Some protecting groups can block two OH groups of a carbohydrate ...

Question

Some protecting groups can block two OH groups of a carbohydrate at the same time. One such group is shown here, protecting the 4-OH and 6-OH groups of b-d-glucose.

(a) What type of functional group is involved in this blocking group? (b) What did glucose react with to form this protected compound? (c) When this blocking group is added to glucose, a new chiral center is formed. Where is it? Draw the stereoisomer that has the other configuration at this chiral center. What is the relationship between these two stereoisomers of the protected compound? (d) Which of the two stereoisomers in part (c) do you expect to be the major product? Why?

Accepted Answer

Hello, everyone today are the following problem. Some protecting groups can protect two hydroxy groups on a carbohydrate. At the same time, one such group protects the hydroxyl groups at carbon four in carbon six of beta de manos as shown below. And then we have the following questions. The first question a says to what functional group does this protective group belong? So this protecting group formed an assay towel Functionality with Carbons four and 6. So we're gonna say that it has a subtle is the derived from an a subtle functional group. And then we have B which states what re Agent is employed in the synthesis of such protected compounds. So if we were to do a retro synthesis, the given structure can be made from two things. one Ben's al to hide. And of course, the sugar which is beta de manos. And so our Benz Benz Aldo hide, it's going to be our ketone connected with a carbon and that carbon, that central carbon is gonna be brought into a final group and an oxygen group. And that is going to stem this structure here. And of course, we have our beta de manos which is the rest of this sugar molecule. So if you go to C C says when this blocking group is coupled with beta D meadows, a new Cairo center is formed, what is its location? So we see here, we have this product, the Cairo center is essentially going to be this carbon right here. This is going to be our new Carol center. This is when the final group is at the equatorial position. However, it's a stereo I summer would have this final group in the actual position. They can get less stable. And that brings us to question D draw the compounds, other stereo and assess which stereo I should be more stable. So we're gonna draw D down here and the Iceman is going to be very similar to the original, we're going to have the sugar here. The main difference and the only difference is the switching of the hydrogen and the final group. So the final group is going to be on top in the actual position and the hydrogen is gonna be in the equatorial position or the sideways position. So we can say that the original stereo Ice hammer is more stable because of that positioning of that final group. And so with that, we have solved the problem overall, I do hope this helped. And until next time.

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