Predict the products and mechanisms of the following reactions...

Question

Predict the products and mechanisms of the following reactions. When more than one product or mechanism is possible, explain which are most likely.

h. 1-bromo-1-methylcyclopentane heated in methanol

Accepted Answer

Hi, everyone. And welcome back in today's video. We are looking at the problem which asks us to determine the mechanism for the following reaction. We have choices among E one E two sn one or SN two. And also the problem states that in the case of multiple products, explain which one is the major product. Now, let's start our analysis as we generally see whenever we are comparing elimination and substitution reactions. First of all, if we look at our substrate, the nature of the substrate is secondary because if you look at this alpha carbon that has the leaving group bonded to it, it has two adjacent neighboring carbon atoms. And therefore, we can say that our substrate is secondary. Number two, we look at the nature of uh look, we look at the nature of our base or our new file. And in this case, it's ethanol and we noticed that it's, it's weak because it's not charged. It's a weak base. Weak nuclei recall that weak nu files preferred to undergo SN one reaction and weak base also preferred to undergo an E one reaction. So we may conclude that in addition to that we also know that substrate substrate is secondary. So these two conditions alone satisfy either SN one or E one reaction. Now essentially condition number three is the fact that we have heat. Notice there's this symbol which stands for heat and heat favors elimination. Notice that this is really good for us because without even knowing the identity of our products, we already said that if heat favors elimination, then elimination wo would produce the major product while the minor product would be produced by a substitution reaction. So we are good to go, we are ready to start drawing our mechanism. And what we're going to do, we are going to start with our substrate. So we are going to draw cyclopse, we are going to add the BROMO substituent pointing up the me, the metal substi pointing down. And now, first of all, we will have a step in which we see the loss of the leaving group to form a carbo ion. Now, during the loss of the leaving group, this bromine essentially leaves notice that we are also using heat. So it makes sense to include that condition and we will be forming a secondary Carbocaine. So let's draw that a secondary carbo is formed. And now essentially because we are primarily interested in an elimination reaction, we may identify our beta hydrogens. Also note that because this is an E one reaction, we are forming a product, a Z of product as opposed to a Hoffman product would be a more substituted alkene. So whenever we are trying to find the beta hygienes that we could eliminate, we are interested in the ones that will form the most stable alkene or the most substituted alkin. One of the beta hygienes is at this position and another beta hydrogen is bonded to that position. Now, here's my question to you. Should I eliminate this hygiene or that hydrogen, which one of the two would form a more stable alkin? And the answer is I would actually remove this hydrogen due to the presence of that methyl substituent. It would give me a more stable alkene. So the removal of this hydrogen would be favored and I'm going to use on a molecule of ethanol because that's the weak base that we have in this mechanism, the lone pair on oxygen would essentially dep proton the structure and we would form a double bond at that position. We may now tell that one of the products is an al. So let's just draw it. We still have that method group. But now this has a play in our geometry due to being SB two hybridized, we may clearly draw that double bond. So this is our first product and we can immediately state that this is our major product. And we already discussed the reason why this is a major product. It's basically because we're using heat and heat favors elimination. So let's label this as our first product for reference. And now let's move on to our minor product because we have a secondary substrate and a weak nuclear file there, a competition between SN one and E one. So we may also draw the mechanism for a substitution reaction if we start with our substrate. The first step is very similar. Essentially, we are adding heat and we have the loss of the leaving group during the loss of the lead group. We are using bromine to form a secondary carbo Keion. So let's draw the carbo canon just as we did before. But now we have to be careful because this is a secondary carbo. We have to always check whether or not we could form a more stable carbo ion. And indeed, because there's a source of hydrogen, a disposition, we may indicate the presence of a hydrogen atom, a disposition. We would expect to see a one comma two hydride shift to form a more stable tertiary carbo. So in this case, we would have one comma two hydride shift to form a more stable carbo ion all that we have to do. Now, we essentially have to show the reaction arrow which represents that shift towards the carbo ion. Now, we are drawing the resultant carbo can, which is now terri having added a hyen atom and disposition. We understand that this carbo will be plainer. So we may essentially add, we may essentially not show any dashes or wedges. It's already planar, we have a positive charge here. And the very last thing is just to understand that the next step, as you recall in a nucleic substitution reaction is the new call for like attack. So if we take a molecule of ethanol, the lump pair on oxygen would just attack the carbo ion. Now, we have two ways to attack it either from the front side or the back side forming a Eric mixture. But that's not really irrelevant because notice there's a line of symmetry making this carbon not chiral in the final product. So we are not worried about serial chemistry and we may simply proceed with our next step. We call this stop anu like attack, stop during the nucleic attack stuff, we are forming a proton to intermediate. So we are going to draw that specifically, we have a metal group and now we will have oxygen directly bonded to that carbon, which will have a formal positive charge still has an ethyl group. And then it has h let's actually expand it because we, it will be important for the mechanism. So in particular, this hydrogen should be shown there's a form of positive charge and we are nearly done. The very final step to warm our product is the de protonation step. We want to get rid of that proton, the protonation. During the de protonation step, we're using another molecule of ethanol. It's a weak base that works for us. So we may draw a molecule of ethanol, the pair on oxygen will be basic and it will essentially capture the proton. And these bonding electrons will be placed in the form of a pair and oxygen neutralizing the negative charge and forming our final product, we are ready to form our final product which but essentially be an Easter. Oh And then we have an ethel group. So we may draw it um in this kind of orientation, we may label this as our minor product. So this is our minor product. And that said we have our products. One of them is major recall that it was produced in an E one reaction and the minor one was produced in an essence to sn one. My apologies reaction. The correct answer to this problem is see and if you have any questions, don't hesitate to leave them down below. Thank you for watching this video.

Predict the products and mechanisms of the following reactions. When more than one product or mechanism is possible, explain which are most likely.
h. 1-bromo-1-methylcyclopentane heated in methanol

Key Concepts

Elimination Reactions

Mechanism Types: E1 vs. E2

Regioselectivity and Stereochemistry

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