Which proton is more acidic, the one on the left or right?

 

My answer would be the proton attached to the N because N is more electronegative than the carbon that the circled H is attached to.

 

I get that the negative charge on the conjugate base (after removing the H on the carbon) is found in an sp orbital, which stabilizes the negative charge more than an sp3 orbital, but isn't it more important that the proton is attached to the more electronegative atom?

 

This question is taken from David Klein's Orgo as a Second Language book.

Edited January 7, 201411 yr by nnath

u r talking about the bond or the nitrogen or protons of the H and N

With that picture, the question that was given to me was "which proton is more acidic?". I'm assuming they are referring to the H that is bonded to the carbon and the H that is bonded to the N. My answer is the H that is bonded to the N, but the correct answer is the H which is bonded to the carbon.

 

I'm asking why is this the case?

It has to do with the stability of the conjugate base and the s character of the terminal alkyne relative to other hydrocarbons. An sp hydridised carbon is said to have 50% character (compared to 33% for sp2 and 25% for sp3), which means that the sp orbital is closer to the nucleus where the electron pair in the orbital experienced more of a stabilising effect. Compared to the conjugate base of an alkene or an alkane, that of an alkyne is much more stable (and by quite large orders of magnitude if you look at their pKa's), making it more acidic.

 

Nitrogenous bases, which are quite strong bases, are typically used to deprotonate alkynes. That should tell you that the conjugate acid of a nitrogenous base would typically be less acidic than that of a terminal alkyne and indeed, I'm pretty sure the pKa's fall between the ranges for alkynes and alkenes.

Thanks hypervalent_iodine. Makes sense.