What substitution products are obtained when each of the follo...

Question

What substitution products are obtained when each of the following compounds is added to a solution of sodium acetate in acetic acid?

a. 2-chloro-2-methyl-3-hexene

b. 3-bromo-1-methylcyclohexene

Accepted Answer

Welcome back, everyone identify the substitution products formed when the compounds below react with sodium ethanoic in Ethanoic acid. Number one, we're given ethyl 55 ethyl five I have re in. So let's go ahead and draw the structure. That would be our first step hept three in, right. So we essentially want to draw a seven member chain. So let's go ahead and do that. We will have 123 or 56, seven carbon atoms, right? Hep three in means that starting with carbon number three, we will have a double bond present between carbons number three and four. And now carbon number five also has an ethyl group attached to it because it says five ethyl five ido meaning there also must be an ido substituent bonded to carbon number five. Now, this is our starting material. And what we want to understand is that this starting material reacts with sodium ethanolate, right? Meaning if we exclude sodium, that is an iron spectator, we're going to use ethanoic as the nhi and we are also using ethanoic acid. So ch three coo H which is a polar product solvent. Now, the starting material is a tertiary alky Halide, the solvent is a polar product solvent, right. So we can perform an S on one substitution reaction. Even though ethanolate is a strong nuclei, it's charged, right. The remaining conditions satisfy an E and one reaction significantly. Let's remember that in the first step of the reaction, we have the loss of the living group in an N one reaction. And in this case, we are forming a stable tertiary resins stabilize carbo can and because it is resonant stabilized, we will be able to get multiple products because first of all, we can shift the pie bond right and obtain another carbo cion. So let's go ahead and throw it well done. So now we're going to shift our pi bond and get our second carbo cion. Now, what is our new profile? In this case? It's Hanno eight. So we simply want to draw a new C like attack, stop, we can remove the arrow or resonance. So we don't get confused. And for our mechanism, we are performing a front side or a backside. E in this case, it doesn't matter because we are not forming any carbon atoms for the first structure. For the second structure. Let's see what we get, we will get four different substituents, meaning we can get two products. Since in an Esselman reaction, we can perform a backside attack or a fronts side attack. So let's go ahead and throw the result on products. Now, for the first carbo cion, we only get a one possible product, right? Because there is no chiral carbon atom formed. Now, for the second carbo cat on, we will get two possible products because we are forming a chiral center. So let's say that we draw Hanno bonded to carbon on a wedge. So we are simply going to say plus en plus in an tumor, meaning the third structure or the third product in this case will have oxygen on a dashed bond because the one that we have thrown is on a watch, right? We can, we can form a erythemic mixture of in an tumors. So those would be our three products for the first reaction. Now let's move on to the second one. For the second one, the starting material is three chloro one methyl cycle one in just as we did before we're going to draw the structure. And in this case, we're given cycled one in. So that means we want to draw a five member grain. Now it says cycle P one in. So we're going to draw a double bond between carbons one and C carbon. Number one has a methyl group and the carbon number three has a chloro substituent. So we have our starting material and we know that it reacts with the same regions is than a in the presence of acidic ass up. OK. So once again, we have a secondary substrate, right, a polar a product solvent which favors an S on one reaction. So we need to think about the loss of the leading group since we're going to produce a secondary resin stabilized carbo Canion. So let's go ahead and illustrate the loss of the leaving group. Let's show the resultant carbo cion formed as well as its resonance form. We're going to shift the pipe band, having shifted the pie bond, we can throw the second possible carbo Canion. OK. And now we understand that we can perform a nucleic attack. Stop. We need to carefully think if we can form any chiral carbon atoms. And yes, we do in each case. So we can draw the products. The first carbo cion can be attacked from both sides. Similarly, the second one can be attacked from both sides. So let's take the first carbo cat I am and replace the positive charge with is an it, we need to choose a wedge or a dash. Let's suppose that we choose a dash, right? Because we can form a mixture of in an tumors. So we will have oxygen bonded to an ethyl group plus it's in an tar, meaning the second product will have a wedge instead of a dash. And now similarly, we can take the second carbo cat I am and replace the positive charge West. Yes. AO it. Now once again, we can form a reca mixture. So let's say that the muscle group is pointing down on a dash. This means that is an, it will be pointing up on a wedge and of course, we're forming an anan tumor. Therefore, we get four products for the second reaction and that would be it for today. Thank you for watching.

What substitution products are obtained when each of the following compounds is added to a solution of sodium acetate in acetic acid?
a. 2-chloro-2-methyl-3-hexene
b. 3-bromo-1-methylcyclohexene

Key Concepts

Nucleophilic Substitution

Electrophilic Reactivity of Alkenes

Acid-Base Chemistry in Solvent Systems

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