Although 2-methyl-1,2-propanediol is an unsymmetrical vicinal ...

Question

Although 2-methyl-1,2-propanediol is an unsymmetrical vicinal diol, only one product is obtained when it is dehydrated in an acidic solution.

b. Why is only one product obtained?

Accepted Answer

Hey, everyone. And welcome back to another video. The dehydration of two third be 33 dimethyl butane 12 dial, a visceral dial in an acidic solution produces only one product. Why is there only one product? Now, first of all, let's draw the structure of that dial. It says that our parent is butane 12 dial. What we're going to do is just draw a four member chain because that's what butane is. And we understand that at positions one and two, we have two alcohol groups. So we're going to add one of them and then the other one in addition to that at position three, we have two metal groups. And finally at position number two, we also have a third beetle group. So we're going to add it as well. Ok. So we got the structure of our alcohol. We can actually redrew it a bit more neatly. So we can clearly see what sort of a dial we have. Let's make our carbon oxygen bond slightly longer. And now we understand that the given dial undergoes the dehydration reaction in an acidic solution. So we are using an acid catalyzed dehydration, which essentially requires us to use sulfuric acid and water. Now, let's remember that an acid catalyzed dehydration is an E one reaction. And because an elimination reaction, E one proceeds via the formation of a carbo ion, it's really important for us to have a stable tertiary carbo ion for an E one elimination. And if we consider the two alcohol groups, one of them is a primary alcohol because it's bonded to a primary carbon, the other one is a tertiary. So we're going to think about the protonation of the tertiary position or basically the tertiary alcohol, which essentially explains partially why there will be only one product. So the first reason is that we are only thinking about a tertiary alcohol, not the primary one, because we would form a very unstable carbo intermediate, which would be a primary carbo ion. If we think mechanistically in the first part of the mechanism, we observe the protonation step where sulfuric acid acts as a proton donor. So let's clearly indicate that stop. And now let's draw the proin intermediate so that we can think about the next step. Now we have two hygienes monitor to oxygen, which makes a very good living group. In the next step, we have the loss of the leaving group since we can form water. So let's draw the resultant stable tertiary carbo Ron intermediate, getting rid of the water molecule, we end up with a tertiary carbo Canion. In the next step, we identify our beta hydrogens. And there are only two beta gens which are equivalent to each other because they're bonded to the same carbon atom. And we're going to use water from the previous step, which will act as a base to form the required pi bond. So let's try the next intermediate. And to do that, we are simply going to remove hydrogen from the structure and introduce a double bond. In addition to that, we will not have any positive charge anymore. We are simply forming a double bond. And now we have to think about it carefully. We have obtained an eno be and we have to recall that enos are not really stable and they actually undergo the kenal totem. Specifically, we can convert our given eno into a more stable carbonyl compound. Let's remember how it's done. We are simply going to use the pair on oxygen to de localize it into a pie bond. And then if we expand our oxygen hydrogen bond, we will understand that the pi bond will essentially capture that proton and we will transfer the oxygen hydrogen electrons anti oxygen in the form of a lone pair. That's how we get a more stable carbonyl compound. And if we think about its nature, we are going to troll our double bond oxygen. We also have to keep in mind that there is an implicit hydrogen. So let's add hydrogen to our carbon group. So we understand that it will be an aldehyde. And then on the other side bonded to the carbonyl group, we have carbon, which now has hydrogen. And in addition to that, there are two tle groups bonded to it. Well done. So we got our product. Now, we can answer our question in full. We clearly see that we get one product and there are two reasons. So we can simply state that the acid catalyze dehydration requires a tertiary alcohol to form a stable tertiary carbo ion just as we discussed before. So even if we proin the primary alcohol, we will not form a stable carbo car. So no acid catalyzed dehydration will take place. And in addition to that, we understand that there are no beta hygienes to eliminate the protein of primary alcohol. Even if the reaction followed an E two pathway. If we carefully analyze the structure at the beta carbon, we don't have any hygienes at all. So even an E E two pathway would not work. And finally, in addition to that, if we think about the structure formed, it doesn't even have an alcohol group anymore because it's the eno to rs into a more stable aldehyde. So there is no more a age group that can participate in the reaction further. That would be our final answer. Thank you for watching. And I hope to see you in the next video.

Although 2-methyl-1,2-propanediol is an unsymmetrical vicinal diol, only one product is obtained when it is dehydrated in an acidic solution.
b. Why is only one product obtained?

Key Concepts

Vicinal Diols

Dehydration Reaction

Carbocation Stability

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