5

2) Singulair is a leukotriene receptor antagonist drug used for the treatment of asthma: Its structure is shown below i) Fill in the missing lone pairs in the struc...

Question

2) Singulair is a leukotriene receptor antagonist drug used for the treatment of asthma: Its structure is shown below i) Fill in the missing lone pairs in the structure below,ii) indicate the hybridization ofatoms 1 - 5 (in the given boxes), and iii) circle and name 3 functional groups. [5 pts](3) OH(2)OH(3)(4)(5)(5)

2) Singulair is a leukotriene receptor antagonist drug used for the treatment of asthma: Its structure is shown below i) Fill in the missing lone pairs in the structure below,ii) indicate the hybridization ofatoms 1 - 5 (in the given boxes), and iii) circle and name 3 functional groups. [5 pts] (3) OH (2) OH (3) (4) (5) (5)



Answers

Albuterol, shown here, is a commonly prescribed asthma medication. For either enantiomer of albuterol, draw a three-dimensional formula using dashes and wedges for bonds that are not in the plane of the paper. Choose a perspective that allows as many carbon atoms as possible to be in the plane of the paper, and show all unshared electron pairs and hydrogen atoms (except those on the methyl groups labeled Me). Specify the ( $R, S$ ) configuration of the enantiomer you drew. (FIGURE CANT COPY)

So we have here is on to H two So we have 12 valence electrons on. We have the following interaction when we have nitrogen sigma interaction and then we have the interactions between our hydrogen and our nitrogen on. We also have our lone pairs. We also have the double bond between the two nitrogen ins That is the pie and P and peace we have Ah Sigma nitrogen sp two h s In the next example we have and to H four So we have a total of 40 in valence electrons. This looks slightly different now. So we have our nitrogen. Thanks Chair Sigma interaction we have our love has there now we have four interactions with one s hydrogen. What we have is N s P three interacting with H one s that is the Sigma. Following this we have C H three n h two 14 Valence Electrons have a carbon interacting in a segment bond with a nitrogen. Nitrogen has a lone pair. It also has two sigma interactions with hydrogen. Carbon has three interactions with hydrogen So we have carbon sp three interacting with hydrogen one s sigma type bonding. We have nitrogen s p three Interacting with hydrogen one s sigma type bonding and we have carbon sp three Interacting with nitrogen SP 34 Sigma type bonding.

So starting with per day we are going to assign the hybridization of the numbered atoms and also give the bond angles surrounding that pattern. So starting with a we have die. It'll either and I already drawn them on how they were given in the text book. And the first atom is this oxygen here. So it has to bonds and two lone pairs of electrons, So does sp three hybridized. And since the electron geometries tetra hydro Um we know that it will be less than one of 9.5 since two lone pairs of electrons will actually push the carbons closer together. And for the second atom is this carbon, It has four bonds. So this Sp three hybridized and as tetra hydro Um electron geometry and shape and has no lone pairs of electrons. So we know that they will all have ideal bond angles of one of 9.5°. Moving on to part B. We have caffeine and the first adam to give the hybridization is this carbon here. So we know that it has two Single bonds and a double bond. So it's sp two hybridized And we'll have bond angles of 120°. The next atom is this carbon also is sp three hybridized and has bond angles of 1 20 and last on the caffeine is this nitrogen, which has two bonds, one single bond, one double bond and a lone pair of electrons Making it sp two hybridized. And we'll have bond angles of slightly less than 120° because that lone pair of electrons again pushes the bonds closer together. Can you get a bent shape for part C. We have a seagull salicylic acid. The first item to give the hybridization of visits oxygen, which has two bonds and two lone pairs. So it is sp three hybridized With bond angles of less than 1.09.5. Next is this oxygen that has one bond, so one double bond in two lone pairs, so does sp two hybridized, and the angle between the carbon and the oxygen Is 1 80°. And this carbon here is sp two hybridized. Inside has three bonds And we'll have bond angles of 120° in party. We have nicotine. This first carbon has four bonds, So it's sp three hybridized. And although the bond angles are 109.5 since etc federal shape Number two is this carbon That it is sp two hybridized. It has three bonds And it's trickle planner shape, so has bond angles of 120. Next is this nitrogen that has to bonds So one double bond, one single bond in a lone pair, So the three regions of electron density making an sp two hybridized, and the long pair of electrons pushes the bombs closer together. Some sort of 120. The bond angle. This bond angle here on the inside Will be slightly less than 120°. Yeah, For number four is this nitrogen that has three bonds and a lone pair of electrons. So it's sp three hybridized and as bond angles a bit less than one of 9.5 because one of the regions of electron density is a lone pair of electrons pushing the bonds closer together and lost. Is this carbon that is sp three hybridized instead of four bonds And the bond angles are one oh 9.5 degrees and says Petra federal shape. And lastly in part E, The F. D. Greene. And the first item is this nitrogen that is three bonds, 1 lone pair. So it's sp three hybridized with bond angles of a bit less than 1 to 9.5 degrees, since it's tetra hydro electron geometry, but there's a long pair of electrons pushing the atoms that's bound to closer together. 2nd is this carbon that has for bonds, So it's sp three hybridized, none of them are lone pairs, so they are all ideal bond angles of one or 9.5. Next is this carbon That is sp two hybridized with bond angles of 120. And lastly, is this oxygen that has to bonds and two lone pairs. So it's sp three hybridized with bond angles of less than 109.5°.

So now we're getting into more complicated molecules. So we look at the skeletal structure thing to give us an idea of how the molecules laid out his first one. Age bonded and I wanted to and spawned its age. But the skeletal structure doesn't tell us necessarily how many bonds we have knows at least one. So Nitrogen wants to form three bonds. It's on ly formed to so far someone to give it a second one Here. Most of the collect can't give the second one the hydrogen. It can only form one Paul. Now we shouldn't forget that alone. Paris Here waken count the number of things around the nitrogen. It's going to be one two three. That makes it F P two hybrids, both of them. You can see that there's symmetric. He could imagine playing a mirror in between them. And because there s p two hybrids, that means that they have a pure metal. It's going to be bonding. And that's where second bond comes from. The Reds are Pete first next one and two h four. We can see the skeletal structure. Looks like I'LL draw over here. Two hydrogen sze bonded to a nitrogen. That's what the H two n means and then the and and then the age to its two more hydrogen, both nitrogen also have a loan pair is because they have five valence electrons. They've only formed three bonds. They also already have all the three bonds that make them happy, which means they're gonna be F p three. We've got four things around their three bonds and bow. Wanna let Trump Air Chris next one. We've got a little bit of, ah, offside molecules who's got each three seats. I like to draw that, starting with the carbon and then adding on the three hydrogen sze and the next one in line is nitrogen, and it's funded to to hide regions. And if we do a quick trick, everything has. The number of bonds that likes hydrogen will have one. Carbon has four nation has three and we just need to add His girl appeared there. This is we've got no double bonds and you were here and no expanded up tests. Everything is F p three

So in this problem, we want to do a hybridization and bonding scheme for the Given Adams. And so the first is C two h two and so will draw are to seize and these Cesaire both going to be sp hybridized and then we'll draw our h is. And so we have a bonding pair at each of these Ch is and those air going to be Sigma bonds between carbons SP, Orbital's and H is S Orbital's. And then, since this central carbon pair is triple bonded, we're going to have two sets of pie bonds. And so we condone draw in the electrons in these pie bonds and then we'll label them. Ah, so these air pie bonds between carbons p orbital's and the other carbons p orbital's. And then we have this Central Sigma bond between the carbons. And so that'll be a Sigma bond between carbons, ESPYs and the other carbons. Espy's so let her be is C two h four and so again, we ever to central carbons. However, this time they are SP two hybridized and so we have our four h is each of them with its own electron pair and so those four orbital's. Those were going to be sigma between carbons SP too, and Hydra Jin's s. And then we still have a double bond in the middle. So as opposed to a we're only going to have one pi bond. Ah, so here we have a pi bond between P Orbital's and one carbon and P orbital's in the other. And then we have Lastly, we have this Sigma bond between carbons SP two and the other carbons SP too. So let her see is C two h six and so will draw are to seize again. This time they have SP three hybridized orbital's and so we can draw in our six h is and then each of them has an electron pair And so those six orbital's are going to be signal orbital's between carbons SP three and hydrogen Sze s And then we just have this last central signal orbital between carbons SP three and another carbons SP three


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