The presence of C=O make the molecule more polar, therefore the interaction between molecules will increase, in other hand the resonance in molecule also increased due to increase the conjugated system by carbonyl, and that make molecule more stable, hence the molecule needs more energy to boil. that's why the Tp is higher.

The structures of the 4 compounds is shown in the attachment. Yes, the -CHO group is polar & usually enhances dipole-dipole interaction but the increase in the boiling points is huge (130.4 for the first ; 88 for the second). Had it been between 20& 50 degrees, I would not have raised my question !

The polarity and the increasing of conjugation act together to increase the b.p

The 4 compounds are aromatic & each has 6 delocalized electrons inside the ring. You know that the Kekule structure of benzene is not the actual one & this fact applies here. If there is conjugation, it will have two effects: i) more stability. ii) ease of detection by U.V.

However, the point emphasized here, is that the increase of Tb between each compound & its derivative is "huge". There must be a reason for this "huge" increase.

I think initially polarization is not as much as in the derivative, initially the molecular is more symmetrical (probably C2V point group) but with carbonyl in, the symmetry is changed form achiral to chiral point group and now its C1h i think and this may be another comparison in addition to S A Abed's justification that is also logical to me.

Molecular symmetry has an effect on the melting or freezing point; more symmetry ==> higher Tm or Tf. The question is about Tbs. Moving from furan to 2-furaldehyde has decreased symmetry & no chirality center has formed. The same applies to moving from pyrrole to 2-pyrrolecarbaldehyde.

I think that you agree with me that something "dramatic" has occurred to the structures upon these moves to account for the "huge" increases of the Tbs. So, what is this dramatic change ?

I wonder if the large increase in bp may have something to do with the tendency of aldehydes to reversibly polymerise to form chains and paraldehyde type rings? This behaviour is particularly well known for formaldehyde and acetaldehyde but also occurs to varying extents in other aldehydes. The process seems to be catalysed by trace amounts of water.

may be this type of polymeric interactions(Stephen Lindell answer) leading to this dramatic change/rise

Dear All,

I kept thinking about this question & in the last two days I got this idea:

Furan & pyrrole each has 3 resonance structures which contribute equally to the actual resonance hybrid structure of each one (similar to benzene). When the carbonyl group is attached to position (2) on each ring, the group will enter into resonance with the ring  adding a 4th resonance structure, at least. Now, this 4th resonance structure is bound to have full ionic positive & negative charges on C & O. Therefore, this will produce 2 new resonance hybrid structures which are far more stable than those of furan alone & pyrrole alone. An ionic character in the resonance hybrid structures will increase the boiling points dramatically.

I wonder if my idea is correct or not. I hope to see one or more comments so as I know if I am thinking in the right direction or not. 

I find the resonance argument very intriguing. If this argument was true, shouldn´t the dihydro- and tetrahydrofurans behave differently? The bp data (below) suggests, however, that a similar effect is operating in these non-aromatic compounds:

Furan 31°C; 2-ethylfuran 96°C; 2-furaldehyde 162°C; 2-furanmethanol 170°C

Dihydrofuran 54°C; 4,5-dihydro-2-ethylfuran 125°C; 4,5-dihydro-2-furancarboxaldehyde 196°C; 4,5 dihydro-2-furanmethanol 200°C

Tetrahydrofuran 65°C; 2-ethyltetrahydrofuran 109°C; tetrahydrofuran-2-carboxaldehyde 161°C; tetrahydrofuran-2.methanol 178°C

In the case of 2-pyrrolecarboxaldehyde, the presence of the N-H and C=O moieties enable intermolecular H-bonding. I suspect that this effect alone may be enough to explain the observed increase in bp?

In the absence of water, no H-bonding is possible in the case of 2-furaldehyde but the C=O bond will be significantly polarised. Perhaps the intermolecular dipolar interaction between the delta-negative O and delta-positive C is enough to explain the observed increase in bp?

 Dear Dr.Stephen Lindell ,  I am really honored by your participation in 3 valuable contributions :

 As for furan & its derivatives, they are aromatic compounds as you know & one may assume resonance between the aromatic ring & the aldehyde or methanol on position (2) which made +ve & -ve charges more prominent (i.e. more than partial).

Dihydrofuran & its derivatives and tetrahydrofuran & its derivatives are definitely no-aromatic. The sharp increases of the boiling points on moving from the "parent" compound to its derivatives is amazing & very intriguing, indeed. This will give me a question about which I shall ponder a lot.

Your second contribution is marvelous. In pyrrole, there is intermolecular H-bonding involving (H..N-H) & in 2-pyrrolecarboxaldehyde there are 2 types of intermolecular H-bonding involving (H..N-H) & (C=O..H-N) which will lead to good increase of boiling point.

Your third contribution is magnificent. Elaborating on your argument: The C=O is highly polarized so there is an enriched delta (+) & delta(-) resulting in more than a polar character (i.e. a character close to ionic but not full ionic).

I think you are the first to give the best answer. No wonder, you work in the giant "Bayer" company as I always describe it to my students in the industrial chemistry course.

 Dear Dr. Matar, Thank you for your kind comments and for posting such an interesting question. The unusually high bp of 2-furaldehyde remains surprising to me. It is only 8°C less than that of  2-furanmethanol, where strong H-bonding interactions are possible.

I think, the incease of boiling points of those heterocycles contaning five membered ring system is firstly due to the hydrogen bonds formed between N-H, and C=O group.