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Chapter 10: Problem 15

Make a new bond between a nucleophile and an electrophile. Reaction of the \(3^{7}\) carbocation (an electrophile) with bromide ion (a nucleophile) gives the alkyl halide. CC(C)(C)CBr CC(C)(Br)CC(C)(C)Br 2-Brome-2-rethylpropane (tert-Butyl bromide)

Short Answer

Expert verified
Question: Describe the reaction between an electrophile and a nucleophile to form an alkyl halide, such as tert-Butyl bromide, and explain its formation and structure. Answer: The reaction between an electrophile (carbocation) and a nucleophile (bromide ion) involves a process called nucleophilic attack, in which the nucleophile donates a pair of electrons to the electrophile. This forms a bond between them, ultimately resulting in an alkyl halide like tert-Butyl bromide. The structure of tert-Butyl bromide is represented by the SMILES notation: CC(C)(Br)CC(C)(C)Br.

Step by step solution

01

Identifying the reactants

The two reactants in this reaction are the carbocation (\(3^7\)) and the bromide ion (Br-). The carbocation acts as the electrophile, which is a molecule that can accept a pair of electrons. The bromide ion (Br-) acts as the nucleophile, a molecule that can donate a pair of electrons.
02

Nucleophilic attack

The nucleophile (bromide ion, Br-) donates a pair of electrons to the electrophile (carbocation), forming a bond between them. As a result, the positively charged carbon atom in the carbocation forms a bond with the negatively charged bromide ion.
03

Formation of the alkyl halide

The bond between the carbocation and the bromide ion forms the desired product, an alkyl halide. In this case, the product is 2-bromo-2-methylpropane, also known as tert-Butyl bromide. Its structure is represented by the SMILES notation: CC(C)(Br)CC(C)(C)Br.
04

Visually representing the product

The final structure of the tert-Butyl bromide after the reaction is as follows: CC(C)(Br)CC(C)(C)Br

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Most popular questions from this chapter

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