[shalikadm]

We learnt that when the polarity of C=O bond increases the magnitude  of intermolecular forces increases so the boiling point increases. As there the polarity is high the magnitude of +δ charge residing on the carbonyl carbon is high. So the reactivity towards neucleophiles such as CN^- increases.

Here's the diagram taken from my note where the reactivity toward CN^- is explained.

It shows that the electron repellant effect by methyl group reduces the polarity of C=O bond which reduces the reactivity towards  CN^-.
But in the section where there is reasoning for BPs , it has showed that the BP of ethanal is greater than that of methanal as the
the polarity of C=O bond increases in ethanal because of electron repellant effect of the methyl group

Why does that electron repellant effect of methyl group decrease the C=O bond polarity in one place and increase the C=O bond polarity in another place...?

[orgopete]

Why does that electron repellant effect of methyl group decrease the C=O bond polarity in one place and increase the C=O bond polarity in another place...?

While that appears to be a contradiction, it assumes the boiling point is solely a function of the C=O bond polarity. That isn't necessarily true, after all, ethane has a higher boiling point than methane. If the interactions created by the methyl group are significant, then they may result in acetaldehyde having a higher bp than formaldehyde while formaldehyde is still the most reactive.

[shalikadm]

While that appears to be a contradiction, it assumes the boiling point is solely a function of the C=O bond polarity. That isn't necessarily true, after all, ethane has a higher boiling point than methane. If the interactions created by the methyl group are significant, then they may result in acetaldehyde having a higher bp than formaldehyde while formaldehyde is still the most reactive.

isn't there a connection between more polar C=O bond and bigger +δ charge residing on carbon?..i got to think like that seeing diagrams

my note describes this increment in BP is mainly because of the polarity of bond rather than molecular weight..this is a part of the note where it describes the effect of C=O bond's polarity.another example given is that acetone having a higher BP than propanal though they are same in M but because electron repelling effect is done by two alkyl groups in acetone which there by increase the polarity of the C=O
which leads to strong intermolecular forces..

do you also find a contradiction here ? is it  a contradiction that everyone find in chemistry ?

still not clear why alkyl group increases the bond polarity when BP is explained and decreases bond polarity when the reactivity of neucleophiles towards the carbonyl carbon is explained..

pardon me if i haven't goy you correctly..please explain this to me..

thanks so much!

[orgopete]

my note describes this increment in BP is mainly because of the polarity of bond rather than molecular weight..this is a part of the note where it describes the effect of C=O bond's polarity.another example given is that acetone having a higher BP than propanal though they are same in M but because electron repelling effect is done by two alkyl groups in acetone which there by increase the polarity of the C=O which leads to strong intermolecular forces..

I don't agree with this. I think the aldehyde C=O is more polar and more reactive than a ketone. I can understand how your notes may say a methyl group would increase the bond polarity in an effort to explain the boiling point only on bond polarity. However, that leads to the apparent contradiction being noted.

still not clear why alkyl group increases the bond polarity when BP is explained and decreases bond polarity when the reactivity of neucleophiles towards the carbonyl carbon is explained..

As just noted, I don't agree that the alkyl groups increase carbonyl polarity for bp and decreases it for reactivity. It is my opinion that the weak intermolecular forces are not well understood. For example, explain what London forces are and how have they been proven? Let us assume for the moment that a methyl group resulted in an increase in a London-type force and not a paradoxical reversal in the bond polarity effect needed to explain the bp. If so, then the same London-type factors that led to the increased bp for acetaldehyde compared to formaldehyde could be repeated in acetone, with two methyl groups. For example, if the methyl groups are proton donors in a London force, they may be expected to have an increased polarity alpha to a carbonyl group compared to the terminal methyl group in propionaldehyde. If so, then acetone could double this effect and further increase the bp.

Note, that I am trying to explain the apparent contradiction. If your book or professor has given the opposite and contradictory effect (and you must use it in your exams), I don't agree with it.

[shalikadm]

Note, that I am trying to explain the apparent contradiction. If your book or professor has given the opposite and contradictory effect (and you must use it in your exams), I don't agree with it.

thank you very much sir..my tutor may have given such explaination because the real phenomenon is beyond our exam level..he may have given the explaination that we must write at our exam which we will get full marks with no risk at all !

[Corribus]

I don't agree with this. I think the aldehyde C=O is more polar and more reactive than a ketone. I can understand how your notes may say a methyl group would increase the bond polarity in an effort to explain the boiling point only on bond polarity.

I always like to turn to real data in these kinds of situations and in this case unfortunately the data does not support your conjecture.

Important point: the permanent dipole moment of acetone is 2.91 D and that of propanal is 2.52 D.  These are fairly consistent for all ketone and aldehydes, respectively.  (For sake of completeness, that of formaldehyde - a "double aldehyde" is 2.29 D, even less polar.)  Probably the most likely cause of this is that a ketone has two fairly rich electron donating aliphatic groups, whereas an aldehyde only has one (and formaldehyde zero).  True there are an equivalent number of CH_n groups on both acetone and propanal, but really it's the number of CHn groups directly adjacent to the carbonyl that count - i.e., and ethyl and methyl have fairly similar electron donating capacity because the second carbon out is too far away to have much of an inductive effect.  Anyway, the long and short of it is that ketones have higher dipole moments, therefore they are more polar, therefore they will tend - all other things being equal - to have higher boiling points (e.g., boiling point of propanal is 46-50 °C and that of acetone is 56 °C).  While all the electrostatic and intermolecular forces will play a role in the boiling point, permanent dipole-dipole interactions are far stronger than London or van der Waals forces, and so any minor differences in these secondary effects in acetone vs. propanal are far outweighed by the large difference in permanent dipole moment between these two species.  Obviously, if you start comparing propanal to something like 2-octanone, where the sheer number of van der Waals forces starts to become very important, then it's no longer a fair comparison.

I think the reason for the apparent contradiction in the OP's post is because he is forgetting that while boiling point is strictly a function of electrostatic forces, reactivity is a function of electrostatic forces and steric considerations.  Ketones simply have less room for reaction, regardless of the electrostatic favorability of a reaction - and formaldehyde, far more reactive than either aldehydes or ketones, have the most room.  In addition, the dipole moment mentioned above is for the whole molecule, not just the carbonyl, which can easily become a point of confusion.  While the presence of two electron donating groups in a ketone paves the way for a large overall molecular dipole moment, it also appears to reduce the positive charge located specifically on the carbonyl carbon.* This would again mean that ketones are less reactive than aldehydes, because reactivity relates to electrostatic forces at a specific point, whereas boiling point considers the electrostatic forces over the entire molecule (and in between them). 

So, to sum, I don't really see a contradiction here.  It's just requires a deeper explanation. 
 
* Now my turn for conjecture.  I puzzled over some other data for a while trying to explain why ketones have higher dipole moments than aldehydes.  I first turned to vibrational frequencies of carbonyls, and aldehydes typically vibrate at lower frequencies than ketones (1730 wavenumber vs. 1715 wavenumber for propanal and acetone).  This might tell us something about the length of the bonds, and hence the polarity, but I came to the realization that it's not helpful because the mass attached to the carbonyl also makes a difference, so that is a confounding factor.  I next turned to bond energies, which shouldn't really depend on the masses at either end of the carbonyl, and found that for the three classes of carbonyls, BDEs are typically in the range of ~712, ~733 and ~754 kJ/mol for formaldehyde, aldehyde and ketone, respectively.  My supposition would be that this translates into a stronger, shorter bond for a ketone versus a longer, weaker bond for aldehyde (and longest, weakest for formaldehyde), which in turn means less polar carbonyl in ketones and more polar carbonyl in aldehydes. Or at least, more pi-bond character in ketone and less in aldehydes.  Contradiction, you might say!  I did, too.  But after thinking about it a while, maybe not.  It is, after all compatible with the idea of extra electron density being fed into the carbonyl carbon of ketones.  This lowers the polarity of the carbonyl itself (because the carbon is less positive) but increases the overall dipole moment because that positive charge has to go somewhere: it is distributed (to some degree) onto the electron donating methyl groups (in the case of acetone).  Dipole moment, after all, is related both to the magnitude of the positive and negative charges - and their degree of separation.  So in the end, orgopete, you are probably right that aldehydes have more polar carbonyls, but ketones, overall, have larger dipole moments, because the charge is separated over large distances.  That's a hand-wavy argument on my part that may not explain everything (if indeed it explains anything - probably it is vulnerable to assault by a determined individual), which is why I left it as a footnote.

[shalikadm]

This means a lot to me. Now I have an idea on how the electron density on carbonyl carbon reduces at the same time keeping a more polar C=O bond in acetone. Thanks a lot Corribus taking time to write a long explaination that solved contradions !

[orgopete]

I always like to turn to real data in these kinds of situations and in this case unfortunately the data does not support your [PW's] conjecture.

Important point: the permanent dipole moment of acetone is 2.91 D and that of propanal is 2.52 D.  These are fairly consistent for all ketone and aldehydes, respectively.  (For sake of completeness, that of formaldehyde - a "double aldehyde" is 2.29 D, even less polar.) 

Let's keep this simple, I was wrong. I didn't even think of using the dipole moments.

I think the reason for the apparent contradiction in the OP's post is because he is forgetting that while boiling point is strictly a function of electrostatic forces, reactivity is a function of electrostatic forces and steric considerations. ... In addition, the dipole moment mentioned above is for the whole molecule, not just the carbonyl, which can easily become a point of confusion.  While the presence of two electron donating groups in a ketone paves the way for a large overall molecular dipole moment, it also appears to reduce the positive charge located specifically on the carbonyl carbon.*
So, to sum, I don't really see a contradiction here.  It's just requires a deeper explanation. 

* Now my turn for conjecture.  I puzzled over some other data for a while trying to explain why ketones have higher dipole moments than aldehydes.  I first turned to vibrational frequencies of carbonyls, and aldehydes typically vibrate at lower frequencies than ketones (1730 wavenumber vs. 1715 wavenumber for propanal and acetone). 

... It is, after all compatible with the idea of extra electron density being fed into the carbonyl carbon of ketones.  This lowers the polarity of the carbonyl itself (because the carbon is less positive) but increases the overall dipole moment because that positive charge has to go somewhere: it is distributed (to some degree) onto the electron donating methyl groups (in the case of acetone).  Dipole moment, after all, is related both to the magnitude of the positive and negative charges - and their degree of separation.  So in the end, orgopete, you are probably right that aldehydes have more polar carbonyls, but ketones, overall, have larger dipole moments, because the charge is separated over large distances.  That's a hand-wavy argument on my part that may not explain everything (if indeed it explains anything - probably it is vulnerable to assault by a determined individual), which is why I left it as a footnote.

Since I was wrong, it is difficult to add anything. I admit I was looking rather narrowly at the effective charge of the carbonyl carbon aspect and thinking the greater the charge, the greater the reactivity. If so, then it becomes predictable that electron donors should decrease reactivity. I had overlooked the simple fact that a compound like dimethylformamide has a very effective electron donor nitrogen. The nitrogen reduces DMF's carbonyl reactivity and the donation increases the molecular dipole and bp (compared to an aldehyde like isobutyraldehyde).

[TheOrganic]

* Now my turn for conjecture.  I puzzled over some other data for a while trying to explain why ketones have higher dipole moments than aldehydes.  I first turned to vibrational frequencies of carbonyls, and aldehydes typically vibrate at lower frequencies than ketones (1730 wavenumber vs. 1715 wavenumber for propanal and acetone).  This might tell us something about the length of the bonds, and hence the polarity, but I came to the realization that it's not helpful because the mass attached to the carbonyl also makes a difference, so that is a confounding factor.  I next turned to bond energies, which shouldn't really depend on the masses at either end of the carbonyl, and found that for the three classes of carbonyls, BDEs are typically in the range of ~712, ~733 and ~754 kJ/mol for formaldehyde, aldehyde and ketone, respectively.  My supposition would be that this translates into a stronger, shorter bond for a ketone versus a longer, weaker bond for aldehyde (and longest, weakest for formaldehyde), which in turn means less polar carbonyl in ketones and more polar carbonyl in aldehydes. Or at least, more pi-bond character in ketone and less in aldehydes.  Contradiction, you might say!  I did, too.  But after thinking about it a while, maybe not.  It is, after all compatible with the idea of extra electron density being fed into the carbonyl carbon of ketones.  This lowers the polarity of the carbonyl itself (because the carbon is less positive) but increases the overall dipole moment because that positive charge has to go somewhere: it is distributed (to some degree) onto the electron donating methyl groups (in the case of acetone).  Dipole moment, after all, is related both to the magnitude of the positive and negative charges - and their degree of separation.  So in the end, orgopete, you are probably right that aldehydes have more polar carbonyls, but ketones, overall, have larger dipole moments, because the charge is separated over large distances.  That's a hand-wavy argument on my part that may not explain everything (if indeed it explains anything - probably it is vulnerable to assault by a determined individual), which is why I left it as a footnote.

I agree to most of your post except the footnote. I do not agree to the part I have highlighted above.

A higher Bond Dissociation Energy ( Homolytic ) might not correspond to a "Stronger" bond between the atoms. Do you think the bond energy excludes the intermolecular forces ? I do not think it does, and in fact the intermolecular bonds will break first, and then the actual bonds between two atoms. You cannot simply have a single molecule and measure it's bond dissociation enthalpy, and hence you cannot really get an idea of how strong the bond actually is from bond dissociation enthalpy of different bonds.

To me, the order of bond dissociation energy given seems consistent, even at first look. As noted by you in the second paragraph of your post, the dipole moment of acetone is greater than propanal. A greater dipole would correspond to stronger intermolecular forces- and for a cluster of molecules - this would result in an overall higher bond dissociation energy. In short, a higher bond dissociation energy results from greater intermolecular forces.

[TheOrganic]

My previous reply was incomplete.

The point that I wanted to make was, simply put: A higher bond dissociation energy does not necessarily correspond to a stronger bond, reason being the intermolecular forces.

However, having higher dipole is certainly not the only criteria for greater intermolecular forces.
Chloroform has a greater dipole than Carbon tetrachloride, but chloroform has a lesser boiling point than Carbon tetrachloride. A higher dipole might result in a stronger dipole-dipole intermolecular bonding. However, in case of CCl₄ and CHCl₃, I think the (net effect of) dispersion forces in CCl₄ are far greater than the dipole dipole interaction in CHCl₃ , which corresponds to the higher B.P. of CCl₄. In fact, Dispersion forces have far greater contribution to intermolecular bonding, than dipole-dipole interactions.