Problem 37 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A ketone 'A' which undergoes haloform reaction gives compound B on reduction. \(\mathrm{B}\) on heating with sulphuric acid gives compound \(\mathrm{C}\), which forms monoozonide D, D on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). Write down the reactions involved.

Step-by-Step Solution

Answer

A is acetophenone, B is ethylbenzene, C is styrene.

Step 1: Analyze Haloform Reaction

The haloform reaction is characteristic of methyl ketones, which have the structure \( \mathrm{RCOCH_3} \). During this reaction, the methyl group is fully halogenated and then leaves as a haloform (like chloroform or iodoform). Therefore, compound \( \mathrm{A} \) must be a methyl ketone.

Step 2: Identify Compound A

Since the final product after the haloform step is acetaldehyde (\( \mathrm{CH_3CHO} \)), and given that the original compound \( \mathrm{A} \) is reduced to \( \mathrm{B} \), we deduce that compound \( \mathrm{A} \) is acetophenone (\( \mathrm{C_6H_5COCH_3} \)). This ketone gives acetaldehyde after the haloform reaction and can be reduced to ethylbenzene (\( \mathrm{C_6H_5CH_2CH_3} \)).

Step 3: Reduction to Compound B

Acetophenone (\( \mathrm{A} \)) is reduced to ethylbenzene (\( \mathrm{B} \)) using a reducing agent such as lithium aluminium hydride or catalytic hydrogenation. The reaction is: \[ \mathrm{C_6H_5COCH_3 \xrightarrow{[H]} C_6H_5CH_2CH_3} \]

Step 4: Dehydration with Sulfuric Acid to Compound C

When ethylbenzene (\( \mathrm{B} \)) is heated with sulfuric acid, it undergoes a dehydration reaction to form ethylbenzene directly. However, as it's already a saturated alkane, dehydration typically involves forming a double bond or losing one of the ethyl hydrogen atoms to form styrene (\( \mathrm{C_6H_5CH=CH_2} \)).

Step 5: Formation of Monoozonide D

Ethylene (\( \mathrm{C} \)), when reacted with ozone forms ozonides. Here, styrene reacts with ozone to form a monoozonide. The structure involves addition across the double bond forming a cyclic ozonide (which is not stable on its own).

Step 6: Hydrolysis of D to Give Only Acetaldehyde

On cleavage and hydrolysis of the ozonide in the presence of zinc dust, compound \( \mathrm{D} \) will give aldehyde products. Since only acetaldehyde is formed, it confirms that compound \( \mathrm{C} \) was styrene. The reaction breaks the double bond and forms two molecules of acetaldehyde: \[ \mathrm{C_6H_5CH=CH_2 + O_3 \rightarrow [C_6H_5CH(O_3)CH_2] \xrightarrow{Zn/H_2O} 2CH_3CHO} \]

Key Concepts

Haloform Reaction

The haloform reaction is a fascinating process in organic chemistry primarily involving methyl ketones. These ketones have the structural formula \( \mathrm{RCOCH_3} \), where \( \mathrm{R} \) is a hydrocarbon group. This reaction entails the complete halogenation of the methyl group in the ketone, leading to the formation of a haloform, such as chloroform \( \mathrm{CHCl_3} \) or iodoform \( \mathrm{CHI_3} \). The essential requirement is the presence of the \( \mathrm{CH_3CO} \) group, which allows the methyl group to transform into a trihalomethane.
The process begins with the enolization of the ketone, which subsequently undergoes halogenation at the methyl group. The fully halogenated methyl group then leaves as a haloform. This reaction enables the cleavage of the carbon-carbon bond adjacent to the carbonyl, resulting in the formation of a carboxylate anion and the haloform. For instance, in the exercise, acetophenone \( \mathrm{C_6H_5COCH_3} \) undergoes haloform reaction resulting in acetaldehyde \( \mathrm{CH_3CHO} \), indicating the identification of compound \( \mathrm{A} \).
Key takeaways:

• Must involve methyl ketone \( \mathrm{RCOCH_3} \).
• Results in trihalomethane (haloform) formation.
• Yields carboxylate anion, which depends on specific methyl ketone used.

Reduction Reactions

Reducing a compound in organic chemistry refers to the addition of hydrogen atoms to it, or the removal of oxygen from it. This reaction is fundamental in converting complex organic molecules into simpler ones, often changing their functional group types.
For this particular solution, acetophenone \( \mathrm{C_6H_5COCH_3} \) is reduced to ethylbenzene \( \mathrm{C_6H_5CH_2CH_3} \). This conversion is typically achieved through agents like lithium aluminium hydride (\( \mathrm{LiAlH_4} \)) or catalytic hydrogenation, which introduces hydrogen to the carbonyl carbon, converting the carbonyl group into an alcohol group, and subsequently into an alkyl group only.
Essential elements of reduction:

• Decreases the oxidation state of the molecule.
• Common reducing agents include hydrides and hydrogen gas.
• A critical reaction in organic synthesis for converting ketones to hydrocarbons.

Dehydration Reactions

Dehydration reactions in organic chemistry involve removing water molecules from alcohols or other compounds to form alkenes or other products. When a compound like ethylbenzene is heated with a strong acid such as sulfuric acid, it can lead to the formation of an alkyne or an alkene.
In the outlined solution, ethylbenzene (\( \mathrm{C_6H_5CH_2CH_3} \)) undergoes a dehydration reaction with sulfuric acid and forms styrene (\( \mathrm{C_6H_5CH=CH_2} \)). This type of reaction often eliminates elements from the original compound, here converting a single bond into a double bond. The transformation involves the removal of a hydrogen atom with an adjacent hydroxyl group, forming the water molecule.
Highlights in dehydration:

• Common catalysts include sulfuric acid and heat.
• Results in the formation of double bonds (alkenes) from alcohols or alkanes.
• A critical step in synthetic organic chemistry, especially in creating alkenes from alcohols.