Alcohols, Phenols and Ethers — Page 2 | JEE Chemistry

Q11

When ethanol is heated with conc.

\mathrm{H}_{2} \mathrm{SO}_{4}

, a gas is produced. The compound formed, when this gas is treated with cold dilute aqueous solution of Baeyer's reagent, is

A formaldehyde

B formic acid

C glycol

D ethanoic acid

Correct Answer

Option C

Solution

\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{\text{ Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\Delta} \mathrm{CH}_{2}=\mathrm{CH}_{2} \frac{\text{ Cold dil solution of }}{\text{ Bayers reagent }} \mathrm{OH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\,(\text{Glycol})

$

Q12

The increasing order of

\mathrm{pK_a}

for the following phenols is (A) 2, 4 - Dinitrophenol (B) 4 - Nitrophenol (C) 2, 4, 5 - Trimethylphenol (D) Phenol (E) 3-Chlorophenol Choose the correct answer from the option given below :

A (A), (E), (B), (D), (C)

B (C), (E), (D), (B), (A)

C (C), (D), (E), (B), (A)

D (A), (B), (E), (D), (C)

Correct Answer

Option D

Solution

Order of acidity for following phenol is A > B > E > D > C. - M and – I increases acidity.

+ M and + I decreases acidity.

$\therefore$ Their pKa value is A < B < E < D < C.

Q13

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Acetal / Ketal is stable in basic medium. Reason R : The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium. In the light of the above statements, choose the correct answer from the options given below :

A A is false but R is true.

B Both A and R are true but R is NOT the correct explanation of A

C Both A and R are true and R is the correct explanation of A

D A is true but R is false.

Correct Answer

Option D

Solution

For Assertion: Acetal and ketals are basically ethers hence they must be stable in basic medium but should break down in acidic medium.

Hence assertion is correct.

For reason: Alkoxide ion (RO–) is not considered a good leaving group hence reason must be false.

Q14

The water gas on reacting with cobalt as a catalyst forms

A Methanoic acid

B Methanol

C Methanal

D Ethanol

Correct Answer

Option B

Solution

Water gas, also known as synthesis gas or "syngas", is a mixture of carbon monoxide (CO) and hydrogen (H2).

When water gas reacts with a catalyst, typically consisting of a combination of cobalt and molybdenum, it can undergo a reaction to form methanol (CH3OH).

This is known as the methanol synthesis reaction and can be written as:

\mathrm{CO(g)} + 2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{3}\mathrm{OH(g)}

This reaction is widely used in industry for the production of methanol, which can be used as a solvent, an antifreeze, a fuel, and as a feedstock for the production of chemicals.

Q15

2-Methyl propyl bromide reacts with

\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}

and gives 'A' whereas on reaction with

\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}

it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :

A

\mathrm{S}_{N} 1, A=

tert-butyl ethyl ether;

\mathrm{S}_{N} 2, B=

iso-butyl ethyl ether

B

\mathrm{S}_{\mathrm{N}} 1, \mathrm{~A}=

tert-butyl ethyl ether;

\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=

2-butyl ethyl ether

C

\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=

iso-butyl ethyl ether;

\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=

tert-butyl ethyl ether

D

\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=

2-butyl ethyl ether;

\mathrm{S}_{\mathrm{N}} 2, \mathrm{~B}=

iso-butyl ethyl ether

Correct Answer

Option C

Solution

2-Methyl propyl bromide (also known as isobutyl bromide) has the formula (CH₃)₂CHCH₂Br.

When it reacts with C₂H₅O⁻ (ethoxide ion), it undergoes an SN2 reaction because ethoxide ion is a strong nucleophile.

The reaction proceeds with a direct exchange of the leaving group (Br⁻) and the nucleophile (C₂H₅O⁻).

The product 'A' would be iso-butyl ethyl ether (CH₃CH₂OCH(CH₃)CH₃).

When it reacts with C₂H₅OH (ethanol), the reaction proceeds via an SN1 mechanism because ethanol is a weak nucleophile.

In this case, the bromide ion leaves first, forming a carbocation intermediate, which is then attacked by the nucleophile (C₂H₅OH).

The product 'B' would be tert-butyl ethyl ether ((CH₃)₃COCH₂CH₃).

So, the correct option is: SN2, A = iso-butyl ethyl ether; SN1, B = tert-butyl ethyl ether

Q16

What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol? (Given molar mass in

\mathrm{g} \mathrm{mol}^{-1}

of

\mathrm{C}, \mathrm{H}, \mathrm{O}, \mathrm{Br}

are

12,1,16,80

respectively )

A 6.0 g

B 10.22 g

C 20.44 g

D 4.0 g

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Moles of phenol }=\frac{2}{94}=0.021 \\ & \therefore \text{ Moles of bromine }=0.021 \times 3=0.064 \\ & \therefore \text{ Mass of bromine }=0.064 \times 160=10.22 \mathrm{~g} \end{aligned}

Q17

Given below are two statements : Statement (I) : The boiling points of alcohols and phenols increase with increase in the number of C -atoms. Statement (II) : The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is false but Statement II is true

B Both Statement I and Statement II are true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

Correct Answer

Option B

Solution

\textbf{Statement (I):}

As the number of carbon atoms in alcohols or phenols increases, the molecule becomes larger, which in turn enhances the van der Waals (dispersion) forces between the molecules.

Stronger intermolecular forces require more energy (i.e., higher temperature) to overcome, leading to higher boiling points.

Therefore, Statement (I) is true.

\textbf{Statement (II):}

Alcohols and phenols have a hydroxyl group (

-OH

) that can form hydrogen bonds, which are significantly stronger than the dipole-dipole interactions typically seen in ethers or the interactions in haloalkanes.

This stronger hydrogen bonding means that alcohols and phenols have higher boiling points compared to ethers and haloalkanes of similar molecular weight.

Hence, Statement (II) is also true.

Based on the above points, the correct answer is: Option B: Both Statement I and Statement II are true.

Q18

Match with . th, (Reactants) (Product)

List - I		List - II
(A)	Phenol, Zn/Δ	(I)	Salicylaldehyde
(B)	Phenol, CHCl3, NaOH, HCl	(II)	Salicylic acid
(C)	Phenol, CO2, NaOH, HCl	(III)	Benzene
(D)	Phenol, Conc. HNO3	(IV)	Picric acid

A (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

B (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

C (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

D (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct Answer

Option B

Solution

Let's analyze each reaction given in List I to identify the correct product from List II.

(A) Phenol, Zn/Δ: When phenol is reacted with zinc dust (Zn) upon heating (Δ), it undergoes a reduction reaction known as the Clemmensen reduction, which results in the removal of the oxygen from the hydroxyl group (-OH) attached to the benzene ring, thus forming benzene.

Hence, (A) corresponds to (III) Benzene.

(B) Phenol, CHCl3, NaOH, HCl: This reaction is known as the Reimer-Tiemann reaction.

Phenol, when reacted with chloroform (CHCl3) in the presence of an aqueous base (NaOH), followed by acidification with HCl, forms salicylaldehyde.

Therefore, (B) corresponds to (I) Salicylaldehyde.

(C) Phenol, CO2, NaOH, HCl: This is the Kolbe-Schmitt reaction.

In this reaction, phenol reacts with carbon dioxide (CO2) under pressure at high temperatures in the presence of sodium hydroxide (NaOH), followed by acidification with HCl, to yield salicylic acid.

Thus, (C) corresponds to (II) Salicylic acid.

(D) Phenol, Conc.

HNO3: The nitration of phenol with concentrated nitric acid (HNO3) produces picric acid (2,4,6-trinitrophenol).

So, (D) corresponds to (IV) Picric acid.

Based on the above reactions, the correct matching would be: (A)-(III) Benzene (B)-(I) Salicylaldehyde (C)-(II) Salicylic acid (D)-(IV) Picric acid Therefore, the correct answer is Option B: (A)-(III), (B)-(I), (C)-(II), (D)-(IV).

Q19

Given below are two statements : Statement I : o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding. Statement II : o-Nitrophenol has high melting due to hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are true

B Statement I is true but Statement II is false

C Statement I is false but Statement II is true

D Both Statement I and Statement II are false

Correct Answer

Option B

Solution

So it is more volatile due to intramolecular H-bonding.

Melting point depends on packing efficiency not on H-bonding thus statement II is false.

Q20

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Butan -1- ol has higher boiling point than ethoxyethane. Reason R: Extensive hydrogen bonding leads to stronger association of molecules. In the light of the above statements, choose the correct answer from the options given below:

A A is false but R is true

B Both A and R are true but R is not the correct explanation of A

C Both A and R are true and R is the correct explanation of A

D A is true but R is false

Correct Answer

Option C

Solution

The boiling point of a substance is dependent on the strength of the intermolecular forces between its molecules.

Stronger intermolecular forces require more energy to overcome, thus resulting in a higher boiling point.

In the case of butan-1-ol and ethoxyethane (diethyl ether), they are both composed of similar numbers of similar atoms, but the type of intermolecular force between the molecules differs due to their functional groups.

Butan-1-ol is an alcohol and has a -OH group, which allows for hydrogen bonding, a strong type of dipole-dipole interaction.

This happens when the hydrogen atom in a polar bond (especially -OH, -NH, -HF) is attracted to some electronegative atom (like O, N, or F) in another molecule.

Because of this, butan-1-ol molecules "stick together" more strongly, requiring more heat energy to separate them and thus exhibiting a higher boiling point.

On the other hand, ethoxyethane, being an ether, does not have the ability to form hydrogen bonds since it lacks the -OH group.

The primary intermolecular forces in ethers are relatively weaker van der Waals forces.

Therefore, it requires less heat energy to break these forces, resulting in a lower boiling point compared to butan-1-ol.

So, both A and R are true and R is indeed the correct explanation for A.