ochem easy question

[phosphorylation]

---

Members don't see this ad.

i know this is really stupid guys, but what the hell is an alpha or beta or whatever carbon...? i tried looking through two texts trying to find it; i know I read it somewere, but I don't know were to look for it. I tried looking through google, but nada... theres a question in kaplan on the first chapter in the review notes, at the review questions that asks which carbons are alpha and beta, etc...can you explain it to me, or can you just tell me were to look for the answer, because kaplan apparently thinks its too easy of a subject to explain.

thanks.


-phospho

 

[justletmein]

whats up dude! alpha carbon is the carbon adjacent to the carbonyl carbon, beta is the carbon adjacent to the alpha carbons..then it goes gamma i believe.. it's in the kaplan stuff! the alpha hydrogens are the hydrogens connected to the alpha carbon, and they're the most acidic too

 

[jk5177]

the reason why they are the most acidic is related to the resonance structure, but I don't remember exactly what happens. Does anyone know?

 

[DrTacoElf]

the reason why they are the most acidic is related to the resonance structure, but I don't remember exactly what happens. Does anyone know?

carbonyl oxygen withdraws electrons from adjacent hydrogens. Most acidic. Also it will stablize any formal charge created.

 

[justletmein]

hey i've got an ochem question now... how come m-nitrophenol is less acidic than o-nitrophenol ?

 

[ranark]

hey i've got an ochem question now... how come m-nitrophenol is less acidic than o-nitrophenol ?

I thought that op directors decreased acidity so wouldnt o-nitrophenol be less acidic.. not sure though.. now i am curious to know.. anyone else have an answer?

 

[justletmein]

I thought that op directors decreased acidity so wouldnt o-nitrophenol be less acidic.. not sure though.. now i am curious to know.. anyone else have an answer?

here's what kaplan says...

This leaves either the m-nitrophenol or o-nitrophenol (choices B and E respectively).
Nitro-substituted aromatic rings are meta directing. This means that the electron density will be higher in the m position than in the o/p position. Since we are looking for low electron density (to stabilize the negative charge) at the hydroxyl group position, choice E would be more acidic than choice B

I don't get it though... isn't it more stable in the meta position? doesnt' that stability make it more acidic? or does that just make the conjugate base more stable? 😕

 

[coodoo]

imagine o-nitrophenol. If the OH group of the phenol loses its proton, it be becomes O-.You can resonate this structure and make it C=O by moving one of the double bonds in the phenyl ring to the Carbon which carries the NO2 group. This makes that carbon negatively charged.

Now look at the structure of NO2. It is N double bonded to one O and single bonded to the next O. If we resonate the lone pair of electrons on the negatively charged carbon to the nitrogen atom we will see this.

C=N-O-
|
O-

For some damn reason the O- keeps showing up on the C. Note that the single bonded o is bonded to the N

You can do this on either the O position or P position, which will give the product that is more resonance stablized. The M-position is not resonance stablized as such

 

[ranark]

Is it because there is an OH on the benzene that the activating groups make them more acidic, or in general do activating groups make all benzene rings more acidic? I heard that deactivating groups make benzene more acidic.

 

[justletmein]

yes, o-nitrophenol is more acidic. Simply, just think of the inductive effects... the NO2 group is closer in the o-position than the m position 😀

 

[phosphorylation]

thanks a lot everyone...im gonna go hit the books again now...