[b]Possible Answer:[/b]
I'm not sure if im right in these.
The first is markovnikov addition
And the second is anti-markovnikov addition.
Could someone draw out the product for me?



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The link above shows different reagents, what is added over the alkene DB, and whether the addition is Markovnikov of not.
If it is Markovnikov the hydrogen that is being added will bind itself to the carbon of the alkene which has the most hydrogens in the first place.


the first reaction is markovnikov.and second one is anti markovnikov because of H2O2. if you perform the reaction with H2O2 the reaction will give you anti-markovnikov addition.

no, no, no, no, NO!

Markovnikov's rule:[quote]When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already[/quote]

In the oxymercuration reaction, the Hg ion adds across the double bond - so both carbons kinda share the charge. When water comes along, it is going to attack the carbon that is most positive - this will be the carbon that is most stable with a positive charge, the most delta positive carbon, whic is the most [i]substituted[/i] carbon. The NaBH[size=8]4[/size] gets rid og the Hg and replaces it with a ydrogen - Markovnikov addition.

In the borane reaction, Boron, an electron deficient species, is acting as the [b]Lewis Acid[/b] The double bond attacks the boron, and does so on the terminal, less substituted carbon, so that a more stable (more substituted) carbocation is formed. This then picks up a hydrogen from the (still unhappy) Boron species, whic is negativly charged. It does this 3 times for each borane. then HOO- (made from HO- and HOOH) comes along ant the negative charge on that attacks the electron deficient Boron. Boron has no hydrogens to get rid of to lose it's newly gained charge, so one of the alkane chains migrates (with retention of configuration) to the first oxygen of BOOH, and kicks off an HO-.
[i]This[/i] happens three times, then Boron is *still* not happy, and gets attacked by HO-, and then complains about having a charge so it kicks out the deprotinated alcohol produce, which becomes the anti-markivnikov product on workup.
(That's all FAR easier to draw out as a mechanism, if you want it explained again)

Basically, look in each case at what is acting as the lewis acid, and how the double bond is added to. In the first the most stable carbocation is attacked by OH, giving you markovnikov overall. In the second you have a migration reaction, so things change a little.

Interestingly, neither of those reactions are "normal" additions across a double bond - certainly two to "just remember"!

(My no, no, no, no, NO! at the beginning was referring to the fact that the anti-markovnikov nature of the second reaction has nothing to do with the fact H2O2 is there - that just furnishes the addition of OH - other functional groups can be added, but they will always add anti-markivnicov because of the Boron, and because it is a MIGRATION reaction)

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