Four pairs of compounds are shown. In each pair, one of the co...

Question

Four pairs of compounds are shown. In each pair, one of the compounds reacts more quickly, or with a more favorable equilibrium constant, than the less conjugated system. In each case, explain the enhanced reactivity.
(d)

(d) Chemical structures comparing acidity, showing one compound with hydroxyl and carbonyl groups is more acidic than another.

Accepted Answer

All right. Hello everyone. So this question says that a pair of compound is shown below when treated with a base compound B reacts more quickly than compound A explain the enhanced acidity of the of compound B compared to compound A. And here we have provided the two structures of compounds A and B A is cyclops, 13 dien, 10 and B is two hydroxy 1677 A tetra hydro five hnd 50. So lets talk about acidity, right? Recall that the strength of an acid is to an extent linked to the conjugate base. Now, the conjugate base just to clarify is the species produced after the acid is dep proton. Now the more stable the conjugate base is the more acidic. That compound is going to be. This is because if the result of dep protonation is more stable, then it's going to be a lot easier to remove that proton. Now, one of the most common ways that a conjugate base is made stable is if it's somehow resonance stabilized. So for both A and B, let's go ahead and detonate the most acidic atom or excuse me, remove the most acidic proton, which would be the one attached to the oxygen of the hydroxy group. So lets start with a. So now I'm just going to go ahead and draw the conjugate base of compound A in which oxygen now has a negative charge after gaining an additional lone pair. Now, this lone pair of electrons is in close proximity to one of the alkenes or one of the P bonds inside of the ring itself. And so here, this means that they are in close enough proximity to each other or resonance to take place. Now, residents describes a movement of electrons from regions of higher electron density to regions of lower electron density. So one of the lone pairs on the oxygen atom that has a negative charge is going to establish a pie bond in between itself and carbon. And that's going to force the pi bond adjacent to it to become a lone pair of electrons on a carpet. So now that lone pair of electrons inside of the ring can move towards the other pyon. And from here, we've completed the set of resonance structures for the conjugate base of compound A. So now let's compare this to the conjugate base of B. Now, before we start drawing the resonance forms of the conjugate base of compound B, there's one thing I do want to point out here because notice how for the conjugate base of compound A in addition to the oxygen atom that had a negative charge. There were two pi bonds that are in conjugation with each other and also with the negatively charged oxygen conjugation. In this context means that there's one single bond separating everything from each other. If we consider the structure of the conjugate base of compound B, there are actually three P bonds this time around and all of them are in conjugation with each other, which means that there will be more resonant structures. So let's go ahead and draw them in much the same way as before the oxygen atom that has a negative charge right now is going to donate a lone pair of electrons and establish a pi bond between itself and carbon. That's going to push the pi bond closest to it to become a lone pair of electrons on carbon, specifically a carbon now inside of the ring. So now that lone pair of electrons on carbon can move towards the other pie bo inside of the ring and move that negative charge to a different carbon atom. So at this point, we once again have three unique resonant structures, but there's also an additional car bundle on the very left side of the structure. This means that one more resonant structure is possible because the low pair of electrons on a carbon atom of the six member ring can move towards the car bal. All right. So here, now that we have all of the resonance forms for both conjugate bases. There are two reasons as to why compound B is more acidic than compound A. This is because there are two reasons why the conjugate base of B is more stable than the conjugate base of A. First of all, there are more resonant structures for the conjugate base of B than there are for the conjugate base of A. And it's also worth mentioning that the conjugate base for compound B features a negative charge on two oxygen atoms which is highly electro negative recall that resonance forms where the negative charge is on an electron atom are more stable because that negative charge is better accommodated by the electron atom. And so with that being said, we've arrived at the final answer which reads as follows. Compound B has an enhanced acidity compared to compound A due to the additional resonance stabilization of its conjugate base. And so with that being said, thank you so very much for watching. And I hope you found this helpful.

Four pairs of compounds are shown. In each pair, one of the compounds reacts more quickly, or with a more favorable equilibrium constant, than the less conjugated system. In each case, explain the enhanced reactivity.
(d)

Key Concepts

Acidity and pKa

Resonance Stabilization

Inductive Effect

Watch next

Four pairs of compounds are shown. In each pair, one of the compounds reacts more quickly, or with a more favorable equilibrium constant, than the less conjugated system. In each case, explain the enhanced reactivity.(d)

Key Concepts

Acidity and pKa

Resonance Stabilization

Inductive Effect

Watch next

Four pairs of compounds are shown. In each pair, one of the compounds reacts more quickly, or with a more favorable equilibrium constant, than the less conjugated system. In each case, explain the enhanced reactivity.
(d)