1-Methylcyclohexene forms two products when it reacts with bro...

Question

1-Methylcyclohexene forms two products when it reacts with bromine in methanol.

a. Draw the mechanism for the formation of the products.

b. Describe their stereochemical relationship—that is, are they enantiomers or diastereomers?

Accepted Answer

All right. Hello everyone. So this question says that two products are formed from the reaction of one ethyl cyclops one E with chlorine and ethanol. Part one says to provide the reaction mechanism for the reaction. And part two determine the stereo chemical relationship of the two products formed. All right. So starting off with the structure of the given, we have one ethyl cyclen one E, that's a five carbon ring with a double bond in between carbons one and two and an ethel or a two carbon substituent on position one. So our studying material is an A K reacting with a halogen, specifically CL two in the presence of an alcohol recall that the product obtained from this combination of reagents is known as a halo hydrant. This means that one carbon of the starting all keen gained chlorine, the halogen and the other carbon gains the oxy group of the alcohol. In this case, an epoxy group or actually a slight correction. This is not a halo hydrant. This is actually a Hallow ether. So, apologies on that note, but as we go through the mechanism, we'll discuss a little bit more about the stereo chemistry of the products. So let's go ahead and get started. So the first step of the mechanism involves the interaction between the keen and the halogen in this case, cl two. So the first thing that happens is an electro pilic addition involving R halogen. All right. So first, the pi electrons of the hay attach themselves to Chine at the same time that chlorine also establishes a bond between itself and carbon. This ultimately breaks the bond between the two chlorine atoms in our molecule of C two. Here, this leads to the formation of a bridged ion intermediate. So the pie bond between both carbons here of the starting alkene breaks. And our intermediate resembles an oxide. But instead of having oxygen in the center of a three member ring, it's chlorine instead. And because of the presence of two bonds, chlorine would actually have a positive charge. So our next step involves nucleic substitution with our alcohol followed by de potent nation of oxygen. So in this case, or in this next step, we'll have ethanol attack, the more substituted carbon of the bridge ion intermediate. This is why this reaction follows Markov Niko's rule. So oxygen behaving as a nuc file attacks, the more substituted carbon of the bridged eye on intermediate, this ultimately breaks one of the bonds between carbon and chlorine. Now this is where stereo chemistry becomes very relevant. So two things to recall here is that this reaction is an example of anti addition, which means that the Al Coxie group and chlorine are going to be on opposite sides. And a mixture of an numer is produced. The reason for this is because the alcohol can attack either above or below the ring. This creates a pair of an tir depending on either possibility. So the first possibility in which the alcohol attacks above the ring would have the a foxy group on the more substituted carbon depicted on a wedge. Whereas chlorine on the other carbon would be depicted on a dash because of the presence of the proton from the alcohol oxygen would currently be positive. So our other possibility is going to switch the orientations shown here in my second in numer, the oxy group would be on a dash which means chlorine would be on a witch. So our last step would then be the depredation of the alcoholic oxygens. This can be done by one molecule of ethanol for each an an tir. So here on the bottom of the screen, I'm drawing out two molecules of ethanol, one for the depot nation of each anant toy. After that proton gets removed, the electrons in between hydrogen and oxygen go back to oxygen. So ultimately, we get our two different products which are once again a numer. So our product and an tir are going to look very similar to our current intermediates except oxygen would be neutral. One thing I didn't mention earlier is that the, the carbon atom that contains the epoxy group also contains the ethyl substituent. The ethyl substituent is going to have opposite stereo chemistry to the oxy group. So if the epoxy group is on a wedge, then the ethel group is going to be on a dash and vice versa. But there you have it. So with that in mind, we have our completed mechanism which explains the formation of the two products hinted at. So with respect to the second part of the question, as stated earlier, the relationship between the two products is that they are in num me. And you can confirm this by rotating one product, 180 degrees. After rotating one product by 180 degrees, you can see that both compounds, both products become non superimposable mirror images. And so with that being said, thank you so very much for watching and I hope you found this hopeful.

1-Methylcyclohexene forms two products when it reacts with bromine in methanol.
a. Draw the mechanism for the formation of the products.
b. Describe their stereochemical relationship—that is, are they enantiomers or diastereomers?

Key Concepts

Electrophilic Addition Mechanism

Stereochemistry

Product Analysis

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