A student wanted to know whether the greater proximity of the ...

Question

A student wanted to know whether the greater proximity of the nucleophile to the C-2 carbon in the transition state is what causes the 1,2-addition product to be formed faster when 1,3-butadiene reacts with HCl. Therefore, she decided to investigate the reaction of 2-methyl-1,3-cyclohexadiene with HCl. Her friend told her that she should use 1-methyl-1,3-cyclohexadiene instead. Should she follow her friend's advice?

Accepted Answer

Hello everyone. Let's do this problem. It says a chemist wanted to investigate whether the faster formation of the 12 edition product in the reaction of 13 B to dine with hydrogen bromide was due to the greater proximity of the nu file to the carbon two carbon in the transition state. Should he investigate the reaction? Using one methyl cyclo Penta, 13 dyne or two methyl cyclo Penta 13 dying. Let's review or recall the different types of products that can be produced in 12 and 14 edition reactions. So there are two types of products, the kinetic product and the thermodynamic product, the kinetic product is the product that forms faster. And as the problem touched on this is due to the proximity effect. So the closer that that nuclear file is to the positive charge in the transition state, the faster it will attack that positive site and form the product. Typically, the kinetic product is formed in the is synonymous with the 12 edition product and it usually has a more stable transition state. OK. The thermodynamic product is the more stable product and you can judge the stability of the product based on how substituted the alkene is. So the more substituted alkene is going to be the more stable product. And conversely, this is typically the 14 edition product. Not always, and it usually has a less stable intermediate or less stable transition state, but again, not always. All right. So let's look at these two possible reactions starting with one methyl cyclo Penta, 13 dying. And let's see how these reactions differ. So let's draw our reactant cyclo Penta 13 dying is a five membered ring with two double bonds. And on carbon one, we have a methyl group. And how do we know which double bond is going to attack that hydrogen bromide? Well, we want to look at which one would form the most stable intermediates. So I'm going to label double bond three A and double bond one B. And I'll show you how we determine which ones we will proceed to form the products with. So if we attack with double bond A, we get the following intermediates, we add the hydrogen to one side and we want the positive charge to be one carbon away from the other double bond so that we can have the resonance structure. So attacking with this double bond gives us a secondary carbo CAD ion and a tertiary carbo CAD ion. If we attack with double bond B, we add the hydrogen where that methyl group is positive positive charge on the carbon next to the double bond, we shift that double bond over towards that positive charge, which gives us two secondary carbo ions. So the first set of carbo canons are more stable because of that tertiary Carbocaine. So that is the pair that we will go with, which means we will attack with double bond A. Does that make sense? So now let me redraw these intermediates and then we will draw our product. So we have added the hydrogen to position one, we have our positive charge on position two. And the product formed from this intermediate would be the product and that product would be three bromo one methyl Cyclo pen with that double bond on carbon number one. So cyclo pen one E and if we shift that double bond to create our other intermediate, that will form the 14 edition product from this intermediate, we have our tertiary carbo CAD ion in which the bromine attacks and that forms three bromo three methyl cyclo pente. OK. Cyclo pen one E. So let's label which product is the kinetic product and which product is the thermodynamic product. So the kinetic product is going to be the 12 edition product because notice that the hydrogen is right next to the carbo CAD on site. And since that bromine was just attached to that hydrogen, it should be floating nearby, right. So that's the proximity effect. So that will be the kinetic product. And for the thermodynamic product, which alkene is more substituted. That would actually be also the 12 edition product, right? Because it has two carbon groups on carbon, one side and one carbon group on the other side. Whereas our other product, the three bromo three methyl product only has one carbon group on either side of the double bond. So, is this a good reaction in order to investigate whether the 12 edition product is faster due to the proximity principle? No, because both the kinetic and thermodynamic product are the same product. So what about two methyl cyclo Penta 13 dying. Let's do the same thing. So we draw our structure five membered ring with two double bonds and a methyl group on carbon two. This time, I won't show the process of selecting which intermediates we are forming, but we are going to attack with the double bond on carbons one in two because it will form the most stable intermediates. So the first intermediate, we add the hydrogen to carbon one, we have that tertiary carbo Canion on carbon two. And that would form the 12 edition product where that bromine is added to the same carbon that the methyl group is on. And to form the 14 edition product, we move that double bond over towards that positive charge, which moves our positive charge over to carbon number four. And the bromine will add there. Hence the 14 edition. OK. So the 12 edition product is going to be our kinetic product and that is three bromo, three methyl cyclo Pentin and then cyclo sorry, three bromo, one methyl cyclo Pentin is going to be our thermodynamic product because it is the more substituted al. So in this reaction, we see that the kinetic product in the thermodynamic product are two different products, two different compounds, right. So that makes this reaction or two methyl cyclo Penta, 13 dyne, the better compound in order to perform this investigation. That's it for this problem. I hope this video is helpful and thanks for watching.

Key Concepts

Nucleophilicity and Electrophilicity

Transition State Theory

Regioselectivity in Electrophilic Additions

Watch next