What is the major elimination product obtained from an E2 reac...

Question

What is the major elimination product obtained from an E2 reaction of each of the following alkyl halides with hydroxide ion?

f.

Accepted Answer

Hey, everyone. And welcome back in today's video. We're going to identify the major product of the E two reaction between the alky halite provided below and the hydroxide ion and we're given two R three S two chloro three methyl pense, that's the starting material and it undergoes an E two reaction just as stated with hydroxide because we are considering a need to reaction with a small base. Let's remember that hydroxide is a really small base. It's not hysterically hindered. That means we are going to favor the of product that will be our major product. And the Z of product is the most substituted alkene. Specifically, if we go back to our starting material, we're going to identify our alpha carbon, which has our living group and the neighboring beta carbons and those beta carbons, they have beta hydrogens, right? So we are going to consider our beta hydrogens and we have to be very careful, we have to remember the anti co planner requirement for an e to reaction. And that means our hydrogen and our living group, they must be in an arrangement. Let's notice. First of all that hygiene on the right is sent to chlorine because they're both pointing down right now. On the left, there is one hydrogen, right? Because we, we don't have any chiral carbon. We actually have our metal group. So we have, we will always have one hydrogen being anti to chlorine. And specifically this is hydrogen, which is on the wedge. So we can definitely eliminate that easily. But the problem is that we are going to form a less substitute at alkene, right? And we want to get a more substitute at alkene. So we want to get a double bond at a different position. So we want to obtain the anti arrangement between our chloro group and hygiene on the right. And to do that, in my opinion, the easiest way is to actually draw the new projection. So if we once again, think about the starting material, right, if we essentially add our metal group, our hydrogen and also our chloral group, let's suppose that we are, we label our carbon atoms as number two and number three. And we are going to draw our implicit hydrogen atom as well, which is winding up on, on carbon number two. And let's suppose that we are looking along our C two C three bond. OK. And now if you want to draw a new one projection, then we are going to draw a circle. OK. And let's assume that our eye is in front of carbon number two. So we have to understand that what we're seeing is simply our metal group pointing up, right. So we have our metal group on the left of our eye, we have our chloral group and on the right pointing up, we have our hydrogen. So we are going to label hydrogen C H three and then chlorine. Now, carbon number three is in the back and we are going to observe three bonds. So let's add them. And what we're seeing is that we have our metal group on the right of our eye. Then on the left, we have our hydrogen and pointing down, we have an ethyl group. So C H two C H three, because our chloral group must be anti to hydrogen. We are simply going to rotate the back carbon slightly or specifically speaking 120 degrees. So if you swap all of these substitutions, we are going to duplicate it for a complete clarity and we're going to get rid of all of our substitutions bonnets to the back carbon, right. We essentially once perform rotation degrees clockwise. So if you do that hydrogen will be anti to chlorine, that's exactly what we want to achieve. The methyl group will be pointing down and then the group will be shifted as well. So we are going to say C H two C H three and now we are nearly done. So if we are using our hyd trioxide, it will attack our beta hydrogen we are going to form our double bond between these two carbon atoms. Of course, we cannot show that very clearly, right? But we're going to form a double bond followed by the loss of the living group. And if we draw the remaining groups, we will pretty much get our final product, which will be our major product. So let's see how it looks like we are removing hydrogen. So let's remove it. We're also removing chlorine and that's it. We now simply want to redraw that alkin in a bond line structure, right? So we simply have to understand that the front carbon and the back carbon, they are bonded via a double bond, right? So we are going to get our double bond and then we understand it on the same side, we have our methyl group on the front carbon, on the back carbon, we have our ethyl substituent and now below the carbon atom that has an metal group also has a metal group and the other carbon which has a metal group has hydrogen bonded to it. So we can actually say that it's implicit, but you can expand it if you wish. So we can just eliminate that because there's no necessity to show it. And we have obtained our major product because it's actually a tri substituted alkene. And as we said, we want to get our states of product instead of the, the other possibility which would give us a substituted alkene or actually to be more accurate, I'm sorry, that would be mono substituted because we would remove one high chain form our double bond. And then followed by the loss of the living group, we will only have one kiel group. So that will be mono substituted Alkin or Hoffman product, which is the minor product. And the major product is the of product, which in this case is a tri substituted alkene. So we got our answer in a green box and that'll be it. Thank you for watching.

What is the major elimination product obtained from an E2 reaction of each of the following alkyl halides with hydroxide ion?
f.

Key Concepts

E2 Reaction Mechanism

Zaitsev's Rule

Sterics and Regioselectivity

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