VITEEE PYQs for Alcohols, Phenols and Ethers with Solutions: Practice VITEEE Previous Year Questions

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Alcohols, Phenols and Ethers is an important topic in the Chemistry section in VITEEE exam. Practising this topic will increase your score overall and make your conceptual grip on VITEEE exam stronger.

This article gives you a full set of VITEEE PYQs for Alcohols, Phenols and Ethers with explanations for effective preparation. Practice of VITEEE Chemistry PYQs including Alcohols, Phenols and Ethers questions regularly will improve accuracy, speed, and confidence in the VITEEE 2026 exam.

VITEEE PYQs for Alcohols, Phenols and Ethers with Solutions

1.
$ p $-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is
- A
- B
- C
- D
View Solution
2.
A substance C$_4$H$_8$O yields on oxidation a compound, C$_4$H$_8$O which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H$_2$SO$_4$ gives C$_4$H$_8$. The structure of the compound is
- CH$_3$COCH$_2$OH
- CH$_3$COOH
- CH$_3$COCH$_2$CH$_3$
- CH$_3$CH$_2$OH
View Solution
3.
Phenol undergoes electrophilic substitution more easily than benzene because
- $ \text{-OH} $ group exhibits +M effect and hence increases the electron density on the $ \alpha- $ and $ \beta- $ positions.
- $ \text{-OH} $ group exhibits $ -M $ effect and hence decreases the electron density on the $ \alpha- $ and $ \beta- $ positions.
- Oxocation is more stable than the carbocation.
- Both (a) and (b)
View Solution
4.
What are $X$ and $Y$ in the following reaction sequence $C _{2} H _{5} OH \ce{->[{ Cl _{2} }]} \underline{X} \ce{->[{ Cl _{2}}]}\underline{Y}$
- $C _{2} H _{5} Cl , CH _{3} CHO$
- $CH _{3} CHO , CH _{3} CO _{2} H$
- $CH _{3} CHO , CCl _{3} CHO$
- $C _{2} H _{5} Cl , CCl _{3} CHO$
View Solution
5.
Phenol undergoes electrophilic substitution more easily than benzene because
- -OH group exhibits +M effect and hence increases the electron density on the o- and p-positions.
- oxocation is more stable than the carbocation
- both (a) and (b)
- -OH group exhibits acidic character
View Solution
6.
What is Z in the following sequence of reactions? \[ \text{Phenol} \xrightarrow{\text{Zn dust}} \text{CH}_6 \xrightarrow{\text{Alkaline KMnO}_4} Z \]
- Benzene
- Toluene
- Benzaldehyde
- Benzoic acid
View Solution
7.
In the reaction, $ C_6H_5OH + NaOH \rightarrow $ (B) $ \rightarrow (C) $, the compound (C) is:
- Benzoic acid
- Salicylaldehyde
- Chlorobenzene
- Salicylic acid
View Solution
8.
Ethylene glycol, on oxidation with periodic acid, gives:
- Oxalic acid
- Glycol
- Formaldehyde
- Glycolic acid
View Solution
9.
When aqueous solution of benzene diazoniumchloride is boiled, the product formed is
- ${C_6H_5CH_2OH}$
- ${C_6H_6 + N_2}$
- ${C_6H_5COOH}$
- ${C_6H_5OH}$
View Solution
10.
$ p $-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is:
- A
- B
- C
- D
View Solution
11.
${A->[K_2Cr_2O_7][dil.H_2SO_4]B->[CH_3MgI][H_2O]CH3}$ the reactant A is
- $CH_3CHOHCH_3$
- $CH_3COCH_3$
- $C_2H_5OH$
- $CH_3COOH$
View Solution
12.
1-Propanol and 2-propanol can be distinguished by
- oxidation with alkaline KMnO$_4$ followed by reaction with Fehling solution
- oxidation with acidic dichromate followed by reaction with Fehling solution
- oxidation by heating with copper followed by reaction with Fehling solution
- oxidation with concentrated H$_2$SO$_4$ followed by reaction with Fehling solution
View Solution
13.
A compound is soluble in conc. H$_2$SO$_4$. It does not decolourise bromine in carbon tetrachloride but is oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning orange solution to blue, green and then opaque. The original compound is
- a primary alcohol
- a tertiary alcohol
- an alkane
- an alkene
View Solution
14.
Compound ‘A’ of molecular formula $ C_6H_10O $ on treatment with Lucas reagent at room temperature gives compound ‘B’. When compound ‘B’ is heated with alcoholic KOH, it gives isobutene. Compound ‘A’ and ‘B’ are respectively:
- 2-methyl-2-propanol and 2-methyl-2-chloropropane
- 2-methyl-1-propanol and 1-chloro-2-methylpropane
- 2-methyl-1-propanol and 2-methyl-2-chloropropane
- butan-2-ol and 2-chlorobutane
View Solution
15.
Anisole is treated with HI under two different conditions. \[ C_6H_5OCH_3 + D \xrightarrow{\text{conc. HI}} A + B \] The nature of A and B will be
- A and B are $ \text{CH}_3 $ and $ \text{C}_6\text{H}_5\text{OH} $, while C and D are $ \text{CH}_3 \text{I} $ and $ \text{C}_6\text{H}_5\text{I} $
- A and B are $ \text{C}_6\text{H}_5\text{OH} $ and $ \text{CH}_3 \text{I} $, while C and D are $ \text{CH}_3\text{I} $ and $ \text{C}_6\text{H}_5\text{OH} $
- A and B are $ \text{C}_6\text{H}_5\text{OH} $ and $ \text{CH}_3 $, while C and D are $ \text{CH}_3\text{I} $ and $ \text{C}_6\text{H}_5\text{I} $
- Both A and B as well as both C and D are $ \text{CH}_3\text{I} $ and $ \text{C}_6\text{H}_5\text{OH} $
View Solution

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VITEEE PYQs for Alcohols, Phenols and Ethers with Solutions: Practice VITEEE Previous Year Questions

VITEEE PYQs for Alcohols, Phenols and Ethers with Solutions

1.
\( p \)-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

2.
A substance C\(_4\)H\(_8\)O yields on oxidation a compound, C\(_4\)H\(_8\)O\(\) which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H\(_2\)SO\(_4\) gives C\(_4\)H\(_8\). The structure of the compound is

3.
Phenol undergoes electrophilic substitution more easily than benzene because

4.
What are $X$ and $Y$ in the following reaction sequence $C _{2} H _{5} OH \ce{->[{ Cl _{2} }]} \underline{X} \ce{->[{ Cl _{2}}]}\underline{Y}$

5.
Phenol undergoes electrophilic substitution more easily than benzene because

6.
What is Z in the following sequence of reactions? \[ \text{Phenol} \xrightarrow{\text{Zn dust}} \text{CH}_6 \xrightarrow{\text{Alkaline KMnO}_4} Z \]

7.
In the reaction, \( C_6H_5OH + NaOH \rightarrow \) (B) \( \rightarrow (C) \), the compound (C) is:

8.
Ethylene glycol, on oxidation with periodic acid, gives:

9.
When aqueous solution of benzene diazoniumchloride is boiled, the product formed is

10.
\( p \)-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is:

11.
\({A->[K_2Cr_2O_7][dil.H_2SO_4]B->[CH_3MgI][H_2O]CH3}\) the reactant A is

12.
1-Propanol and 2-propanol can be distinguished by

13.
A compound is soluble in conc. H\(_2\)SO\(_4\). It does not decolourise bromine in carbon tetrachloride but is oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning orange solution to blue, green and then opaque. The original compound is

14.
Compound ‘A’ of molecular formula \( C_6H_10O \) on treatment with Lucas reagent at room temperature gives compound ‘B’. When compound ‘B’ is heated with alcoholic KOH, it gives isobutene. Compound ‘A’ and ‘B’ are respectively:

15.
Anisole is treated with HI under two different conditions. \[ C_6H_5OCH_3 + D \xrightarrow{\text{conc. HI}} A + B \] The nature of A and B will be

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VITEEE PYQs for Alcohols, Phenols and Ethers with Solutions

1.\( p \)-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

2.A substance C\(_4\)H\(_8\)O yields on oxidation a compound, C\(_4\)H\(_8\)O\(\) which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H\(_2\)SO\(_4\) gives C\(_4\)H\(_8\). The structure of the compound is

3.Phenol undergoes electrophilic substitution more easily than benzene because

4.What are $X$ and $Y$ in the following reaction sequence $C _{2} H _{5} OH \ce{->[{ Cl _{2} }]} \underline{X} \ce{->[{ Cl _{2}}]}\underline{Y}$

5.Phenol undergoes electrophilic substitution more easily than benzene because

6.What is Z in the following sequence of reactions? \[ \text{Phenol} \xrightarrow{\text{Zn dust}} \text{CH}_6 \xrightarrow{\text{Alkaline KMnO}_4} Z \]

7.In the reaction, \( C_6H_5OH + NaOH \rightarrow \) (B) \( \rightarrow (C) \), the compound (C) is:

8.Ethylene glycol, on oxidation with periodic acid, gives:

9.When aqueous solution of benzene diazoniumchloride is boiled, the product formed is

10.\( p \)-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is:

11.\({A->[K_2Cr_2O_7][dil.H_2SO_4]B->[CH_3MgI][H_2O]CH3}\) the reactant A is

12.1-Propanol and 2-propanol can be distinguished by

13.A compound is soluble in conc. H\(_2\)SO\(_4\). It does not decolourise bromine in carbon tetrachloride but is oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning orange solution to blue, green and then opaque. The original compound is

14.Compound ‘A’ of molecular formula \( C_6H_10O \) on treatment with Lucas reagent at room temperature gives compound ‘B’. When compound ‘B’ is heated with alcoholic KOH, it gives isobutene. Compound ‘A’ and ‘B’ are respectively:

15.Anisole is treated with HI under two different conditions. \[ C_6H_5OCH_3 + D \xrightarrow{\text{conc. HI}} A + B \] The nature of A and B will be

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1.
\( p \)-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is

2.
A substance C\(_4\)H\(_8\)O yields on oxidation a compound, C\(_4\)H\(_8\)O\(\) which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H\(_2\)SO\(_4\) gives C\(_4\)H\(_8\). The structure of the compound is

3.
Phenol undergoes electrophilic substitution more easily than benzene because

4.
What are $X$ and $Y$ in the following reaction sequence $C _{2} H _{5} OH \ce{->[{ Cl _{2} }]} \underline{X} \ce{->[{ Cl _{2}}]}\underline{Y}$

5.
Phenol undergoes electrophilic substitution more easily than benzene because

6.
What is Z in the following sequence of reactions? \[ \text{Phenol} \xrightarrow{\text{Zn dust}} \text{CH}_6 \xrightarrow{\text{Alkaline KMnO}_4} Z \]

7.
In the reaction, \( C_6H_5OH + NaOH \rightarrow \) (B) \( \rightarrow (C) \), the compound (C) is:

8.
Ethylene glycol, on oxidation with periodic acid, gives:

9.
When aqueous solution of benzene diazoniumchloride is boiled, the product formed is

10.
\( p \)-cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is:

11.
\({A->[K_2Cr_2O_7][dil.H_2SO_4]B->[CH_3MgI][H_2O]CH3}\) the reactant A is

12.
1-Propanol and 2-propanol can be distinguished by

13.
A compound is soluble in conc. H\(_2\)SO\(_4\). It does not decolourise bromine in carbon tetrachloride but is oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning orange solution to blue, green and then opaque. The original compound is

14.
Compound ‘A’ of molecular formula \( C_6H_10O \) on treatment with Lucas reagent at room temperature gives compound ‘B’. When compound ‘B’ is heated with alcoholic KOH, it gives isobutene. Compound ‘A’ and ‘B’ are respectively:

15.
Anisole is treated with HI under two different conditions. \[ C_6H_5OCH_3 + D \xrightarrow{\text{conc. HI}} A + B \] The nature of A and B will be