When (1-bromoethyl)cyclohexane is heated in methanol for an ex...

Question

When (1-bromoethyl)cyclohexane is heated in methanol for an extended period of time, five products result: two ethers and three alkenes. Predict which of the three alkenes is the major elimination product.

Accepted Answer

All right. Hello everyone. So this question says that extended heating of two bromo 11 dimethyl cyclops in the presence of methanol yields seven products including four ethers and three alkenes determine the all keen that is the major product of the elimination reaction. So here are starting material is two bromo 11 dimethyl cyclops. That's a five carbon ring to start off with two methyl substituents on the same carbon of the rink. In addition to that next to carbon one here with two methyl substituents is a bromine substituent attached to carbon two of the ring. So this Ocu Halide is reacting with methanol or CH 30. Now, based on the descriptions of the products given we have substitution and elimination products here recall that substitution is a reaction in which a leaving group. In this case, bromine is replaced by the nucleo. In this case, methanol because of that, bromine is replaced with the meth oxy group of methanol to create our ether products. On the other hand, in elimination, the leaving group is lost and a double bond is subsequently produced. Hence the all keen products that were discussed previously. Now, substitution is of course, going to be present in this reaction mixture. But because this question is focusing on elimination, that's what we're going to discuss further. So it's worth mentioning here that our base methanol is neutral, which means it's considered to be quote unquote weak as a base. And this is done in the presence of extended heating recall that heat is a catalyst of ionization. So if we go ahead and we redraw our starting material on the bottom of the screen here, in this case, heat is a catalyst of ionization because it allows for the bond between carbon and bromine to break, thereby producing a carbo cion intermediate. So because of this carbo on intermediate and also the use of a neutral base in this reaction, this is going to be an example of an E one mechanism. So now that we've created a carbo cion recall that rearrangement is always going to be a concern here because if a given carbo cion can rearrange itself to obtain a more stable position, it's going to do precisely that. And here we have a secondary carbo cion. That's right next to a ordinary carbon containing two methyl substituents. This ordinary carbon is going to allow four rearrangement to occur specifically a metal shift because one of our methyl groups here can migrate towards the secondary carbon, thereby creating a tertiary carl cion where the quad carbon initially was. I'm rearranging this mechanism a little bit because even though bromide is going to form. It's it's not going to be involved in this reaction further. So just for the sake of space, but going back to our previous discussion, the reason why multiple s are formed is because really we have two different carbos, one before and one after rearrangement. However, the major elimination product is going to stem from the more stable rearranged carbo cion recall that elimination is also described as beta elimination in which a hydrogen from the carbon atom. Next to wait. Let me correct. That previous statement, beta elimination describes the removal of a hydrogen attached to a carbon next to the carbo cion to form an ke that carbon is known as the beta carbon and the hydrogen removed is therefore a beta hydrogen. So if we consider the un rearranged secondary Carbel cion, really there's only one beta carbon that has hydrogens to remove and that would be the Betta carbon to the right of our secondary Carbo Kia. This is because our ordinary carbon here does not have any hydrogens to remove. So here water which is behaving at our base here or sorry. Methanol. Methanol is in solution is going to go ahead and remove this beta hydrogen which creates one possible or K namely, it creates 33 dimethyl cyclen one E. So our remaining two keen products stem from this rearranged tertiary carbocation intermediate. So let's consider how many beta hydrogens can be removed relative to our tertiary Rebo K on here now focusing our attention on the rearranged tertiary carbo cion intermediate. We can remove beta hydrogens from the adjacent primary and tertiary carbons. So to draw out these possible products, we can go ahead and draw double bonds in between the tertiary cruel cion and either position. So on one hand, we have one methyl, two methylene cyclops. And on the other hand, we have 12 dimethyl cyclen one E. So to understand which is going to be the major product. Here, we have to recall Zito's rule because our base methanol is relatively small. The more substituted, keen referred to here as the sites of product is going to be the major elimination product. So to understand how substituted and all keen happens to be, we have to consider how many or groups are attached to both carbons of that double bond. And in this case, we can see that 12 dimethyl cyclen one E is an example of a tetra substituted all king because notice how both sp two carbons of the double bond each have two R group substituents attached to them. So there you have it, 12 dimethyl cyclo one in is the most substituted akin and therefore the major elimination product. And with that being said, thank you so very much for watching. And I hope you found this helpful

When (1-bromoethyl)cyclohexane is heated in methanol for an extended period of time, five products result: two ethers and three alkenes. Predict which of the three alkenes is the major elimination product.

Key Concepts

Elimination Reactions

Zaitsev's Rule

Regioselectivity

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