Question

Print CalculatorPeriodic Table Question 4 of 8 presented by saptinig Learring Complete the mechanism for the acid-catalyzed racemization of the ketone below by adding any missing atoms, bonds, charges, nonbonding electron pairs, and curved arrows (forward reaction only). Hint Previous & Give Up & View Solution Check Answer NextExit

Answer 1

The complete mechanism is shown below:

Firstly, we have the ketone in an acidic medium (due to the presence of the hydronium ion), which acts as a base (because its oxygen from the keto group C=O, is an electronegative element, with free electron pairs), deprotonating an acidic hydrogen from the hydronium ion, being positively charged, said oxygen, while the electron pair of the deprotonated acidic hydrogen, is located in the oxygen atom of the hydronium ion and stabilizes it. After that, a resonance movement occurs, where the protonated oxygen atom of the ketone will be stabilized by the electron pair pi, since this charge makes it unstable; and in this way, it stabilizes. Simply, the dipole moment moves towards it, because it is more electronegative than carbon. Once this occurs, the carbon atom of the keto group is positively charged (carbocation), and the hydrogen atom adjacent to it (which forms the sigma sp₃-s bond with the carbon atom adjacent to the carbocation), goes to seek to stabilize this positive charge by hyperconjugation (approaching it); where said charge causes this compound to behave like an acid; while the water molecule (because it is amphoteric), acts as a base, deprotonating said adjacent hydrogen so that the sigma bond forms a pi bond with the carbocation and stabilizes it; and for this reason, the water molecule remains protonated. Next, the pair of electrons of the pi bond will be located in the carbon from which it came, in such a way that a pair of electrons of oxygen stabilizes the electronic deficiency that is generated by the previous displacement (the oxygen being positively charged); and the pair of electrons that was located in the carbon, will deprotonate a hydrogen acid of the hydronium ion formed in the previous step, generating a sigma carbon hydrogen bond. Now, this protonated oxygen generates an acidic behavior of the species, so that its bonding hydrogen will be deprotonated by water, which acts as a base, so that finally the electron pair will be located in the oxygen and stabilize; and the water will be protonated, leaving the ketone that we had initially in the acid medium. In the image, all electron movements, bonds, breaks, free electron pairs, formal charges (positive and negative), Lowry-Bronsted acid-base attacks and the missing stretch of the ketone that was absent in the steps are shown end of the mechanism.