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Chapter 27: Problem 86

When \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\) is added to \(\mathrm{CH}_{3} \mathrm{OH}\) at room temperature, the major product is \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\) and a minor product is \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .\) Write out the mechanisms for the reactions leading to these products and use curved arrows to show the movement of electrons.

Short Answer

Expert verified
The major product results from a SN2 reaction where \(\mathrm{CH}_{3} \mathrm{OH}\) acts as a nucleophile, attacking the carbon in \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\), resulting \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\). The minor product comes from an elimination mechanism, where a $\beta -$hydrogen is removed along with the leaving group to form a double bond, forming $\mathrm{CH}_{3}\mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .$ Curved arrow notation shows the flow of electrons.

Step by step solution

01

Mechanism for the Major Product

The major product is due to a Nucleophilic Substitution reaction. \(\mathrm{CH}_{3} \mathrm{OH}\) acts as a nucleophile and attacks the \(\mathrm{CBr}\), of \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\) while \(\mathrm{Br}^{-}\) leaves, resulting in the product \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\).
02

Mechanism for the Minor Product

The minor product is due to an Elimination reaction. In this reaction, a \(\beta-\)hydrogen is removed along with \(\mathrm{Br}^{-}\) to form a double bond in $\mathrm{CH}_{3}\mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .$ Despite being less predominant due to room temperature conditions, this reaction does result in a minor product.
03

Showing Movement of Electrons

Curved arrow notation is used to show the movement of electrons during these reaction mechanisms. An arrow from \(\mathrm{CH}_{3} \mathrm{OH}\) to the Carbon atom of \(\mathrm{CBr}\) will depict the nucleophilic attack for step 1. In step 2, an arrow from the \(\beta-\)hydrogen to the adjacent Carbon followed by an arrow from the Carbon-Br bond to \(\mathrm{Br}\) will show the elimination process.

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