PROBLEM 7.61
Question
Draw a stepwise mechanism for the following reaction that illustrates how two substitution products are formed. Explain why 1-bromohex-2-ene reacts rapidly with a weak nucleophile (CH3OH) under reaction conditions, even though it is a alkyl halide.
Step-by-Step Solution
VerifiedAnswer
The 1-bromo-hex-2-ene is a primary alkyl halide, but it reacts with the weak nucleophile under the SN1 reaction conditions because of the formation of the resonance stabilized carbocation.
The two substitution products are formed by the following reactions:
The nucleophilic attack on carbon 1
The nucleophilic attack on carbon 2
The SN1 mechanism involves the following two steps:
- The formation of a carbocation intermediate: The leaving group moves from the alkyl halide and leads to the formation of an intermediate carbocation.
- The carbocation intermediate is planar in structure, due to which the nucleophile can attack either side of the carbocation. This results in the formation of a racemic mixture of the products.
The SN1 reactions are always favored by the polar protic solvents such as water, ethanol, etc.
The reaction of 1-bromo-hex-2-ene with ethanol takes place in the following way:
The above reaction proceeds in the following steps:
- The removal of the leaving group: The bromide ion leaves the alkyl halide and results in the formation of a carbocation which is resonance stabilized.
Formation of resonance stabilized carbocation
The 1-bromo-hex-2-ene is a primary alkyl halide, but it reacts with the weak nucleophile under the SN1 reaction conditions because of the formation of the resonance stabilized carbocation.
- The attack of the nucleophile: The attack of the nucleophile can result in the formation of two substitution products because the nucleophile is able to attack two different carbon atoms.
The nucleophilic attack on the carbon atoms is shown below:
The nucleophilic attack on carbon 1
The nucleophilic attack on carbon 2