Identify D:

The correct answer is option 1.

Let us analyze the reaction sequence thoroughly. The initial reaction in the question is prop-1-ene. So, the steps of the reaction will be as follows:

Step 1: \(Prop-1-ene \ \underset{H^+}{\overset{C_6H_6}{\longrightarrow}} A\)

This step represents Friedel-Crafts alkylation, where prop-1-ene (\(CH_2=CH-CH_3\)) reacts with benzene (\(C_6H_6\)) in the presence of an acid catalyst (\(H^+\)).  The propene undergoes protonation on the double bond, forming a carbocation. The carbocation generated is a secondary carbocation . Benzene then attacks this carbocation, undergoing electrophilic aromatic substitution to form isopropylbenzene (commonly known as cumene), \(C_6H_5-CH(CH_3)_2\).

So, compound A is cumene.

Step 2: \(A \ \overset{KMnO_4 / H^+ }{\longrightarrow} B\)

In this step, cumene undergoes strong oxidation with potassium permanganate (\(KMnO_4\)) in acidic conditions. \(KMnO_4\) is a powerful oxidizing agent, and when it reacts with side chains on aromatic rings, it typically oxidizes the entire alkyl group, regardless of its size or branching, into a carboxyl group (-COOH). Cumene \(C_6H_5-CH(CH_3)_2\) is oxidized to benzoic acid \(C_6H_5-COOH\).

So, compound B is benzoic acid.

Step 3: \(B \ \overset{NH_3 / \Delta }{\longrightarrow} C\)

In this step, benzoic acid is heated with ammonia (\(NH_3\)). The reaction between a carboxylic acid and ammonia under heat leads to the formation of an amide. Benzoic acid (\(C_6H_5-COOH\)) reacts with ammonia to form **benzamide** (\(C_6H_5-CONH_2\)) by releasing a molecule of water.

So, compound C is benzamide.

Step 4: \(C \ \overset{KOBr }{\longrightarrow} D\)

This step involves the Hoffmann bromamide degradation reaction. In this reaction, an amide reacts with bromine in a basic medium (here potassium hypobromite, \(KOBr\)) to produce a primary amine with one less carbon atom than the amide. Benzamide (\(C_6H_5-CONH_2\)) reacts with \(KOBr\) and undergoes a rearrangement that results in the formation of aniline (\(C_6H_5-NH_2\)).

So, compound D is aniline.

Thus, the correct answer is  Aniline.

The overall reaction can be represented as:

This reaction sequence illustrates the transformation of an alkene (propene) to a primary aromatic amine (aniline) through a series of well-known organic reactions: Friedel-Crafts alkylation, oxidation, formation of amide, and the Hoffmann degradation.