Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following sequence of reactions \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\text { Ethanolic } \mathrm{KOH}}{\longrightarrow}(\mathrm{X}) \stackrel{\mathrm{Br}_{2}}{\longrightarrow}(\mathrm{Y})\) \(\frac{\mathrm{NaNH}_{2}}{\mathrm{liq} \cdot \mathrm{NH}_{3}}(\mathrm{Z})\) The end product \((Z)\) is (a) propane (b) propyne (c) propene (d) propan-2-al

Step-by-Step Solution

Answer

The end product \((Z)\) is propyne (b).

Step 1: Identify the First Reaction

The first reaction involves 2-bromo propane \((\text{CH}_3)_2\text{CHBr}\) with ethanolic KOH. This is a dehydrohalogenation reaction which eliminates HBr and forms an alkene. The product \((X)\) is propene \((\text{CH}_3\text{CH}=\text{CH}_2)\).

Step 2: Determine the Second Reaction

The second reaction involves the addition of Br2 to propene \((X)\). Br2 adds across the double bond to form 1,2-dibromopropane \((\text{CH}_3\text{CHBrCH}_2\text{Br})\), giving us product \((Y)\).

Step 3: Analyze the Final Reaction

The final step involves treating 1,2-dibromopropane \((Y)\) with sodium amide \((\text{NaNH}_2)\) in liquid ammonia \((\text{NH}_3)\). This is a dehydrohalogenation reaction which removes both bromine atoms and forms a triple bond, resulting in propyne \((\text{CH} \equiv \text{C-CH}_3)\), which is our final product \((Z)\).

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a chemical reaction that involves the elimination of a hydrogen halide (such as HBr). This process is significant in organic chemistry as it converts halogen-containing compounds into alkenes or alkynes. In the given exercise, 2-bromo propane reacts with ethanolic potassium hydroxide (KOH). During this reaction, the hydrogen and the bromine atom from adjacent carbon atoms are removed.
This elimination step leads to the formation of water, a bromide ion, and an alkene. The process is characterized by the use of strong bases, such as KOH, to facilitate the removal of the hydrogen ion (proton), while the halogen leaves as a halide ion.
In summary, dehydrohalogenation is a vital reaction type that assists in forming double and triple bonds from halogenated hydrocarbons.

Alkene Formation

Alkene formation is a central process in organic reactions, particularly when transforming saturated alkanes into unsaturated alkenes through elimination reactions. After dehydrohalogenation, the resulting compound in the exercise is propene (\( \text{CH}_3\text{CH} = \text{CH}_2 \)).
Alkenes, also known as olefins, are characterized by the presence of one or more carbon-carbon double bonds, which offer distinct reactivity compared to alkanes. The double bond is a site of high electron density, making alkenes more reactive toward electrophilic addition reactions.
Understanding alkene formation is crucial as it serves as an intermediate step for further chemical transformations, such as bromine addition.

Bromine Addition

The addition of bromine to an alkene, like propene in the exercise, is a classic example of an electrophilic addition reaction. In this process, bromine molecules approach the double bond of propene, which is electron-rich, allowing the bromine to form a cyclic bromonium ion intermediate.
This reaction leads to the creation of 1,2-dibromopropane (\( \text{CH}_3\text{CHBrCH}_2\text{Br} \)), as the bromine atoms add across the double bond. The bromine atoms add anti (opposite sides) from each other creating a di-bromo product. This step saturates the molecule, eliminating the double bond.

• Electrophilic addition is significant as it transforms alkenes into more functional compounds.
• Also, it allows for multiple types of reactions to follow due to the presence of added functional groups.

Triple Bond Formation

Finally, creating a triple bond from a di-substituted alkane involves a two-step dehydrohalogenation process. In the exercise, 1,2-dibromopropane undergoes another phase of dehydrohalogenation via treatment with sodium amide (\( \text{NaNH}_2 \)) in liquid ammonia.
This reaction leads to the removal of two halogen atoms and two hydrogen atoms, forming propyne (\( \text{CH} \equiv \text{C-CH}_3 \)). Sodium amide is a potent base that not only promotes the elimination of hydrogen but also encourages the formation of a more stable alkyne from an alkene.

• The formation of a triple bond requires substantial energy, evident from the use of bases like \( \text{NaNH}_2 \).
• Triple bonds in alkynes are shorter and stronger than double bonds, making them unique chemical entities with distinct properties.

Understanding these transformation steps is essential in organic synthesis, allowing the creation of diverse, complex molecules from simpler starting materials.