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Question

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Answers

Treatment of propionaldehyde with dil. $\mathrm{NaOH}$ solution gives (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$ (b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}$ (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$

Okay. This problem is essentially asking Why are different? Esther's more reactive with an a average than others. So here we have, um this compound which looks like this this is my Esther. It is a metal estate. And then we're going to have our other states as well. We're gonna have one which has an ethnic group attached to it, one that has a isopropyl attached to it, and then one that has a turn bugle group attached to it. Okay. And I drew these all in order of, um, decreasing rectory, meaning that this one is more active than this one and so on and so forth. Okay, so the reason behind all this, why any which is considered more reactive with this one and it is with this one is because we have the presence of this metal group compared to the presence of this turbulent group. So don't forget that Noh or my hydroxide ion has to, um, add to my carbon. If I just only have a metal group there, then there's not a whole lot of interference between my alcohol coming in and attacking my carbon. Whereas with my tribunal group, let's a have a hydroxide on all the way over here. It'll be a lot harder for my hydroxide ion to come in with the presence of this tribunal group because this actually has 1 to 300 is attached to it. 2300 attention to it, etcetera. So this is a very bulky group and might be prohibiting the hydroxide I had to come in. And that goes the same for all of these as well, because this is a little bit more bulky than this, and this is a little bit more bulky than this, and so on.

As we all know that where A whenever at C H. O. Including Canada a reaction it is oxidized and the other, I'll be hide it radioed. So I'm just writing the re accent CS three C. See http. CST CH two plus sc at you in potential Annie you edge will give the final compounded C S. T. See see http see http CH 20 h plus at c O n E. Therefore according to the given option option B. Of seven BH. Correct answered.

No the question is what will be the product to see. Okay so here we have criminal paranoid broken all is reacted with any which so this is the best and as a parent so that's two mobile gets eliminate and we will get phen oxide and they're welcome. Thank you. And further it is reacting with the CS three CS 2 I oh minus will attack on the carbon and then we will get OC two H 5 here will no further then and I see it so things are reducing metal that will reduce nitro into amenable. Well C2 H five here is that a miracle right? This molecule will react further. She has trees fields theater So here from Excel will gets eliminate and we will get Most repeat that's five and it's sealed PhD so Kay Thompson will be deep.

This is the answer to chapter 20 to problem number 47 fromthe Smith Organic chemistry textbook. Ah, this problem asks us to draw the product formed when fennel acetic acid is treated with the tree agent. And we're told that with summary agents, no reaction occurs. Ah, and there are 12 of these. So I'm just gonna go ahead and get started because this is gonna take a little bit of time. Okay. S o for a Ah, we have sodium bicarbonate. Uh, that's gonna act as base. Um, and we have a car back so weak acid that we're starting with. Oh, and I drew fennel acetic acid at the top of the screen here. Rather than redraw the aromatic ring 12 times, I'm just gonna use a pH to abbreviate it. Since the ring itself is not participating in any of these reactions. Okay, s O sodium bicarb here is gonna act as a base. And so we're gonna get an asset base reaction and will de protein, eight hour carb oxalic acid. Ah, and we'll get the sodium ion as the counter I on here. Okay, s so that's a for be, um, again be Is a base. Ah, and so the exact same thing is gonna happen as in a we will de protein eight. Ah, and we'll have the sodium ion as our counter ion. Okay. Ah, see, we're using signing chloride. Remember, signing chloride, um, takes a carb oxalic acid to the acid chlorides, and so we'll have an acid chloride for a product there. And C D is just salt table salt. There's gonna be no reaction here. Okay, s so then e um and actually, let me move these up a little bit. Okay? So in e our using one equivalent of ammonia. Ah, and so that is going to just act as a base. And so again, we're gonna get D prone nation. Ah, and no further reaction. So Deep Throat Nation, and then we'll have the ammonia my eye on as the counter ion. Okay. Ah, for F. We're using ammonia, followed by heat. Um, and so that is actually going to drive this reaction. And so, rather than just a pro nation here, we're actually going to make the A mite. Okay, so there we go. There's f moving on to the next page here. Fergie were using methanol and then sulfuric acid. Um, and so we're going to get the Esther product here on DSO Since we're using methanol will get the methyl ester for H. We're gonna try to do the same thing, but instead of using acid, we're gonna use base. Um, and so that's not going to work. Ah, and so we will end up just deep resonating. Okay. For I AA were using sodium hydroxide, um, followed by on acid chloride. And so, uh, this is going to create the mixed and hydride. So we d pro Nate. Ah, and then our deep throat nated oxygen is able to act as nuclear file Ah, and kick out the chlorine. Um, in the acid chloride that were using is the second re agent in this sequence. And so we end up making the next anhydride. So then, for j r. Using Mesilla mean ah, in D c c. And remember D c. C. Is gonna act as a dehydrating agent here, and so that is going to allow us to form the A mine. Instance for using meth will mean our product is going to be metal. A my okay for kay, we're gonna use. Sign your chloride. Um, and then we're gonna use problem meid. Um and so the final chloride is gonna make the acid chloride. Um, And then, uh, the pro chloride is going to, um, add to that. And so we end up with the A mine. So 123 Okay. Ah, and s o l is going to be similar. Um, but we're not using a mean, so we're not gonna get in a mine. Ah, we're using an alcohol. And so we'll get an Esther. So the Esther will be the Pro Bowl. Esther. Okay, so there we go. Um, yeah. So that's all of the products to these reactions. Ah, and that's the answer to chapter 20 to problem number 47.


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