An important protecting group developed specifically for polyhydr...

Question

An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyldisiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule.

(a) The TIPDS group is somewhat hindered around the Si atoms by the isopropyl groups. Which OH is more likely to react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen. (b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final product with two oxygens protected. (c) The unprotected hydroxy group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with deprotection with Bu4NF.

Accepted Answer

Hello, everyone. Today we have the following problem. Tetra Isopropyl die Silo Oxon Hill or Tip DS is an important protective group that was created specifically for poly hydroxy substances like sugar. Tip Ds can shield two alcohol groups of a molecule. Then we have the following questions. A says the isopropyl groups around the silicon atoms hinder the tip DS group slightly which hydroxy group of one meth oxy alpha D Rubino Fury knows has a higher chance of reacting with T tip DS chloride first draw the product with one oxygen attached to the tip DS group. So essentially the isopropyl or the unhindered hydroxy group is going to be attached first and that is going to be this here. And so how that's going to look is we're going to have the following. So it's gonna be bound to see that oxygen there. And then we're gonna draw like normal, we'll have our silicon touch the oxygen touched with our silicon. And then that's bounded by two for isopropyl groups And the chlorine and in the first silicon is joined by two isopropyl groups as well. B says the next oxygen nearest to it reached is reached once the tip DS group has been connected to the first oxygen, draw the final product when the tip DS attaches to two oxygen's. So we're actually gonna draw this down here. So we're going to retain our original sugar molecule. Only our second oxygen is going to now essentially be connected. And then see says, note that the ox, now that the oxygen's are shielded, the unprotected hydroxy group can undergo reactions. All the final product is a result of treating the protected one meth oxy alpha D Arabian Arabian off your nose from B with Tulsa chloride period and then sodium burrow hydrate. So If we treat this second product here with one tussle chloride in peru emitting py for short and then sodium bromide, we're going to get the following product. And then after that, we have to treat it with borough hydride and a water molecule. And when we do that, we're essentially going to get just the remaining product, which is our sugar with a browning. And with that, we have solved the problem overall, I hope this helped. And until next time.

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