [76.6%C, 6.38%H, vapour density 47] (i)CO2,NaOH,120,high pressure (ii)H3O+ Compound (y) (Major Product) Compound (y) develops characteristic colour with neutral FeCl3 solution. Identify the INCORRECT statement from the following for the above sequence.

Based on empirical data for (x): Molar Mass = 2 x Vapour Density = 2 x 47 = 94~g/mol. C = 76.6% (76.6/12) approx 6.38 H = 6.38% (6.38/1) approx 6.38 O = 17.02% (17.02/16) approx 1.06 Ratio C:H:O approx 6:6:1 C6H6O (Phenol). Compound (x) is Phenol. It reacts with CO2/NaOH via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y). (1) Both Phenol and Salicylic acid dissolve in NaOH. (True) (2) Salicylic acid (y) has a -COOH group and dissolves in NaHCO3, evolving CO2 gas. (True) (3) Salicylic acid (y) is much more acidic than Phenol (x) due to the presence of the carboxylic acid group and intramolecular hydrogen bonding in its conjugate base. Therefore, the statement "x is more acidic than y" is INCORRECT. (4) Being highly aromatic compounds, both burn with a sooty flame. (True) The reaction sequence Phenol + CO2/NaOH arrow Salicylic ; acid is a staple (Kolbe's Reaction). Always test functional group properties: Phenols dissolve in NaOH but not NaHCO3, while Carboxylic acids dissolve in both. Class 12 Chemistry: Alcohols Phenols and Ethers Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Consider the following reaction sequence Compound (x) [76.6%C, 6.38%H, vapour density 47] (i)CO2,NaOH,120,high pressure (ii)H3O+ Compound (y) (Major Product) Compound (y) develops characteristic colou

Based on empirical data for (x): Molar Mass = 2 x Vapour Density = 2 x 47 = 94~g/mol. C = 76.6% (76.6/12) approx 6.38 H = 6.38% (6.38/1) approx 6.38 O = 17.02% (17.02/16) approx 1.06 Ratio C:H:O approx 6:6:1 C6H6O (Phenol). Compound (x) is Phenol. It reacts with CO2/NaOH via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y). (1) Both Phenol and Salicylic acid dissolve in NaOH. (True) (2) Salicylic acid (y) has a -COOH group and dissolves in NaHCO3, evolving CO2 gas.

Consider the following reaction sequence Compound (x) [76.6%C, 6.38%H, vapour density 47] (i)CO2,NaOH,120,high pressure…

Year	2026	2025	2024	Total
Questions	7	5	11	23

Q63 jee_main_2024_29_january_evening pKa of Phenols

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 :

A. A-I, B-II, C-III, D-IV
B. A-IV, B-I, C-II, D-III
C. A-III, B-IV, C-I, D-II
D. A-II, B-I, C-IV, D-III

Solution

Related Formula

pKₐ = - ₁₀ Kₐ

Higher acidity Higher Kₐ Lower pKₐ

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

Q jee_main_2024_27_jan_morning Acidity Trends in Phenols

The ascending order of acidity of -OH group in the following compounds is: Choose the correct answer from the options given below:

Acidity compounds structural forms for Q72 - JEE Main 2024 Morning — Structures of molecules P, Q, R, S assigned letters A through E.

A. $(A) < (D) < (C) < (B) < (E)$ $(A) < (D) < (C) < (B) < (E)$
B. $(C) < (A) < (D) < (B) < (E)$ $(C) < (A) < (D) < (B) < (E)$
C. $(C) < (D) < (B) < (A) < (E)$ $(C) < (D) < (B) < (A) < (E)$
D. $(A) < (C) < (D) < (B) < (E)$ $(A) < (C) < (D) < (B) < (E)$

Solution

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.

Acidity progression schematic overview for Q72 - JEE Main 2024 Morning — Structures of molecules P, Q, R, S assigned letters A through E.

Step 1: Order verification

Bu-OH (A) < p-methoxyphenol (C) < phenol (D) < p-nitrophenol (B) < 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

Q71 jee_main_2024_27_jan_morning Acidity of Phenols and Lucas Test

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 :

A. Statement I is true but Statement II is false
B. Both Statement I and Statement II are true
C. Both Statement I and Statement II are false
D. Statement I is false but Statement II is true

Solution

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 ( $conc. HCl + anhydrous ZnCl₂$ ) 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

Q jee_main_2024_29_jan_morning Cleavage of C-O Bond in Ethers

The major product(P) in the following reaction is

Cleavage of C-O Bond in Ethers diagram for Q67 - JEE Main 2024 Morning — Reaction showing an aromatic ether with a vinyl group reacting with excess HBr.

A. ""
B. ""
C. ""
D. ""

Solution

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

Q jee_main_2024_30_january_evening Reactions of Phenols

Salicylaldehyde is synthesized from phenol, when reacted with

A.
B. $CO₂ , NaOH$ $\text{CO}_2 \text{, NaOH}$
C. $CCl₄ , NaOH$ $\text{CCl}_4 \text{, NaOH}$
D. $HCCl₃ , NaOH$ $\text{HCCl}_3 \text{, NaOH}$

Solution

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.

Reimer Tiemann reaction mechanism diagram for Q62 - JEE Main 2024 Evening

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

Solution & Explanation

Core Logic

Step 1: Identifying Compounds

Step 2: Checking Options

Pattern Recognition

Chapter Mix

Reference Study Guides

More Alcohols Phenols and Ethers Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Assembly

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Order verification

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Detailed Mechanism

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Alcohols Phenols and Ethers Questions — jee_main_2026_28_january_morning

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