In an E1 reaction, I understand that the base reacts with the hydrogen of the carbon which is next to the positive charged carbon of the cation (of an haloalkane), because this hydrogen is supposed to be acidic. Why is that? And again, in an E2 reaction, why is the hydrogen that is attacked by the base, acidic? I am not sure how the polarity of the halogen affects the situtation.
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asked May 26, 2016 at 22:15
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1 Answer
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After the first step of the reäction (i.e., loss of the halogen), you have the carbocation. You could draw a resonance structure of this: push the electrons from the C–H bond into making a new C–C bond, leaving you with an alkene and a nearby proton (note the conservation of charge, moving the + from a carbon to the hydrogen). This is like the reverse of an addition of a hydrohalic acid to an alkene, whereïn you'd have the double bond attack a proton, then have the halogen attack the carbocation. Either way, you've drawn a very reasonable* resonance structure showing an alkene and H+. That H+ is pretty much the definition of a Brønsted acid, so that would be a hand-waving explanation of how to see this hydrogen as being particularly acidic.
For the E2 example, we can't use the same method because we can't show this resonance structure (well, we kind of can if you want to get into a really tedious discussion that would be highly debated). You'd probably want to use the argument about halogens being inductively electron withdrawing, which pulls electron density away from nearby bonds. This results in less electron density existing in the C–H bond, making the bond longer, and slightly increasing the positive charge on the hydrogen. Since Brønsted acid theory defines the acid as being H+, the more we can show a partial positive charge on the hydrogen, the more Brønsted-acidic we can argue it to be.
* Note that I mentioned drawing a "reasonable" resonance structure. That's important, because we could technically show a resonance structure of benzene as having a carbon with a negative charge and a hydrogen with a positive charge, but that wouldn't be an argument for that hydrogen being particularly acidic, as it's a horrible resonance structure to draw.
It certainly is more complicated than this, but I feel like this highlights some of the most important effects without oversimplifying to the point of making anything explicitly wrong or creating a bunch of exceptions.
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$\begingroup$ For E1: So we can say that the hydrogen of the carbon next to the positive charged carbon is acidic because the electrons of the C–H bond tend to balance the positive charge of the cation? For E2: I still don't understand why the C–H bond next to the halogen bonded carbon sholud be more affected than the hydrogens attached to halogen bonded carbon... $\endgroup$ May 26, 2016 at 23:03
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1$\begingroup$ For E1: That's a good way of describing it, yes. Since C<sup>+</sup> is far less stable than H<sup>+</sup>, it would be favourable to move the two electrons over from the C–H bond to make a new C–C bond, since doing so would transfer that positive charge from the C to the H. While we may draw the intermediate as a carbocation, there is already quite a bit of electron density that has moved from C–H over to make a new C–C bond, making that H quite a bit closer to H<sup>+</sup>. $\endgroup$ May 26, 2016 at 23:22
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1$\begingroup$ For E2: The halogen certainly does affect the hydrogens on the same (geminal) carbon. There are a few ways of explaining why the base wouldn't get involved with one of those hydrogens, so I'll choose one. If you did reprotonate there, you'd end up removing both a proton and the halide from the same carbon. This would make a carbene, which is quite reactive, both nucleophilically and electrophilically. The process would likely just reverse. This can be done, though! For example, you can use potassium tert-butoxide on CHCl<sub>3</sub> to make a :CCl<sub>2</sub> carbene. $\endgroup$ May 26, 2016 at 23:26
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$\begingroup$ Ok I think I got it. So in E2, the halogen affects the bonds nearby making the hydrogens bonded to the carbons next to the geminal carbon little, yet enough acidic for a strong base to come and get them? $\endgroup$ May 27, 2016 at 15:55
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1$\begingroup$ That sounds right to me. There are other factors that complicate things, but that's true for anything in science. For example, depending on your molecule, you may have other hydrogens that are more acidic, so a base would want to deprotonate at the more acidic sites first, BUT if there's no good leaving group next to that carbon, then an elimination won't happen, and the process would likely reverse. You also need to use a base that is small enough and strong enough to deprotonate where you want it to. But what you said is still correct (I'm just stressing that other factors exist). $\endgroup$ May 27, 2016 at 17:51