Write the chemical equation for the following: (a) Hydration of propene in presence of an acid
(b) Reaction between Ethyl bromide and C2H5ONa
(c) Reaction between Dimethyl ether and Hydrogen iodide
Official Solution
Correct Option: (1)
(a) The hydration of propene in acidic medium follows Markovnikov’s rule. The –OH group attaches to the more substituted carbon to form 2-propanol. (b) The reaction between ethyl bromide and sodium ethoxide is a classic Williamson ether synthesis, producing diethyl ether. (c) Dimethyl ether reacts with hydrogen iodide (HI), cleaving the ether to form methyl iodide and methanol. This is an example of ether cleavage using hydrogen halides.
02
PYQ 2023
medium
chemistryID: cbse-cla
Why is sulphuric acid not used during the reactions of alcohols with KI?
Official Solution
Correct Option: (1)
Step 1: In the reaction of alcohols with KI, HI is required as a nucleophile to replace the hydroxyl group. Step 2: When sulphuric acid (H SO ) is used, it oxidizes iodide ions ( ) to iodine ( ), reducing the amount of HI available for the reaction. Step 3: Hence, sulphuric acid is avoided, and phosphoric acid (H PO ) is preferred instead, as it does not oxidize iodide.
03
PYQ 2023
medium
chemistryID: cbse-cla
Choose the compound which is more acidic than phenol:
1
2
ethanol
3
4
Official Solution
Correct Option: (1)
Step 1: Phenol is weakly acidic because the phenoxide ion formed after deprotonation is stabilized by resonance. Step 2: An electron-withdrawing group increases acidity by stabilizing the negative charge on the oxygen atom. The nitro group ( ) is a strong electron-withdrawing group due to its (mesomeric) and (inductive) effects. Step 3: In , the nitro group is ortho to the hydroxyl group, so it withdraws electron density both through the ring and by induction. This additional stabilization makes the conjugate base (phenoxide ion) more stable, thereby increasing acidity compared to phenol. Step 4: Ethanol is less acidic than phenol because it lacks resonance stabilization of the alkoxide ion. Methyl and methoxy groups are electron-donating due to their and effects, which increase electron density and thus reduce acidity relative to phenol.
04
PYQ 2023
hard
chemistryID: cbse-cla
Assertion (A): Phenol gives 2,4,6-trinitrophenol on treatment with concentrated and concentrated . Reason (R): –OH group in phenol is -directing.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is \textit{not} the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option: (3)
Step 1: The Assertion is correct. Phenol reacts with concentrated nitric acid in the presence of sulfuric acid to form 2,4,6-trinitrophenol (picric acid). Step 2: The Reason is incorrect. The –OH group in phenol activates the benzene ring and is an -directing group due to its +M effect, which increases electron density at the ortho and para positions. Step 3: Therefore, while the Assertion is true, the Reason is false.
05
PYQ 2024
medium
chemistryID: cbse-cla
Ethanol on heating with conc. H SO at 413 K gives:
1
2
3
4
Official Solution
Correct Option: (3)
When ethanol is heated with conc. sulfuric acid (H SO ) at 413 K, it undergoes an intermolecular dehydration reaction, leading to the formation of diethyl ether ( ): - This reaction occurs at lower temperature (413 K), favoring ether formation.
- At higher temperatures (443 K), ethanol undergoes elimination to form ethene ( ).
06
PYQ 2024
medium
chemistryID: cbse-cla
Dehydration of tertiary alcohols with Cu at 573 K gives:
1
Alkyne
2
Alkene
3
Aldehyde
4
Ketone
Official Solution
Correct Option: (2)
Tertiary alcohols undergo dehydration when heated with copper (Cu) at 573 K, leading to the formation of alkenes via the E1 (elimination) mechanism. The reaction follows: This method is used in industrial applications for producing alkenes from alcohols.
07
PYQ 2024
medium
chemistryID: cbse-cla
Phenol on reaction with aqueous bromine at room temperature gives:
1
2-bromophenol
2
3-bromophenol
3
4-bromophenol
4
2,4,6-tribromophenol
Official Solution
Correct Option: (4)
Phenol is highly reactive toward electrophilic aromatic substitution due to the electron-donating effect of the -OH group. When reacted with bromine water (Br in H O), substitution occurs at the ortho (2), para (4), and ortho (6) positions, forming 2,4,6-tribromophenol, a white precipitate.
08
PYQ 2024
medium
chemistryID: cbse-cla
Consider the following reaction: The major product obtained is:
1
2
3
4
Official Solution
Correct Option: (2)
When benzyl alcohol (C H CH OH) reacts with thionyl chloride (SOCl ), the hydroxyl group (OH) is replaced by a chlorine atom, resulting in the formation of benzyl chloride (C H CH Cl). This is a typical reaction for alcohols with SOCl , where the alcohol undergoes nucleophilic substitution. Thus, the major product is benzyl chloride (C H CH Cl).
09
PYQ 2024
medium
chemistryID: cbse-cla
CH -O-CH when treated with excess HI gives:
1
CH -OH + CH -I
2
2CH -OH
3
2CH -I
4
CH -I + CH
Official Solution
Correct Option: (3)
The reaction of ether (CH -O-CH ) with excess hydroiodic acid (HI) results in the cleavage of the ether bond. This cleavage produces methyl iodide (CH -I) and methyl alcohol (CH -OH). However, with excess HI, both ether bonds are cleaved, producing 2 moles of CH -I and CH -OH. Thus, the correct answer is (C), where two moles of CH -I are formed.
10
PYQ 2024
medium
chemistryID: cbse-cla
Match the reagents required for the given reactions: I. Oxidation of primary alcohols to aldehydes (p)
II. Butan-2-one to Butan-2-ol (q) 85% phosphoric acid at 440 K
III. Bromination of Phenol to 2, 4, 6- (r) PCC Tribromophenol
IV. Dehydration of propan-2-ol to (s) Bromine water propene
1
I — (r), II — (p), III — (s), IV — (q)
2
I — (q), II — (r), III — (p), IV — (s)
3
I — (s), II — (q), III — (p), IV — (r)
4
I — (p), II — (s), III — (r), IV — (q)
Official Solution
Correct Option: (1)
- The oxidation of primary alcohols to aldehydes is done using PCC Tribromophenol.
- Butan-2-one is reduced to Butan-2-ol using .
- Bromination of Phenol to 2, 4, 6-tribromophenol is done using Bromine water propene.
- Dehydration of propan-2-ol leads to propene when treated with 85% phosphoric acid at 440 K
11
PYQ 2024
medium
chemistryID: cbse-cla
Assertion (A): p-nitrophenol is less acidic than phenol. Reason (R): Nitro group is electron-withdrawing and helps in the stabilization of p-nitrophenoxide ion.
1
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option: (4)
- Assertion (A) is incorrect because p-nitrophenol is actually more acidic than phenol, due to the electron-withdrawing nature of the nitro group, which stabilizes the conjugate base (p-nitrophenoxide ion).
- Reason (R) is true because the nitro group, being electron-withdrawing, helps in the stabilization of the conjugate base, thereby increasing the acidity of p-nitrophenol.
Step 1: The electron-withdrawing nitro group stabilizes the conjugate base, making p-nitrophenol more acidic than phenol.
12
PYQ 2024
medium
chemistryID: cbse-cla
Nucleophilic addition of Grignard reagent to ketones followed by hydrolysis with dilute acids forms:
1
Alkene
2
Primary alcohol
3
Tertiary alcohol
4
Secondary alcohol
Official Solution
Correct Option: (3)
The reaction of a Grignard reagent (RMgX) with a ketone results in the formation of a tertiary alcohol. The mechanism involves the nucleophilic attack of the Grignard reagent on the carbonyl carbon of the ketone, followed by hydrolysis to yield a tertiary alcohol. This is because the Grignard reagent adds a second alkyl group to the carbonyl carbon, resulting in a tertiary alcohol.
Mechanism:
Grignard reagent attacks the carbonyl carbon of the ketone to form a tetrahedral intermediate.
Hydrolysis of the intermediate yields a tertiary alcohol.
13
PYQ 2024
medium
chemistryID: cbse-cla
Write the reactions involved in the following: (i) Reimer-Tiemann reaction (ii) Kolbe’s reaction
Official Solution
Correct Option: (1)
(i) Reimer-Tiemann Reaction: The Reimer-Tiemann reaction involves the introduction of a formyl group (–CHO) to the aromatic ring using chloroform ( ) and sodium hydroxide ( ): (ii) Kolbe’s Reaction: This reaction involves the carboxylation of sodium phenoxide with carbon dioxide under high pressure, yielding salicylic acid:
14
PYQ 2024
medium
chemistryID: cbse-cla
Give the equations of reactions for the preparation of: (any three)(a) Phenol from chlorobenzene (b) Salicyaldehyde from phenol (c) 2-Methoxyacetophenone from anisole
Official Solution
Correct Option: (1)
(a) The Wurtz-Fittig reaction is used here. Chlorobenzene (C H Cl) undergoes nucleophilic substitution with NaOH under high temperature and pressure (623K, 300 atm), followed by acidic hydrolysis, to form phenol.
(b) Salicyaldehyde from phenol
(b) Phenol (C H OH) undergoes haloform reaction with CHCl and NaOH to produce salicylic acid intermediate, which then undergoes acid hydrolysis to yield salicyaldehyde. (c) Anisole (C H OCH ) reacts with acetyl chloride (CH COCl) in the presence of anhydrous AlCl to form 2-methoxyacetophenone. (d) Picric acid from phenol Phenol is nitrated with concentrated nitric acid (HNO ) to produce picric acid (2,4,6-trinitrophenol).
15
PYQ 2024
medium
chemistryID: cbse-cla
Write the mechanism of the following reaction:
Official Solution
Correct Option: (1)
The mechanism for the dehydration of ethanol (CH3CH2OH) to ethene (CH2=CH2) under acidic conditions involves the following steps:
Step 1: Formation of protonated alcohol (Fast step)
Ethanol (CH3CH2OH) reacts with the proton (H+) to form the protonated alcohol (ethyl oxonium ion).
CH3CH2OH + H+ → CH3CH2OH+
Step 2: Formation of carbocation (Slow step)
The protonated alcohol undergoes the loss of water to form a carbocation, which is the rate-determining step.
CH3CH2OH+ → CH3C+ + H2O
Step 3: Formation of ethene by elimination of a proton (Fast step)
The carbocation formed in Step 2 loses a proton to form ethene.
Overall reaction:
CH3C+CH2 → C2H4 (Ethene) + H+
16
PYQ 2024
medium
chemistryID: cbse-cla
Assertion (A): Boiling point of butan-1-ol is higher than that of butan-1-amine.Reason (R): Being more polar, butan-1-ol forms stronger intermolecular hydrogen bonds as compared to butan-1-amine.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option: (1)
Hydrogen bonding is stronger in alcohols ( group) than in amines ( group) due to the higher electronegativity of oxygen compared to nitrogen. As a result, butan-1-ol exhibits stronger hydrogen bonding and thus a higher boiling point.
17
PYQ 2024
medium
chemistryID: cbse-cla
Assertion (A): Phenol on reaction with aqueous NaOH gives sodium phenoxide. Reason (R): This reaction supports the acidic nature of phenol.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option: (1)
Phenol reacts with NaOH to form sodium phenoxide, confirming its weakly acidic nature: This reaction occurs because phenol can donate a proton ( ), showing its acidic behavior.
18
PYQ 2025
hard
chemistryID: cbse-cla
Can sodium ethoxide and t-butyl chloride be used for the preparation of t-butyl ethyl ether? Give suitable explanation. Justify your answer by suggesting the appropriate starting material required for preparation of t-butyl ethyl ether.
Official Solution
Correct Option: (1)
Sodium ethoxide (C H ONa) and t-butyl chloride (C H Cl) can indeed be used to prepare t-butyl ethyl ether through a nucleophilic substitution reaction. Sodium ethoxide, being a strong nucleophile, can replace the chlorine atom in t-butyl chloride, leading to the formation of t-butyl ethyl ether. The reaction is as follows: The appropriate starting materials for the synthesis of t-butyl ethyl ether are t-butyl chloride and sodium ethoxide.
19
PYQ 2025
medium
chemistryID: cbse-cla
The reaction -OH + Na RO Na + H (g) suggests that alcohols are:
1
Acidic
2
Basic
3
Neutral
4
Amphoteric
Official Solution
Correct Option: (2)
The reaction shows that alcohols react with sodium metal to form alkoxide ions (RO ) and hydrogen gas. This indicates that alcohols are basic in nature, as they are able to donate a proton (H ) in the reaction.
20
PYQ 2025
hard
chemistryID: cbse-cla
Arrange the following compounds in increasing order of their boiling points:
1
(ii) < (i) < (iii)
2
(iii) < (i) < (ii)
3
(i) < (ii) < (iii)
4
(iii) < (ii) < (i)
Official Solution
Correct Option: (2)
Understanding Factors Affecting Boiling Point
1. Molecular Weight (Size): Larger molecules generally have higher boiling points due to increased van der Waals forces.
2. Branching: Branching decreases the surface area for intermolecular interactions, leading to lower boiling points.
3. Polarity: Polar molecules have stronger dipole-dipole interactions, resulting in higher boiling points than nonpolar molecules of similar size.
Analyzing the Compounds
(i) 2-bromo-3-methylbutane: This molecule has branching.
(ii) 1-bromobutane: This is a straight-chain alkyl halide.
(iii) 2-bromo-2-methylpropane: This molecule has significant branching around the bromine atom.
Determining the Order
1. Molecular Weight: All three compounds have a similar molecular weight due to the presence of bromine and a 4-carbon alkyl chain. However, compound (ii) has a slightly higher molecular weight as it is a straight-chain.
2. Branching: Branching significantly impacts boiling points. Compound (iii) has the most branching, causing it to have the lowest boiling point, followed by compound (i). Compound (ii) is straight-chained.
Therefore, the order of increasing boiling point should be: (iii) < (i) < (ii).
Answer: The correct answer is (B).
21
PYQ 2025
easy
chemistryID: cbse-cla
Give a simple chemical test to distinguish between butan-1-ol and butan-2-ol.
Official Solution
Correct Option: (1)
A simple chemical test to distinguish between butan-1-ol and butan-2-ol is the oxidation test using potassium dichromate ( ) in acidic medium. Butan-1-ol will oxidize to butanoic acid, while butan-2-ol will oxidize to acetone. Butan-1-ol: On oxidation, it gives a carboxylic acid, butanoic acid. Butan-2-ol: On oxidation, it gives a ketone, acetone. Another test is the reaction with iodine and NaOH: Butan-1-ol gives a yellow precipitate of iodoform when treated with iodine and NaOH. Butan-2-ol also gives a yellow precipitate of iodoform, but this reaction can differentiate based on the intensity of the reaction.
22
PYQ 2025
medium
chemistryID: cbse-cla
What will be formed after oxidation of secondary alcohol with chromic anhydride (CrO3)?
1
Aldehyde
2
Ketone
3
Carboxylic acid
4
Ester
Official Solution
Correct Option: (2)
When a secondary alcohol is oxidized with chromic anhydride (CrO3), it undergoes oxidation to form a ketone. Secondary alcohols do not oxidize to aldehydes or carboxylic acids. Step 2: Thus, the correct answer is option (2).
23
PYQ 2025
medium
chemistryID: cbse-cla
Alkenes are formed by heating alcohols with conc. H SO . The first step in the reaction is:
1
formation of carbocation
2
formation of ester
3
protonation of alcohol molecule
4
elimination of water
Official Solution
Correct Option: (3)
When alcohols are heated with concentrated H SO , the first step involves the protonation of the alcohol molecule, which leads to the formation of a carbocation. This carbocation can then undergo elimination of water to form an alkene. Therefore, the correct first step is the protonation of the alcohol molecule.
24
PYQ 2025
hard
chemistryID: cbse-cla
At low temperature, phenol reacts with Br in CS to form:
1
2,4,6-Tribromophenol
2
p-Bromophenol
3
o-and p-Bromophenol
4
2,4,4-Dibromophenol
Official Solution
Correct Option: (1)
At low temperatures, phenol reacts with bromine in CS to form a tribrominated product at the 2, 4, and 6 positions due to the activating effect of the hydroxyl group.
25
PYQ 2025
hard
chemistryID: cbse-cla
Assertion (A): A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation. Reason (R): o-nitrophenol is steam volatile due to intermolecular hydrogen bonding.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option: (1)
Since o-nitrophenol has hydrogen bonding, it is more volatile and thus can be separated from p-nitrophenol by steam distillation. The assertion is correct, and the reason is valid as well. Intermolecular hydrogen bonding in o-nitrophenol contributes to its steam volatility.
26
PYQ 2025
medium
chemistryID: cbse-cla
In the reaction R–OH + HCl RCl + H O, what is the correct order of reactivity of alcohols?
1
1 2 3
2
2 1 3
3
3 1 2
4
1 2 3
Official Solution
Correct Option: (1)
Lucas reagent (HCl + ZnCl ) reacts fastest with tertiary alcohols due to the formation of stable carbocations, so the order is 1 2 3 .
27
PYQ 2025
medium
chemistryID: cbse-cla
The reaction suggests that phenol is :
1
Basic
2
Neutral
3
Acidic
4
Amphoteric
Official Solution
Correct Option: (3)
Phenol reacts with strong bases like NaOH to form phenoxide ion and liberate H gas. This shows phenol is acidic enough to donate a proton to OH .
28
PYQ 2025
medium
chemistryID: cbse-cla
Assertion (A): The boiling point of ethanol is higher than that of methoxymethane. Reason (R): There is intramolecular hydrogen bonding in ethanol.
1
Assertion (A) is true, but Reason (R) is false.
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option: (1)
To solve this question, let's analyze the assertion and reason provided.
Assertion (A): The boiling point of ethanol is higher than that of methoxymethane.
Reason (R): There is intramolecular hydrogen bonding in ethanol.
1. Boiling Points of Ethanol and Methoxymethane: Ethanol (C2H5OH) has a higher boiling point compared to methoxymethane (CH3OCH3). The reason for this difference is primarily due to the presence of hydrogen bonding in ethanol.
2. Hydrogen Bonding in Ethanol: Ethanol exhibits intermolecular hydrogen bonding between the -OH group of one ethanol molecule and the hydrogen of another ethanol molecule. This strong intermolecular force requires more energy to break, resulting in a higher boiling point for ethanol compared to methoxymethane.
3. Evaluation of Assertion (A): The assertion that ethanol has a higher boiling point than methoxymethane is correct. Methoxymethane, a simple ether, does not exhibit hydrogen bonding, resulting in a lower boiling point compared to ethanol.
4. Evaluation of Reason (R): The reason provided is not entirely accurate. While it is true that ethanol has hydrogen bonding, it is intermolecular hydrogen bonding, not intramolecular hydrogen bonding, that influences the boiling point. Therefore, the reason is incorrect.
5. Conclusion: The assertion is correct, but the reason is incorrect. Ethanol has a higher boiling point due to intermolecular hydrogen bonding, not intramolecular hydrogen bonding.
Final Answer: The correct option is that the assertion is true, but the reason is false.
29
PYQ 2025
hard
chemistryID: cbse-cla
The conversion of phenol to salicylic acid can be accomplished by
1
Reimer-Tiemann reaction
2
Friedel-Crafts reaction
3
Kolbe reaction
4
Coupling reaction
Official Solution
Correct Option: (1)
The Reimer-Tiemann reaction is used to convert phenols to salicylic acid. In this reaction, phenol is treated with chloroform and a base to form salicylic acid. Step 2: Thus, the correct answer is option (1).
30
PYQ 2025
medium
chemistryID: cbse-cla
What will be formed after oxidation reaction of secondary alcohol with chromic anhydride (CrO )?
1
Aldehyde
2
Ketone
3
Carboxylic acid
4
Ester
Official Solution
Correct Option: (2)
When a secondary alcohol reacts with chromic anhydride (CrO ), it undergoes oxidation to form a ketone. This is a characteristic reaction for secondary alcohols where the hydroxyl group is oxidized to a carbonyl group.
31
PYQ 2025
medium
chemistryID: cbse-cla
The conversion of phenol to salicylic acid can be accomplished by
1
Reimer-Tiemann reaction
2
Friedel-Crafts reaction
3
Kolbe reaction
4
Coupling reaction
Official Solution
Correct Option: (1)
The conversion of phenol to salicylic acid is typically achieved by the Reimer-Tiemann reaction, where phenol undergoes electrophilic substitution in the presence of chloroform and a base (usually sodium hydroxide) to form salicylic acid.
32
PYQ 2025
medium
chemistryID: cbse-cla
Phenol reacts with Bromine (Br ) water to form:
1
2,4-dibromophenol
2
2-bromophenol
3
2,4,6-tribromophenol
4
4-bromophenol
Official Solution
Correct Option: (3)
Phenol activates the benzene ring towards electrophilic substitution, resulting in bromination at ortho and para positions, giving 2,4,6-tribromophenol.
33
PYQ 2025
medium
chemistryID: cbse-cla
Assertion (A): The bond angle C-O-C in ethers is slightly greater than tetrahedral angle.
Reason (R): This is because of the repulsive interaction between the two bulky alkyl groups.
Official Solution
Correct Option: (1)
The bond angle in ethers is slightly greater than the tetrahedral angle due to the lone pairs of electrons on the oxygen atom, which push the bonding pairs away, increasing the bond angle. However, the reason given regarding the repulsive interaction between bulky alkyl groups is incorrect; the main factor is the lone pair-bond pair repulsion.
34
PYQ 2025
medium
chemistryID: cbse-cla
How will you convert the following: (i) Toluene to benzoic acid (ii) Ethanol to propan-2-ol (iii) Propanal to 2-hydroxy propanoic acid
Official Solution
Correct Option: (1)
(i) Toluene to Benzoic acid: Toluene can be oxidized to benzoic acid using an oxidizing agent like potassium permanganate (KMnO ) or chromic acid (H CrO ). The methyl group (-CH ) is oxidized to a carboxyl group (-COOH), forming benzoic acid. (ii) Ethanol to Propan-2-ol: Ethanol can be converted to propan-2-ol by catalytic hydrogenation using hydrogen (H ) in the presence of a catalyst such as palladium (Pd). This reaction adds a hydrogen atom to the ethanol molecule at the second carbon, yielding propan-2-ol. (iii) Propanal to 2-hydroxy propanoic acid: Propanal can be oxidized to 2-hydroxy propanoic acid by using an oxidizing agent like potassium permanganate (KMnO ). The aldehyde group (-CHO) is oxidized to a carboxyl group (-COOH), and a hydroxyl group (-OH) is introduced at the second carbon via hydrolysis.
35
PYQ 2025
medium
chemistryID: cbse-cla
Arrange the following in increasing order of acid strength: phenol, ethanol, water.
Official Solution
Correct Option: (1)
To arrange phenol, ethanol, and water in increasing order of acid strength, we need to look at their ability to donate a proton (H ). The greater the ability to donate a proton, the stronger the acid: - Phenol is a weak acid, but it is stronger than ethanol due to the resonance stabilization of the phenoxide ion after deprotonation. - Ethanol is weaker than phenol as the ethoxide ion is less stabilized than the phenoxide ion. - Water is the weakest acid because it does not have any resonance stabilization after losing a proton. Thus, the order of increasing acid strength is:
36
PYQ 2025
medium
chemistryID: cbse-cla
Give a simple chemical test to distinguish between butan-1-ol and butan-2-ol.
Official Solution
Correct Option: (1)
A simple test to distinguish between butan-1-ol and butan-2-ol is the use of a Lucas test. In this test, Lucas reagent (concentrated HCl with ZnCl ) is added to the alcohol: - Butan-1-ol (a primary alcohol) reacts slowly to form an alkyl chloride. - Butan-2-ol (a secondary alcohol) reacts more quickly to form an alkyl chloride. The reaction rate can help distinguish the two alcohols. Butan-2-ol will show turbidity more rapidly than butan-1-ol.
37
PYQ 2025
medium
chemistryID: cbse-cla
Write chemical equations of the following reactions: (i) Phenol is treated with conc. HNO (ii) Propene is treated with B H followed by oxidation by H O /OH (iii) Sodium t-butoxide is treated with CH Cl
Official Solution
Correct Option: (1)
(i) Phenol reacts with concentrated nitric acid (HNO ) to give a mixture of nitrophenols. The major product is 2,4,6-nitrophenol (picric acid). The reaction is: (ii) Propene reacts with diborane (B H ) followed by oxidation with hydrogen peroxide (H O ) in the presence of hydroxide ions (OH ) to form propan-2-ol. The reaction is: (iii) Sodium t-butoxide (CH C(OH)CH C O) reacts with methyl chloride (CH Cl) to form isobutyl methyl ether. The reaction is:
38
PYQ 2025
medium
chemistryID: cbse-cla
An organic compound ‘A’, molecular formula C H O oxidises with CrO to form a compound ‘B’. Compound ‘B’ on warming with iodine and aqueous solution of NaOH gives a yellow precipitate of compound ‘C’. When compound ‘A’ is heated with conc. H SO at 413 K gives a compound ‘D’, which on reaction with excess HI gives compound ‘E’. Identify compounds ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ and write chemical equations involved.
Official Solution
Correct Option: (1)
- Compound ‘A’ is phenol (C H O), which reacts with CrO to form compound ‘B’, which is quinone (C H O ). - Compound ‘B’ reacts with iodine and aqueous NaOH to form compound ‘C’, which is iodoquinone (C H I O ). - Compound ‘A’ (phenol) when heated with conc. H SO at 413 K gives compound ‘D’, which is bisphenol (C H OH), through sulfonation. - Compound ‘D’ reacts with excess HI to form compound ‘E’, which is 2-iodophenol (C H IOH). \text{The chemical equations involved are:}
39
PYQ 2025
medium
chemistryID: cbse-cla
Give the IUPAC name of above mentioned ether.
Official Solution
Correct Option: (1)
The IUPAC name of t-butyl ethyl ether is 2-methoxypropane. The ether consists of an ethyl group (C H ) attached to a t-butyl group (C H ). Therefore, the correct name is 2-methoxypropane.
40
PYQ 2025
medium
chemistryID: cbse-cla
Write the IUPAC name of the product formed in the Reimer-Tiemann reaction.
Official Solution
Correct Option: (1)
Reimer-Tiemann Reaction
The Reimer-Tiemann reaction is a method used for the ortho-formylation of phenols. It involves the introduction of a formyl group (-CHO) to the aromatic ring of phenol using chloroform (CHCl₃) in the presence of a strong base, typically sodium hydroxide (NaOH).
General Reaction:
The general reaction can be represented as:
Product Formed:
The product of the Reimer-Tiemann reaction is 2-formylphenol, also known as ortho-hydroxybenzaldehyde.
IUPAC Name:
The IUPAC name of the product formed in the Reimer-Tiemann reaction is 2-formylphenol.
Structure of the Product:
In this reaction, the formyl group (-CHO) is added to the ortho position (adjacent to the hydroxyl group -OH) on the benzene ring.
Conclusion:
Therefore, the IUPAC name of the product formed in the Reimer-Tiemann reaction is 2-formylphenol.
41
PYQ 2025
easy
chemistryID: cbse-cla
Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of the OH group attached to the benzene ring. Since the OH group increases the electron density more to the ortho- and para- positions, one of the examples of the aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols. Answer the following questions:
Official Solution
Correct Option: (1)
42
PYQ 2025
medium
chemistryID: cbse-cla
Write chemical equations of the following reactions: (i) Phenol is treated with conc. HNO3. (ii) Propene is treated with B2H6 followed by oxidation by H2O2/OH−. (iii) Sodium t-butoxide is treated with CH3Cl.
Official Solution
Correct Option: (1)
1. Phenol treated with conc. : Phenol reacts with concentrated nitric acid ( ) to form a mixture of nitrophenols. The reaction is as follows: 2. Propene treated with followed by oxidation by : The hydroboration-oxidation reaction of propene results in the formation of an alcohol: Followed by oxidation: 3. Sodium t-butoxide treated with : Sodium t-butoxide undergoes nucleophilic substitution with methyl chloride to form t-butyl methyl ether:
43
PYQ 2026
medium
chemistryID: cbse-cla
on reaction with HI gives
1
and
2
and
3
and
4
and
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept: Reaction of ethers with HI depends on the nature of the alkyl groups. If one of the alkyl groups is tertiary ( ), the reaction follows an mechanism. Step 2: Detailed Explanation: The ether is tert-butyl ethyl ether. Upon protonation by HI, the molecule is . Because the tert-butyl group can form a highly stable carbocation , the C-O bond breaks via . The iodide ion ( ) then attacks this stable carbocation to form (tert-butyl iodide). The other product formed is ethanol ( ). Step 3: Final Answer: The products are and .
44
PYQ 2026
easy
chemistryID: cbse-cla
Ethers are prepared by the dehydration of alcohols in presence of protic acids at 413 K. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves attack of an alkoxide ion on a primary alkyl halide. If tertiary alkyl halide is used, elimination reaction occurs and alkene is formed instead of ether.
C–O bond in ethers are cleaved under drastic conditions with excess of HI. When unsymmetrical ethers react with HI, the alkyl halide is formed from the smaller alkyl group. If one of the alkyl groups is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary carbocation is more stable than primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl–oxygen bond due to more stable alkyl–oxygen bond.
The order of reactivity of hydrogen halides is:
Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring and directs the incoming group to ortho and para positions.
Official Solution
Correct Option: (1)
45
PYQ 2026
medium
chemistryID: cbse-cla
Write the reaction involved in Friedel-Crafts acylation of anisole.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: Friedel-Crafts acylation introduces an acyl group ( ) into the benzene ring using an acid chloride and a Lewis acid catalyst. Step 2: Detailed Explanation: Anisole ( ) is an activated aromatic ether. The methoxy group is ortho-para directing. Reaction with acetyl chloride ( ) in the presence of anhydrous yields a mixture of ortho and para products. The para-isomer (4-Methoxyacetophenone) is the major product due to less steric hindrance. Chemical Equation:
Step 3: Final Answer: The major product is 4-Methoxyacetophenone.
46
PYQ 2026
medium
chemistryID: cbse-cla
Write the reaction involved in Kolbe’s reaction.
Official Solution
Correct Option: (1)
Step 1: Conceptual Overview:
Kolbe's reaction involves the electrophilic substitution of the phenoxide ion by carbon dioxide, leading to the formation of a hydroxy acid. This reaction is commonly used to synthesize salicylic acid, which is the precursor to aspirin. Step 2: Detailed Explanation:
To begin the reaction, phenol is first converted into sodium phenoxide by reacting phenol with sodium hydroxide (NaOH).
Next, sodium phenoxide undergoes a reaction with carbon dioxide ( ) under conditions of 400K temperature and 4-7 atm pressure. This reaction leads to the formation of an intermediate compound, sodium salicylate.
The final step involves acidifying the sodium salicylate with a proton source ( ) to yield the desired product, 2-hydroxybenzoic acid, also known as salicylic acid.
The overall chemical equation for Kolbe's reaction is as follows:
Step 3: Final Conclusion:
The product formed in Kolbe's reaction is salicylic acid, which is an important compound in the pharmaceutical industry.
47
PYQ 2026
medium
chemistryID: cbse-cla
Write the reaction involved in Reimer-Tiemann reaction.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The Reimer-Tiemann reaction is used to introduce an aldehyde group ( ) at the ortho position of the phenol ring. Step 2: Detailed Explanation: When phenol is treated with chloroform ( ) in the presence of sodium hydroxide, a substituted benzaldehyde (salicylaldehyde) is formed. The intermediate is a dichlorocarbene ( ) which acts as an electrophile. Chemical Equation:
The main product is 2-Hydroxybenzaldehyde (Salicylaldehyde). Step 3: Final Answer: The reaction yields salicylaldehyde as the major product.
48
PYQ 2026
medium
chemistryID: cbse-cla
Assertion (A) : The presence of group in phenols directs the incoming group to ortho and para positions. Reason (R) : group in phenols activates the aromatic ring towards electrophilic substitution reaction.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: Phenol molecules undergo electrophilic aromatic substitution very easily due to the electron-releasing nature of the hydroxyl group. Step 2: Detailed Explanation: The group has lone pairs on oxygen which are in conjugation with the benzene ring. By the resonance effect ( effect), it increases the electron density on the ring, especially at the ortho and para positions. This high electron density makes these positions more reactive towards electrophiles (Activation). Since the density is highest at ortho and para, the electrophile attacks there, making the group o,p-directing. Thus, both statements are true and R explains why substitution occurs at o/p positions. Step 3: Final Answer: Code (A) is correct.
49
PYQ 2026
hard
chemistryID: cbse-cla
Which of the following Grignard Reagent will be used to prepare cyclohexylmethanol when treated with methanal?
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept: Reaction of a Grignard reagent ( ) with methanal ( ) followed by acid hydrolysis always yields a primary alcohol with one extra carbon atom ( ). Step 2: Detailed Explanation: The target product is cyclohexylmethanol, which has the structure: (Cyclohexyl ring) . From the general reaction, the part comes from the methanal ( ). The group must therefore be the cyclohexyl group. Thus, the Grignard reagent required is Cyclohexylmagnesium bromide. Step 3: Final Answer: The correct reagent is Cyclohexylmagnesium bromide.
50
PYQ 2026
medium
chemistryID: cbse-cla
Identify the correct increasing order of boiling points of the given compounds:
1
Propan-1-ol butan-1-ol butan-2-ol pentan-1-ol
2
Pentan-1-ol butan-1-ol butan-2-ol propan-1-ol
3
Propan-1-ol butan-2-ol butan-1-ol pentan-1-ol
4
Butan-1-ol butan-2-ol propan-1-ol pentan-1-ol
Official Solution
Correct Option: (3)
Concept: Boiling points of alcohols depend on:
Hydrogen bonding (present in all alcohols)
Molecular mass (higher mass higher boiling point)
Branching (more branching lower boiling point due to less surface area)
Step 1: Compare carbon chain length. Higher carbon chain stronger van der Waals forces higher boiling point. So:
Step 2: Compare branching among butanol isomers. Butan-1-ol = primary alcohol (less branched) Butan-2-ol = secondary alcohol (more branched) More branching lower boiling point. So:
Step 3: Combine both trends.
