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

Year	2026	2025	2024	Total
Questions	7	5	8	20

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

Q jee_main_2024_30_january_evening Preparation of Alcohols

Product A and B formed in the following set of reactions are:

Preparation of Alcohols diagram for Q66 - JEE Main 2024 Evening — The diagram shows a reaction pathway for synthesizing product A and B.

A. $A = CH₂OH , B = CH₂OH$
B. $A = CH₃OH , B = CH₃OH$
C. $A = CH₂OH , B = CH₃OH$
D. $A = CH₃OH , B = CH₃OH$

Solution

Core Logic

The given substrate is an alkene attached to a benzene ring (styrene derivative).

Reaction 1: Acid-catalyzed hydration ( $H^+ / H₂O$ $H^+ / H_2O$) This proceeds via Markovnikov's rule. The proton attacks the alkene to form the most stable carbocation (benzyllic and secondary). Water then attacks this carbocation to yield the alcohol at the substituted position (product A: $Ph-CH(OH)-CH₃$ $Ph-CH(OH)-CH_3$).

Reaction 2: Hydroboration-Oxidation ( $B₂H₆$ $B_2H_6$, then $H₂O₂/NaOH$ $H_2O_2/NaOH$) This proceeds via anti-Markovnikov addition of water across the double bond without carbocation rearrangement. The $OH$ $OH$ group adds to the less substituted carbon atom of the alkene (product B: $Ph-CH₂-CH₂-OH$ $Ph-CH_2-CH_2-OH$).

Reaction pathway for products A and B diagram for Q66 - JEE Main 2024 Evening — The diagram shows a reaction pathway for synthesizing product A and B.

Pattern Recognition

$H^+/H₂O$ $H^+/H_2O$ = Markovnikov hydration (carbocation intermediate). $B₂H₆ / H₂O₂, OH^-$ $B_2H_6 / H_2O_2, OH^-$ = Anti-Markovnikov hydration (concerted, no rearrangement).

Chapter Mix

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

Q69 jee_main_2024_31_jan_morning Chemical Reactions of Alcohols

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Alcohols react both as nucleophiles and electrophiles. Reason R: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen. In the light of the above statements, choose the correct answer from the options given below:

A. $A is false but R is true.$ $\text{A is false but R is true.}$
B. $A is true but R is false.$ $\text{A is true but R is false.}$
C. $Both A and R are true and R is the correct explanation of A.$
D. $Both A and R are true but R is NOT the correct explanation of A.$

Solution

Core Logic

Assertion (A) is correct. Alcohols can act as nucleophiles (in reactions involving O-H bond cleavage) and as electrophiles (in reactions involving C-O bond cleavage after protonation).

Reason (R) is correct. Alcohols react with active metals to form alkoxides and $H₂$ $H_2$ gas.

However, Reason (R) only explains the acidic nature of the O-H bond (which relates to nucleophilic behavior when broken) but does not explain the electrophilic behavior (C-O cleavage). Thus, (R) is not the correct explanation for (A).

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Q74 jee_main_2024_31_jan_morning Acidity of Phenols

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A:

pKₐ

value of phenol is 10.0 while that of ethanol is 15.9. Reason R: Ethanol is stronger acid than phenol. In the light of the above statements, choose the correct answer from the options given below:

A. $A is true but R is false.$ $\text{A is true but R is false.}$
B. $A is false but R is true.$ $\text{A is false but R is true.}$
C. $Both A and R are true and R is the correct explanation of A.$
D. $Both A and R are true but R is NOT the correct explanation of A.$

Solution

Core Logic

Assertion A is true: Phenol has a $pKₐ$ $pK_a$ of around 10.0, and ethanol has a $pKₐ$ $pK_a$ of around 15.9.

Reason R is false: A lower $pKₐ$ $pK_a$ value implies higher acidity. Therefore, phenol is a stronger acid than ethanol. This is because the conjugate base of phenol (phenoxide ion) is resonance-stabilized, whereas the ethoxide ion is destabilized by the +I effect of the ethyl group.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

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 — Page 4

Solution

Core Logic

Step 1: Detailed Mechanism

Pattern Recognition

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Solution

Core Logic

Pattern Recognition

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Solution

Core Logic

Pattern Recognition

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Solution

Core Logic

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Solution

Core Logic

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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 — Page 4

Solution

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Step 1: Detailed Mechanism

Pattern Recognition

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Solution

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Pattern Recognition

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