Alcohols, Phenols and Ethers — Page 4 | NEET Chemistry

Q31

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A:

\mathrm{pK}_{\mathrm{a}}

value of phenol is 10.0 while that of ethanol is 15.9 . Reason R: Ethanol is stronger acid than phenol. 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

\mathrm{A}

and

\mathrm{R}

are true and

\mathrm{R}

is the correct explanation of

\mathrm{A}

.

D

A

is true but

R

is false.

Correct Answer

Option D

Solution

Phenol is more acidic than ethanol because conjugate base of phenoxide is more stable than ethoxide.

Q32

Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolyzed in presence of an acid results

A Salicylic acid

B Benzene-1,2-diol

C 2-Hydroxybenzaldehyde

D Benzene-1,3-diol

Correct Answer

Option C

Solution

It is Reimer Tiemann Reaction

Q33

Given below are two statements : Statement I : Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent. Statement II : Sodium metal can be used to dry diethyl ether and not ethyl alcohol. In the light of given statements. choose the correct answer from the options given below

A Statement I is false but Statement II are true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Both Statement I and Statement II are False

Correct Answer

Option C

Solution

St-I - St-I is correct because both given ether are soluble in water $\rightarrow$ Di ethyl ether and butan-1-ol are miscible to almost same extent i.e., 7.5 and 9 gm per 100 ml water due to H -bonding St-II : - St.

II is also correct because sodium metal is not used with ethyl alcohol as $\mathrm{H}_2$ gas release with ethyl a below

Q34

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})

$

Q35

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.

Q36

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.

Q37

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.

Q38

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

Q39

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}

Q40

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.