Show how you would add Grignard reagent to acid chloride or es...

Question

Show how you would add Grignard reagent to acid chloride or ester to synthesize the following alcohols.

a. Ph₃C–OH

Accepted Answer

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem. Show how you would synthesize the following alcohol by adding green yard region to an asset chloride or Esther. Now, if you look at the given alcohol, we first will understand that this is a tertiary alcohol because the carbon atom that is bonded to the alcohol group has three alcohol substitute prints and they're all ethnic groups. Right? So let's keep that in mind. Now, the problem suggests to use either and acid chloride or an ester for our sentences. Let's see why these to undergo a granite reaction. Pretty much in an exactly same manner. Recall that if we're using some L kill chain, let's suppose this is R one R two. Let's label this one as our three. And suppose that we have our four bonnet, two MG B. R. That's the granite region. Right? So essentially what happens if you look at these two structures whenever we have that nuclear attack at the carbon ill, we are basically breaking the pi bond. But due to the fact that there's a good leaving group nearby, the double bond can get restored and then we will have the loss of the leaving group to first of all use the first equivalent and form a key tone. Right? So we are going to form a key tone first, where now we have our three on the left are four on the right. And now the second equivalent can be used to produce that final tertiary alcohol. So if the second equivalent is used, we understand that our four MGB r. That same granite region can attack one more time. And then when we are forming that bond, there's no more possibility to restore the carbon in a bond because in particular, we already have four bonds total and no good leaving groups around. So we can say that our final product here would be first of all that Alcock side and ion which is At a tertiary position. Right? We have our three R four R and then we're essentially protein eating it in an asset to form our final alcohol. And then we are getting our tertiary alcohol. Now, if you look at this scheme in particular, you will understand that it doesn't really matter whether we use an acid chloride or an ester, why they essentially have good leaving groups each. So the mechanism for an ester looks the same for the purposes of this problem. Let's choose an acid chloride. And we will perform our analysis based on the given scheme. Notice that there are two equivalents of the Green Yard Regent added number one over here. And then we're adding the second equivalent during our next edition, because there are two equivalents of the Green Yard Regent added. We notice that these two are our al kiel suspicions that come from the green region and our goal is to look at our tertiary alcohol forms and identify to exactly same substitutions. However, we understand that all these are ethel substations and all of them are the same meaning we can just say that because we see two equivalent substations to ethyl groups. Our Green yard region would be ethel magnesium bromide. So basically ch three ces two MGB R. That's the granting of region we want to use. Right? So we understand that. Okay, we have assigned our two ethyl groups right to the Green yard region. Now, this implies that the remaining L kill chain would give us our starting material. We previously said that our three in this case is the substitution bondage that carbonell group. So if we change our three to an ethyl group, we will get our assets Glory that we are going to use. So let's label this is our Grenier region, right? Because this will be important for us. And now the world last thing we're going to do, we're going to think about it this way. We have already eliminated these two ethnic groups. And the remaining ethyl group corresponds to our substitution R. Three which is bonded to the carbon group of the acid chloride. So we need to draw an ethyl group. Let's start by drawing the carbon group itself. Okay, so that be our carbon group. Glory in And then we are signing two more carbon atoms. Because as you notice we have 12 more. That's our al kill chain bonded to that carbon. So that would be 12. So that's our starting material. We can just label this acid chloride is our starting material And then we will just look at the overall synthesis. So we have determined our starting regions and this is how the synthesis can be performed. Now we are using our assets chloride. So that would be pro pion l chloride. And as we previously sad based on our scheme, we need to use two equivalents of the Green Yard Region. So step number one is to add the equivalence of ethyl magnesium bromide. And then when we produce that, Alcock set an eye on, we essentially want to protein ate it right? This would give us our tertiary alcohol. So there will be we ethel groups and that will be it. That's how final synthesis. Now, once again, if you want to think about whether or not that works, you can just refer back to the mechanism, right? We are going to use the Green reagent attack the carbon ill group wants to add that first ethel substitution over here. Right then the carbon group would be restored because chlorine can easily leave and adding the second equivalent would add another ethyl group followed by acidic hydrolyzed as we would protein ate our elk oxide and iron and as a result we would form our tertiary alcohol. So we can say that this is our final reaction scheme based on the reaction scheme. We can say that the correct answer to this multiple choice question is B However, if you have any questions, don't have States, leave your comment down below in the comment section and thank you for watching.

Show how you would add Grignard reagent to acid chloride or ester to synthesize the following alcohols.
a. Ph₃C–OH

Key Concepts

Grignard Reagents

Nucleophilic Addition

Hydrolysis of the Intermediate

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