Questions: Determine a stepwise mechanism for the following reaction that illustrates why two substitution products are formed. Explain why 1-bromohex-2-ene reacts rapidly with a weak nucleophile (CH3OH) under SN1 reaction conditions, even though it is a 1° alkyl halide.

Determine a stepwise mechanism for the following reaction that illustrates why two substitution products are formed. Explain why 1-bromohex-2-ene reacts rapidly with a weak nucleophile (CH3OH) under SN1 reaction conditions, even though it is a 1° alkyl halide.
Transcript text: Determine a stepwise mechanism for the following reaction that illustrates why two substitution products are formed. Explain why l-bromohex-2-ene reacts rapidly with a weak nucleophile $\left(\mathrm{CH}_{3} \mathrm{OH}\right)$ under $\mathrm{S}_{\mathrm{N}} \mathrm{l}$ reaction conditions, even though it is a $1^{\circ}$ alkyl halide.
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Solution

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Solution Steps

Step 1: Identify the Reaction Mechanism

The reaction involves 1-bromohex-2-ene reacting with methanol (CH3OH) under SN1 reaction conditions. The first step in an SN1 mechanism is the formation of a carbocation by the departure of the leaving group (Br-).

Step 2: Formation of the Carbocation

The bromine atom leaves, forming a carbocation at the second carbon: \[ \text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_2^+ + \text{Br}^- \]

Step 3: Nucleophilic Attack

Methanol (CH3OH) acts as a nucleophile and attacks the carbocation, leading to the formation of the product: \[ \text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_2^+ + \text{CH}_3\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_2\text{OCH}_3 + \text{H}^+ \]

Final Answer

The first step in the reaction proceeds according to the SN1 mechanism, where the bromine atom leaves, forming a carbocation.

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