Give the products that would be obtained from the reaction of the following compounds with sodiu

Give the products that would be obtained from the reaction...

Key Concepts

Role of Sodium Amide as a Strong Base

Sodium amide is a powerful base that is extensively used to induce dehydrohalogenation reactions by abstracting protons adjacent to leaving groups. Its high basicity is crucial for initiating the elimination process, particularly in substrates that can form multiple bonds. Understanding its reactivity is key in predicting the formation of unsaturated products such as alkenes or alkynes.

Formation of Alkynes via Double Elimination

Many organic reactions involving dihalide substrates with sodium amide in liquid ammonia result in the formation of alkynes through a double elimination process. In these reactions, two molecules of hydrogen halide are removed, leading to the establishment of a carbon–carbon triple bond. This method is a pivotal synthetic route for alkyne production in organic chemistry.

Use of Liquid Ammonia as a Solvent

Liquid ammonia serves as an excellent solvent for reactions involving strong bases like sodium amide. It stabilizes reactive intermediates and solvates ions effectively, thereby facilitating smooth elimination processes. This concept is essential in organic synthesis, especially when controlling the reaction conditions for dehydrohalogenation reactions.

Elimination Reactions

Elimination reactions are processes in organic chemistry where atoms or groups are removed from adjacent carbons, resulting in the formation of a double or triple bond. When a strong base like sodium amide is used, the mechanism typically follows an E2 pathway, with the simultaneous removal of a proton and a leaving group. This concept is fundamental when predicting the outcome of reactions involving halogenated compounds under basic conditions.

Transcript

00:01 We're going to see how paraclorotolulene will react with some sodium amide in liquid ammonia.

00:13 Okay, so that we could have here is a benzene intermediate, a benzine intermediate, okay, followed by some amine substitution.

00:29 Okay, that can either take place on either side of this.

00:33 Here where the benzene alken would have been.

00:41 So we can either get para substitution or we can get meta substitution as well.

00:51 Okay, since after, you know, we formed the benzine intermediate, we can get addition of the amine on either side, since both sides, you know, are reactive.

01:06 So the exact same thing will happen here.

01:11 Okay, this bromophenyl ethene.

01:18 Okay, so when we treat it with sodium amine, you will get formation of an alkyne intermediate, most likely at this position, as well as the other side.

01:40 So there will be two benzine intermediates here.

01:47 Okay, so then now that will yield quite a bit of products here.

01:55 So the first one is the most obvious one, where we just directly, it looks like the direct, you know, position, the direct substitution.

02:07 But it's also possible to have the ortho and the para.

02:12 Here plus the para here.

02:22 Okay.

02:24 And that's because both sides of that bromine are, you know, available and non -equivalent.

02:35 And both can get, you know, benzene formation.

02:40 Okay, but as we can see, this one will be in a larger ratio compared to the rest...