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Chapter 10: Problem 17

Malve a new bond between a nucleophile and an electrophile and simultaneously break a bond to give stable molecules or ions. Nucleophilic displacement of MEO by Br gives the bromoalkane. ONCCCBr

Short Answer

Expert verified
Answer: The product formed after the nucleophilic substitution reaction is a bromoalkane with the structure BrCH2CH2CH2ON.

Step by step solution

01

Identify the nucleophile and electrophile

In this reaction, the nucleophile is the bromide ion (Br-), which has a high electron density and is attracted to the positive centers of molecules. The electrophile is the carbon atom bonded to the methoxy group (MEO) in the given molecule (ONCCCBr). The electrophile is electron-poor and can accept an electron pair from a nucleophile.
02

Understand the mechanism of nucleophilic substitution reaction

Nucleophilic substitution reaction takes place via two mechanisms: SN1 and SN2. In this exercise, the reaction is assumed to be an SN2-type mechanism, which is a direct one-step reaction where the nucleophile can easily attack the electrophilic carbon center, simultaneously expelling the leaving group. This mechanism usually occurs in primary substrates.
03

Attack of the nucleophile

The bromide ion (Br-) attacks the electrophilic carbon atom, which is attached to the methoxy group (MEO) in the given molecule. This leads to the formation of a new bond between the carbon atom and the bromide ion (Br-).
04

Leaving group departure

While the new bond between the carbon atom and the bromide ion is being formed, the bond between the carbon atom and the methoxy group (MEO) simultaneously breaks. The methoxy group acts as the leaving group and exits the molecule, forming a stable methoxide ion (CH3O-).
05

Formation of the bromoalkane

After the nucleophilic substitution reaction, a bromoalkane is formed. The structure of the new molecule can be represented as follows: BrCH2CH2CH2ON. In summary, the reaction proceeds via the SN2 mechanism, where the bromide ion (Br-) attacks the electrophilic carbon atom, forming a new bond. The methoxy group (MEO) acts as the leaving group, and a bromoalkane is formed.

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Most popular questions from this chapter

When \((R)-2\) butanol is left standing in aqueous acid, it slowly loses its optical activity. Account for this observation.

Break bonds to give stable molecules or ions. Redistribution of valence electrons within the cyclic periodate gives \(\mathrm{HIO}_{3}\) and two carbomyl groups. A result of this electron redistribution is an oxidation of the organic component and a reduction of the iodine-containing component. \(\mathrm{HUO}_{3}\) (lodic acid)

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