Consider the open-chain structure for D-gulose in Question $13.51$.a. Draw the Fischer projection and name the product formed by the reduction of D-gulose.b. Draw t...

Draw the product formed from an aldol reaction with the given starting material(s) using $^-$OH, H$_2$O.

a. $(\mathrm{CH}_3)_2\mathrm{CHCHO}$ only

b. $(\mathrm{CH}_3)_2\mathrm{CHCHO} + \mathrm{CH}_2 = 0$

c. $\mathrm{C}_6\mathrm{H}_5\mathrm{CHO} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}$

d. $(\mathrm{CH}_3\mathrm{CH}_2)_2\mathrm{C}=\mathrm{O}$ only

And asked us to determine the structures. Structure a Structure B structure A is in D. Aldo, Texas, which means we'll have six carbons, one carbon at the top. 234 i six Carbon at the bottom. The D configuration means we have the hydroxy group on the rights, and then the eldest means we have an alga hide functional group at the top, and this is the generic structure of a D LDA Hyksos structure B. As an LDA pintos, we will have five carbons, carbon one at the top, 234 and 1/5 carbon right at the bottom. We will maintain the D configuration and the Aldo Hide Function Group at the top, and this is a generic structure of an elder pantos. We also know that one structure be undergo the Colonial Fisher synthesis. We make structure a so you just look at the structures. You'll see that in Elda Pintos differs from an elder hacksaws, and then elder Texas has one extra carbon. So therefore we know that the Colony Fisher synthesis we'll extend the carbon chain of a sugar by one carbon. Additionally, we know that one structure A has reacted with sodium borough hydride, we make an optically inactive albuterol. What that means is we maintain the six carbons and an elder toll will have come all alcohol functional groups, optically inactive means if you were to draw a line in the middle of the compound so right you're here. We will see that this top half will mirror the bottom half. And since we know this is a D sugar, this group up here must maintain the same configuration. We also know that the oxidation of compound B will form an optically active all dark acid. Small, dark acid. Again. Compound B has five carbons. No dark acid has, uh, has carb oxalic acid groups at the top. And at the bottom. We know that structure B is a D sugar. That does not change. And we also know since it is optically active if you were to draw a line right through the middle, you would have different configurations at the top Carbon this carbon right here versus this other carbon. So we know. But this, like structure me, we'll have hydroxy group on that side. We would have now dropped the complete structure of structure and structure. Would be structure A Remember that it's a D yellow Hyksos 56 carbons India Sugar Who have the outrage group optically inactive around all the hydrogen, of course, was around the all the high group as well. Should be the complete structure of structure a structure B remembering that it's optically inactive. Well, the same. We'll have five carbons. Well, the same Alba High Group. We'll have the same configuration here with the D sugar, but then it's optically inactive. We will have the hydroxy group at the top on the other side.

Asses to grow The Hayworth projection from the Fischer projection. So first, I would better the tallow Pierre and knows. So there are certain rules that already mentioned in the textbook and that we're just gonna apply that rooms. So in order to draw Ah Hayworth projection off. This is a paranoid. So it's six member ring so far star from drawing a six member ring with the with oxygen at the up right corner off the ring. Excuse. My draw always looks good. Bet. Okay. And so, in order to make things easier, I label the number over here, and we can follow it. So this is gonna be number one, two, three oh, five. And a siege to always will be number six. So, um, fossa bone is the better, and so better. I mean, he always group will be up. An alpha means the always group at the him. Our star position will be there. So this is a better so age will be here Next. Um, it's a d ah more osaka. Right. Which means that the siege to a wage will be up all right. And now the next parked will be growing thehe wage for number. Come on before 32 So the, um rule is if the a wage on the right side it would be down. And if on the left side will be up and all three off 234 over here on the left side. So they should be up, away, away. Ohh! I didn't take it too, chin. OK, so using the Seine Ah, roll. Let's draw the Hey Whoa! Hey, Walls Projection of better the mountain of Mount No, paranoid. So again, starting with a six member ring with oxygen at the rye Up Carmen. And this is a better. So again, better What is, uh, always up again? We keep the label. One, two, three. Well, five. And number six will be up because it's a better. I'm sorry. Is that a the ah de manos? So see a stew. Oh, it will be appeared. And now we need to right now. The ohh position for 23 and four and again left up right now. So these two, two and three left should be up, and this one's dead. Now let us fill in the hydrogen hydrogen attuned chin. Have it all right. Ah, far de Gallo. Octo Pyra knows so again is the six member ring. Start from labeling one too. 3456 And Starwood growing a six member ring. Oxygen on the ride Top Kona, huh? OK, it's the offer again. Alfa with mean that awaits. You will be there. And I would keep the label. 123456 Against a store, It will be up. The car is he's a d mortal. Second rate now to three for a wage on the right side, that means now a wage on the left side That means up and that our staff would be put in hydrogen Hye jin hydrogen had to. So this one will be five member ring Furano ce But the rules is basically the same. So 12345 And we start with growing a five member dream Sam role. If it's Alfa, then always should be down. And the glucose csto it will be a and because, uh, okay, that's still a boy. One, 2345 again, if he always in this case two and three on the rights I that should be there and the last step would be put in hydrogen. So the finals one over here is again is Ah ah! Five members ring and we label from here. So is the Alfa, which meant that always now that also mean the seas to away. He's upset me. Soc doorway is over here. These are on the left side, which means that it should be up and that we just pulled in hydrogen.

Action on outdoor pintos A of 12345 days You can see here oxidized by nitric acid Thio die acid, which is inactive optically so because it is a d outdoor pendel. So this always group has to be here, and so b is optically active and you can see that there's a plan of symmetry going through right here. So which they use to possible structure of the this to always have to be here and this. Oi, either be here are on this side. And so, which again? Late to flip back to two possible structure of a which is Ohh! Over here, Over here. Over here. Uh, they are all in one side now. Um, so with this, let's go on this one and two. And I rewrite them here, wanting to sew with key down. You fish us. And this is basically at one more cab on into the chain and generate compound C and d soc inlay A plumber's, um so going with the first possibility, you could have seen which a wage off the side and the witch from the side. Either one going through the second possibility you can have sea away from the side. Indeed. Hoid on that side. And so I rewrite. Ah, the structure over here. So you knowthis are our first possibility and this is our second possibility. So another piece of information is when you oxidize the sea and the product see is optically active and me is optically inactive. So when you oxidized, bossy Ohh There's a plan of symmetry that going through the middle off the molecule right here. So you could see that in this pairs they are This one is optically active and this one is also optically active. This bear, this one is optically inactive and this one is optically active. And that's why um see have to be this. Come over here and the is this company over here? So which Ah, give us. Um, the final answer for the correct a is this Well compound over here. And B is this compound over here

Some were submitting these three actions to mild oxidizing engines. Ah, so what that's going to do for this 1st 1 is what that typically does is convert uh, alcohols into carbon nails. Ah, So for this 1st 1 the reaction we're going to get is when we mildly oxidize this, we're gonna end up with a key tone. Ah, and that's true for any secondary alcohol, right? So if the next one, obviously we're starting with this out of hide on the one thing to remember is that when we're oxidizing and how to hide where we're going to end up with is a complex that look acid, which is why it's very dangerous to use a strong oxidizing agent on a primary alcohol because more than likely, you're just gonna accidentally oxidize it twice. All right, so what we're gonna end up with is an alcohol in here instead of just the age making that car books that, like acid, right in the last one again is another secondary alcohol, meaning that when we make this product, basically what we're going to get is something that just looks like that