For each compound,
1. name the functional group.

Question

For each compound,

1. name the functional group.

2. show what compound(s) result from complete hydrolysis.

(c)

Accepted Answer

Welcome back everyone name the functional group present in the following compounds and show the products of their complete hydrolyzed sis. Now let's start by looking at the functional group in this compound notice that there is one carbon atom specifically here that is bonded to two elk. Oxy groups right on the left, we have one auction atom that is bonded to carbon on the right we have another oxygen bonded to carbon. Let's recall that whenever you have two elk oxy groups wanted to exactly the same carbon atom. Let's suppose that you're analyzing O. R. One, then you have R. T. O. Over here. And let's suppose here you have to substitute prints. Perhaps they're just other alcohol groups R. Three and R four. Now whenever you have these two elk oxy groups wanted to the same carbon atom you call this as an acid. I'll so we can say that the given structure because it has that carbon atom with two elk oxy groups bonded to it actually that would correspond to an A saddle. And this would be our first answer to this question. Now the world, last thing we want to do is just draw the products of the complete hydrolyzed this of this? S settle and we will remember the steps how to do so. So first we need to recall that we are using an asset right? So we are going to read through our asset. Okay and now let's just try that adjacent meth oxy group O ch three. The welfare step is the protein nation. Stop, let's recall that oxygen has long pairs and it can get easily protein ated. So as long as we use our asset, let's suppose that we use this lone parent this oxygen because we have two elk. Oxy groups will repeat that same step. But for the worry beginning it's just sufficient to show that we are protein ating it using hydro knee. Um So we're going to use H 30 plus. Let's get a bit more space. So this is our hydro knee. Um which represents strongly acidic conditions. Now we're going to use equilibrium arrows in each step because these steps are reversible celeste at equilibrium arrows, let's clearly indicate that here we are using acidic conditions. Let's draw the result and intermediate. So that would be our five member ring. Let's add our meth oxy group. And then we understand that this oxygen is protein ated. So we have that oxygen hydrogen bond. There will be a former positive charge and oxygen. And from here we're ready to think what happens next. So basically because oxygen has a positive charge, it's easy to remove it in the form of a leaving group. This is a cyclic structure. And actually, if you notice the adjacent oxygen atom would essentially use its lone pair in the formation of a pi bond, followed by the cleavage of the adjacent carbon oxygen bond. And this would give us our next intermediate. So let's try it. Now we notice clearly that we are opening our ring and that's really important to understand because we're doing so it would make sense to enumerate our carbon atoms. So 1234. Okay, and now here we have our meth oxy group. Now let's draw our chain. So we may begin by drawing four carbon atoms at position one. We're going to create our alcohol group. And now at position four, it'll be over here. We clearly notice that we have hydrogen atom, right? So there's hydrogen which is implicit let's expand it. And now we also have our carbon steel group. So that would be double bonded carbon C oxygen which still has that metal group. Right? So we're going to add that metal group Oxen would have a formal positive charge here because we have three bonds. And from here we're ready to think what happens next. So in the next stop, we have our water molecule formed over here. Let's not forget to include that additional era, which represents that. We are essentially adding a proton and then removing the water molecule. And this water molecule can essentially attack that the electra Philip position. So we're going to use water. Now the lump erin auction would essentially attack the electrolytic position. We would break our pi bond and then we would form the next intermediate. From here, we are ready to draw another set of equilibrium arrows. Let's redraw that structure. We understand that at this position we will have two fragments, one of them will be our meth oxy group O ch three. Right? And now on the other and we will have our oh age age, right, that's our water molecule. But now in this case we have auction with a formal positive charge because auction forms three bonds. We want to make sure that we d protein ate it now. So we're going to use another molecule of water for that D protein nation. Stop. Let's see that. So another molecule of water acts as a proton except er All right, we're going to get our hemi acid out because now we will have an elk oxy group and an alcohol group and it's the same carbon atom. So now let's retry that hemi acid out. We have our O. H. And we have our meth oxy group O CH three. Now from here it's really important to understand that we can have another protein nation step we have acidic conditions between these two. Remember that anne Alcock C group can accept a proton more easily due to the fact that there's an inductive effect adjacent to that auction. Notice that there is this metal group as opposed to hygiene. So it's donating electrons to oxygen and therefore oxygen becomes more basic here than here. And essentially we're going to use that lone pair hydro nian will donate a proton. Okay and now we are essentially protein eating it. So let's see what the outcome is. We're going to do another set of equilibrium arrows. Now we still have our alcohol. And now this O. C. Three got protein ated. So we're going to show that oxygen hydrogen bond with a former positive charge on oxygen. And from here because once again we got a positive charge and oxygen, it's easy to get rid of this fragment from the structure. We can use the lone pair on oxygen adjacency that meth oxy group or in particular, this is already methanol to be specific. Right, we just have it positively charged. So we can use the lone parent oxygen to essentially form a pi bond followed by the loss of the leaving group. So we're going to form methanol and then in this step we still have a protein ated form of our carbon group. So let's see how that looks like we have our ohh and you'll notice how we are forming that double bond oxygen still contains that proton. So it's possibly charged as a formal positive charge. And from here we can use another molecule of water to D. Protein ate it. And that would give us our final products. Let's keep in mind that one of the other products would be must And also let's add that ch 30. H. Right, we got read a bit from the original fragment and now here we're going to use water to dip protein it oxygen and form our Aldo hide. So we are ready to draw our final products from here, we are just going to indicate our final equilibrium arrows and throw the final product. That'd be our hydroxy substitution. And now we are getting our al to hide. Well done. So basically we can say that the products of this high troll is this would be these two hour al to hide and methanol. In addition to that, one part of our problem asked us what kind of functional group we were dealing with at the beginning. We set aside so we may also conclude that one of the answers that he may wish to remember was acid. Also let's label this as well. We have three answers. For the first part we said that the original functional group was style because it contained two elk oxy groups bonded to the same carbon atom. And then when we looked at this extensive hydrolyzed this mechanism, we said that we are forming two products. One of them is our alcohol, which is methanol and the other products our al hide with our hydroxy substitution. The correct answer to the multiple truest question is a However, if you have any questions, don't hesitate to leave a comment down below and thank you for watching

For each compound,
1. name the functional group.
2. show what compound(s) result from complete hydrolysis.
(c)

Key Concepts

Functional Groups

Hydrolysis

Ethers

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