Question

What are ‘X’ and ‘Y’ respectively in the following reactions?

• A.
• B.
• C.
• D.

Hint:

Electrophilic Aromatic Substitution \& Reduction Pathways.

• Electron-donating groups like -OCH\textsubscript3 are ortho/para directing.
• In Friedel–Crafts acylation, the acyl group gets added to the para position if it is sterically and electronically favored.
• Clemmensen reduction or similar conditions reduce the acyl group (-COCH\textsubscript3) to an alkyl group (-CH\textsubscript3) on the aromatic ring.
• Ensure orientation and group positions are carefully analyzed when dealing with substituted benzene compounds.

Answer 1

The Correct Option is B

The given reaction involves two steps:

• Step 1: A compound (X) undergoes Friedel–Crafts acylation using \( \mathrm{CH_3COCl} \) and anhydrous \( \mathrm{AlCl_3} \) to introduce an acetyl group on the aromatic ring.
• Step 2: The ketone group is reduced using \( \mathrm{CH_2OH} \) (likely via Clemmensen reduction using \( \mathrm{Zn(Hg)} \)), producing an alkylated aromatic compound (Y).

Now let’s analyze the structure of X. The compound has a bromine and a methoxy group attached to a benzene ring. These are ortho/para directing groups and favor substitution at the para position.

The acetylation happens at the para position relative to \( -\mathrm{OCH_3} \) due to electronic effects, and this ketone group is then reduced to a \( -\mathrm{CH_3} \) group.

Hence:

\[ \text{X = } \mathrm{Br{-}C_6H_4{-}OCH_3}, \quad \text{Y = } \mathrm{CH_3{-}C_6H_4{-}OCH_3} \]

This corresponds to option (2).

Answer 2

To determine compounds 'X' and 'Y' respectively in the given set of reactions, let's analyze the transformations step-by-step:

Step 1: Electrophilic Aromatic Substitution (Bromination)

The starting compound is anisole (methoxybenzene), which contains a methoxy group (-OCH₃) attached to a benzene ring. The methoxy group is an electron-donating group (EDG) and activates the benzene ring towards electrophilic aromatic substitution by increasing the electron density on the ring, especially at the ortho and para positions relative to itself. When anisole reacts with bromine (Br₂) in the presence of acetic acid (CH₃COOH), bromination occurs at the positions activated by the methoxy group. Because of steric hindrance, the para-product is usually the major product. Therefore, the major product X is para-bromoanisole or 4-bromoanisole.

Step 2: Reaction with Hydroiodic Acid (HI)

The second step involves heating the compound X with hydroiodic acid (HI). Anisole derivatives, when treated with HI at high temperature, undergo cleavage of the C–O bond in the ether group. The general reaction is: (Ar-OCH₃) + HI → Ar-OH + CH₃I In this case, the methoxy group (-OCH₃) is cleaved, resulting in the formation of phenol (Ar–OH) and methyl iodide (CH₃I). So, applying this to compound X (which is para-bromoanisole), the product Y will be para-bromophenol (4-bromophenol) and methyl iodide. However, according to the figure and the given options in the question, the bromination step occurs first, and then HI causes cleavage. But in the reverse order shown in the original question, the bromination happens on anisole first to form 4-bromoanisole, and then HI cleaves it to give phenol and CH₃I (the bromine is no longer present as the reaction seems to be with plain anisole). So, correctly aligning with the figure and transformations: - The major product after bromination of anisole is: X = 4-Bromoanisole - The product Y after reaction with hot HI is: Y = Phenol and Methyl iodide (CH₃I)

Final Answer:

X = 4-Bromoanisole, Y = Phenol + CH₃I