Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: qw098 on April 03, 2011, 11:50:02 AM
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Hey Guys,
I am doing a problem on carbonyl reactivity. I have attached a PDF of my question. I am pretty sure about my answer a), b), c) and e) but I am unsure about d). How does carbonyl reactivity change if the carbonyl is in a cage? My thought is that the bigger the cage, the less the steric hinderance so the more reactive the carbonyl would be in nucleophilic addition?
Could you guys please take a look at d) and tell me if my rationale is correct? Meaning that the bigger cage would be of highest reactivity in nucleophilic addition?
Thanks!
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c) is not correct.
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How is c) not correct?
CF3 is an electron withdrawing group therefore the center molecule will be the most reactive because the electron density on the carbon will be pulled away making the carbon bonded to oxygen very delta positive therefore it will be the most reactive under nucleophilic addition.
Is my rationale incorrect?
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Your logic for c) is correct, but is there any reason why a CCl3 group would render a carbonyl more reactive than a CF3 group?
Also, e) is incorrect. For this just try drawing resonance forms where the substituent is involved and you'll see why.
For d), think about hybridization and bond angles.
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Your logic for c) is correct, but is there any reason why a CCl3 group would render a carbonyl more reactive than a CF3 group?
I could agree with the original poster on this one, but I don't KNOW the correct answer. The acidity of CX3H increases F<Cl<Br or I. CF3 groups don't fall off readily like a haloform reaction, so I could think the CCl3 group should be more reactive. Can anyone post data for this problem?
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CF3 is more EWing than CCl3 to alpha carbon but less so to itself like in CX3H.
in d) smaller cyclics are more nucleophilic like alkenes so cyclopropanone is most reactive
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CF3 is more EWing than CCl3 to alpha carbon but less so to itself like in CX3H.
Can you give me some data that shows this?
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pKa of CF3COOH is 0.23, pKa of CCl3COOH is 0.77.
Sort of shows that the CF3 group has a greater inductive effect.
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pKa of CF3COOH is 0.23, pKa of CCl3COOH is 0.77.
Sort of shows that the CF3 group has a greater inductive effect.
I concede fluoroacetic acids are the poster example for this argument. However, hydroxyacetic acid (3.83) is less acidic than iodoacetic acid (3.18) also. Does that mean iodine is a stronger electron withdrawing group than OH? If you compare Hammett sigma values, HX acidities, R3CX solvolysis, fluorine is less electron withdrawing than the other halogens. This is consistent with the acidity of the halomethanes I cited earlier. It appears that a general trend is developing consistent with why the haloform reaction fails with fluorine, namely because it is not as strong of an electron withdrawing group.
I also agree that the professor could have used the acidity argument to assume the trifluoroacetone to be more reactive than the trifluoro isomer. I would like to see would be some real data to determine whether it can be experimentally verified.
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I agree that fluorine on its own seems to have less of an electron-withdrawing effect than the heavier halogens. One explanation I was given is that (for aromatic systems at least), the pi overlap is much better with fluorine, so the inductive/resonance effects almost nearly cancel each other out. Personally I don't like this explanation much, but it's plausible I suppose.
The funny thing is, if you compare Hammett sigma values for CF3 vs CCl3, the trend is reversed:
CF3: σm = 0.43, σp = 0.54
CCl3: σm = 0.40, σp = 0.46
Granted the difference is not too significant, but it is there, particularly for σp
(Data taken from Chem. Rev. 1991, 165-195; relevant values are on p168, CF3 = entry 70, CCl3 = entry 66)
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@nox, good point. I cannot argue with data. Since these are the groups being discussed in the example, it may prove determinant.
I am not ready to concede the point completely however. I argue the haloform reaction or why CF3 (seemingly) is not lost as readily as the other halogens is due to the acidity of CX3H. The exchange rate is slowest for CF3H, (Hine, J., Burski, N.W., Hine, M., Langford, P.B.; J. Am. Chem. Soc.; 79, 1406 (1957)).
If anyone has followed my opinions on electronegativity, I completely disagree with the scientific merit of this concept. The fact that the reaction
HF :rarrow: H• + F•
is more energetic than predicted from the contributions of the homolytic cleavage of hydrogen and fluorine should make fluorine a better electron withdrawing group than chlorine, bromine, or iodine is wrong. I argue these are different reactions and different bonding elements that are being broken.
Attempts to justify the electronegativity arguments to other chemistry does lead to contradictory results. Although the homolytic bond strength of HF is greater then HCl, for example, the heterolytic bond strength is the opposite. They are just not the same reaction. I argue the heterolytic bond strength is an excellent measure of electron withdrawing properties. I argue that acidity measurements provide an excellent scale to measure heterolytic bond strengths or electron withdrawing properties. While fluorine is a fine electron withdrawing group, it is less electron withdrawing than the other halogens. I think this is born out in many reactions. I think if you were unaware of electronegativity and all of your knowledge came from the laboratory, being familiar with the reactivity of the different halo compounds, you would naturally think that iodide is the most electron withdrawing and fluorine the least.
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@nox, good point. I cannot argue with data. Since these are the groups being discussed in the example, it may prove determinant.
I am not ready to concede the point completely however. I argue the haloform reaction or why CF3 (seemingly) is not lost as readily as the other halogens is due to the acidity of CX3H. The exchange rate is slowest for CF3H, (Hine, J., Burski, N.W., Hine, M., Langford, P.B.; J. Am. Chem. Soc.; 79, 1406 (1957)).
If anyone has followed my opinions on electronegativity, I completely disagree with the scientific merit of this concept. The fact that the reaction
HF :rarrow: H• + F•
is more energetic than predicted from the contributions of the homolytic cleavage of hydrogen and fluorine should make fluorine a better electron withdrawing group than chlorine, bromine, or iodine is wrong. I argue these are different reactions and different bonding elements that are being broken.
Attempts to justify the electronegativity arguments to other chemistry does lead to contradictory results. Although the homolytic bond strength of HF is greater then HCl, for example, the heterolytic bond strength is the opposite. They are just not the same reaction. I argue the heterolytic bond strength is an excellent measure of electron withdrawing properties. I argue that acidity measurements provide an excellent scale to measure heterolytic bond strengths or electron withdrawing properties. While fluorine is a fine electron withdrawing group, it is less electron withdrawing than the other halogens. I think this is born out in many reactions. I think if you were unaware of electronegativity and all of your knowledge came from the laboratory, being familiar with the reactivity of the different halo compounds, you would naturally think that iodide is the most electron withdrawing and fluorine the least.
I would argue that the reason CF3 groups are not viable substrates for haloform reactions is not due to their diminished electronegativity in comparison with their other halogen counterparts but a function of their small size onto which they can distribute the negative charge. The formal negative charge lies on the carbon (in terms of CF3-) but the actual electron density lies in the halogens.
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I would argue that the reason CF3 groups are not viable substrates for haloform reactions is not due to their diminished electronegativity in comparison with their other halogen counterparts but a function of their small size onto which they can distribute the negative charge. The formal negative charge lies on the carbon (in terms of CF3-) but the actual electron density lies in the halogens.
I'm not sure I am understanding this. Even though CBr3H is a stronger acid than CF3H, CF3- is not ??.
The formal negative charge lies on the carbon? Formal charge? Protons are positive and electrons are negative, no? The actual electron density lies on the halogens?? Iodine is the most electron dense of the halogens, right? Fluorine is a gas. It is the lightest of the halogens. Its electrons extend the furtherest from the nuclear proton field, that is why fluorine holds its proton the strongest and it is the least acidic. Iodine is the opposite.
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I could be wrong posting this, as I am the least educated person here. I though another reason HF was a weak acid (HF being an analogue to the molecules you're speaking of) was because if it does adopt the negative charge, the high electronegativity and the relatively small size of its electron shell overall, concentrates the electron density to an unstable extent. with CF3- there isn't much hyperconjugation. Is it wrong to assume that even though the charge is technically on C, the fluorine extreme polarity will spread the charge densely over themselves leaving a positivish charge on C... Overall a rather unstable molecule. Br, although electronegative in its own respect, has a large shell such that even if it does pull the charge unto itself, the overwhelming electron density surrounding carbon will stabilize it. No postive effect due to polarity would hinder the stability of the ion.
Its a long shot idea, and I know that technically the charge should be on carbon, however, it seems to make sense to me, that fluorine is so electronegative that it might actually invoke and opposite effect on carbon? With the hydrogen bonded to carbon, there is at least some electrons stuck in place around C.
Out of the box idea, feel free to shred me!
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Again, I argue the problem arises by trying to explain why, when fluorine bonds are broken homolytically, they are much stronger than expected from the contributions of fluorine itself. It has been suggested by Pauling that bonds have covalent and ionic character and that it is fluorine's ionic character that increases its bond strength.
I do not agree with this. I do not believe this has been proven in any manner. If fluoride had high ionic content in a compound like HF, then its resistance to ionization virtually disproves itself. Pauling's calculations of electronegativity are not consistent with an additive principle of covalent and ionic content. He argues that opposite ionic content can only increase bond strength, yet it is well known that metal hydrides are weaker.
Except for the haloacetic acids, virtually all examples of electron withdrawing properties, acidity, reactivity, etc. point to the well known order I>Br>Cl>>F. Because of the Pauling electronegativity table, there are all sorts of rationalizations of why common sense fails to hold and how fluorine is not be a potent electron withdrawing group. To wit, the charge is more concentrated in fluorine, therefore it is a weaker acid. The bulk of atomic volume is the space occupied by the electrons. A low electron density (mostly empty) results in a low density atom. If the electrons are packed into a tighter space, that is more electrons per unit volume, it will have a higher electron density. If an electron's volume were constant, the density of atoms were vary with the neutron/proton ratio, or not very much. The reality is the nuclear charge of iodine is much greater than fluorine, its electrons are concentrated in a smaller volume per electron, it is a dense solid. Fluorine is a gas. If you took a measure, you would find more electrons in a smaller space around iodine. If electrons were negative, then the actual charge around iodine is larger. That isn't the entire story however. Iodine's nuclear proton field is larger and can repel a proton as well. The acidity is really a function of both fields, the local electron pair and the nuclear proton field. The closer iodine's electrons are to its nucleus, the closer and the more strongly will a proton feel the effects of the nuclear proton field as well.
I argue acidity is an inverse square result. Iodine, with more protons, pulls its electrons in more. Fluorine, with fewer protons does not. They reach out further. They react like a boxer, the one with the longer reach will hit you first. If you apply this simple principle, then HF, with its shorter bonds and electrons closer to its nucleus, is more acidic than H2O, and in turn more acidic than NH3 and more acidic than CH4. CH4 has the longest bonds. It has the fewest protons in its nucleus.
Acidity is a measure of heterolytic bond strength or about electron withdrawing properties. Electronegativity has been derived from bond strength data. It is actually a measure of homolytic bond strength. It is unfortunate that electronegativity arguments are as pervasive as they are. In my class, I simply divided electron withdrawing properties from electronegativity properties. You use electron withdrawing properties in the lab and electronegativity on the exam.
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Off topic
The people I would like to be critical of are the reviewers of Pauling's 1932 paper. We don't think about how something is thought to be true in science. Science is not a level playing field. Bigger reputations can get funding more easily than someone yet to prove themselves. A weak paper from a well regarded chemist can be accepted for publication more easily as well. However, to paraphrase myself, no matter how many times something is repeated, it will not be any more true than its original proof or proofs.
I think I have an explanation for the homolytic bond strength data for why fluorine bonds are stronger than expected. This is what Pauling was trying to achieve with electronegativity. However it becomes rather more complicated. It is in part a challenge to explain why the bonds of elemental fluorine as much weaker than expected. If you look at bond strengths, you will find the bonds are becoming stronger in the order I<Br<Cl and weaker in the order C>N>O. What is going to happen with fluorine and why? I just don't think it is because the electrons surrounding fluorine are more negative (ionic) than the electrons surrounding any other atom.
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Although I understand (I think) most of what you're saying orgopete, I do not quite get the comparison of the density of the electrons around the nucleus of Iodide vs Fluorine. I get the vast difference in number, and the greater positive charge pulling ect.... My question is what role/effects do the valence electrons play differently than the descending closed shells. Does the radius of the over all shell have any role to play? Overall this is some dense information to absorb, especially since I've always been taught that only the valence electrons really matter, with exception to the "lightly touched" idea of sheer electron density being able to stabilize a true charge just by being in proximity. I'm gonna pop out some books on the subject before I post again, I just don't quite get what it matters if in comparison ratio-wise iodide's electrons are held closer to the nucleus when they're in such a further out shell.... The true distance of bonding electrons is further away from the nucleus. Also what proportion does the positive field increase with proton density increase? Maybe the distance and positive increase at different rates??? allowing for a still predictable, however, less intuitive bond strength as electron mass increases proportionally to proton mass increase?
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I need to take a step back on electron density. This is what I had thought. If you take the bulk density of the halogens (solid or liquid forms) and adjust for the number of electrons, then it appears as though the electron density increases as you move down the periodic table. However, if you take an atomic radius and calculate the volume per electron, then the density is reasonably consistent. It doesn't matter whether it is fluorine or iodine.
In the atomic model that I have been thinking about, the atomic properties emanate rather more simply from Coulomb's Law. As the nuclear charge is increased, the proton field grows proportionally with the charge. As the distance increases, the field decreases with the inverse square of the distance. In this model, distance is more important than (formal) charge. That is, as one moves across the periodic table from C to F, the bonds become shorter, consistent with the nuclear charge effect. NH3 is more basic than H2O (or HF) because the electrons of oxygen are closer to the nucleus of oxygen. Ammonia is also more basic than chloride even though chloride has an excess of electrons. It isn't the net charge that matters because the charge of electrons is constant, minus one. The difference is the distance or rather the field of the electrons. The further they are from the nuclear proton field, the greater their attraction for protons. In this model, electrons have microscopic fields consistent with the orbitals they exist in. Their fields are dependent on the distance from the nucleus.
As a consequence, in this model, compounds like HF or H2O can form intermolecular bonds consistent with the tetrahedral orientation of their orbitals. Compounds like NaF are in cubic close packing crystals in which no orbital orientation is maintained and is consistent with a larger separation between the ions. This greater distance is also consistent with weak ionic bonds. In this case, weak and ionic are redundant. Bonds are the result of a Coulombic attraction and weak is a function of distance. When we refer to ionic bonds, we mean weak bonds. A greater separation seems rational as sodium and fluoride have the same number of electrons surrounding each nucleus. A Coulombic repulsion at short distances is present. However, a differential in charges creates an ionic attraction which in this case is qualitatively different from the short range attraction making covalent bonds.