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Progunta 32 Sin responder nDe entre las sigulentes reacciones, la que muestra producto correcto es NOz Nz CHCIPuntua comoWarcar IpreguntaAICI;Cl-C= -CH3 AICI}NHz CI...

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Progunta 32 Sin responder nDe entre las sigulentes reacciones, la que muestra producto correcto es NOz Nz CHCIPuntua comoWarcar IpreguntaAICI;Cl-C= -CH3 AICI}NHz CI-C-CHa"CH3 NHzAICHd.1,"CH3CH;

Progunta 32 Sin responder n De entre las sigulentes reacciones, la que muestra producto correcto es NOz Nz CHCI Puntua como Warcar Ipregunta AICI; Cl-C= -CH3 AICI} NHz CI-C-CHa "CH3 NHz AICH d.1, "CH3 CH;



Answers

$\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \frac{\text { diethyl ether }}{\text { Anhyd. } \mathrm{AlCl}_{3}}>\mathrm{A}$ Product $A$ is (a) $\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}$ (b) $\mathrm{CH}_{3} \mathrm{COOCH}_{3}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOC}_{2} \mathrm{H}_{5}$ (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$

The order of the strength of hydrogen born order strength of hydrogen bond is and it and listen to it. Mm. Just listen efforts have Its strength is 13 kg julie Parmalee. It is 18 and it is 40 km from all closing mood. So as the difference in the electronic negativity increases, the hydrogen born will also increase. So we can write AS L. To IAN that is electronic activity difference increases. Then strength of it wasn't born will also increase. So we see that option. Option A option A is correct. Option is correct.

In this problem, I can write DDX and head see STD CH two CS three in Potenza, one equal land In pageants of one equivalent land. We are to HIV will give the The agent at CS three ch ch t d br and this in but then so MG or italy will give the components C h t see it, see http. Mdbr and this in emergence of duty. Or We'll give the final compounded CH three CHED see http. So according to the option options. See it correct here. Option C It's correct. I hope you understand these solutions.

For this question. We're gonna draw the sister transpired, Rose, for this particular, uh okay. And so we're just gonna start off by drawing one of the ice mers naming it and then finding out with the other one has to be The first thing I'm gonna do is from my butt off. And so coming off this carbon, I need, uh, basically what hydra I need to know. So I put in my and my coming off of the other side. There is a high fashion. There's a medical group, so it doesn't matter what I draw first. So this has to be one of the Iceman's, and if we look at what side everything is on So basically, um, for this particular car. But the biggest argument method that's on bottom side, this carbon, the biggest are your business, but also on the bottom side. So these two are Groups are the biggest argues for each of their respective parent, Harbin's. And so that actually makes this this this and contrast is really easy to draw a trans. I saw what you figured out. One of them. We just need to keep one side the same, so I'm gonna keep the left side the same. And I just need to flip right side. So now my biggest argue on this carbon If I must come here because our group on this carbon is my It is over here. And so these two groups, my biggest artists, are no longer on the same side that makes this transit.

In this problem, I'm writing the reaction. Just look at it carefully. See us afford Plus to K C N will react to give, get to and so forth. Plus see you CN two to see you. CN two will react to give to see you CNN plus CNN too. Finally see you CNN Plus today K C N will react to give jittery. See you CN food. So the answer for this problem is option B. Option B. It correct answer.


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