Identify the three products formed when 2-bromo-2-methylpropan...

Question

Identify the three products formed when 2-bromo-2-methylpropane is dissolved in a mixture of 80% ethanol and 20% water.

Accepted Answer

Hey, everyone. And welcome back in today's video, we're going to draw the three products formed in a reaction between to Idaho to metal propane and the solution of ethanol. Let's begin solving this problem by drawing a three member chain which represents propane in that position too. We are adding our metal substitution and our idol substitution. The structure is treated with ethanol. Sees three sees to wage. That's our alcohol. And actually we have an agreed solution which also implies that there's water present and that solution besides the molecules of ethanol, let's analyze our material. So we have a tertiary substrate because there's a good leaving group, right? That's our substrate iodine is our leaving group. The reason why it's tertiary is because the carbon atom that responded to iodine has three neighbors. Now thinking about our other region which is ethanol. First of all, it's not charged, it's neutral. So we can state that it must be a week new profile. And also it's a weak base because oxygen does not have a negative charge, right? It's basically bonded to hydrogen. It has two lone pairs, no excess amount of loan pairs that would make it negatively charged and therefore a strong base. So we have a weak nuclear file because it's not charged in a weak base, pretty much because of the same reason. So we have to rule out what kind of a reaction that would be A tertiary substrate would always undergo an SN one reaction. Now, a weak nuclear file and a weak base, that essential implies that if we have a weak nuclear file, It supports an SN one reaction, a weak base With support and E one reaction. So we will have a mixture of products, tertiary substrates will also undergo E one reactions perfectly fine. So that means we will have products of SN one and E one reactions. Let's think how an SN one reaction would take place. But actually, the first step in each reaction would be exactly the same, we would form a stable tertiary carbo carry on. So we can afford the loss of the leaving group and the formation of a stable tertiary carbo Karelian. Now thinking about SN one products, this car Bull Karen can be attacked by two nuclear files. One of them is ethanol itself. So the lone parent oxygen would attack the positively charged carbon. We would form a new carbon, oxygen bond oxygen is still protein ated, right? So we have a formal positive charge on oxygen. So another molecule of ethanol, Kenji's D protein to give us our final product. So that means our final product for this one would correspond to a three member chain position. So we are adding our methyl group at the same position. We are adding our with oxy substitution. Oh Cease to C three. Wonderful. So we got our first product to more products are required. So going back to our carb. Okay Ryan, because it's an acquittal solution of ethanol. As we previously discussed, the other nuclear file is water and the lone pair in oxygen from water can attack it as well. So that gives us another product. First of all, we are creating our carbon oxygen bond. So we have oxygen with two hye Jin's still meaning it has a positive charge. And of course, another molecule of water can essentially d protein it that to produce and alcohol. So let's draw that as our next product We can through our three member chain, add a metal group at position two. And at that same position, we are now going to add oh age, that's how we get our third, be eternal. And finally, our third product must be an elimination product because we are done with our two nuclear files. So now we are going to think about ethanol acting as a base. So going back to our cardboard Calyon, we are going to redraw it. Once again, we have to think about our beta protons and they are all equivalent, there are three metal groups. So we can show one of the hydrogen atoms at any metal group. And now here we can use ethanol as our base that removes the proton. Then we are forming a pi bond out of the carbon hydrogen electrons. And we are getting our elkin Which would correspond to our 3rd product. So to draw our third product, we are just going to draw a three member chain, Add a methyl group at position two and add a double bond between Carbons one and 2. And those will be our three products. We can essentially label them as our final products. We may recall that we had to substitution products because we had too weak nuclear files and a tertiary substrate. One of the nuclear files was ethanol, the other one was water and then we got one elimination product because ethanol is also weak base and it allowed us to form an L king based on our answers. We can conclude that the correct choice to this multiple choice question is a. However, if you have any questions don't have states, leave a comment down below in the comments section and thank you for watching.

Identify the three products formed when 2-bromo-2-methylpropane is dissolved in a mixture of 80% ethanol and 20% water.

Key Concepts

Nucleophilic Substitution

Ethanol as a Solvent

Elimination Reactions

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