[darkv0id]

Out of the keto-enol tatomers of acetoacetic acid ester, which has lower volatility?
I think it should be the enol form, considering the intermolecular H-bonding due to the presence of -OH groups in the enol form.

However, my textbook says quite the opposite. It says that the intramolecular H-bonding in the enol will limit intermolecular interaction, increasing the BP vis-a-vis  the keto form.

I'm rather confounded. Aren't electrostatic interactions between two charges (or partial charges) independent of their interactions with other charges?

[discodermolide]

Out of the keto-enol tatomers of acetoacetic acid ester, which has lower volatility?
I think it should be the enol form, considering the intermolecular H-bonding due to the presence of -OH groups in the enol form.

However, my textbook says quite the opposite. It says that the intramolecular H-bonding in the enol will limit intermolecular interaction, increasing the BP vis-a-vis  the keto form.

I'm rather confounded. Aren't electrostatic interactions between two charges (or partial charges) independent of their interactions with other charges?

They are tautomers and as such in equilibrium in the liquid phase and the gas phase. As such I do not think you can speak of a property such as volatility or indeed BPt. the keto form will always be in equilibrium with the enol tautomer.
Hypothetically speaking if the could exist separately then I might think that the enol had the higher BPt.

[orgopete]

I cannot verify whether the bp should indeed be lower or not. However, I did a search with ChemSpider.com for 2-butanone and 2-methoxypropene, the same molecular formula. Obviously, 2-methoxypropene cannot hydrogen bond, but it also will not have the polarity of a carbonyl group. 2-Butanone has a bp of ~80°C while methoxypropene is 36°C. I should assume that an intramolecular hydrogen bond should diminish intermolecular hydrogen bonding and thus decrease the boiling point.