Draw the predominate structures of Leu at each of the indicate...

Question

Draw the predominate structures of Leu at each of the indicated sections ( 1, 2, 3) on its titration curve.

Accepted Answer

So this practice problem wants us to draw the predominant structure of loosen at each of its indicated sections on the Thai tray shin curve. And so the sections are one which is here. We have section two, which is over here on the Thai tradition curve, and we'll draw that structure here. And then we have section three of the Thai tray shin curve which corresponds down here, and we'll draw that structure here. Now, before we actually get to drawing the structures, we need to notice that there are these two PK. So we've got one PK over here showing us PK one and we've got a second PK over here showed us p k two and then over here in the middle. What we have is the P I of Alan in, or the ISO electric point of Allan E. And so up here in this first region that we're gonna draw the structuring we're referring to. Essentially this highlighted region here and this highlighted region corresponds with the pH value on the Y axis that air higher than the P k A for both of the P. K s and recalled that when the pH is higher than the PK that the conjugate base predominates. And so loosen on. Lee has to ionized able groups the amino group in the car Boxer group And both of those ionized herbal groups are gonna be, uh, in their conjugate base form. So the conjugate base form of the amino group is just a NH two, which is not charged. So we'll leave it like that and then we can draw in the car boxing group and the conjugate base form of the car Boxer Group is going to be a car Boxley n i n, which is negatively charged and then recall that loosens our group so loosens our group is a loose extension of veiling. So it just has an extra ch two in comparison availing so it's just gonna look like, uh, this so that is losing structure at this particular region of its tight tracing curve. So now let's move on to the next region and the next region is showing pointing right at here, which is the same location as Theis Electric Point. So really, you can think of, uh, this whole second region as being this region and between all the PK is because the predominant structure is gonna be the same at this in this entire region. And so it's gonna be the region where the p I is. So we know that the overall charge has to be neutral in this green region. And so, uh, basically, what you'll see is that in this green region here, this is, uh, lower than the peak A A for the amino group, which means that the peak a uh, which means that the amino group is going to be in its conjugate acid form and the conjugate acid form of the Amino group is NH three with a positive charge. And so this region here is actually higher than the PK of the car boxing group. And when the pH is higher than the car boxing group P. K, that means that the car boxing group is gonna be in its conjugate base form and so we can go ahead and draw the car boxing group and its conjugate base form. And the conjugate base form, of course, is just gonna be the car Boxley an eye on. And then we can, of course, filling our our group, which is again non ionized able. So it will be ch two ch and two metal groups coming off. And so this is the predominant structure here, and it's net charge is going to be zero. And so keep that in mind here we can keep track of the net charges below. So the net charge zero for this structure and for this structure over here, the net charge is negative one. So that was up here in this region. The net charge was negative. One and the net charge and this other region here is zero. So now let's look at the third region, which is down here in this region down here, and notice that this region corresponds with pH values that are both below the both PK. So PK one m p k. Two. So this region is below both PKs and when the pH is below the P. K. That means that the conjugate acid is gonna predominate for both of those ionized herbal groups. And so that means that the amino group is gonna be in its kinda get acid form which is pro donated like that with the charge and then the car boxes group is also going to be protein ated. Tow the car. Boxer group is gonna be in its O H form, which is neutral. And so again, we can draw in the our group and the our group is gonna be just a ch two with the CH and two lines coming off for the method groups. And then the overall net charge at this structure is just going to be positive one. And so down here, we can put a positive one. Help us remember the charge. So again, as the pH starts to increase, notice that as the pH increases that the charge of the predominant structure decreases. So there's this inverse relationship between the ph of the solution and the charge off the predominant structure. And so that concludes this practice problem, and I'll see you guys in our next one.

Draw the predominate structures of Leu at each of the indicated sections ( 1, 2, 3) on its titration curve.

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