Question

Consider the following sequences of reactions: \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3} \frac{(\mathrm{i}) \mathrm{BH}_{3}}{\text { (ii) } \mathrm{CH}_{3} \mathrm{COOH}}(\mathrm{A}) \frac{\mathrm{dil} \mathrm{KMnO}_{4}}{\text { cold }} \longrightarrow(\mathrm{B})\) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3} \stackrel{\mathrm{Na}, \mathrm{NH}_{3}(l)}{\longrightarrow}(\mathrm{C}) \frac{(\mathrm{i}) \mathrm{OsO}_{4}}{\text { (ii) } \mathrm{Na}_{2} \mathrm{SO}_{3}, \mathrm{H}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{D})\) The products \((\mathrm{A}),(\mathrm{B}),(\mathrm{C})\) and \((\mathrm{D})\) are respectively (A) cis-but-2-ene, meso-butane-2,3-diol, trans-but-2-ene and ( t)-butane- 2,3 -diol (B) trans-but-2-ene, ( \(\pm\) )-butane-2,3-diol, cis-but-2-ene and meso-butane- 2,3 -diol (C) cis-but-2-ene, \((\pm)\)-butane- 2,3 -diol, trans-but- 2 -ene and meso- butane- 2,3 -diol (D) trans-but-2-ene, \((\pm)\)-butane-2,3-diol, trans-but-2-ene and \((\pm)\)-butane- 2,3 -diol

Short answer

The correct products for this sequence of reactions are (B) trans-but-2-ene, (±)-butane-2,3-diol, cis-but-2-ene, and meso-butane-2,3-diol.

Step-by-step solution

Identifying initial reactant and first reaction

The initial reactant given is CH₃C≡CCH₃, which is an alkyne (but-2-yne). The first reaction given is the addition of BH₃ followed by CH₃COOH.

Determining product A

The first reaction is the addition of BH₃ to but-2-yne, followed by CH₃COOH. This reaction involves hydroboration-oxidation, which converts the alkyne to an alkene with anti-Markovnikov addition and syn stereochemistry. The product A formed is cis-but-2-ene.

Determining product B

Product A (cis-but-2-ene) is then treated with dilute KMnO₄, which is a cold and mild oxidation. This reaction leads to the formation of a diol on the alkene group without breaking the carbon-carbon bond. The product B is (±)-butane-2,3-diol.

Identifying new initial reactant and reaction

The new initial reactant given is the same alkyne - but-2-yne (CH₃C≡CCH₃). The reaction given involves sodium (Na) in liquid ammonia (NH₃(l)).

Determining product C

The reaction of but-2-yne with sodium in liquid ammonia is a Birch reduction, which will convert the alkyne into a trans-alkene. The product C formed is trans-but-2-ene.

Determining product D

Product C (trans-but-2-ene) is treated with OsO₄ followed by Na₂SO₃ and H₂O. This reaction involves the addition of two OH groups across the double bond (alkene) while retaining the carbon-carbon bond. The reaction follows syn stereochemistry, resulting in the formation of a meso compound. The product D is meso-butane-2,3-diol.

Matching the products to the given options

Based on our analysis, the products A, B, C, and D are: - A: cis-but-2-ene - B: (±)-butane-2,3-diol - C: trans-but-2-ene - D: meso-butane-2,3-diol Comparing these products with the given options, we find that the correct option is (B) trans-but-2-ene, (±)-butane-2,3-diol, cis-but-2-ene and meso-butane-2,3-diol.

Key concepts

Alkyne Reactions

Alkyne chemistry is a cornerstone of organic chemistry, with the reactions of alkynes offering diverse pathways to a variety of complex molecules. Alkynes, characterized by their carbon-carbon triple bonds, are known for their unique reactivity patterns due to the bond’s high energy.

One common reaction of alkynes is hydroboration-oxidation, a two-step process that transforms the alkyne into an alkene with anti-Markovnikov addition. In the context of the exercise, but-2-yne reacts with BH₃ to form cis-but-2-ene, showcasing the predictable and valuable change in saturation level from an alkyne to an alkene. Additionally, alkenes can be further transformed via reactions such as hydrogenation, halogenation, and hydroxylation, each rooted in the diverse reactivity of alkynes.

Understanding the outcomes of these reactions is critical for students, particularly in advanced competitive exams like the JEE Advanced. The application of these reactions can lead to the synthesis of a wide array of structures, highlighting the importance of mastering alkyne chemistry for a solid foundation in organic synthesis.

Stereochemistry

Stereochemistry is the study of the three-dimensional arrangement of atoms within molecules and the impact of this arrangement on reaction outcomes and properties. In the educational context, it is essential for students to comprehend the concept of isomerism, which includes cis-trans isomers and chiral molecules.

For instance, when discussing the hydroboration-oxidation of but-2-yne in the given problem, it is crucial to note that the reaction not only changes the chemical group but also introduces stereochemistry with the addition occurring from the same side (syn addition). Similarly, when transforming trans-but-2-ene into a diol with OsO₄, followed by Na₂SO₃ and H₂O, the reaction follows syn stereochemistry and leads to meso-butane-2,3-diol, emphasizing the conservation of the geometric configuration of the alkene.

A clear understanding of these stereochemical outcomes is paramount, particularly when students are tackling complex molecules with multiple chiral centers. Knowing the difference between chiral and achiral compounds, like the meso compound produced, aids in predicting the physical and chemical behaviors of these molecules.

Hydroboration-Oxidation

Hydroboration-oxidation is a two-step chemical reaction mechanism taking center stage in the realm of alkyne and alkene chemistry. What is pivotal for students to grasp is that this reaction is a gateway to synthesize alcohols from unsaturated carbon compounds in an anti-Markovnikov fashion, where the hydroxyl group attaches to the less substituted carbon atom.

Initially, in hydroboration, a boron atom from BH₃ adds to the less substituted carbon of the triple or double bond. Simultaneously, a hydrogen atom attaches to the other carbon involved in the multiple bonds, which ultimately results in an alkylborane intermediate. In the oxidation step, typically using a weak acid like CH₃COOH, the boron is replaced with a hydroxyl group, forming the final alcohol.

This reaction is stereospecific, yielding syn addition products, meaning the new atoms attach to the same side of the formerly unsaturated molecule. This characteristic is integral for predicting the stereochemical outcome, such as the production of cis-but-2-ene from but-2-yne in the JEE Advanced problem discussed. Understanding hydroboration-oxidation is essential for students, not only for exams but also for practical organic synthesis, where the construction of complex, stereochemically-rich molecules is commonplace.