Question

The synthesis of 3 -octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are A) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) B) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) C) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\) D) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\)

Short answer

The correct starting materials for the synthesis of 3-octyne are option B, \(BrCH_2CH_2CH_3\) and \(CH_3CH_2CH_2C\equiv CH\).

Step-by-step solution

Identify the Target Molecule - 3-Octyne

Understand that 3-octyne is an 8-carbon alkyne with the triple bond located between the third and fourth carbons. It has the formula \(C_8H_{14}\) with the structure \(CH_3(CH_2)_2C\equiv C(CH_2)_3CH_3\).

Determine the Required Alkyne and Alkyl Halide

To synthesize 3-octyne, we need a 5-carbon alkyne and a 3-carbon alkyl halide or vice versa. The alkyne must have the triple bond at the correct position to yield the triple bond at the third carbon in the final product.

Match the Starting Materials with the Structure of 3-Octyne

Scrutinize each pair of starting materials. The correct alkyne should have either 2 or 5 carbons to give us the correct length of 8 carbons when combined with the other moiety. The bromoalkane must add to the correct carbon next to the triple bond in the alkyne to give the triple bond in the third position in the final product.

Select the Correct Option

Option B has the bromoalkane \(BrCH_2CH_2CH_3\) and alkyne \(CH_3CH_2CH_2C\equiv CH\). When these react, the alkyne carbon that bonds to the bromoalkane is the third one, resulting in the formation of 3-octyne. Thus, option B yields \(CH_3CH_2CH_2C\equiv C(CH_2)_2CH_3\), which is the correct structure.

Key concepts

Synthesis of Alkynes

The synthesis of alkynes is an important topic in organic chemistry, particularly for students preparing for competitive exams like JEE Advanced. Alkynes are hydrocarbons with at least one carbon-carbon triple bond and follow the general formula \( C_nH_{2n-2} \). One common method of synthesizing alkynes is by coupling two smaller fragments, such as an alkyne and an alkyl halide, in the presence of a strong base like sodium amide (NaNH2).

The reaction typically involves the deprotonation of the terminal alkyne by the base to form a highly reactive acetylide anion, which then undergoes a nucleophilic substitution reaction with an alkyl halide. The challenge lies in ensuring that the triple bond is located at the correct position in the final alkyne product, which requires careful selection of the starting materials.

For instance, the given exercise demonstrates the synthesis of 3-octyne, an eight-carbon alkyne with the triple bond between the third and fourth carbons. It requires a 5-carbon alkyne with the triple bond at one end and a 3-carbon alkyl halide. Students should visualize the reaction and understand how the length and position of the triple bond are maintained throughout the synthesis process.

Alkyl Halide Reactions

Alkyl halide reactions are a cornerstone in organic chemistry education, and mastering them is crucial for success in exams like JEE Advanced. These compounds contain a halogen atom covalently bonded to an aliphatic carbon chain, and they are known for their reactivity due to the polar nature of the carbon-halogen bond.

For the synthesis of alkynes, the most relevant reaction of alkyl halides is the nucleophilic substitution. This occurs when a nucleophile, such as an alkoxide or acetylide ion, displaces the halogen. The result is the formation of a new carbon-carbon bond. In our problem, the nucleophilic species is the deprotonated alkyne (acetylide ion), which attacks the carbon atom bearing the halogen in the alkyl halide.

Understanding the specifics of this reaction is essential for students, including reaction mechanisms like SN2 or SN1, the role of base and solvent, and the concept of primary, secondary, and tertiary carbon atoms, which influence the rate and outcome of the reaction.

JEE Advanced Organic Chemistry

Preparing for the JEE Advanced requires a deep understanding of intricate concepts in organic chemistry, such as the synthesis of alkynes and reactions involving alkyl halides. The exam not only tests a student’s knowledge of reactions and mechanisms but also their ability to apply this knowledge to problem-solving.

In our exercise, we use a reaction between an alkyne and an alkyl halide, which exemplifies the type of synthetic strategy questions frequently encountered in the JEE Advanced. Students must identify the starting materials and predict the structure of the product. Focus should be placed on recognizing the reagents, reaction conditions, and the stereochemistry of the reaction when applicable.

To excel, students should frequently practice problems, understand the underlying principles, and apply those principles to novel situations. Emphasizing concepts like acidity of alkynes, base strength, nucleophilicity, and the stability of intermediates can help in achieving a higher level of proficiency in organic chemistry for competitive exams.