Suggest explanations for each of the following observations: a. The equilibrium constant for cya

step 1 Hello everyone. Today we're doing chapter 11 Palm 66. And this poem asks us to draw stepwise mec

Suggest explanations for each of the following observations:...

Key Concepts

Substituent Effects in Nucleophilic Addition Reactions

This concept deals with how different groups attached to a carbonyl compound influence both its electrophilicity and the stability of the addition product. Electron?donating or withdrawing substituents, as well as steric hindrance, can alter the favorability of nucleophilic attack and the resulting adduct’s energy, thereby affecting equilibrium constants such as in cyanohydrin formation.

Neighboring Group Participation in Reaction Mechanisms

Neighboring group participation refers to the assistance provided by an adjacent functional group that can stabilize a developing charge or provide an alternative reaction pathway, thus accelerating a reaction. This effect is particularly significant in processes like ester hydrolysis where intramolecular interactions can lead to a dramatically increased reaction rate.

Electronic Effects and Leaving Group Ability in Elimination Reactions

This concept involves the role of electronic distribution and the intrinsic stability of the leaving group in determining the rate of an elimination process. The ease with which a leaving group departs, and the stabilization of the developing negative charge in the transition state, are key factors that influence the overall speed of base?catalyzed elimination reactions.

Hydration Equilibria and Carbonyl Oxygen Exchange Mechanisms

The rate of isotopic exchange of a carbonyl oxygen often depends on the reversible hydration of the carbonyl group. Factors such as resonance stabilization, conjugation, and the electrophilicity of the carbonyl carbon affect the formation and stability of the hydrate intermediate, thereby modulating the exchange rate of the oxygen atom in processes like isotopic labeling.

Transcript

0:00 Hello everyone.

00:01 Today we're doing chapter 11 palm 66.

00:04 And this poem asks us to draw stepwise mechanism from these following two reactions and explain why the more stable 2 butine is isomerized to less stable 1 butine.

00:15 But under the similar conditions, we get the 2 .5 diomethal 3 hexyl forming a 2 .5 diomethyme.

00:22 2 .3 hexadeene.

00:24 So we don't actually isomerize the alkyne.

00:27 We actually go down to two different.

00:30 Alken bonds.

00:31 So let's go over the mechanism for the first one.

00:33 So from the first problem here we see that we're reacting with potassium amy so potassium is just a counter ion doesn't really do anything but the real key here is the nh2 which is a strong base and some sort of equilibrium with ammonia.

00:57 So we see here that now what would we have here? well we have carbon with four bonds carbon with four bonds this carbon has some protons so we know that bases can take away protons so let's do that let's use this base to take away one of these protons making a bond you make a bond you make a bond you break a bond and those sigma bond electrons will go to form a carb an ion so we formed a carb anion negative charge we have still two protons on this carbon but carbon is not happy to have electrons not happy to to be negatively charged, it would like to be neutral.

01:42 So we will use those two electrons to form a new pie bond, making a bond.

01:45 When you make a bond, you break a bond.

01:47 So one of these pie bonds, all this alkyne will break to form a new carbon out.

02:00 Now we have another alkyne right next to it.

02:04 Then we have those two electrons and formal negative charge here.

02:07 So once again, carbon is not happy to be negatively charged.

02:11 And if you remember, when we first used our base to pick up a proton, we still have that base with its proton.

02:17 So it's conjugate acid.

02:20 It's actually a neutral now.

02:23 So it's nh2 with h.

02:24 So it's ammonia.

02:26 So this carbon ion can actually pick up that proton from our conjugate acid to reform our basic catalyst.

02:32 So remember you always have to reform your catalyst.

02:34 So we know that somehow we would have to reform nh2.

02:42 And now we reform our basic catalyst as well.

02:45 So we can do this once over again.

02:49 So this base can pick up this proton making a new carb anion.

03:03 So now this carbon has two electrons here, but remember carbon doesn't like to be anionic, so it's going to make a new double bond here with his lone pair electrons.

03:16 When you make a bond, you break a bond.

03:17 So now you made a new carb anion.

03:22 So let me draw that over here.

03:25 So now you made three bonds here, and now you reduce that to alkan.

03:32 And then this has formal negative charge.

03:36 And then you also pronated your base to its congeate acid.

03:40 So we know that this anionic carbon can pick up that proton to reform your basic catalyst.

03:47 So you always have to reform your basic catalyst.

03:50 And now we form the product in the question stem...