Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: ve_90 on March 16, 2025, 11:58:23 AM
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When I add HCl to the triple bond of 5-methyl-2-hexyne, is it correct to form two secondary carbocations and proceed with the reaction mechanism through both, since they have the same stability? Can the same reasoning be applied to the subsequent attack on the double bond? Finally, I obtain three products: 2,3-dichloro-5-methylhexane, 2,2-dichloro-5-methylhexane, and 3,3-dichloro-5-methylhexane. Is it correct to say that 2,3-dichloro-5-methylhexane is the major product? It indeed results from the nucleophilic attack of the chloride ion on two carbocations (so, two of the four carbocation intermediates).
Thank you for your response!
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According Markovinkow you will get mainly the 3,3 - Di-Bromoproduct. H will add to the C with most H- atoms already.
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Regarding Markovnikov's rule, when I add hydrogen to the triple bond (and then to the double bond), in all cases I get two equally stable secondary carbocations (because I don’t have one carbon less substituted than the other; both are equally substituted). Maybe I’m wrong, but I don’t consider Cl as a substituent. When I say “less substituted,” I mean in terms of carbon chains. Maybe I’m wrong? If not, I don’t understand your conclusion.
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Why you dont consider Cl.
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You are right, now I have properly understood Markovnikov's rule, which states: "when an asymmetric reagent adds to an asymmetric alkene, the electropositive part of the reagent attaches to the carbon atom of the triple/double bond that is bonded to the greatest number of hydrogen atoms."
To summarize:
In the first step, the hydrogen adds to either of the two carbons involved in the triple bond, since both do not have any hydrogen atoms attached. Two secondary carbocations are formed, and both are equally stable. The addition of the chloride ion leads to the formation of 3-chloro-5-methyl-2-ene and 2-chloro-5-methyl-2-ene.
Hydrogen adds to the double bond of both alkenes: a total of 4 secondary carbocations are formed, but the ones that follow Markovnikov's rule are those in which the positive charge is located on the carbon that is bonded to the chlorine.
As a result, 4 products are formed, but the ones that are produced in higher quantities in accordance with Markovnikov's rule are 3,3-dichloro-5-methylhexane and 2,2-dichloro-5-methylhexane.
Now I can say I have a clear understanding? Thank you for your feedback.
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Yes correct.