Which one of the following, with HBr will give a phenol?

Year	2026	2025	2024	Total
Questions	7	5	8	20

Q53 jee_main_2026_22_january_morning Reimer Tiemann Reaction

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:

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

Solution

Related Formula

Reimer-Tiemann Reaction: Phenol $CHCl₃, KOH$ $\xrightarrow{CHCl_3, KOH}$ o-hydroxybenzaldehyde (Salicylaldehyde) as major product.

Core Logic

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.

Reimer-Tiemann reaction mechanism diagram

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.

Step 1: Final Conclusion

Therefore, Statement I is false but Statement II is true.

Pattern Recognition

Always remember: Ortho-isomers capable of intramolecular H-bonding are steam volatile. Para-isomers exhibit intermolecular H-bonding and are not steam volatile.

Chapter Mix

Class 12 Chemistry: Alcohols Phenols and Ethers

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.

B.

C.

D.

E.

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. $Option 4$ $\text{Option 4}$

Solution

Core Logic

We work backwards. Both (Q) and (R) give a positive iodoform test (yellow precipitate with NaOI). Since (Q) and (R) are obtained by treating alkyl iodides with aq. NaOH, they must be alcohols.

Alcohols that give a positive iodoform test must possess the structural unit $CH₃-CH(OH)-$ $CH_3-CH(OH)-$. Therefore, both alkyl iodides must contain the $CH₃-CH(I)-$ $CH_3-CH(I)-$ group.

The precursor mixed ether (P) was treated with excess concentrated HI to yield these two alkyl iodides. For both sides of the ether to produce iodides capable of forming secondary alcohols with terminal methyls upon hydrolysis, the ether must be composed of two branches that look like $-CH(CH₃)-$ $-CH(CH_3)-$.

Let's evaluate the correct option (1): Structure (P) is an ether linking a sec-butyl group and a secondary-pentyl group (or similar structure). Excess HI cleaves the ether to give two secondary alkyl iodides. Hydrolysis with aq. NaOH converts both alkyl iodides into secondary alcohols. Both secondary alcohols have a methyl group directly adjacent to the hydroxyl-bearing carbon, fully satisfying the iodoform test requirement.

Pattern Recognition

Excess hot HI cleaves ethers ( $R-O-R'$ $R-O-R'$) into two moles of alkyl iodides ( $R-I$ $R-I$ and $R'-I$ $R'-I$). Only secondary alcohols with a methyl group or ethanol itself give positive iodoform tests. Look for an ether containing $CH₃-CH(O-)-$ $CH_3-CH(O-)-$ branching on BOTH sides.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q65 jee_main_2026_24_january_morning Acetylation of Hydroxyl Groups

A hydroxy compound (X) with molar mass

122 g mol⁻¹

is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of

290 g mol⁻¹

. The number of hydroxyl groups present in compound (X) is :

A. $3$ $3$
B. $5$ $5$
C. $2$ $2$
D. $4$ $4$

Solution

Core Logic

During the acetylation of an alcohol, an $-H$ $-\text{H}$ atom (molar mass = 1) is replaced by an acetyl group ( $-COCH₃$ $-\text{COCH}_3$, molar mass = 43). Therefore, the net increase in molar mass per hydroxyl group acetylated is:

Δ M = 43 - 1 = 42 g/mol

Given: Molar mass of starting compound (X) = $122 g/mol$ $122 \text{ g/mol}$ Molar mass of completely acetylated product = $290 g/mol$ $290 \text{ g/mol}$

Total increase in molar mass:

290 - 122 = 168 g/mol

Step 1: Calculate Number of Hydroxyl Groups

Number of -OH groups = Total Mass IncreaseIncrease per group = (168)/(42) = 4

Acetylation mechanism showing mass increase

Pattern Recognition

Acetylation replaces $H$ $H$ with $COCH₃$ $COCH_3$. Always divide the mass difference by 42 to find the number of $-OH$ $-OH$ (or $-NH₂$ $-NH_2$) groups.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Biomolecules

Q68 jee_main_2026_24_january_morning Reactions with Active Metals and Bicarbonate

Consider the following two reactions A and B. (A)

Phenol Na Main Product + gas (x)

(B)

Benzoic acid NaHCO₃ Main Product + gas (y)

Numerical value of [molar mass of x+ molar mass of y] is

A. $4$ $4$
B. $88$ $88$
C. $46$ $46$
D. $160$ $160$

Solution

Core Logic

Reaction (A): Phenol + Sodium metal Phenol reacts with active metals like Sodium to form sodium phenoxide and evolve hydrogen gas.

2 C₆H₅OH + 2 Na arrow 2 C₆H₅O^-Na^+ + H₂

Gas (x) is $H₂$ $H_2$. Molar mass of $H₂ = 2 g/mol$ $H_2 = 2 \text{ g/mol}$.

Reaction (B): Benzoic acid + Sodium bicarbonate Benzoic acid is a sufficiently strong acid to react with weak bases like $NaHCO₃$ $NaHCO_3$, undergoing decarboxylation to evolve carbon dioxide gas.

C₆H₅COOH + NaHCO₃ arrow C₆H₅COO^-Na^+ + H₂O + CO₂

Gas (y) is $CO₂$ $CO_2$. Molar mass of $CO₂ = 44 g/mol$ $CO_2 = 44 \text{ g/mol}$.

Step 1: Final Calculation

Sum of molar mass = M(H₂) + M(CO₂)

Sum = 2 + 44 = 46

Pattern Recognition

Active metals + acids/alcohols = $H₂$ $H_2$ gas. Bicarbonates + acids (stronger than carbonic acid) = $CO₂$ $CO_2$ effervescence.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Which one of the following, with HBr will give a phenol?

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Alcohols, Phenols and Ethers Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Calculate Number of Hydroxyl Groups

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

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