Acidity inductive effect

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chiddler

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I'm self studying, so i'm probably going to be spamming this forum for the next 2 months or so 👍

Which is the most acidic?

OH-CH(F)-R
OH-CH(Cl)-R
OH-CH(Br)-R

It's with bromine because the conjugate base is most stable. The increased stability is attributed to the reduced electronegativity of bromine which allows more negative charge to remain on the oxygen.

Am I correct?

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HBr is a stronger acid than HF due to the large size of Br. But in this case, since the hydrogen is not directly attached to the electronegative atom, I believe the structure with Fluorine is the most acidic one since F is the most electronegative out of all.
(I'm not 100% sure on this one but that would be my guess) 😎
 
HBr is a stronger acid than HF due to the large size of Br. But in this case, since the hydrogen is not directly attached to the electronegative atom, I believe the structure with Fluorine is the most acidic one since F is the most electronegative out of all.
(I'm not 100% sure on this one but that would be my guess) 😎

hmm why do you say fluorine would make it the most acidic?

i mean can you please elaborate how F's increased electronegativity would contribute to increased acidity?
 
HBr is a stronger acid than HF due to the large size of Br. But in this case, since the hydrogen is not directly attached to the electronegative atom, I believe the structure with Fluorine is the most acidic one since F is the most electronegative out of all.
(I'm not 100% sure on this one but that would be my guess) 😎

I think you're right. Fluorine should be the most acidic would be my choice.
 
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the electrons get left on the oxygen. the most electronegative element of the three you listed is fluorine. thus, it has the strongest inductive pull and can best stabilize the oxygen atom's extra electrons.

well explained. thanks very much!
 
hmm why do you say fluorine would make it the most acidic?

i mean can you please elaborate how F's increased electronegativity would contribute to increased acidity?

The size of Br compared to F affects acidity because the larger the atom the more polarizable it is.

As to the electronegativity of F, it stabilizes the conjugate base better than Br does since it is less electronegative.

So in reality you can make arguments for either being the stronger acid. Best thing to do is look up the pKa and understand the reasons why it is so.


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The size of Br compared to F affects acidity because the larger the atom the more polarizable it is.

As to the electronegativity of F, it stabilizes the conjugate base better than Br does since it is less electronegative.

So in reality you can make arguments for either being the stronger acid. Best thing to do is look up the pKa and understand the reasons why it is so.


Sent from my medical tricorder using Tapatalk

you're right but i think the polarizability argument applies to haloacids. since the H isn't bound directly to the halogen, the second part of what you wrote is more applicable.
 
you're right but i think the polarizability argument applies to haloacids. since the H isn't bound directly to the halogen, the second part of what you wrote is more applicable.


Correct. HF < HCl < HBr < HI and CH2FOH > CH2ClOH > CH2BrOH > CH2IOH
Another way to think of/explain why CH2FOH > CH2IOH is that the electronegative F "steals" electrons from the Oxygen, creating a partial positive on the Oxygen, and the Oxygen "steals" the electron from the bond with the Hydrogen making it very easy for the acid to donate that Hydrogen == more acidic hydrogen, more acidic compound.

Edit: Acidic hydrogen = hydrogen that is easily lost. I don't remember if I got this from my orgo teacher, or from TBR (I think TBR), but you can use this thought process to remember why HI is the strongest haloacid. Iodide has a huge electron cloud, and fluorine has a small electron cloud. They both have 7 valence electrons, so it would be harder for a hydrogen to "find" an electron in Iodine's huge cloud, thus making it harder to bond, and easier for HI to donate its proton.
 
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