Question

Identify the products \(\mathrm{X}\) and \(\mathrm{Y}\) in the following reaction $$ \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{Br}_{2}}{\longrightarrow} \mathrm{X} \frac{\text { (i) } \mathrm{NaNH}_{2}(3 \text { equivalents) }}{\text { (ii) } \mathrm{CH}_{3} 1} \mathrm{Y} $$

Short answer

Answer: The structures of the products are as follows: Product X: C₆H₅-CHBr-CH₂Br (formed after reaction with bromine) Product Y: C₆H₅-C≡C-NHCH₃ (formed after reactions with sodium amide and methyl iodide)

Step-by-step solution

Reaction of alkene with bromine

Firstly, we need to determine the product X formed by the reaction between the alkene (C\(_{6}\)H\(_{5}\)-CH=CH\(_{2}\)) and bromine (Br\(_{2}\)). Bromine reacts with alkenes to form a vicinal dibromoalkane, via a halogenation reaction. In this case, the product X is: $$ \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CHBr}-\mathrm{CH}_{2}\mathrm{Br} $$

Reaction of product X with sodium amide

The product X reacts with 3 equivalents of sodium amide (NaNH\(_{2}\)). Sodium amide is a strong base and is a good nucleophile, so it will deprotonate the most acidic hydrogens present. Since there are 3 equivalents, it will deprotonate the alpha-carbon hydrogens (adjacent to the phenyl group) one at a time, ultimately forming a triple bond between the carbon atoms: $$ \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}\equiv\mathrm{C}-\mathrm{NH}_{2} $$

Reaction with methyl iodide

Next, the resulting compound reacts with methyl iodide (CH\(_{3}\)I). The amine (NH\(_{2}\)) group in the compound will react with the methyl iodide in a nucleophilic substitution reaction, adding a methyl group to the nitrogen atom. This gives us the final product Y: $$ \mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}\equiv\mathrm{C}-\mathrm{NHCH}_{3} $$ Throughout this process, we have seen how the alkene is first brominated, then undergoes deprotonation and formation of a triple bond, ultimately reacting with methyl iodide to form an N-methylated amine. By following these steps, we have determined the structures of products X and Y.

Key concepts

Halogenation Reaction

The halogenation reaction is a type of chemical reaction where halogens, such as bromine (Br₂), react with another substance, often an alkene or an alkane, to form a halogenated product. In the presence of an alkene, like in our exercise, bromine can react to form a vicinal dibromoalkane. This happens through a mechanism where the electrons in the double bond of the alkene attack the electron-deficient bromine molecule, forming a bromonium ion intermediate.

This intermediate then quickly reacts with a bromide ion (Br^-) that comes from another molecule of bromine, leading to the addition of bromine atoms across the double bond. The reaction is stereospecific, typically yielding a trans addition, which means the bromine atoms are added on opposite sides of the alkene plane, resulting in a more stable configuration. Notably, this process doesn't require a catalyst, and the bromine provides both a visual and mechanistic indication of the reaction’s progression, as the color of bromine dissipates upon successful halogenation.

Nucleophilic Substitution Reaction

Nucleophilic substitution reactions are fundamental in organic chemistry. They occur when a nucleophile, a species with a free electron pair, donates its electrons to an electrophile, a positively charged or electron-deficient atom, to form a new chemical bond. In our problem, the nucleophilic substitution occurs during the reaction of a compound with methyl iodide (CH₃I).

The nucleophile in this case is an amine, specifically the NH₂ group attached to the alkyne. Amines are quite reactive due to the lone pair of electrons on nitrogen. When CH₃I is introduced, the lone pair of electrons from the nitrogen atom attacks the positively charged carbon in the CH₃I, which is connected to the iodine, a good leaving group. As a result, the iodine is displaced, and the nitrogen forms a new bond with the methyl group, yielding the N-methylated product. This type of reaction is key in forming a variety of organic molecules with diverse functionalities.

Alkyne Synthesis

Alkyne synthesis is often achieved by the deprotonation of an alkane or alkene to form an acetylide anion, followed by a coupling reaction. In the exercise's context, the synthesis of an alkyne begins with the introduction of a strong base, like sodium amide (NaNH₂), to a dibromoalkane compound. This strong base can remove acidic hydrogen atoms from carbons adjacent to the halogenated carbons in the dibromoalkane, which are acidic due to the electron-withdrawing nature of the adjacent bromine atoms.

As the base deprotonates the molecule sequentially, it induces the formation of a carbon-carbon triple bond, characteristic of alkynes. This step-wise elimination of hydrogen bromide (HBr) is essential for creating the structure of the alkyne. It is worth noting that the regioselectivity of the base in choosing the hydrogens to remove shows the importance of understanding reactivity trends in organic synthesis. Once the alkyne is formed, it undergoes further reactions, like the nucleophilic substitution mentioned previously, to yield a variety of functionalized organic compounds.