Cytosine, uracil, and guanine have tautomeric forms with aromatic...

Question

Cytosine, uracil, and guanine have tautomeric forms with aromatic hydroxy groups. Draw these tautomeric forms.

Accepted Answer

Welcome back, everyone draw the taut tome forms of adenine and thymine with aromatic hydroxy and amino groups. We're going to start with the recollection that adenine and thymine are representative as complementary bases that are essential to the structure of DNA. Recall that these two bond with one another via hydrogen bonding. Now, next, we'll focus on the keyword tato meic. Now recall the word tatom Americanism recall that Tatom Mars describes the ability of a structure to inter convert between multiple structural forms known as constitutional isomers. And this will occur via a mechanism that will involve the movement and transfer of protons and electrons. So let's start with the structure. The simplified structure of adenine recall that it has a fused ring structure as follows. And because this again is going to show the conversion of this structure to a tumeric product, this will be required requiring sorry an equilibrium arrow reaction and sorry an equilibrium reaction, arrow double head arrow. So now let's go ahead and recognize a very crucial group present in the structure which we observe is where the nitrogen is bonded to the carbon of the ring. And that nitrogen has two bonds to hydrogen and a lone parent itself. So recall that, that represents a group, which we recall is a amino or a mean group. Recall that the cyclic amino containing structure of adenine is going to taize to an Iine product. We're going to need to show the mechanism of how that occurs. So let's start with the lone pair on our nitrogen in our amine group and shifting that to form a pi bond with carbon of the rink that is then therefore going to cause the adjacent pi bond in our structure to shift as an additional lone pair on this aerial nitrogen or nitrogen, a part of the ring. And that is therefore going to then create a potential for a negative charge on this nitrogen. And to mend that or avoid it, we're going to be able to utilize the original lone pair present on this aerial nitrogen to capture the hydrogen from our amine group, which will therefore cause the remainder bond to shift as a replacement lone pair on this nitrogen of our amine. So as a result of this movement of electrons and ultimately protons, in a sense, we're going to form the following structure where we can now observe the presence of our iine group where nitrogen has now a pi bond shared to the carbon of the ring and one bond to hydrogen. So from this mechanism, we can understand that the cyclic amino containing structure of a sorry Adenine Tatom rises to form a cyclic amine Toomer product. And now we're going to think of stability. So, recall that and sorry, this is actually missing a bond here to carbon. OK. So with that corrected, now, let's recall that the more favorable structural form is actually going to be our reactant form of adenine. And so we're going to make the left hand half arrow of our equilibrium go longer towards the left or reactant side. This can be explained by the recollection of the fact that when we've got cyclic pie bonds with between atoms such as nitrogen and carbon, in which we have two that we can observe in this structure, this reactant structure of adenine that is going to enhance the stability of the molecule as a whole because of the stronger kalumba attractions due to the smaller radius of those atoms in the pyon nitrogen and carbon. Whereas in our amine product or amine Toomer products, we have a nitrogen carbon p bond that is technically exocytic and just one nitrogen carbon py bond, which is endocytic. And so ultimately, as we stated, the reactant form is going to be favored containing more endocytic nitrogen carbon py bonds which enhance the stability of this reactant Toomer form of adenine. So with this understood, our first answer is going to be the structure that we've created for this, I mean Toomer product. So now let's move on to the second base considered thymine that the prompt mentions and recall that it contains the following structure, which is actually just a single ring. So we'll start with the identification of another second type of essential group that we can observe in the structure where we've got a carbon, which is connected to a carbon and also with a bond to a nitrogen with just one bond to hydrogen. So this is going to be recognizable as a group which we recognize as a primary amide. And ultimately, it is a cyclic amide structure that we have of thymine. With this recognition, we will start with the recollection that thymine is going to taught to marise to form an en all product. In order to form an en all product, we're going to need the oxygen in our carbonyl of the amide group to have a bond to hydrogen. And so, in order to form that bond, recall that we can utilize a hydrogen capture step where we can use one of the or a set of the electrons in the pi bond to capture this hydrogen atom from our nitrogen in the ring, which is going to in turn shift its remaining bond as a lone pair on this nitrogen, that's going to create the potential of a negative charge. So we can shift then the original lone pair on this nitrogen as a pi bond and as an adjacent pi bond, which will cause this pi bond to have to shift to this position on the ring and now writing out our equilibrium half arrows and forming our product, we will have the following structure in which now we have our en all Coomer product of thymine. Now recognize that equilibrium is actually going to favor the product side. So we'll have a longer half arrow going towards the right. And that is due to the fact that we have an endo cyclic nitrogen carbon pi bond signifying significant calumba attraction, strengthening this bond and overall the stability of this molecule because it is endocytic, it is an endocytic nitrogen carbon pi bond. In our reactant structure of thymine. We did not observe any endo cyclic nitrogen, carbon py bonds. And therefore, again, the product form is favored, which is our en all products that we've exemplified here where we now observe oxygen with a bond to hydrogen. And that is actually going to complete this prompt as our two final answers where we've got the highlighted structures for the Toomer products of adenine and thymine respectively. I hope that this made sense. And let us know if you have any questions.

Cytosine, uracil, and guanine have tautomeric forms with aromatic hydroxy groups. Draw these tautomeric forms.

Key Concepts

Tautomerism

Aromaticity

Hydroxy Groups

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