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Identify the most acidic hydrogen atom in the compound below:

$$\ce{NC-CH2-CH2-CH2-OH}$$

My teacher said that the $\ce{H}$ connected to $\ce{O}$ is the most acidic since $\ce{O}$ is more electronegative than carbon.

However, I see that the $\ce{H}$ connected to the carbon adjacent to cyanide must be more acidic since the conjugate base is stabilized by resonance, contrarily to that connected to $\ce{O}$.

Isn't resonance the most significant criterion in determining the stability of a conjugate base? Or does the electronegativity of the donor atom matter more?

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    $\begingroup$ Electronegativity nearly always matters more, unless there is a lot of resonance (think malonates). My other suggestion is to stop thinking that one can simplify it into an "X matters more than Y" or "Z is the most important thing" situation that you can use to predict acidities. There will always be exceptions, and the only way to figure it out is to look up the data. In this case, alcohols have pKa ~15, and nitriles ~30. Thus alcohols are more acidic. You can figure out your own rules once you have enough experience with the hard numbers to know what's right and wrong. Not before. $\endgroup$ – orthocresol Sep 18 at 16:22

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