Answer the question on the basis of passage given below:

Strong oxidizing agents, oxidize toluene and its derivatives to benzoic acid. However, it is possible to stop the oxidation at the aldehyde stage with suitable reagents that convert methyl group to an intermediate that is difficult to oxidize further.

Match List I with List II

Choose the correct answer from the options given below:

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

The correct answer is option 2. A-III, B-I, C-IV, D-II.

Let us break down each pair in detail to understand the correct matches between List I and List II.

(A) Benzyl alcohol and phenol : (III) Blue litmus:

Benzyl alcohol has the structure \(C_6H_5CH_2OH\). It consists of a phenyl ring \((C_6H_5)\) attached to a -\(CH_2OH\) group. The hydroxyl group (-OH) attached to the \(-CH_2-\) group does not significantly affect the benzene ring's electronic nature, which means benzyl alcohol behaves similarly to simple alcohols like ethanol.

Alcohols, including benzyl alcohol, are generally neutral substances. They don't readily donate or accept protons \((H^+)\) in water to any significant extent. Since benzyl alcohol doesn't ionize in water to produce \(H^+\) or \(OH^-\) ions, it does not make the solution acidic or basic.

Litmus paper is an indicator that tests whether a solution is acidic or basic. Blue litmus paper turns red in acidic conditions (pH < 7). Benzyl alcohol, being neutral, does not change the pH of the solution significantly. Therefore, it will not change the color of blue litmus paper; it will remain blue.

Phenol has the structure \(C_6H_5OH\). The hydroxyl group \((-OH)\) is directly attached to the benzene ring, which influences the acidity of phenol.

Phenol is considered a weak acid because the hydroxyl group can lose a proton \((H^+)\), forming the phenoxide ion \((C_6H_5O^-)\). The resonance stabilization of the phenoxide ion makes it more likely for phenol to donate a proton compared to alcohols like benzyl alcohol. When phenol is dissolved in water, it partially ionizes, releasing \(H^+\) ions, which increases the acidity of the solution.

Because phenol can increase the concentration of \(H^+\) ions in a solution, it makes the solution acidic. Blue litmus paper, when exposed to an acidic environment (where \(H^+\) ions are present), will turn red. Therefore, phenol will turn blue litmus paper red.

(B) Phenol and alcohol : (I) Neutral \(FeCl_3\):

When phenol and alcohol are treated with neutral ferric chloride \((FeCl_3)\), they exhibit different reactions due to the distinct chemical properties of the hydroxyl groups attached to the aromatic ring in phenol versus the aliphatic chain in alcohols. 

Phenol has an \(-OH\) group directly attached to the aromatic ring. This makes the hydroxyl group more reactive due to resonance stabilization when the hydrogen from the \(-OH\) is lost.

When phenol reacts with neutral \(FeCl_3\) solution, a complex is formed between the \(Fe^{3+}\) ions and the phenoxide ion \((C_6H_5O^-)\), which results from phenol losing a proton \((H^+)\). This reaction typically produces a colored complex, often a violet, purple, or green color, depending on the specific structure of the phenol derivative.

The colored complex indicates the presence of phenolic groups.

Alcohols do not have the same resonance stabilization as phenols because the \(-OH\) group is attached to an alkyl chain, not an aromatic ring. As a result, simple alcohols, including benzyl alcohol, do not react with \(FeCl_3\) to form a colored complex. The solution typically remains yellow or shows no significant color change, indicating the absence of phenolic groups.

Phenol reacts with neutral \(FeCl_3\) to form a colored complex (usually violet, purple, or green), confirming the presence of a phenolic hydroxyl group. Alcohol (like benzyl alcohol) does not react with neutral \(FeCl_3\) to form a colored complex, indicating the absence of phenolic hydroxyl groups.

This test is commonly used to distinguish between phenols and alcohols.

(C) Secondary butyl alcohol and isobutyl alcohol : (IV) Iodoform test:

The Iodoform test is used to identify compounds that contain either a methyl ketone group or a secondary alcohol with the structure. The test works by detecting the presence of a \( \text{-CH(OH)-CH}_3 \) group (as in secondary alcohols) or a \( \text{CH}_3\text{CO-} \) group (as in methyl ketones), which can be oxidized to produce iodoform (CHI₃), a yellow precipitate.

Secondary butyl alcohol has a \( \text{-CH(OH)-CH}_3 \) group attached to the carbon, which meets the criteria for the Iodoform test. Upon oxidation with iodine in the presence of a base, 2-butanol can form acetone (\( \text{CH}_3\text{COCH}_3 \)), which has the necessary \( \text{CH}_3\text{CO-} \) group. This results in the formation of a yellow precipitate of iodoform \((CHI_3)\). So, the compounds gives positive iodoform test. 

Isobutyl alcohol does not have the \( \text{-CH(OH)-CH}_3 \) group or the \( \text{CH}_3\text{CO-} \) group in its structure. It is a primary alcohol, and the hydroxyl group is attached to a carbon that is not adjacent to a methyl group. Since isobutyl alcohol lacks the necessary structural features, it does not undergo the Iodoform reaction. o, the compound gives negative iodoform test.

Secondary Butyl Alcohol (2-butanol) will give a positive Iodoform test, resulting in a yellow precipitate of iodoform. Isobutyl Alcohol (2-methyl-1-propanol) will give a negative Iodoform test because it does not have the structural requirements to produce iodoform.

(D) Propan-1-ol and propan-2-ol : (II) Lucas reagent:

The Lucas test is used to distinguish between primary, secondary, and tertiary alcohols based on their reactivity with Lucas reagent, which is a solution of zinc chloride (\( \text{ZnCl}_2 \)) in concentrated hydrochloric acid (\( \text{HCl} \)). The test is based on the formation of an alkyl chloride, which is insoluble in water and forms a cloudy solution. When an alcohol reacts with Lucas reagent, the hydroxyl group (\(-OH\)) is replaced by a chloride ion (\(-Cl\)), forming an alkyl chloride.The rate of this reaction depends on the stability of the carbocation intermediate formed during the reaction:

Tertiary alcohols react almost immediately, forming a cloudy solution quickly (within seconds to minutes). Secondary alcohols react more slowly, forming a cloudy solution within a few minutes. Primary alcohols react very slowly or not at all, taking a very long time (hours) to form a cloudy solution, if at all.

Primary alcohols like propan-1-ol form a relatively unstable carbocation intermediate during the reaction with Lucas reagent. As a result, the reaction is very slow, and it takes a long time for any cloudiness (indicating the formation of the alkyl chloride) to appear, often several hours or not at all. Propan-1-ol (1-propanol), a primary alcohol, will give a negative Lucas test, with no immediate cloudiness.

Secondary alcohols like propan-2-ol form a more stable carbocation intermediate compared to primary alcohols. The reaction with Lucas reagent is faster than with primary alcohols, leading to the formation of cloudiness (alkyl chloride) within a few minutes. Propan-2-ol (isopropanol), a secondary alcohol, will give a positive Lucas test, with cloudiness forming within a few minutes.