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You have Propanoic Acid ($\ce{CH3CH2CO-O-H}$) so an $\ce{OH}$ on the end You have Ethanol ($\ce{CH3CH2-OH}$) so an $\ce{OH}$ on the end

When you combine the two for a condensation reaction you have the $\ce{OH}$ group from 1 reactant and the $\ce H$ group from the other that joins to release a water molecule correct?

So which of these groups loses its $\ce{OH}$ group and which group loses its $\ce H$ group?

How can i determine this is im guessing it may have something to do with strength of bonding and shielding but can somebody please explain i cant find the answer

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Acid loses its OH and alcohol loses its H, that's the way it is. Esterification mechanism

Long story short, oxygen from alcohol's $\ce{-OH}$ attacks the carbon from the acid's $\ce{-COOH}$. That carbon, being linked to two oxygens, carries significant positive charge, so this is much more likely than the other way around.

Also, it has been proven rigorously with the use of isotope-labeled compounds. Say, you take some acid with "normal" oxygen ($^{16}\ce O$) and alcohol with $^{18}\ce O$, and behold, the ester comes out with one $^{16}\ce O$ and one $^{18}\ce O$, so the latter must have come from alcohol, so it is the oxygen from acid that was lost in the form of water.

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