Hey, Students, I am going to show in my post some chemical reaction in each reaction box, place the best reagent and conditions from the list below images.

All chemical reactions I show, how you can solve easily and best way. In each reaction box place the best reagent and conditions from the list below, I am describing in detail.

You are not finding your’ chemical reactions’ problem please ask your question in the comment box. I am trying to best answer as soon as possible. So let’s go to the problem.

In each reaction box, place the best reagent and conditions from the list below: 3-Hexanone Exclusive Final Product.

In each reaction box, place the best reagent and conditions from the list below: 3-Hexanone Exclusive Final Product.

The first thing we notice regarding. This Synthesis is that we are beginning with a reagent that contains two carbons. But our final product contains six carbons. So of course we have to find a way to add four total additional carbons on to our starting reagent.

What we’re going to do is actually begin by examining a couple of reagents from this pool  And one of those reagents is this very strong base known as sodium amide and then another reagent. 

 That’s going to come in handy is this reagent here so this is basically Ethyl iodide. Now why are these reagents going to be useful to us well let’s take a look at what we’re starting with a very simple alkene.

If we add the reagent sodium amide first what we want to do is understand that this is an ionic compound so the sodium has a positive charge and the amide is a negatively charged ion  And the amide ion is a very strong base.  

It has a lone pair on the nitrogen and what happens is that very strong base is able to de protonate the terminal hydrogen here on the alkene.  So it’s basically going to rip that hydrogen off these two electrons here will stay with the carbon so what we end up forming is a negative acetylide ion basically it’s a triple bonded carbon.

 That contains a negative charge on it and so that’s the result of treating our alkyne with that strong base. Then we said it would be useful to include the ethyl iodide but why is that well let’s take a look let’s draw the ethyl iodide.

We have a ch3 bonded to a ch2 and then bonded to an iodide. Iodide is a very good leaving group. So this sets us up for a classic sn2 reaction where. 

The negatively charged carbon will attack that carbon right there the reason it attacks that carbon is that it’s bonded to the very electronegative iodine.  That gives that carbon right there a little of positive charge because basically its electrons are being tugged away from it.

So that’s why we attack there and then at the same time the iodide will leave so what ends up happening.  If we kind of come down maybe over here is we add these extra two carbons to the molecule, so we might colour them to emphasize that we are bonded to the ch2 followed by the ch3 and then the iodide has left.

 So this is nice because now instead of having two carbons on our molecule we have a total of

 Four and so you might guess well since we need a grand total of six we would run this reaction again  and that’s exactly right. 

 So if we do this two-step mechanism again where we add the sodium amide and then the ethyl iodide.

 What’s going to happen is this terminal hydrogen over here will be removed will form the lone pair of electrons and a negative carbon.  There and that’s going to come over and attack at that same carbon and the iodide is going to leave again so what ends up happening Is that we place an additional two carbons.

On the left side of the molecule we’ll kind of colour them differently here to emphasize that they are different carbons and so after four steps one two three four we have achieved a six carbon molecule.

 So that’s excellent we’re getting there but of course we need to create a ketone we have to some.

 How introduce a double bond to oxygen within the molecule and also lose the triple bonds so it doesn’t seem like.  That’s feasible, but we recall in this chapter that we have learned about a hydration  reaction involving these reagents right here.

 So you have water sulphuric acid and then this mercury ii sulfate.  So let’s go down below now this is a mechanism you probably.  Don’t need to know most textbooks sort of just explain.

 The process in passing what you really need to know is the outcome. So let’s talk about the outcome these reagents are going to be used to hydrate a triple bond.  Now hydrate simply means to add a hydrogen at one carbon and an o h at the other carbon of the triple bond.

 Now in this case we have a symmetrical alkane this is very important for us to understand.

 So it doesn’t really matter so to speak whether we add the hydrogen to this carbon and the oh to that carbon or vice versa and the reason again is that This is a symmetrical alkane.

 So let’s choose the case that we just drew so what we’re going to have here if we kind of redraw it is a ch3 ch2 we’re going to give that carbon the hydrogen.

 The neighbouring carbon is going to get the oh group. Then we have the ch2 ch3 here now the triple bond is no longer a triple bond in this case because if it were you would have

 Carbons that possess five bonds.

 So what happens is it’s reduced down to a double bond but this molecule is actually very unstable this is what we call an Ecole because it contains a double bond.

 All because it has a hydroxyl group in it  and an enol is an unstable molecule so again you probably learned in this chapter.

That an enol will actually rearrange or if you want to be fancy it will tautomerize  and it will tautomerize or rearrange into a ketone.  In this case so basically what happens is this double bond.

You can sort of think of it as coming up here to form a double bond right there and then this hydrogen will migrate over to that carbon. There that is certainly not the mechanism but that’s basically the end result it’s sort of the punch line of the rearrangement.

 So what happens is you end up with that carbon that had that hydrogen and then.  As we said the hydrogen of the hydroxyl sort of migrates over to that carbon here.

That pi bond you can think of it as migrating upward so you’re going to form to colour coded a little.

 There’s your oxygen and then you have the ch2, ch3.  So in essence we are left with the product.  If we write this in a skeletal structure we have the six carbons and then at this carbon here we have the double bond to oxygen so that does work.

 I hope you notice that, if we were to do this the other way this is a very important point. That if we back up and instead of putting the hydrogen and the oh there we sort of reversed it.

 We noted that wouldn’t matter because it’s a symmetrical alkene let’s understand why that doesn’t matter.  So you would form this initially it would tautomerize or rearrange and when it rearranges in that case you would end up with the double bond to oxygen here and the hydrogen.

 There but notice this is the same molecule this is still one two three four five, six carbons long and at the third carbon.  If we number it in that fashion you have the carbonyl group, so we would have the six carbons and then you would have your carbonyl group. 

 Here it might look different but it’s the same thing again you still have the carbonyl group at carbon 3. 

 Just like we had the carbonyl group over here at carbon 3.  So that’s why this synthesis works because whether you add the h and the oh in this fashion or  in reverse fashion as we did earlier you’re still going to get basically three hexa none is basically.

 The name of that molecule and so let’s go back and just put into the answer boxes what we

 Discovered here so step one was the sodium amide. Step two was the ethyl iodide then we did a second round of that, so we had step three sodium amide again.

 Then we have the ethyl iodide and then finally we had added those reagents in here. That hydrated the triple bond and there is indeed the answer

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