Alcohols, Phenols and Ethers — Page 4 | JEE Chemistry

Q31

Hydrolysis of which compound will give carbolic acid?

A Cumene

B Benzenediazonium chloride

C Benzal chloride

D Ethylene glycol ketal

Correct Answer

Option B

Solution

Phenol, is known as Carbolic acid.

Diazonium salt are hydrolysed to phonols.

Benzal chloride on hydrolysis gives benzaldehyde

Q32

Suitable reaction condition for preparation of Methyl phenyl ether is

A Benzene, MeBr

B

\mathrm{Ph{O^\Theta }N{a^ \oplus }}

, MeBr

C Ph - Br,

\mathrm{Me{O^\Theta }N{a^ \oplus }}

D

\mathrm{Ph{O^\Theta }N{a^ \oplus }}

, MeOH

Correct Answer

Option B

Solution

$\mathrm{PhONa}+\mathrm{MeBr} \rightarrow \mathrm{PhOMe}+\mathrm{NaBr}$ Williamson's Synthesis

Q33

Phenolic group can be identified by a positive:

A Tollen's test

B Phthalein dye test

C Carbylamine test

D Lucas test

Correct Answer

Option B

Solution

The correct answer for identifying the phenolic group is Option B, the Phthalein dye test.

Option A: Tollen's test - This test is used to identify aldehydes.

Aldehydes are oxidized to carboxylic acids by the Tollen's reagent (ammoniacal silver nitrate), and this reaction results in a characteristic silver mirror or a black precipitate of silver metal, which is not relevant to phenolic compounds.

Option B: Phthalein dye test - The phenolic group can be identified using the phthalein dye test.

In this test, phenols react with phthalic anhydride in the presence of a dehydrating agent, typically concentrated sulfuric acid, to form phthalein dyes.

A common example of a phthalein dye that can be formed in this manner is phenolphthalein, which is synthesized using phenol and phthalic anhydride.

These dyes exhibit color changes in different pH environments, which is a characteristic helpful in identifying phenolic compounds.

Option C: Carbylamine test - The carbylamine test, also known as the isocyanide test, is specific to primary amines.

In this reaction, primary amines are heated with chloroform and an alkali base, typically potassium hydroxide, to produce a foul-smelling isocyanide or carbylamine.

Since phenolic compounds do not have an amine group, they will not give a positive carbylamine test.

Option D: Lucas test - The Lucas test is used to classify alcohols of low molecular weight based on their reactivity with Lucas reagent, a solution of zinc chloride in concentrated hydrochloric acid.

The test differentiates between primary, secondary, and tertiary alcohols.

Phenols, which have an -OH group bonded to an aromatic ring, do not react in the same way as aliphatic alcohols in the Lucas test, and therefore, this test is not used to identify phenolic groups.

Thus, the test that is specifically used to identify a phenolic group is the Phthalein dye test.

Q34

Given below are two statements : Statement I : Phenols are weakly acidic. Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water. Choose the most appropriate option :

A Both Statement I and Statement II are correct.

B Both Statement I and Statement II are incorrect.

C Statement I is correct but Statement II is incorrect.

D Statement I is incorrect but Statement II is correct.

Correct Answer

Option C

Solution

Phenol are weakly acidic. Phenol is more acidic than alcohol & H2O statement (I) is correct. (II) is incorrect.

Q35

Which one the following compounds will readily react with dilute

\mathrm{NaOH}

?

A

\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}

B

\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}

C

\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}

D

\left(\mathrm{CH}_3\right)_3 \mathrm{COH}

Correct Answer

Option C

Solution

To determine which of the given compounds will readily react with dilute

\mathrm{NaOH}

, we need to understand the chemical reactivity of these compounds towards bases like sodium hydroxide (

\mathrm{NaOH}

). Here's a brief overview of each compound's reactivity towards

\mathrm{NaOH}

: Option A:

\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}

(Benzyl Alcohol) The presence of a benzyl group (a phenyl group attached to a CH2 group) adjacent to the hydroxyl group can somewhat increase the acidity of the hydroxyl hydrogen.

However, benzyl alcohol is still not significantly acidic to react vigorously with a weak base like dilute

\mathrm{NaOH}

.

Nonetheless, under certain conditions, it might undergo reactions, but not as readily as an acidic hydrogen-containing compound would.

Option B:

\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}

(Ethanol) Ethanol is a simple alcohol with no acidic hydrogen atoms that would react with

\mathrm{NaOH}

. The hydroxyl group in ethanol is not sufficiently acidic to deprotonate in the presence of a base like

\mathrm{NaOH}

, making it unreactive in this context. Option C:

\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}

(Phenol) Phenol contains a hydroxyl group directly attached to an aromatic ring.

This structural feature significantly increases the acidity of the hydroxyl hydrogen compared to alcohols.

The reason behind this is the stabilization of the phenoxide ion (the conjugate base) through resonance within the aromatic ring.

As a result, phenol can react with a weak base like

\mathrm{NaOH}

to form sodium phenoxide and water.

\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} + \mathrm{NaOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^- \mathrm{Na}^+ + \mathrm{H}_2\mathrm{O}

Option D:

\left(\mathrm{CH}_3\right)_3 \mathrm{COH}

(tert-Butyl Alcohol) tert-Butyl alcohol, being a tertiary alcohol, does not have a hydrogen atom on the carbon bearing the hydroxyl group that is acidic enough to react with

\mathrm{NaOH}

.

Tertiary alcohols, in general, are resistant to deprotonation because of the steric hindrance around the hydroxyl group and the lack of an easily removable hydrogen atom connected to the carbon with the -OH group.

Given the choices, the correct answer is Option C:

\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}

(Phenol), as it will readily react with dilute

\mathrm{NaOH}

due to the increased acidity of its hydroxyl group caused by the resonance stability of the phenoxide ion formed in the reaction.

Q36

An ethar is more volatile than an alcohol having the same molecular formula. This is due to

A alcohols having resonance structures

B inter-molecular hydrogen bonding in ethers

C inter-molecular hydrogen bonding in alcohols

D dipole characters of ethers

Correct Answer

Option C

Solution

Alcohol and ether are isomer with each other.

So, with same molecular formula we can make ether as well as alcohol.

For ex, With molecular formula C2H6O (1)

\,\,\,

alcohol will be CH3CH2 OH (2)

\,\,\,

ether will be CH3 $-$ O $-$ CH3 In Alcohol there is hydrogen bond and in Ether there is Van der walls force of attraction.

We know that H bond is stronger bond than van der walls force of attraction as the atoms of alcohol are strongly attached with each other by hydrogen bonding so tendency of vaporization of alcohol is less compared to ether.

In alcohol inter-molecular hydrogen bonding look like this -

Q37

Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives

A 2,4,6-trinitrobenzene

B o-nitrophenol

C p-nitrophenol

D nitrobenzene

Correct Answer

Option B

Solution

Phenol on reaction with conc.

{H_2}S{O_4}

gives a mixture of

o

- and

p

- products (i.e.,

- S{O_3}H

group, occupies

o

-,

p

- position). At room temperature

o

-product is more stable, which on treatment with conc.

HN{O_3}

will yield

o

-nitrophenol.

Q38

Phenyl magnesium bromide reacts with methanol to give

A a mixture of anisole and Mg(OH)Br

B a mixture of benzene and Mg(OMe)Br

C a mixture of toluene and Mg(OH)Br

D a mixture of phenol and Mg(Me)Br

Correct Answer

Option B

Solution

C{H_3}OH + {C_6}{H_5}MgBr\overset{\,}\longrightarrow C{H_3}O.MgBr + {C_6}{H_6}

v

Q39

From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is

A 2–Butanol

B 2–Methylpropan–2–ol

C 2–Methylpropanol

D 1–Butanol

Correct Answer

Option B

Solution

Tertiary alcohols react fastest with conc.

HCl

and anhydrous

ZnC{l_2}

(lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation. Mechanism

Q40

Common name of Benzene - 1,2 - diol is -

A catechol

B quinol

C o-cresol

D resorcinol

Correct Answer

Option A

Solution

Benzene-1, 2-diol is also called catechol