Question

\(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{Br}_{2} \rightarrow(\mathrm{A}) \stackrel{\mathrm{NaNH}_{2}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{H}^{+}}{\longrightarrow}(\mathrm{C}) .\) The structure of (C) is (a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (c) \(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\)

Short answer

Answer: The structure of compound (C) is \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\).

Step-by-step solution

Identify the structure of compound (A) by reacting the given compound with Br₂

To identify compound (A), we need to react \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) with \(\mathrm{Br}_{2}\). This reaction is an electrophilic addition reaction where one Br atom adds across the double bond of the given compound. The structure of compound (A) will be \(\mathrm{CH}_{3}-\mathrm{CHBr}-\mathrm{CH}_{2}\mathrm{Br}\).

Identify the structure of compound (B) by reacting compound (A) with NaNH₂

The second reaction involves treating compound (A) (\(\mathrm{CH}_{3}-\mathrm{CHBr}-\mathrm{CH}_{2}\mathrm{Br}\)) with \(\mathrm{NaNH}_{2}\). This is a dehydrohalogenation reaction, where two HX molecules are eliminated in the presence of strong base. NaNH₂ will selectively remove the HX from the more subtituted side of the compound. In this case, it will remove the hydrogen from the carbon between the two Br atoms. The structure of compound (B) will be \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH} + \mathrm{2Br}^{-} + \mathrm{2Na}^{+} + 2 \mathrm{NH}_{2}^{-}\).

Identify the structure of compound (C) by reacting compound (B) with H⁺

The third reaction involves treating compound (B) (\(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\)) with \(\mathrm{H}^{+}\). This is an acid-base reaction, in which the \(\mathrm{CH}^{-}\) in compound (B) will abstract a proton (H⁺) from the acid present in the reaction. The result is the triple bond remaining intact and the formation of compound (C). So, the structure of compound (C) will be \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\). When comparing our final structure for compound (C) to the answer choices provided in the problem, we see that option (a) matches our result. Therefore, the correct choice is: (a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\)

Key concepts

Electrophilic Addition Reaction

The Electrophilic Addition Reaction is a fundamental type of reaction that occurs in organic chemistry, particularly with compounds that have double or triple bonds, like alkenes and alkynes.

An electrophile is an electron-deficient atom or molecule that is attracted to an electron-rich center, where it accepts a pair of electrons to form a new chemical bond. In the case of alkenes like owrap(CH}_3\right)-\text{CH}=\text{CH}_2\right)\(, the double bond is electron-rich and readily reacts with electrophiles. When a diatomic bromine (\text{Br}_2\right)\) molecule approaches the double bond, one of the bromine atoms acts as an electrophile. The bromine atom accepts the electron pair from the double bond to form a bromonium ion intermediate, and the other bromine atom bonds to the other carbon atom of the original double bond, resulting in a vicinal dibromide compound.

Understanding how these additions occur helps students predict reaction products and mechanism pathways in organic synthesis. This reaction is crucial for modifying molecular structures and forms the basis for more complex chemical transformations.

Dehydrohalogenation Reaction

A Dehydrohalogenation Reaction is a type of elimination reaction where a hydrogen halide (\text{HX}\right)\(, such as hydrogen bromide (\text{HBr}\right)\), is removed from an alkyl halide. This reaction typically requires the presence of a base, such as sodium amide (\text{NaNH}_2\right)\(, which can abstract a proton (\text{H}^+\right)\) from the substrate.

In the provided exercise, when compound (A), a vicinal dibromide, is treated with a strong base like \text{NaNH}_2\right)\(, the base removes a hydrogen atom from the carbon atom next to the one carrying the bromine atom (halogen). This removal creates a double bond, and with the elimination of a second \text{HBr}\right)\), a triple bond is formed, resulting in an alkyne. Students should note that \text{NaNH}_2\right)\( is strong enough to induce the elimination of two equivalents of \text{HBr}\right)\), thereby transforming a dihalide to an alkyne in a process termed double dehydrohalogenation.

This type of reaction is significant for creating alkenes and alkynes, which can serve as key intermediates for further chemical synthesis. It exemplifies the importance of understanding how various reagents and conditions can drastically alter molecular structure.

Acid-Base Reaction

Acid-Base Reactions are pervasive in chemistry and central to numerous processes in both organic and inorganic chemistry. They involve the transfer of a proton (\text{H}^+\right)\( from an acid to a base. According to the Br\text{\text{o}}nsted-Lowry theory, an acid is a proton donor, while a base is a proton acceptor.

In the final step of the exercise, compound (B) (an alkyne) reacts with a proton source (\text{H}^+\right)\). The negatively charged carbon in the alkyne (\text{CH}^−\right)$ is highly nucleophilic and readily accepts a proton to form the more stable hydrocarbon, compound (C). This is an acid-base reaction where the alkyne acts as a base and the acid provides a proton to the carbon with excess electrons.

Students should recognize the acid-base reaction not only as a simple proton transfer but also as a way to adjust the charges on molecules, affect their stability, and alter their reactivity. These reactions are also crucial in biochemistry, as they are part of amino acid side-chain interactions, enzyme activities, and cellular processes like buffering systems.