Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product. Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation. In the light of the above statements, choose the correct answer from the options given below:

Reimer-Tiemann Reaction: Phenol CHCl3, KOH o-hydroxybenzaldehyde (Salicylaldehyde) as major product. Statement I claims that p-hydroxybenzaldehyde is the major product. This is incorrect. In the Reimer-Tiemann reaction, ortho-hydroxybenzaldehyde (salicylaldehyde) is the major product due to the stabilization of the intermediate and product via intramolecular hydrogen bonding. Statement II states that the mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be separated by steam distillation. This is true. Ortho-hydroxybenzaldehyde forms intramolecular hydrogen bonds, making it more volatile (steam volatile), whereas the para-isomer forms intermolecular hydrogen bonds, increasing its boiling point and making it non-steam volatile. Therefore, Statement I is false but Statement II is true. Always remember: Ortho-isomers capable of intramolecular H-bonding are steam volatile. Para-isomers exhibit intermolecular H-bonding and are not steam volatile. Class 12 Chemistry: Alcohols Phenols and Ethers

Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydr

Reimer-Tiemann Reaction: Phenol CHCl3, KOH o-hydroxybenzaldehyde (Salicylaldehyde) as major product. Statement I claims that p-hydroxybenzaldehyde is the major product. This is incorrect. In the Reimer-Tiemann reaction, ortho-hydroxybenzaldehyde (salicylaldehyde) is the major product due to the stabilization of the intermediate and product via intramolecular hydrogen bonding. Statement II states that the mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be separated by steam distill

Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product. Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation. In the light of the above statements, choose the correct answer from the options given below:

Reimer-Tiemann Reaction: Phenol CHCl3, KOH o-hydroxybenzaldehyde (Salicylaldehyde) as major product. Statement I claims that p-hydroxybenzaldehyde is the major product. This is incorrect. In the Reimer-Tiemann reaction, ortho-hydroxybenzaldehyde (salicylaldehyde) is the major product due to the stabilization of the intermediate and product via intramolecular hydrogen bonding. Statement II states that the mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be separated by steam distillation. This is true. Ortho-hydroxybenzaldehyde forms intramolecular hydrogen bonds, making it more volatile (steam volatile), whereas the para-isomer forms intermolecular hydrogen bonds, increasing its boiling point and making it non-steam volatile. Therefore, Statement I is false but Statement II is true. Always remember: Ortho-isomers capable of intramolecular H-bonding are steam volatile. Para-isomers exhibit intermolecular H-bonding and are not steam volatile. Class 12 Chemistry: Alcohols Phenols and Ethers

Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydr

Reimer-Tiemann Reaction: Phenol CHCl3, KOH o-hydroxybenzaldehyde (Salicylaldehyde) as major product. Statement I claims that p-hydroxybenzaldehyde is the major product. This is incorrect. In the Reimer-Tiemann reaction, ortho-hydroxybenzaldehyde (salicylaldehyde) is the major product due to the stabilization of the intermediate and product via intramolecular hydrogen bonding. Statement II states that the mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be separated by steam distill

Given below are two statements: Statement I: Phenol on treatment with CHCl3/aq. KOH under refluxing condition, followed…

Year	2026	2025	2024	Total
Questions	7	5	11	23

More Alcohols Phenols and Ethers Previous-Year Questions

Q69 jee_main_2026_22_january_evening Synthesis of 3,3-Dimethyl-2-butanol

3,3-Dimethyl-2-butanol cannot be prepared by: A.

Reaction paths scheme A for Q69 - JEE Main 2026 Evening — Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.

Choose the correct answer from the options given below:

A. B only
B. B and E only
C. B and C only
D. B, C and E only

Solution

Related Formula

Acid-catalyzed hydration: Carbocation formation arrow Ethyl/Methyl Shift

Oxymercuration-Demercuration: Markovnikov addition without carbocation rearrangement

Core Logic

Step 1: Evaluate Route B: Acid catalyzed hydration of 3,3-dimethyl-1-butene involves carbocation formation followed by 1,2-methyl shift to give 2,3-dimethyl-2-butanol as major product instead of 3,3-dimethyl-2-butanol.

Step 2: Evaluate Route E: Oxymercuration-demercuration or specific hydration route E fails to give the targeted alcohol structural framework.

Hence, 3,3-Dimethyl-2-butanol CANNOT be prepared by routes B and E.

Carbocation rearrangement mechanism for Q69 - JEE Main 2026 Evening — Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.

Pattern Recognition

Sees: Carbocation rearrangement during acid hydration. Shortcut: Acid-catalyzed hydration of $(CH₃)₃C-CH=CH₂$ $(CH_3)_3C-CH=CH_2$ undergoes methyl shift yielding 2,3-dimethyl-2-butanol, failing to produce 3,3-dimethyl-2-butanol.

Chapter Mix

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

Q66 jee_main_2026_23_january_evening Reactions of Ethers

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):

Reactions of Ethers diagram for Q66 - JEE Main 2026 Evening — A schematic diagram showing the transformation sequence from ether (P) to products (Q) and (R) giving a yellow precipitate.

A. $Option 1$ $\text{Option 1}$
B. $Option 2$ $\text{Option 2}$
C. $Option 3$ $\text{Option 3}$
D. $Option 4$ $\text{Option 4}$