Suppose we have some optically pure (R)-2-butyl acetate that has ...

Question

Suppose we have some optically pure (R)-2-butyl acetate that has been 'labeled' with the heavy O-18 isotope at one oxygen atom as shown.

(b) Repeat part (a) for the acid-catalyzed hydrolysis of this compound.

Accepted Answer

Hello, everyone. So let's consider the fallen compound that has labeled with this 0 18 isotope of oxygen. We have to write a plausible mechanism for the hydraulic sis of this compound under acidic conditions. So under acidic conditions, we will never have a negative charge, will only have positive charges in our molecule. So the reaction starts with the presentation of the carbon Neil group. So let's first of course, throughout the starting material. So again, we have this six member ring here with alternating pi bonds and this double bound oxygen then are labeled 0 here and the rest of our chain with a methyl group sticking up. Alright. So again, we're talking about the pronation of the carbon Neo group. So we have our H plus here. So of course, we need to go ahead and add in our lone pairs. So this lone pair will go ahead and grab this H plus to continue our reaction. Again, we're drawing basically everything the same of our story material except for the pollinated carbonell group. So double bound. Oh alrighty. So now that we have this oxygen attached, we only have one long pair. So basically we have a positive charge on this oxygen. Now, the next step is to attack of the watermelon keel on the promenade carbondale group to form a tetrahedron intermediate. And then a proton transfer then takes place between the water oxygen atom and the like oxy oxygen. So then we have our H20 coming in. Of course, we're always using our lone pairs. So this oxygen here will go ahead and tack this slightly positive carbon. And then one of the pipelines will go ahead and break and form a lone pair on the carbon Neil oxygen. So this will go ahead and form. Let's see. So we have again, this untouched six member ring with alternating pi bonds. And it will happen. Of course, again, we have a tetrahedron intermediate. So we will have an O H group will have Basically an H20 group. And then this will be a positive charge on this oxygen and continuing our chain. We have this 018 Here drawing in the lone pairs and the rest of the molecule with this metal group here. And what will happen from here is that the lone pairs from this labeled oxygen is going to go ahead and attack and grab a proton. Then this bond between the O and H will go ahead and break to form lone pairs on the oxygen right over here. And this is going to remain untouched. So what this will yield is, let's draw this like. So again, the six member ring is not touched. So what will yield is going to be now that we have our hydrogen from our water being taken, then we have 20 H groups, then we have our labeled auction having one proton. This will of course gain a positive formal charge and then the rest of our chain. So step number four is that one of the two H group pushes its electrons toward the carbon atom to expel the alcohol forming a protein ated car box silic acid. So again, using the long pairs here, we're saying that the lone pair from the alcohol will go ahead and make a double bond here and expelling this right over here, including our labeled O H T isotope. So this will go ahead and yield again, we have our six member dring here. Alternating Pi bonds will have two different products. So double bound oxygen and R H. So we still have this positive charge. And then, oh H at the end, and will also yield the O H group here. But this one is labeled with 0 and this method group as well. So the loss of a proton from the Carbondale group produces the final carb oxalic acid product. So then from our product here, it's actually kind of redraw this. So we do have a bond with the hydrogen. And of course, we need our lone pairs. So water comes back again. And what the H U L will do is act as a base grab this proton here and the O H bond right here. Let's see. So we don't have this long pair. Actually, this will go ahead and break to form a double bond or a lone pair for this oxygen. So then our final products, if I go ahead and scroll down for more space, what we will get is going to be again, of course, our two products well, gets The six member ring here than our double bound oxygen O H group. And by product being H 30 plus. So our products for this, of course, is just this right over here. But this whole entire thing is going to be our plausible mechanism for the hydraulic sis of the compound under acidic conditions.

Suppose we have some optically pure (R)-2-butyl acetate that has been 'labeled' with the heavy O-18 isotope at one oxygen atom as shown. (b) Repeat part (a) for the acid-catalyzed hydrolysis of this compound.

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Suppose we have some optically pure (R)-2-butyl acetate that has been 'labeled' with the heavy O-18 isotope at one oxygen atom as shown.
(b) Repeat part (a) for the acid-catalyzed hydrolysis of this compound.