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Aujenmgplr cm cduqeMALMLIUNEReturn to BlackboardOrganic Chemistry; sxstem announccmaM5Devstsyninetic rouleconud-el /-r-ntemmethylhekanal:I0 1) NaOEt; 2) BH-THF; J) ...

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Aujenmgplr cm cduqeMALMLIUNEReturn to BlackboardOrganic Chemistry; sxstem announccmaM5Devstsyninetic rouleconud-el /-r-ntemmethylhekanal:I0 1) NaOEt; 2) BH-THF; J) HzOar NaOH 0 1} NBS; hv; ?) BHi-THF; 3) H,Oz NaOH1) Bta; Giocon NaNh ; 3) Hzo; 4) 1 HCI; ROOR heat1) NaMt ; 2) CH,M; 3) HC; = ROOR;4B;Tess MaNtz; 9) mo;0,; 5) #oMartt ; 3) #;0; disumylboran €H-OkOh")M;;

Aujenmgplr cm cduqeMALMLIUNE Return to Blackboard Organic Chemistry; sxstem announccmaM 5 Devst syninetic roule conud -el /-r-ntem methylhekanal: I0 1) NaOEt; 2) BH-THF; J) HzOar NaOH 0 1} NBS; hv; ?) BHi-THF; 3) H,Oz NaOH 1) Bta; Giocon NaNh ; 3) Hzo; 4) 1 HCI; ROOR heat 1) NaMt ; 2) CH,M; 3) HC; = ROOR; 4B; Tess MaNtz; 9) mo; 0,; 5) #o Martt ; 3) #;0; disumylboran € H-OkOh ")M;;



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$\mathrm{FeCr}_{2} \mathrm{O}_{4}$ (Chromite) is converted to Cr by following steps: Chromite $\stackrel{1}{\sim} \mathrm{NaC} \mathrm{O}_{4} \stackrel{11}{\longrightarrow} \mathrm{Cr}_{2} \mathrm{O}_{3} \stackrel{\mathrm{III}}{\longrightarrow} \mathrm{Cr}$ Reagenrs in 1, II and III srep might be Srep I Srep II Srep III (a) NaOH/air, $\Delta$ C $C$ (b) $\mathrm{Na}_{-} \mathrm{CO}_{5} /$ air, $\Delta \quad \mathrm{C}, \Delta$ (c) Conc.H $\mathrm{SO}_{4}, \Delta \quad \mathrm{NH}_{4} \mathrm{Cl}, \Delta$ $C, \Delta$ (d) $\mathrm{NaO} 11 /$ air, $\Delta$ $\mathrm{C}, \Delta$ $\mathrm{C}, \Delta$

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.

The ocean is correct problematic. So we have one degree alcohol is reacting with that and we are H2 so far. And we are left. Who will give up? I mean we are someone will give degree al Qaeda attached to the boat. Then first we will get this thinking not the same thing we will get. It does help if you know any PR is reacting and it was very electable attacks. We're -1 attack here. Then living there, We are here. 50s. Thank you. So they will go to that be Three degree alcohol is reacting with 100. Let's see it. Then we will get Uh huh. They start to real cold. Real cold is reacting with the PBR three. Then we will get become So he will go to that S We will go through that. I believe in ghosts are civil. Goes without c saving gift. He and they will give job. So this is the correct column match that we have to identify. Okay.

Well the question is sodium. I drove. I dropped sleep. I think I grow a score pick absorb moisture when exposed to the atmosphere. I student placed a quarter of anyway. On the last a few years later, he noticed that the color was covered. A white solid, identify the white solid. We need to identify, identify divide solid. So sodium hydroxide as we know this is hi gyroscopic material. This is the hydra skah pick compound which absorbs the moisture then exposed to the atmosphere so we can write a. It wasn't to any elect less. C. 02 the atmosphere atmosphere. This will give who Here to see you three less Will not give care to 03 since it is too anyway. So it will give between you U. N. O. H. Plus your two will give an A. To C. +03 which is solid. So this is what which looks like a white solid. Yeah, we can see that. Hobson. She is correct. And to see you.

Hello portion is full of magic. Okay so we have to master column and the first two molecules there is Mhm. We will get this by one will break and also to trust will come here and this will add on the bench you drink so we will get vengeance and here will be ch Ch one or twice. This is what you mean. Okay I said Robin Benji to ask and for telegraph reaction and electrically substitution reactions. And that the molecule we will get after the elimination of SCM we will get this molecule this this and all. See you this molecule. So it is a certain woman reaction and it is a dreamer but okay because the only shorten human reaction in the sea molecule. Yeah it is a middle class reaction and electro filic substitution reactions and in this molecule we will get sandy sending sms. So electoral Filic substitution reaction and the delicate after the action. Also. So to remote demand reaction. Not shuttlecraft. It is a remote demand reaction. The multi condition is this. So our comparative. Okay


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