Talk:Markovnikov's rule

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I have watched this article bounce back and forth between which carbon the halide will go. Before someone changes it back, please re-read and re-check yourself - the halide will go to the carbon with the MOST substituents, and the proton will go to the carbon with the LEAST substituents. The opposite is true for the Anit-Markovnikov rule. Umdelt 06:45, 20 February 2007 (UTC)[reply]

I am uncomfortable with the anthropomorphic wording of the following two phrases:

the positive charge wants to be in the most stable form

the radical prefers to be in the center of the molecule

A molecule cannot "want" or "prefer" anything. Reactions occur because of underling physical properties. Unfortunately, I do not feel that I am qualified to write those phrases with more appropriate terms.--Mikebrand 04:03, 12 June 2006 (UTC)[reply]

I hear educators use words like that all the time - I guess it's hard not to anthropomorphise a little. Anyway, it looks like this has been dealt with now. Richard001 08:52, 21 March 2007 (UTC)[reply]

This needs to be checked: "The rule states that with the addition of a protic acid such as H-X to an alkene, the acid hydrogen (H) becomes attached to the carbon atom with the greatest number of hydrogens, and the halide (X) group becomes attached to the carbon with the fewest hydrogens.[3] In other words the negative part of the addendum attaches itself to the carbon atom with the fewest hydrogen atoms."

The reference link states "In the addition of HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents." Not what is written up above. —Preceding unsigned comment added by Apilgrim135 (talk • contribs) 06:27, 8 December 2009 (UTC)[reply]

I have "In the addition of HX to an alkene, the hydrogen atom bonds to the alkene carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents." Organic Chemistry, 6th Edition, by John McMurry. Section 6.9, page 187. —Preceding unsigned comment added by Apilgrim135 (talk • contribs) 06:05, 8 December 2009 (UTC)[reply]

• We will go with McMurry V8rik (talk) 17:11, 8 December 2009 (UTC)[reply]

Because Markovnikov's rule is taught in the undergraduate context of hydrating alcohols, you end up with the assumption that it refers to where the hydroxyl ends up. This isn't the case, as is obvious from the actual statement of the rule: it's a question of where the cation ends up, and that's at the more stabilized location. In hydroboration, the cation ends up at the more stabilized location; unlike in straight-up hydration or in oxymercuration-demercuration, the first reagent to add contains a hydride; as such, the cation gets itself a hydrogen, the less substituted position gets the boron, and the rest is history.

I also brought this up in the talk section of hydroboration-oxidation. Everyday847 (talk) 02:22, 7 March 2012 (UTC)[reply]

You're confusing the cause (carbocation-intermediate stability, which dictates where the nucleophile attacks) with the observed result (which end winds up with the electronegative atom at the end of the process). The formulation of the rule focuses on the empirical result. Was the mechanism or C+ stability-difference as the underlying cause even known in the 1860s? The mechanism and structural details are obviously the cause (and always are) but I've never seen a textbook tie "Markovnikov's rule" to the cause of the observed result. On the contrary, definitions of the rule as given such as those in McMurry's Organic Chemistry, "In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon wiyh more alkyl substituents." seem common. Then the hydroboration–oxidation reaction is described as "a complementary method that yields the non-Markovnikov product". It's therefore

verifiably correct

to describe that reaction with that terminology.

And the cause of this reversal is at least in part steric (large B cluster at less hindered end). I hope you are not suggesting (but your wording says you are) that the hydroboration–oxidation reaction involves a C+ intermediate that then undergoes hydride attack (similar to how HBr involves C+ intermediate with bromide attack). That's very wrong. DMacks (talk) 03:17, 7 March 2012 (UTC)[reply]

The opening paragraph of the article should say what Markovnikov's rule actually states. Not that it "describes the outcome of reactions". — Preceding unsigned comment added by 69.180.17.136 (talk) 23:05, 26 February 2017 (UTC)[reply]