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JEE • CHEMISTRY • ALCOHOLS & PHENOLS & ETHERS

Alcohols & phenols & ethers practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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121 Medium
121 of 193
Analyzing +4 | -1
Assertion (A): 7.5 g7.5 \mathrm{~g} of ethoxyethane and 9 g9 \mathrm{~g} of butan-1-ol is miscible to almost same extent while pentane is essentially immiscible with water.

Reason (R): Alcohols and ethers can form hydrogen bonds with the water molecule.
View answer and explanation
Correct option: A
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is the correct explanation of A\mathrm{A}.
Option A (Both A and R are true and R is the correct explanation of A) is correct because the back bonding in BF3BF_3 makes it a weaker Lewis acid compared to BCl3BCl_3.
122 Easy
122 of 193
Remembering +4 | -1
Saponification means hydrolysis of an ester with:
View answer and explanation
Correct option: D
NaOH\mathrm{NaOH}
Option D (NaOH) is correct because saponification is the hydrolysis of an ester using a strong base like NaOH to produce a soap (salt of a fatty acid) and alcohol.
123 Hard
123 of 193
Understanding +4 | -1
Name the reagent used to convert bromoethane to ethoxy ethane
View answer and explanation
Correct option: A
Sodium ethoxide
Option A (Sodium ethoxide) is correct because bromoethane reacts with sodium ethoxide via Williamson ether synthesis to form ethoxyethane.
124 Medium
124 of 193
Remembering +4 | -1
Chlorex which is a good solvent for aromatic impurities is:
View answer and explanation
Correct option: B
Dichlorodiethyl ether
Option B (Dichlorodiethyl ether) is correct because Chlorex, a good solvent for aromatic impurities, is dichlorodiethyl ether.
125 Medium
125 of 193
Analyzing +4 | -1
An alcohol on oxidation is found to give CH3COOHC{H_3}COOH and CH3CH2COOHC{H_3}C{H_2}COOH The alcohol is:
View answer and explanation
Correct option: D
CH3CH(OH)CH2CH2CH3C{H_3}CH(OH)C{H_2}C{H_2}C{H_3}
Option D (CH3CH(OH)CH2CH2CH3C{H_3}CH(OH)C{H_2}C{H_2}C{H_3}) is correct because its oxidation leads to the formation of both acetic acid and propanoic acid via bond cleavage.
126 Medium
126 of 193
Remembering +4 | -1
Ethers on hydrolysis yield
View answer and explanation
Correct option: A
Alcohol
Option A (Alcohol) is correct because ethers, upon hydrolysis, yield alcohols.
127 Medium
127 of 193
Understanding +4 | -1
The reaction of CH3OC2H5\mathrm{C}{\mathrm{H}_3}\mathrm{O}{\mathrm{C}_2}{\mathrm{H}_5} with HI\mathrm{HI} gives:
View answer and explanation
Correct option: C
CH3I+C2H5OH\mathrm{C}{\mathrm{H}_3}\mathrm{I} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}
Option C (CH3I+C2H5OH\mathrm{C}{\mathrm{H}_3}\mathrm{I} + {\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}) is correct because the reaction of an ether with HI cleaves the ether bond, with the smaller alkyl group typically forming the iodide and the larger forming the alcohol.
128 Medium
128 of 193
Analyzing +4 | -1
For which of the following parameters the structural isomers C2H5OH{C_2}{H_5}OH and CH3OCH3C{H_3}OC{H_3} would be expected to have the same values? (Assume ideal behaviour)
View answer and explanation
Correct option: D
Gaseous densities at the same temperature and pressure
Option D (Gaseous densities at the same temperature and pressure) is correct because the ideal gas law relates density to molar mass, which is the same for both isomers.
129 Easy
129 of 193
Understanding +4 | -1
Ethyl iodide on treatment with dry Ag2O\mathrm{A}{\mathrm{g}_2}\mathrm{O} will yield:
View answer and explanation
Correct option: B
Diethyl ether
Option B (Diethyl ether) is correct because ethyl iodide reacts with dry Ag2O\mathrm{A}{\mathrm{g}_2}\mathrm{O} to produce diethyl ether.
130 Easy
130 of 193
Understanding +4 | -1
Phenol is weakly acidic but does not react with NaHCO3\mathrm{NaHC}{\mathrm{O}_3} like carboxylic acids hence:
View answer and explanation
Correct option: A
Phenol is weaker than carbonic acid
Option A (Phenol is weaker than carbonic acid) is correct because phenol does not react with sodium bicarbonate (NaHCO3\mathrm{NaHC}{\mathrm{O}_3}), indicating it's a weaker acid than carbonic acid, which is formed when NaHCO3\mathrm{NaHC}{\mathrm{O}_3} reacts with an acid.
131 Medium
131 of 193
Analyzing +4 | -1
The most acidic compound among following is:
View answer and explanation
Correct option: D
CH3−CO∣∣−OHC{H_3} - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - OH
Option D is correct because carboxylic acids are more acidic than other given compounds due to resonance stabilization of the carboxylate ion.
132 Medium
132 of 193
Analyzing +4 | -1
Phenol→(ii)   CO2(i)  NaOH(A)→H+/H2O(B)→Ac2O(C){\text{Phenol}}\xrightarrow[{\left( {{\text{ii}}} \right)\,\,\,C{O_2}}]{{\left( {\text{i}} \right)\,\,NaOH}}\left( A \right)\xrightarrow{{{H^ + }/{H_2}O}}\left( B \right)\xrightarrow{{A{c_2}O}}\left( C \right)
In this reaction, identify the incorrect statement?
View answer and explanation
Correct option: D
CC is a paracetamol
Option D (CC is a paracetamol) is correct because the final product CC is aspirin (O-acetoxybenzoic acid), not paracetamol.
133 Easy
133 of 193
Understanding +4 | -1
The reaction of Lucas reagent is fastest with
View answer and explanation
Correct option: D
(CH3)3COH\left({CH}_3\right)_3 {COH}
Option D ((CH3)3COH) is correct because tertiary alcohols react fastest with Lucas reagent due to the formation of a stable tertiary carbocation.
134 Medium
134 of 193
Remembering +4 | -1
Lucas reagent is
View answer and explanation
Correct option: A
Conc. HCI\mathrm{HCI} and anhydrous ZnCI2\mathrm{ZnC}{\mathrm{I}_2}
Option A (Conc. HCI\mathrm{HCI} and anhydrous ZnCI2\mathrm{ZnC}{\mathrm{I}_2}) is correct because Lucas reagent is a solution of concentrated hydrochloric acid and anhydrous zinc chloride.
135 Medium
135 of 193
Remembering +4 | -1
Phenol have ____ acidic strength than alcohols:
View answer and explanation
Correct option: B
Greater
The correct answer is B (Greater) because phenols are more acidic than alcohols due to the stabilization of the phenoxide ion by resonance.
136 Medium
136 of 193
Application +4 | -1
What is the product when C6H5−CH2−NH2C_6H_5 - CH_2 - NH_2 reacts with nitrous acid?
View answer and explanation
Correct option: C
C6H5−CH2−OH{C_6}{H_5} - C{H_2} - OH
Option C (C6H5−CH2−OH{C_6}{H_5} - C{H_2} - OH) is correct because primary amines react with nitrous acid to form diazonium salts, which decompose to form alcohols.
137 Medium
137 of 193
Understanding +4 | -1
To distinguish between salicylic acid and phenol one can use
View answer and explanation
Correct option: A
NaHCO3\mathrm{NaHC}{\mathrm{O}_3} solution
Option A (NaHCO3\mathrm{NaHC}{\mathrm{O}_3} solution) is correct because salicylic acid reacts with NaHCO3NaHCO_3 to produce CO2CO_2 effervescence, while phenol does not.
138 Easy
138 of 193
Application +4 | -1
The product C in the following sequence of reaction,
C2H5Br→NaOH  (aq)A→NaB→CH3IC{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Br}\xrightarrow{{\mathrm{NaOH\;}\left( {aq} \right)}}A\mathop \to \limits^{\mathrm{Na}} B\xrightarrow{{\mathrm{C}{\mathrm{H}_3}\mathrm{I}}}C is:
View answer and explanation
Correct option: C
Methyl ethyl ether
Option C (Methyl ethyl ether) is correct because the sequence of reactions leads to the formation of an ether via Williamson ether synthesis.
139 Easy
139 of 193
Analyzing +4 | -1
If the boiling point of ethanol (molecular weight=46) is 78∘C^ \circ C, what is the boiling point of diethyl ether?
(molecular weight=74)
View answer and explanation
Correct option: D
34∘C^ \circ C
Option D (34 ∘C^\circ C) is correct because diethyl ether lacks hydrogen bonding, resulting in a significantly lower boiling point compared to ethanol, despite having a higher molecular weight.
140 Easy
140 of 193
Remembering +4 | -1
Proof spirit contains about:
View answer and explanation
Correct option: A
48% alcohol by weight
Option A (48% alcohol by weight) is correct because proof spirit contains approximately 48% alcohol by weight or 57.1% by volume.
141 Medium
141 of 193
Understanding +4 | -1
Phenol→ Zinc Distillation (A)→ Conc.  HNO3Conc. H2SO4 at 60∘C(B){\text{Phenol}}\xrightarrow{{{\text{ Zinc Distillation }}}}(A)\xrightarrow{{{\text{ }}\begin{array}{cccccccccccccccccccc} {{\text{Conc}}{\text{.}}\,\,HN{O_3}} \\ {{\text{Conc}}{\text{. }}{H_2}S{O_4}{\text{ at }}{{60}^\circ }{\text{C}}} \end{array}}}\left( B \right)

In the above reaction, compounds (A)(A) and (B)(B) are:
View answer and explanation
Correct option: D
benzene and nitrobenzene
Option D (benzene and nitrobenzene) is correct because phenol on zinc distillation gives benzene, which on nitration with conc. HNO3HNO_3 and conc. H2SO4H_2SO_4 at 60°C gives nitrobenzene.
142 Medium
142 of 193
Remembering +4 | -1
Purity of ether before using it as anaesthetic agent is tested by:
View answer and explanation
Correct option: A
KI+\mathrm{KI} + starch
Option A is correct because the presence of peroxides in ether, which can form upon exposure to air, is detected using KI and starch, leading to the formation of iodine and a blue color.
143 Medium
143 of 193
Analyzing +4 | -1
Which of the following can be used to distinguish between phenol and cyclohexanol
View answer and explanation
Correct option: B
NaOHNaOH
Option B (NaOHNaOH) is correct because phenol reacts with NaOH to form a salt soluble in water, while cyclohexanol does not react.
144 Medium
144 of 193
Analyzing +4 | -1
For the molecule
CH3CH(OH)CH(OH)CH(OH)CH3{CH}_3 {CH}({OH}) {CH}({OH}) {CH}({OH}) {CH}_3 total number of stereoisomers and meso form respectively are
View answer and explanation
Correct option: D
4,2
Option B (2-methylpentane) is correct because it has the most non-equivalent hydrogen atoms, leading to the maximum number of monochloro derivatives.
145 Medium
145 of 193
Analyzing +4 | -1
Ph− C∣HOH∣−CH3Ph - \mathop {\mathop {{\text{ }}C}\limits_{\begin{array}{cccccccccccccccccccc} | \\ H \end{array}} }\limits^{\begin{array}{cccccccccccccccccccc} {OH} \\ | \end{array}} - C{H_3}

can be prepared from which of the following combinations?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because Grignard reagents can react with aldehydes and ketones, and reduction of ketones can also yield the desired alcohol.
146 Easy
146 of 193
Application +4 | -1
The number of isomeric alcohols of formula C4H10O{\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O} is:
View answer and explanation
Correct option: B
4
Option B (4) is correct because there are four structural isomers of alcohols with the formula C4H10O: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol.
147 Medium
147 of 193
Understanding +4 | -1
The alcohol which does not give a stable compound on dehydration is
View answer and explanation
Correct option: B
methyl alcohol
Option B (methyl alcohol) is correct because dehydration of methyl alcohol leads to the formation of unstable carbene.
148 Medium
148 of 193
Analyzing +4 | -1
Which of the following sequence of reagent is the good means to furnish the conversion?

R−CH2OH→R−CH2−CH2−NH2\mathrm{R}-\mathrm{CH}_2 \mathrm{OH} \rightarrow \mathrm{R}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2
View answer and explanation
Correct option: B
SOCl2,NaCN,H2/Ni\mathrm{SOCl}_2, \mathrm{NaCN}, \mathrm{H}_2 / \mathrm{Ni}
Option B is correct because the sequence of reagents SOCl₂, NaCN, and H₂/Ni effectively converts the alcohol to an amine with two additional carbon atoms.
149 Easy
149 of 193
Remembering +4 | -1
Carbolic acid is
View answer and explanation
Correct option: D
C6H5OH{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH}
Option D (C6H5OH{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH}) is correct because carbolic acid is the common name for phenol, which has the formula C6H5OH{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{OH}.
150 Medium
150 of 193
Application +4 | -1
HOOC−CH2−CH2−COOC2H5→1 equivalentBH3/THFA. HOOC - C{H_2} - C{H_2} - COO{C_2}{H_5}\xrightarrow[{1 \text{ equivalent}}]{{B{H_3}/THF}}A.{\text{ }}

A will be
View answer and explanation
Correct option: A
CH2∣CH2OH−CH2−COOC2H5\mathop {C{H_2}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {C{H_2}OH} \end{array}} - C{H_2} - COO{C_2}{H_5}
Option A is correct because BH3BH_3 reduces the carboxylic acid group to an alcohol, while the ester group remains unaffected.
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