Problem 172 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C}-\mathrm{CH}_{3} \quad \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{P}\) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{Q}\)( \(\mathrm{P}\) ) and \((\mathrm{Q})\) are respectively (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CHOH}-\mathrm{CH}_{3}\) and \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHOH}-\mathrm{CH}_{3}\) and \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CHOH}-\mathrm{CH}_{3}\) and \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHOH}-\mathrm{CH}_{3}\) and \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

Option (b) is correct: \( \mathrm{P} = \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHOH}-\mathrm{CH}_{3} \) and \( \mathrm{Q} = \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH} \).

Step 1: Understanding LiAlH4 Reaction

Lithium aluminum hydride (LiAlH extsubscript{4}) is a strong reducing agent commonly used to reduce carbonyl groups in compounds. It is effective in converting aldehydes and ketones into alcohols. In this exercise, LiAlH extsubscript{4} will reduce the carbonyl groups in the given compounds.

Step 2: Analyzing the Reaction for Compound 1

For the first compound, \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C}-\mathrm{CH}_{3} \), the structure indicates the presence of a carbonyl group. However, as presented, it seems the compound formula is incomplete. Assuming the carbon after the double bond is an aldehyde, the reduction will convert this carbonyl group to a primary alcohol.

Step 3: Determining Product P

If we infer the compound correctly, \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CHO} \), then the reduction with LiAlH\textsubscript{4} will produce \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH} \). Given the exercise wording, assume it's most analogous unless there is a structural detail implied but not visible.

Step 4: Analyzing the Reaction for Compound 2

The second compound \( \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \) is a conjugated aldehyde. LiAlH\textsubscript{4} will reduce the \( \mathrm{CHO} \) group to \( \mathrm{CH}_{2}\mathrm{OH} \), yielding \( \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH} \).

Step 5: Determining Product Q

The reaction implies the reduction of the aldehyde without affecting the double bond beyond, producing an allylic alcohol form with the existing alkene.

Step 6: Identify Correct Pair for P and Q

From the determined transformations, \( \mathrm{P} \) could potentially resemble (a) or (d), while \( \mathrm{Q} \) represents option (b) or (d) as \( \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH} \). Therefore, association with \( \mathrm{P} \) suggests answer (b) and (d); however, considering the reduction observed matches explicitly given conditions without further backbone reduction.

Step 7: Select the Correct Answer

Based on standard reactions and structural assumptions, \( \mathrm{P} \) and \( \mathrm{Q} \) connect with answer choice (b).

Key Concepts

LiAlH4 as a Reducing Agent

Lithium aluminum hydride, commonly abbreviated as LiAlH₄, is a powerful reducing agent extensively used in organic chemistry. This compound is particularly popular for its ability to reduce carbonyl groups, such as those found in aldehydes and ketones, into alcohols. The magic of LiAlH₄ lies in its high reactivity, as it can effectively donate hydride ions to the carbonyl carbon, facilitating the reduction process.

• Hydride donor: LiAlH₄ provides hydride ions (H⁻), which are crucial in reducing carbonyl compounds.
• Sensitivity: It is highly reactive and must be handled with care, often under an inert atmosphere.
• Solvent choice: Typically, it is used in anhydrous diethyl ether or tetrahydrofuran (THF) to prevent reaction with moisture.

Students often encounter LiAlH₄ in reactions where complete reduction of various functional groups is required, further transforming the nature of the compounds involved.

Reduction of Carbonyl Compounds

Carbonyl compounds are pivotal in organic chemistry, characterized by their carbon-oxygen double bond, present in aldehydes and ketones. When a reducing agent like LiAlH₄ is introduced, the carbonyl carbon, which is electrophilic, interacts with the nucleophilic hydride ions. This reaction allows the conversion of the C=O double bond to a C-O single bond, forming an alcohol.

• Aldehydes reduction: When reduced, aldehydes yield primary alcohols.
• Ketones reduction: The reduction of ketones results in secondary alcohols, as they have two carbon-containing groups attached.
• Reaction specificity: LiAlH₄ reduces carbonyl groups but leaves other functional groups like double bonds untouched. This selectivity is crucial for targeted syntheses.

Reduction reactions are pivotal, allowing chemists to transform and synthesize complex molecules efficiently.

Formation of Alcohols from Aldehydes and Ketones

The transformation of aldehydes and ketones into alcohols is a cornerstone reaction in organic chemistry. When LiAlH₄ acts as a hydride donor, the aldehyde or ketone accepts the hydride ion at the carbonyl carbon, reducing the double bond to create an alcohol.

• Primary alcohols: These form when an aldehyde is reduced, typically exhibiting the presence of one alkyl group attached to the alcohol.
• Secondary alcohols: These result from the reduction of ketones, containing two alkyl groups linked to the central carbon.
• Reaction context: In examining the provided exercise, one can predict products like alcohols from conjugated and standard carbonyl compounds, emphasizing LiAlH₄'s utility in transformation.

Understanding these principles expands one's capability to predict molecular transformations, playing a vital role in synthetic organic chemistry.