Predict the major products of the following reactions, includi...

Question

Predict the major products of the following reactions, including stereochemistry where appropriate.

(h) cyclooctylmethanol + CH₃CH₂MgBr

(i) potassium tert-butoxide + methyl iodide

(j) sodium methoxide + tert-butyl iodide

Accepted Answer

Hey everyone and welcome back to another video. In today's lesson, we're going to learn how to solve the following problem, determine the identity of products obtained in each of the reactions shown below and include stereo chemistry in your answers. Reaction number one. The reaction takes place between an alcohol because we have an O. H. Group bond, it's an L. Kill chain and the green equation which is ethyl magnesium bromide due to the presence of that L kill group bonded magnesium. This is our green equation. Now let's recall that the gradient region is strongly basic. So if we think about this reaction generally, we see granite regions reacting with carbon compounds or night trials to form alcohols. But what if we have that acidic proton monetary oxygen? We call that Green Earth regions are strongly basic because there's polarization towards carbon right here in that carbon magnesium bond, Since magnesium is electoral positive due to the fact that carbon is polarized negatively, it can act as a base. So the bonding electrons here might essentially attack that hygiene and abstracted, followed by the formation of the elk oxide and iron. So one of our products here would be Alcock side itself. We may of course indicate the presence of magnesium bromide to stabilize the charge and indicate that there's an overall neutrality in that solution. So positively charged, negatively charged. Right? This is one of our products and if you look at the second product, this ethnic group abstracted hydrogen forming anything. So our next product would be anything. C three, C three. Right? So we can say that these are our products for reaction. No one you will get reaction number two, we have cyclops insane oxygen. It's very important to understand that sodium is an ion spectator, right? This is a sold. But keep in mind that from general chemistry, we know that sodium sodium separates from an any ionic compound because all of sodium souls are soluble. Right? So there's a dissociation process and sodium becomes a non spectator, which doesn't have any effect on the outcome of the reaction. So the major part here is the elk oxide and iron itself that reacts with methyl iodide. We understand that this is a strong base strong nuclear file because there's a negative charge on the oxygen and we have a metal substrate. Right? So, definitely that would be an essence. To reaction, we have a good leaving group, a metal substrate and a strong nuclear power. So all of these conditions satisfy an SN two reaction in an essence to reaction the lone parent auction attacks the electrical a carbon atom followed by the loss of the leaving group. And as you notice, we are forming an ether, we're going to draw the easter formed O. c. three. Right? And of course, if you wish, you can add the presence of sodium iodide, right? Because we had iodide as the leaving group and then we would add sodium over here to indicate that charge stabilization or overall neutrality. So that be our overall equation, we can say that the products for reaction number two would be these two. The main product is either itself sodium iodide is an inorganic product right now, if you look at reaction number three, we are starting with 3rd beetle bromide. Let's draw it which reacts with sodium ether oxide. So we're going to drool with oxide itself. Okay, so let's not worry about sodium so far. Of course we can add it. Now. Let's think about this reaction. First of all, the substrate that we're seeing here is a tertiary substrate because broom in which is a good leaving group responded to a carbon atom that has three neighbors. So the substrate is tertiary, meaning it would generally participate in sN one reactions if it was substitution. But the nuclear file itself is negatively charged oxygen, which is a strong nuclear file. It's not molecular meaning sN one. It's not something we can perform here. Right? But this is not only a strong nuclear file, it's also a strong base. East oxide is a strong base as well. And if we think about that strong base, it can definitely participate in an E two reaction. It doesn't matter if our substrate is tertiary or not because that is a strong base, it can form an elimination product. So if we draw our elimination product and that's exactly what we're observing here. That would be an E two mechanism, mainly because the substrate is there Suri, which satisfies us in one, but the nuclear file does not satisfy us in one. So because the strong base it implies that it must be an E two reaction. If we think about our beta hydrants, there are three equivalent metal groups. So we can just show one beta hygiene because all of beta high genes are exactly the same. And there's only one possibility to form that product. We would attack that hydrogen form a double bond and then we would have the loss of the leaving group. So one of the structures that we would get here would be an AL king. So we draw our ALC in. We don't understand that there will be an alcohol form. So let's draw ethanol and then sodium bromide. Right? Based on the given ions. So that's sodium, iron, bromine and iron has been formed. So we can say an A. B. R. This is our reaction scheme, of course the main product is just the alcan itself because we're interested in the conversion of the substrate into something else. So, if you look at these three reactions, we understand that reaction. Number one was a basic acid based reaction because we had an acidic proton reacting reacting with a basic remote region reaction. Number two was an SN two reaction because we had a strong based nuclear file and the substrate was a metal substrate. Right? So this really favors and listen to reaction. And reaction number three was an elimination reaction to you too because we had a strong based, strong nuclear file. But the substrate was third sherry, meaning this must be exclusively an E two reaction. Now we may conclude that the correct answer to this multiple choice question is D. However, if you have any questions, don't have states, leave your comments down below in the comments section and thank you for watching.

Predict the major products of the following reactions, including stereochemistry where appropriate.
(h) cyclooctylmethanol + CH₃CH₂MgBr
(i) potassium tert-butoxide + methyl iodide
(j) sodium methoxide + tert-butyl iodide

Key Concepts

Grignard Reagents

Elimination Reactions

Nucleophilic Substitution

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