Alcohols, Phenols and Ethers

$ \text { Reaction given Alcohol } $

$ \text { (A) } \mathrm{CH}_4+\mathrm{O}_2 \xrightarrow[\Delta]{\mathrm{Mo}_2 \mathrm{O}_3} \mathrm{HCHO} $

$ \text { (B) } 2 \mathrm{CH}_3 \mathrm{CH}_3+3 \mathrm{O}_2 \xrightarrow[\Delta]{\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Mn}} \mathrm{CH}_3 \mathrm{COOH} $

$ \text { (D) } 2 \mathrm{CH}_4+\mathrm{O}_2 \xrightarrow[523 \mathrm{~K}, 100 \text { atm }]{\mathrm{Cu}} \mathrm{CH}_3-\mathrm{OH} $

$ \text { And. } \rightarrow \quad C, D $

$ \text { II, IV & V are secondary alcohol. } $

No. of OH groups $=\frac{290-122}{42}=4$

(can be separated by steam distillation)

Total number of $\pi $ and lone pair electrons in final product = 18

Compound $X$ is phenolphthalein (an acid–base indicator).

In basic solution, phenolphthalein first shows a pink colour.

But when $\mathrm{NaOH}$ is in excess (very strong base), phenolphthalein changes to a form that is colourless.

So, the colour of compound $X$ in excess $\mathrm{NaOH}$ is colourless.

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

Relation with HBr to give phenol

(1)

The reaction involves two steps,

$\star$) The lone pair of oxygen atom attacks on the hydrogen of HBr.

$\star$) The nucleophile Br$^-$ attacks on the methyl group which lead to the formation of phenol along with methyl bromide.

Phenol is formed from this compound on reaction with HBr.

(2)

No phenol formation in this case because alkyl chain is attached to the benzene ring and on addition of H-Br, either H$^+$ or Br$^-$ get attached to this chain and no $-$OH group is formed as it is directly bonded to the benzene ring.

(3)

No phenol formation in this case. $-$OH group is formed as

So, phenol product is formed in case of option (1) compound only.

Rate of (a) is faster than (b) because of its intramolecular substitution.

$\textbf{Statement (I):}$

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.

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

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.

Methoxy group is o-, p-directing. Nitration of methoxy benzene will produce p-isomer as the major product due to steric crowding in o-isomers. Bromination of p-nitroanisole (A) with excess of $\mathrm{Br}_2$ will result in disubstitution to give 2, 5-dibromo-4-nitroanisole. (B) as the major product.

Carbene is intermediate.

Intermediate given in option (1) is produced.

The above reaction is not possible as due to resonance $\mathrm{C}-\mathrm{O}$ bond has partial double bond character. So, bond strength will increase and substitution will be less likely.

Presence of Lewis acid promotes the reaction due to ease in formation of electrophile.

Boiling point order will be :

(B) < (A) < (C) < (D) due to H bonding in (D) and dipole interactions in (A) and (C).

Benzene-1, 2-diol is also called catechol

(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.

(A)-(III) Benzene
(B)-(I) Salicylaldehyde
(C)-(II) Salicylic acid
(D)-(IV) Picric acid

Assertion A is that alcohols can act as both nucleophiles and electrophiles. This is indeed true.

Alcohols can act as nucleophiles because they have a polar O-H bond; the oxygen atom has a pair of nonbonding electrons that can be donated to electrophiles. For example, in the presence of a strong acid such as hydrochloric acid, the lone pair on the oxygen can attack the positively charged hydrogen atom to form an oxonium ion ($R-OH_2^+$), which then acts as a good leaving group in substitution reactions.

Alcohols can also behave as electrophiles. An example of an alcohol acting as an electrophile is when it reacts with a strong base, such as sodium hydride (NaH), to undergo deprotonation forming an alkoxide ion ($R-O^-$). The alkoxide ion is a strong nucleophile, but the original alcohol molecule in the presence of a base acts as an electrophile due to the acidity of the hydroxyl hydrogen.

Reason R posits that alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen. This is also true. When alcohols react with active metals, the alcohol acts as an acid, donating a hydrogen ion ($H^+$) to form hydrogen gas ($H_2$), while the remaining alkoxide ion ($R-O^-$) bonds with the metal cation ($M^+$, where M is an active metal like sodium) to form the alkoxide salt ($MOR$).

Now, considering the above explanations, the relationship between Assertion A and Reason R needs to be evaluated:

While both Assertion A and Reason R are indeed true, Reason R is not the correct explanation of Assertion A. Reason R describes a reaction (alcohol with active metal) where the alcohol functions only as an acid (not as a nucleophile or an electrophile). Hence, Reason R does not explain how alcohols can act as both nucleophiles and electrophiles, which is what Assertion A states.

Therefore, the correct answer is:

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

It is Reimer Tiemann Reaction

$\begin{aligned} & \text { Ethanol } \rightarrow 15.9 \\ & \text { Phenol } \rightarrow 10 \\ & \text { M-Nitrophenol } \rightarrow 8.3 \\ & \text { P-Nitrophenol } \rightarrow 7.1 \end{aligned}$

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.

(A) $\rightarrow$ Kolbe Schmidt Reaction

(B) $\rightarrow$ Reimer Tiemann Reaction

(C) $\rightarrow$ Oxidation of phenol to p-benzoquinone

(D) $\rightarrow \mathrm{PhOH}+\mathrm{NaOH} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{PhO}^{-}$

Acidic strength

Ethanol give lucas test after long time

Statement (I) $\rightarrow$ correct

Statement (II) $\rightarrow$ incorrect