Alcohols Phenols and Ethers Previous Year Questions — JEE Main Chemistry
20 past-year Alcohols Phenols and Ethers questions from JEE Main (Chemistry) — page 3.
Q1746 (2024)
Product A and B formed in the following set of reactions are:
{{IMG1}}
- $\text{A = } \mathrm{CH_2OH} \text{, B = } \mathrm{CH_2OH}$
- $\text{A = } \mathrm{CH_3OH} \text{, B = } \mathrm{CH_3OH}$
- $\text{A = } \mathrm{CH_2OH} \text{, B = } \mathrm{CH_3OH}$
- $\text{A = } \mathrm{CH_3OH} \text{, B = } \mathrm{CH_3OH}$
### Core Logic
The given substrate is an alkene attached to a benzene ring (styrene derivative).
Reaction 1: Acid-catalyzed hydration ($H^+ / H_2O$)
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: $Ph-CH(OH)-CH_3$).
Reaction 2: Hydroboration-Oxidation ($B_2H_6$, then $H_2O_2/NaOH$)
This proceeds via anti-Markovnikov addition of water across the double bond without carbocation rearrangement. The $OH$ group adds to the less substituted carbon atom of the alkene (product B: $Ph-CH_2-CH_2-OH$).
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### Pattern Recognition
$H^+/H_2O$ = Markovnikov hydration (carbocation intermediate).
$B_2H_6 / H_2O_2, OH^-$ = Anti-Markovnikov hydration (concerted, no rearrangement).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols Phenols and Ethers
Class 11 Chemistry: Hydrocarbons
Q69 (2024)
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:
- $\text{A is false but R is true.}$
- $\text{A is true but R is false.}$
- $\text{Both A and R are true and R is the correct explanation of A.}$
- $\text{Both A and R are true but R is NOT the correct explanation of A.}$
### 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 $H_2$ 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).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q74 (2024)
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: $pK_a$ 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:
- $\text{A is true but R is false.}$
- $\text{A is false but R is true.}$
- $\text{Both A and R are true and R is the correct explanation of A.}$
- $\text{Both A and R are true but R is NOT the correct explanation of A.}$
### Core Logic
Assertion A is true: Phenol has a $pK_a$ of around 10.0, and ethanol has a $pK_a$ of around 15.9.
Reason R is false: A lower $pK_a$ 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q53 (2026)
Given below are two statements:
Statement I: Phenol on treatment with $CHCl_{3}/aq$. 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:
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are true}$
- $\text{Statement I is false but Statement II is true}$
### Related Formula
Reimer-Tiemann Reaction: Phenol $\xrightarrow{CHCl_3, KOH}$ 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.
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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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols Phenols and Ethers
Q69 (2026)
3,3-Dimethyl-2-butanol cannot be prepared by:
A. {{IMG1}}
B. {{IMG2}}
C. {{IMG3}}
D. {{IMG4}}
E. {{IMG5}}
Choose the correct answer from the options given below:
- B only
- B and E only
- B and C only
- B, C and E only
### Related Formula
$$\text{Acid-catalyzed hydration: Carbocation formation } \rightarrow \text{Ethyl/Methyl Shift}$$
$$\text{Oxymercuration-Demercuration: Markovnikov addition without carbocation rearrangement}$$
### 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.
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{{SOL_IMG2}}
{{SOL_IMG3}}
{{SOL_IMG4}}
### Pattern Recognition
Sees: Carbocation rearrangement during acid hydration.
Shortcut: Acid-catalyzed hydration of $(CH_3)_3C-CH=CH_2$ undergoes methyl shift yielding 2,3-dimethyl-2-butanol, failing to produce 3,3-dimethyl-2-butanol.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 11 Chemistry: Hydrocarbons
