Draw the most stable carbocation intermediate and the product of the reaction of hex-1-ene with HCl.
Understand the Problem
The question asks to draw the most stable carbocation intermediate and the final product of the reaction between hex-1-ene and HCl. This involves understanding electrophilic addition to alkenes and the formation of carbocations.
Answer
Carbocation Intermediate: $CH_3CH_2CH_2CH_2CH^+CH_3$ Product: $CH_3CH_2CH_2CH_2CHClCH_3$
Answer for screen readers
Carbocation Intermediate: $CH_3CH_2CH_2CH_2CH^+CH_3$
Product: $CH_3CH_2CH_2CH_2CHClCH_3$
Steps to Solve
- Draw the hex-1-ene molecule
Draw the structure of hex-1-ene, which is a six-carbon chain with a double bond between the first and second carbon atoms. The formula is $CH_3CH_2CH_2CH_2CH=CH_2$.
- Protonation of the alkene
The first step in the reaction of hex-1-ene with HCl is the protonation of the double bond. The double bond acts as a nucleophile and attacks the proton ($H^+$) from HCl. This results in the formation of a carbocation intermediate and a chloride ion ($Cl^-$).
- Determine the most stable carbocation
There are two possible carbocations that can be formed: a primary carbocation (on the first carbon) or a secondary carbocation (on the second carbon). Secondary carbocations are more stable than primary carbocations due to hyperconjugation and inductive effects. Therefore, the most stable carbocation is the one formed on the second carbon atom.
- Draw the carbocation intermediate
The most stable carbocation intermediate will have the positive charge on the second carbon atom. The structure is $CH_3CH_2CH_2CH_2CH^+CH_3$
- Chloride ion attack
The chloride ion ($Cl^-$) acts as a nucleophile and attacks the carbocation to form the final product. The chloride ion attacks the positively charged carbon.
- Draw the final product
The final product is formed when the chloride ion bonds to the second carbon atom. The product is 2-chlorohexane, with the structure $CH_3CH_2CH_2CH_2CHClCH_3$.
Carbocation Intermediate: $CH_3CH_2CH_2CH_2CH^+CH_3$
Product: $CH_3CH_2CH_2CH_2CHClCH_3$
More Information
The reaction of hex-1-ene with HCl follows Markovnikov's rule, which states that in the addition of a protic acid HX to an alkene, the hydrogen atom (H) becomes attached to the carbon with more hydrogen substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents. In other words, the hydrogen adds to the carbon with more hydrogens, and the halogen adds to the carbon with more alkyl groups. This favors the formation of the more stable, more substituted carbocation intermediate!
Tips
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Forming the less stable carbocation: A common mistake would be to form the primary carbocation instead of the more stable, secondary carbocation. Remember that the more substituted carbocation is more stable.
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Incorrect product: Forgetting to have the chloride ion attack the correct carbon leading to an incorrect product.
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