general chemistry - acidity and electron withdrawing group (EK problem)

[Monkeymaniac]

So I thought that when you attach an electron withdrawing group (say electronegative halogens) to alpha carbon of a carboxylic acid, this would spread out the electron density on the O of hyroxyl group, leading to the better stablization of the acid's conjugate base, effectively increasing the acidity of the Hydroxyl Hydrogen.

But while solving EK general chemsitry question

The addition of an electorn withdrawing gorup to the alpha carbon of a carboxylic acid will:
A. increase the acidity of the proton by making the O-H bond more polar.
B. increase the acidity of the proton by making the O-H bond stronger.
C. decrease the acidity of the proton by making the O-H bond more polar.
D. decrease the acidity of the proton by stablizing the conjugate acid.

The solution says A is true, which is completely absurd. Electrons are withdraw from the Hydroxyl O, decreasing the electron density difference between O and H of OH, making the bond less polar. The answer, in my opinoin, should be "increase the acidity of the proton by stablizing the conjugate acid."

Any ideas? and can anyone please recommend a better general chemistry book to study with? Thanks.

---

Members don't see this ad.

 

[G1SG2]

So I thought that when you attach an electron withdrawing group (say electronegative halogens) to alpha carbon of a carboxylic acid, this would spread out the electron density on the O of hyroxyl group, leading to the better stablization of the acid's conjugate base, effectively increasing the acidity of the Hydroxyl Hydrogen.

But while solving EK general chemsitry question

The addition of an electorn withdrawing gorup to the alpha carbon of a carboxylic acid will:
A. increase the acidity of the proton by making the O-H bond more polar.
B. increase the acidity of the proton by making the O-H bond stronger.
C. decrease the acidity of the proton by making the O-H bond more polar.
D. decrease the acidity of the proton by stablizing the conjugate acid.

The solution says A is true, which is completely absurd. Electrons are withdraw from the Hydroxyl O, decreasing the electron density difference between O and H of OH, making the bond less polar. The answer, in my opinoin, should be "increase the acidity of the proton by stablizing the conjugate acid."

Any ideas? and can anyone please recommend a better general chemistry book to study with? Thanks.

I agree. But look at it this way-from EK's perspective, they might mean that the EWG withdraws electron density from the oxygen, making the oxygen less negative, or more positive, and causing repulsion or DECREASING attraction between the now more positive O and the partial positive H, lengthening and weakening the O-H bond, and thus increasing acidity.

As for your second question: Any ideas? and can anyone please recommend a better general chemistry book to study with?

Holy crap, get TBR chemistry book. Best chem review ever, hands down.

 

[Monkeymaniac]

the EWG withdraws electron density from the oxygen, making the oxygen less negative, or more positive, and causing repulsion or DECREASING attraction between the now more positive O and the partial positive H, lengthening and weakening the O-H bond, and thus increasing acidity.

Thank you for your comments. I see. But does that mean a polar bond in general is stronger than a non-polar bond (say C-H bond)? I think a better question would be how is a bond strengh effected by the polarity of the bond?

 

[G1SG2]

Thank you for your comments. I see. But does that mean a polar bond in general is stronger than a non-polar bond (say C-H bond)? I think a better question would be how is a bond strengh effected by the polarity of the bond?

Well, the more polar a bond, the weaker it is because there is a greater charge separation. The dipole moment which measure's a bond's polarity is given by p=q*r where q=charge and r=distance. Coloumb's law of electrostatic attraction (or repulsion) states that F=q1q1/r^2, or the force of attraction/repulsion is inversely proportional to the square of the distance between the two charges. The MORE POLAR a bond is, the more separated it is/the greater the charge separation (p=q*r), and thus the WEAKER the electrostatic force between them (F=1/r^2). Thus, bond strength and bond polarity are inversely proportional to each other.

Now, in terms of acidity, we can then say that the more polar a bond is, the weaker it is, and that consequently makes it all the more acidic (as acidity involves the dissociation of a proton).

 

[boaz]

Acidity actually does not depend on polarity. H-F is more polar than H-I yet H-I is a MUCH stronger acid. Acidity has only to do with relative stabilities of products vs. reactants. Since I- is much more stable than F-, delta-G for the HI acid/base reaction is much more negative than that for HF.
Same goes for electron withdrawing groups on a carbon alpha to COOH. The product of the acid/base rxn is COO-. If there is an electronegative group nearby, it can withdraw some of the negative charge, spreading it out, and thus stabilizing it. This means it is a relatively more stable prduct than it would be without electron withdrawing group.
You should know this from orgo.

 

[G1SG2]

Acidity actually does not depend on polarity. H-F is more polar than H-I yet H-I is a MUCH stronger acid. Acidity has only to do with relative stabilities of products vs. reactants. Since I- is much more stable than F-, delta-G for the HI acid/base reaction is much more negative than that for HF.
Same goes for electron withdrawing groups on a carbon alpha to COOH. The product of the acid/base rxn is COO-. If there is an electronegative group nearby, it can withdraw some of the negative charge, spreading it out, and thus stabilizing it. This means it is a relatively more stable prduct than it would be without electron withdrawing group.
You should know this from orgo.

No, acidity does have to do with polarity, although it may be relatively less important when there are other factors to be taken into consideration.

For example, suppose we're comparing the relative acidity of the following compounds:

H-CH3 (pka 48), H-NH2 (pka 38), H-OH (pka 15.7), H-F (pka 3.2).

Clearly, HF is a stronger acid than CH3-H. The reaction H-A-->H+ + A- requires the breaking of the H-A bond. The MORE POLARIZED this bond is due to differences in electronegativity, the WEAKER it is, and thus the STRONGER the acid.

In the situation you mentioned, stability of the conjugate base overrides polarity (I- distributing it's negative charge over a larger surface area than F-). Stability of the conjugate base makes the acid stronger, which is why HI is a stronger acid than HF.

Bond polarity is definitely important to consider when comparing the relative acidity of compounds, especially when comparing elements within the same period. But like I said, its relatively less important in the face of other factors (such as stability of the conjugate base).

A good rule of thumb would be to consider polarity when comparing compounds with atoms in the same period/row in the periodic table. If they are in different periods/rows (as in HF and HI), then consider the atomic/ionic radii in terms of stability of the conjugate base.

 

[boaz]

Don't confuse polarity with acidity. It just so happens that in general polarity is correlated with acidity but it is not the reason for acidity. Acidity is a qualitative way of talking about magnitude of reaction constant, K. The larger K is, the more acidic it is. K, in turn, depends upon delta-G. Delta-G depends upon relative thermodynamic stability of products vs. reactants. Picture the typical reaction diagram. Delta-G does NOT depend upon bond polarity. A polar bond might be a good indicator that the product will be able to stabilize the negative charge, and it might tell us something about the kinetics of the reaction, but nowhere does it come into the calculations of delta-G.

 

[G1SG2]

Don't confuse polarity with acidity. It just so happens that in general polarity is correlated with acidity but it is not the reason for acidity. Acidity is a qualitative way of talking about magnitude of reaction constant, K. The larger K is, the more acidic it is. K, in turn, depends upon delta-G. Delta-G depends upon relative thermodynamic stability of products vs. reactants. Picture the typical reaction diagram. Delta-G does NOT depend upon bond polarity. A polar bond might be a good indicator that the product will be able to stabilize the negative charge, and it might tell us something about the kinetics of the reaction, but nowhere does it come into the calculations of delta-G.

Like I said, bond polarity comes into play when one is considering the relative dissociation/acidity of the compounds, but is relatively less important when considering other factors. But it certainly does matter and can help in solving problems. You're correct when you say it's not the reason for acidity, of course. But it matters.

 

[boaz]

If you look back at the OP's opening remark, you'll notice that it's the bond polarity thing that was confusing him/her. I also used to have that problem until I realized that bond polarity is basically irrelevant.

I recommend the OP completely disregard bond polarity when comparing acidity and just focus on stability of the products.

 

[G1SG2]

If you look back at the OP's opening remark, you'll notice that it's the bond polarity thing that was confusing him/her. I also used to have that problem until I realized that bond polarity is basically irrelevant.

I recommend the OP completely disregard bond polarity when comparing acidity and just focus on stability of the products.

And if you look back a few posts, you'll see that the OP wanted to know how bond strength and ultimately acidity is affected by polarity.

OP, don't disregard polarity. I believe you should be smart when considering different factors, and you should know when one thing is more important than the other (for instance, in HI vs HF, stability of the conjugate base outweighs polarity, but as you can see in the example I provided you with, polarity/electronegativity clearly does matter). But, to each his/her own, of course.