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(36 pts) Fill in the missing products reagents, Make sure include stereo chemistry when necessany. there reachon, write NO REACTIONShow Major Product(s) OnlyAICI;Sh...

Question

(36 pts) Fill in the missing products reagents, Make sure include stereo chemistry when necessany. there reachon, write NO REACTIONShow Major Product(s) OnlyAICI;Show Major Product(s) OnlyHzN-NHzKOHShow Major Product(s) OnlyAICI;

(36 pts) Fill in the missing products reagents, Make sure include stereo chemistry when necessany. there reachon, write NO REACTION Show Major Product(s) Only AICI; Show Major Product(s) Only HzN-NHz KOH Show Major Product(s) Only AICI;



Answers

Draw the product formed when pentanal (CH $\left._{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\right)$ is treated with each reagent. With some reagents, no reaction occurs.

We will solve this problem from hydrogen and the S block elements were certain reactions were given for which we have to suggest the product off the reaction. So the first reaction is the reaction between sodium peroxide and carbon dioxide. When Sardenberg oxygen carbon X that react, they will produce sodium carbonate that is any to see you three and oxygen. The second reaction is the reaction between sodium oxide and water. Sodium oxide reacts with water to produce sodium hydroxide. That is any which the third reaction is a reaction between potassium hydroxide and sulfuric acid. This reaction is a reaction between a base on an acid and is known as neutralization reaction, in which salt and water will produce that is que two s 04 on its two are produced. The next reaction is a reaction between petition hydroxide and hydrogen iodide, which reacts to produce potassium iodide and water. The next reaction is the reaction of oxygen with sodium in the presence of liquid ammonia. On this will produce sodium super oxide. That is any photo, and the last reaction is production between sodium and chlorine. This will produce sodium chloride, that is any seal. So these are the products off the given chemical equations

Problem. 89 here is giving us this massive fat molecule. And it's asking for us to react it with K O H and give the products. Okay, so, um, we have hiss. What? What we have here is a, um a try Asia, legless rok. And whenever we're reacting, try a So glycerol with a base, it makes a split like so. Okay, so probably not try that, but it makes a split. And this molecule and what it does is it gives it it gives us back our components for a try. Ace of bliss. A religious one goes for a molecule and three fatty acids cape. So let's go ahead and split this molecule. So we're splitting it. Right? So now we're gonna have hydrogen sze on thes um, oxygen's right. So these oxygen's we're going to be crabbing hydrogen. Okay, so now we have our glycerol and three fatty acids. So let's go ahead and Neymar products here. So we have a glycerol molecule. Okay. And here we have we have on 18 carbon fatty acid. Gaby got 16 c h two's in. We got two more carbons in here, So 18 carbon, um, fatty acid would know double bonds is gonna be steri gas it. Right. So this is just information that you could find in your book. Okay, so that's a stereo guy said next, we have a fatty acid with one double bond. Okay, so that makes it our Oleic acid. Like acid. Okay. And finally, here we have this, um, fatty acid with three double bonds. So it got has got three double bundt, which makes it our Lena Lennox acid Glen. No Lennox acid. Okay, so there we go, that that's all our products.

All right. So we're given to to die substitute Ben Zines again with some schemes. Uh, and we just got to predict where the sight of reactions going to happen. What's positions that substitution gonna happen at so keep in mind is that it's not the reaction. Um, the reaction of the next substitution reaction that's going to dictate where, um, where the substance mint goes, it's what's already on the rings. Right? So in this case, you can look at this disorder Dobro Motel. You mean, um and you can say, OK, so they're both Ortho para directors. Um, the Alcoa Group is a stronger director than the bro mean, although they're both Ortho para right. So you probably said the Method group would want to direct here and here. The broom Ian would want to direct here and here. But as I said, the Method group is the stronger director. So you probably favour substitution at either of the sites that the methyl group is directing you to. So really, the Method group is the major director here. Um, and we're doing a Friedel crafts calculation. So we have ethyl chloride with aluminum chloride as the Lewis acid um, catalysts. So, um, you're going to get a mixture of, uh, this product, and then you're going to get a mixture. The other component is likely going to be the one that's parent to That's your adding and Ethel Group. Right. So between these two products, I would say you're probably going to have a small R small excess of this one on because that Ethel Group is gonna have a fridge rotation. Um, and you see some Starik blocking here, Um, where you have the ftf of group C. H. Is and the method groups ch is air colliding. Um, it's probably gonna favor the power position substitution while still being a mixture of Ortho power. You're probably gonna favor the PARA as the major products among the major products, right? Something else. The text doesn't really say this, but you know that, um, that the products of Al Kalay shins are more active than the starting material because that Alka group is going to activate Um, so just as a fun exercise, you might even see, um, some products of let's say this Paris substantive prowess. Now the V positions, um, here and here are gonna be more activated because now you have NFL Group, um, directed to those sites so you might even see, um, for example, it's gonna be pretty hysterically encumbered, but you might even see Ah, some Tetris substituted benzene form if you get a second alkaloid. A second calculation on the already alkaloid instead of products because you know that that can happen with calculations. So just a thought experiment. Um, right. So you'll probably get a mixture of those products with the metal group Ortho para directing, um, over the broom in and then similar to this situation. Right. So you're going to still see those same sites being activated, but instead of a NALC elation, Now we're doing a nitrate shin, so you're gonna have either the Ortho less likely because you have that, you know, to those Eno bonds are gonna be in the way of the ch is here, Right? So, um, you probably see a mixture of Ortho para with the side excess of the power of products. So, um, those are your two products with respect to this substrate moving on rights. And now we have a different war. Now we have, um you have this meth OC or film a fox, he tell you Mean, um, so you can do the same things here, right? So they're both Ortho parrot directing. So you're gonna want the Method Group will be directing to these two positions or thrown parrot to the Method group while Harrell use Green. The meth Oxy group is going to be directing to these two positions. So the meth Oxy group is a much stronger activator than the method group is activating. So you're probably not going to see um, your CV Ortho Power products of the where the meth Oxy is directing those next substitue INTs going. But remember them. The foxy group is probably gonna have some Starik effects hindering this Ortho position slightly giving probably a slight excess of that of that power of products. Um, so we have the same mixtures here, So you have You have a methyl group. You have the meth oxy, so you might see eth elation at the Ortho position. Plus, uh, no Thanakorn. It's a metal. There you go. And you might see some power products again. You might see some Polly substitution ins from that scheme, but also your to nutrition products. We're gonna be the same positions. So you might see an Ortho nitrate Ah nitro group or you'll see a Parametric group again, probably preferring that power position. Do hysterics, they go.

This question asked us to draw the products from each of these reactions, assuming that there's an equivalent amount of each region. So we have a 1 to 1 ratio here. So for part A, we have two different double bonds that we can react with in the way that we wanted to. The side which wanted to use is based on the carpet. Katyn. So the first step in this reaction is the double bond grabbing the hydrogen, and then we're left over with a car pool. Cattle. So if we were to use this elevon down here, we would have a secondary couple cat eye on each side. If you were to use this one up here, we could get a tertiary Carbo Catomine. So what, we're gonna dio we're gonna grab the age. But that's double bond. Um, and that's going to give us an intermediate Hey, trying seven member drinks, um, an intermediate carbo, can I? And that looks like this. So it's hard for a share of carbon Carradine. And then the only other stuff we have is to attack the carbo cabin with the, uh, roaming nuclear file. So that gives us this product So this, uh, double bond hasn't been touched, And then we have my muscle group and are roaming on the same carbon. So that is part A. And then in part B, we also have to Elkins to choose from. So again, we're gonna look at the possible possible carbon Catalonians. So from this top, one of the best car broke down and we could get is right here, and that's a secondary. But if we use this one, we can get a tertiary right here. So it's what we're gonna do. Grab the hydrogen. Could go, bro. Mean that one. Don't think tush. And then our car broke out. I and it's right there on the tertiary carbon. And then the broening comes into the tax it. So our product looks like this and there is our brought me. It's a lot of party


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