Predict the major products of dehydration catalyzed by sulfuri...

Question

Predict the major products of dehydration catalyzed by sulfuric acid.

(a) hexan-1-ol

(b) hexan-2-ol

(c) pentan-3-ol

Accepted Answer

Welcome back everyone show the major products for the acid catalyst dehydration of the following alcohols. Because we are dealing with the asset catalyzed dehydration. We are interested in an E. One mechanism because it basically explains us how the reaction takes place And recalling the principles of an E one mechanism. We will be able to predict our products. Now starting with a we have our alcohol and the world fair stopped in an E. One reaction or the acid catalyzed dehydration reaction. In this case we are using a strong asset such as sulfuric acid. In the general case we can just depict hydro knee um itself and the lone parent oxygen gets protein ated producing a water molecule as a byproduct. So we control the protein ated intermediate. We can also label this step as the protein ation step and the protein ated intermediate that reform essentially has oxygen with a positive charge three bonds. Now the next step is the loss of the leaving group. Notice that we have it was a molecule that is possibly charged, so it's a good leaving group. We can just remove it from the structure. This step is the loss of the leaving group and we are forming a secondary carbo Koran. When this carbon Karen is formed, we will actually produce the actual elimination reaction. We have to identify our beta hydrogen if you notice there's a plane of symmetry. So the two beta hydrogen surrounding that, that carbon Karen are exactly the same or equivalent to each other. And we can use the water molecule produced to eliminate any of the two. It doesn't matter which one you choose, but suppose we go with the left one and the lump, iron oxygen attacks ha jin forming an Nl keen. We just our final product. Well done. That's our final product and that's the only product we could have formed in this reaction. We may troll the product for a. Now let's move on to be for B. We will essentially re draw the alcohol. The first step is the protein nation step. We already know how it takes place. So we can use a shorthand notation and indicate that we're using H plus, which is hydro ni um We are protein eating our alcohol which is done followed by the loss of the leaving group and the formation of the carbo Karelian. However, this is a primary carbon Karen and we could make a more stable carbo Karen notice that there is a source of hydrogen. So if we undergo a one comet to hydrate shift, we will form a more stable secondary carbo cast iron. We can indicate that now we have a more stable secondary carbo Karen. And now we are ready to eliminate let's find our beta hydrogen. There are two beta hydrogen surrounding the car bo katan. The question is which one of the two will we choose to get our major product and keeping in mind that we wish to form a more stable al keen. We will go with the left hydrogen atom. In other words, we will eliminate as hydrogen, the lone pair on water will attack it forming a double bond and forming our major product. Notice that if I actually had removed this hygiene, I would have won a less stable al keen because that's that's actually the structure that I would have got. It's mono substituted as opposed to a di substituted elkin that I got. So simona substituted Elkin is less stable and therefore this would be the major product. Well done. I'm ready to show this is my final answer for B. So I made troll the owl came form and finally let's look at sea for C. We are given the alcohol. We already know that according to E one mechanism. The first step is the protein nation stop and we are forming the protein ated intermediate, followed by the loss of the living group. So we are forming a secondary carbo carry on. In this case we cannot make it more stable. There are no ways to make it tertiary one. So we are good with that. Now, similarly to the previous problem, we wish to indicate which two beta high jeans we could remove these are our two beta hydrants. And again, we'll go with this one because that would produce us with a more stable di substituted elkin exposes. Have ruled that's what we are going to do here. So using water that's our base, we are capturing that proton and producing a double bond to form our final product. And now this is our final product, which is a major product. Notice that in this reaction in particular, we could have formed the cIS eyes Marvet because there's a free bond rotation, but trans versions of all kinds are more stable and therefore we have formed a trans I. Zimmer, so we did not worry about any bond rotation to indicate any minor products. We are ready to show. That is our final product for part C. Now, which is exactly the same structure that we got for B. These three are our products for A B and C, respectively. And the correct answer for this question is the if you have any questions, feel free to leave them down below and thank you for watching.

Predict the major products of dehydration catalyzed by sulfuric acid.
(a) hexan-1-ol
(b) hexan-2-ol
(c) pentan-3-ol

Key Concepts

Dehydration Reaction

Carbocation Stability

Markovnikov's Rule

Watch next