Alcohols Phenols and Ethers Previous Year Questions — JEE Main Chemistry
20 past-year Alcohols Phenols and Ethers questions from JEE Main (Chemistry) — page 4.
Q66 (2026)
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):{{IMG1}}
- $\text{Option 1}$
- $\text{Option 2}$
- $\text{Option 3}$
- $\text{Option 4}$
### 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 $CH_3-CH(OH)-$. Therefore, both alkyl iodides must contain the $CH_3-CH(I)-$ 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 $-CH(CH_3)-$.
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 ($R-O-R'$) into two moles of alkyl iodides ($R-I$ and $R'-I$). Only secondary alcohols with a methyl group or ethanol itself give positive iodoform tests. Look for an ether containing $CH_3-CH(O-)-$ branching on BOTH sides.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q65 (2026)
A hydroxy compound (X) with molar mass $122 \text{ g mol}^{-1}$ 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 $290 \text{ g mol}^{-1}$. The number of hydroxyl groups present in compound (X) is :
- $3$
- $5$
- $2$
- $4$
### Core Logic
During the acetylation of an alcohol, an $-\text{H}$ atom (molar mass = 1) is replaced by an acetyl group ($-\text{COCH}_3$, molar mass = 43).
Therefore, the net increase in molar mass per hydroxyl group acetylated is:
$$\Delta M = 43 - 1 = 42 \text{ g/mol}$$
Given:
Molar mass of starting compound (X) = $122 \text{ g/mol}$
Molar mass of completely acetylated product = $290 \text{ g/mol}$
Total increase in molar mass:
$$290 - 122 = 168 \text{ g/mol}$$
### Step 1: Calculate Number of Hydroxyl Groups
$$\text{Number of } -\text{OH groups} = \frac{\text{Total Mass Increase}}{\text{Increase per group}} = \frac{168}{42} = 4$$
{{SOL_IMG1}}
{{SOL_IMG2}}
### Pattern Recognition
Acetylation replaces $H$ with $COCH_3$. Always divide the mass difference by 42 to find the number of $-OH$ (or $-NH_2$) groups.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Biomolecules
Q68 (2026)
Consider the following two reactions A and B.
(A) $\text{Phenol} \xrightarrow{\text{Na}} \text{Main Product} + \text{gas (x)}$
(B) $\text{Benzoic acid} \xrightarrow{\text{NaHCO}_3} \text{Main Product} + \text{gas (y)}$
Numerical value of [molar mass of x+ molar mass of y] is
- $4$
- $88$
- $46$
- $160$
### Core Logic
Reaction (A): Phenol + Sodium metal
Phenol reacts with active metals like Sodium to form sodium phenoxide and evolve hydrogen gas.
$$2 C_6H_5OH + 2 Na \rightarrow 2 C_6H_5O^-Na^+ + H_2 \uparrow$$
Gas (x) is $H_2$. Molar mass of $H_2 = 2 \text{ g/mol}$.
Reaction (B): Benzoic acid + Sodium bicarbonate
Benzoic acid is a sufficiently strong acid to react with weak bases like $NaHCO_3$, undergoing decarboxylation to evolve carbon dioxide gas.
$$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COO^-Na^+ + H_2O + CO_2 \uparrow$$
Gas (y) is $CO_2$. Molar mass of $CO_2 = 44 \text{ g/mol}$.
### Step 1: Final Calculation
$$\text{Sum of molar mass} = M(H_2) + M(CO_2)$$
$$\text{Sum} = 2 + 44 = 46$$
### Pattern Recognition
Active metals + acids/alcohols = $H_2$ gas. Bicarbonates + acids (stronger than carbonic acid) = $CO_2$ effervescence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q66 (2026)
From the following, how many compounds contain at least one secondary alcohol?
(I) {{IMG1}}
(II) {{IMG2}}
(III) {{IMG3}}
(IV) {{IMG4}}
(V) {{IMG5}}
(VI) {{IMG6}}
Choose the correct answer from the options given below:
- Five
- Three
- Four
- two
### Core Logic
A secondary alcohol features a hydroxyl group (-OH) attached to a carbon atom that is bonded to two other carbon atoms ($2^{\circ}$ carbon).
Evaluating the structures based on visual analysis:
Compounds II, IV, and V contain at least one hydroxyl group on a $2^{\circ}$ carbon.
Therefore, 3 compounds feature a secondary alcohol.
### Pattern Recognition
Spot the carbon carrying the -OH group. Count the C-C bonds attached to it. 1 = Primary, 2 = Secondary, 3 = Tertiary. Phenols are aromatic and distinct from secondary alcohols.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q54 (2026)
Consider the following reaction sequence
Compound (x) [76.6%C, 6.38%H, vapour density 47] $\xrightarrow{\text{(i)}\mathrm{CO}_{2},\mathrm{NaOH},120^{\circ}\mathrm{C},\text{high pressure}} \xrightarrow{\text{(ii)}\mathrm{H}_{3}\mathrm{O}^{+}}$ Compound (y) (Major Product)
Compound (y) develops characteristic colour with neutral $\mathrm{FeCl}_3$ solution.
Identify the INCORRECT statement from the following for the above sequence.
- Both compounds x and y will dissolve in NaOH.
- Compound y will dissolve in $\mathrm{NaHCO}_3$ and evolve a gas.
- Compound x is more acidic than compound y.
- Both compounds x and y will burn with sooty flame.
### Core Logic
Based on empirical data for (x):
$\text{Molar Mass} = 2 \times \text{Vapour Density} = 2 \times 47 = 94\mathrm{~g/mol}$.
$\text{C} = 76.6\% \implies (76.6/12) \approx 6.38$
$\text{H} = 6.38\% \implies (6.38/1) \approx 6.38$
$\text{O} = 17.02\% \implies (17.02/16) \approx 1.06$
Ratio $\text{C:H:O} \approx 6:6:1 \implies \mathrm{C_6H_6O}$ (Phenol).
### Step 1: Identifying Compounds
Compound (x) is Phenol. It reacts with $\mathrm{CO}_2/\mathrm{NaOH}$ via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y).
{{SOL_IMG1}}
### Step 2: Checking Options
(1) Both Phenol and Salicylic acid dissolve in $\mathrm{NaOH}$. (True)
(2) Salicylic acid (y) has a $-COOH$ group and dissolves in $\mathrm{NaHCO}_3$, evolving $\mathrm{CO}_2$ 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)
### Pattern Recognition
The reaction sequence $\mathrm{Phenol} + \mathrm{CO_2/NaOH} \rightarrow \mathrm{Salicylic \; acid}$ is a staple (Kolbe's Reaction). Always test functional group properties: Phenols dissolve in $\mathrm{NaOH}$ but not $\mathrm{NaHCO}_3$, while Carboxylic acids dissolve in both.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Alcohols Phenols and Ethers
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
