5

QueSTION7The prodability Inut =umed may gel EIe A in Muth: 102608 Jna Shjier In"y Ect E'Jde A in Kutns 10230.65 Wnjl i {ne Probub Iity Ihat 48/0'4 on...

Question

QueSTION7The prodability Inut =umed may gel EIe A in Muth: 102608 Jna Shjier In"y Ect E'Jde A in Kutns 10230.65 Wnjl i {ne Probub Iity Ihat 48/0'4 one 0f them mill sore Ejoe4075 0.85 0.020,93

QueSTION7 The prodability Inut =umed may gel EIe A in Muth: 102608 Jna Shjier In"y Ect E'Jde A in Kutns 10230.65 Wnjl i {ne Probub Iity Ihat 48/0'4 one 0f them mill sore Ejoe4 075 0.85 0.02 0,93



Answers

Manoalide, a sesterterpenoid isolated from the Pacific marine sponge $Luffariella$ veriabilis by Scheuer and co-workers at the University of Hawai'i at M$\bar{a}$noa, has anti-inflammatory, analgesic, and antifungal properties. Find the isoprene units in manoalide.

Lim Kok Granuloma Veneer IAM is a sexually transmitted infection that is generally asymptomatic when it is symptomatic. It is usually mild and involves mostly regional lymph node swelling. Even though it is asymptomatic or mildly symptomatic. It is highly contagious and is very common on college campuses. It is caused by Cira Brewers of chlamydia, true commenters.

I have even so in this question they asked the second in addition energy of the carbon nitrogen option. And flooring atoms are such as is carbon far far greater than nitrogen far far greater than flooring for greater than carbon bees. Flooring greater than option greater than nitrogen greater than carbon C. Is carbon greater than night auction greater than nitrogen greater than deep. Please oxygen greater than floating greater than nitrogen greater than got a button so yeah. Option dis correct. Yeah. Option beads curry nine days, 9 days the second ionization energy of carbon nitrogen option flooring are yeah, The 2nd Innovation and Adultery of carbon nitrogen option and flooring are in the older Yes, yeah auction greater than flooring which is greater than nitro ocean. This is greater than carbon. Yeah. Yeah. Yeah. Yeah. Yeah. Mhm. Half fill two P three sub shell of a plus is highly stable, highly stable then stable compared to compared two to be force sub selloff. Yes. Yeah. Of a plus sold. I used to of folks who is young will be larger than f minus. Yeah. No Jeff then eight minus. Thanks lot.

This is the answer to Chapter 30. Problem number 32 from the Smith Organic Chemistry textbook. In this problem asks us to draw a mechanism for this cycle ization reaction between foreigner sealed I phosphate and I So Penton Neil Penton he'll die phosphate. Um And then it asks us to comment on some unusual features of this reaction. So to start drawing foreign steel die phosphate here, So the first thing that's gonna happen, Um, And remember, we've seen this before in other problems in this chapter. So the dye phosphate group, and far too sealed by phosphate, is willing to leave. So the electrons will go to the oxygen, um, and will form carbo cat eye on. Right. So So after the dye phosphate leaves, we're here where we have that carbo cat eye on. Um And so at that point, that's when the ice Oh, Penton, he'll die phosphate eyes going to come in and react. So here is our S o pente, Neil die phosphate. And, uh, the electrons in this double bond will attack the corporal cat eye on that re formed in the first step. And so that, uh, will give us this intermediate. So give me a moment to draw it out. Oh, and actually, I just noticed that I made a mistake in the previous intermediate. So this double bond should not be here. It should instead be here. So I apologize for that. Uh, and so back to drawing this intermediate. So here is the product of, uh, this addition. Let me just particularly note that nitrogen there on. Of course, we have a carbo cat eye on here now, um, and so it's important to note that nitrogen there because that's the nitrogen that a base in solution. And we're just gonna represent base with a capital B. They're so base is gonna come, um, and grab that proton. Grab that hydrogen. Um, and the electrons will revert to that bond to make it a double bond. Um, and so product here, eyes. Now, um, this So basically all that we've done is, uh, if you look at our starting for in a sealed by phosphate, we've basically just extended it by one unit. Right? So it looks like we just just tacked on another repeating sub unit. Another another one of these, um, and in effect, that's that's what's happened so far. So, um, what we need to do now is redraw this in such a way that it's easy to visualize the cycle ization that's going to occur here. So in order to do that, um, you can start at one end here. Just just bear with me. This made take me a moment to draw. Oops. Okay. And then here's our ropp or die phosphate. Okay, so the obviously the point of redrawing it like this, uh oops. And I dropped a metal group. There we go. Okay, so the point of redrawing it like this is, um just because it makes it much easier to show this cycle ization that's going to occur. So once again, the O. P. P. Group is very willing to leave, so we can just show that bond breaking the electrons going to that oxygen there. Um and so what will happen next? Um, is that you know what? I'm not even gonna redraw this entire thing. So here's what I'll do. So there's one terminus, then the other terminus is going to look like this now, so we'll do this, and we can use this sort of squiggle curvy line connecting these just to show that there's a whole ring going on over here and that cuts down on time having to redraw the ring, making mistakes and drawing rang. And then right. So nothing, nothing from this portion of the molecule here has changed it all and is really affected by the rest of this reaction at all. So we might as well just leave it out, and I will redraw it for the final product. Ah, but until then, there's there's not really any point. Um, and so we now have a carbo cat eye on here. And very similarly to the first reaction in this mechanism, this double bond is going to attack this carbo cat eye on. And so the product that is going to be this Ah, and again, I'm going to draw this simplified version that is ugly, right? So I'm actually now, of course, that closed the ring so and again, we can just draw squiggle there, um, and so will draw, actually, and draw it this way. Uh, Proton there, hydrogen there. Um And then, of course, we will have a carbo cat eye on here. Um, and so, uh, what will happen now again, Exactly. Has happened in. The first reaction in this sequence is that, um base here will show up. Snag this hydrogen. These electrons will revert to their to reform the double bond. Um, you know, unfortunately, it looks like I'm gonna have to draw, uh, my product on a new page. So? So there we go. There's the final product. Um, flecks of bean. Right. So, uh, so the other part of this question that were asked is what is unusual about this cycle? Ization reaction and what's unusual about this cycle? Ization reaction. Is this right here where I'm gonna show you in red. So a corporate cat ion has formed here in the mechanism that I've drawn. Um, and that's a secondary carbo cata. Uh, whereas ah, corporal cat I own could form uh, elsewhere. Our team make a, uh, tertiary carbo chi on. So if if instead of of that corbeau cat eye on we formed Ah, Corbeau cat eye on, uh right. So if instead of that, if instead of that one we formed the carbo cat eye on here, um so this the ring wouldn't close here. It would close here instead, Um, and that would be a tertiary carbo cat eye on. And, of course, tertiary carbo cat ions are more stable, but it would result in a different ring to it would result in a 14 member grain instead of 15 members. Um, so, uh, the the reason for for this the reason that this is able to happen is that, um you know, in reality, uh, the electrons here are sort of bouncing back and forth. So each of these positions has somewhat of a cargo cat eye on by the difference being when the carbo cat eye on is at this position, the position it's actually in it allows us to close the string down and make this ring. Um, which for whatever reason, is preferential. Ah, and so that's what's unusual is that we would expect this to to form a tertiary carbo cat eye on, or at least to rearrange to a tertiary carbo cat. Eye on over. That's not what we see. Um, and of course, the result. A CZ, I said, Is this molecule flecks of being? Um and so that's the answer to Chapter 30. Problem number 32

In this problem in this problem Sino hydro rain. So I know. Hi dream of CS three CH. Oh, I know. I do know CS three. She at your really yield bill? Yeah, I will you'd lactic ation. I will gild like the cash in according to the option of Cindy H of 10 b. H., correct answer for this problem.


Similar Solved Questions

5 answers
EntndFnne(W-JnMn Trip *711ComaciDAUAHIcuyre nETOALIH"cor Plllin " ZT"IPFZ f 0o4iP7r cosin prr)40tanranen IDJ ecoitaFarhlzhenanvux-Hoarllncuon upiniol[[email protected] l0temanlo(II =6" 0<*<W(U[VM(1, WOrMhdepookMTC
Entnd Fnne (W-JnMn Trip *711 Comaci DAUAHI cuyre nETOALI H"cor Plllin " ZT"IPFZ f 0o4iP7r cosin prr) 40tanranen IDJ e coita Farhlzh enanvux-Hoarllncuon upiniol [[email protected] l0temanlo (II = 6" 0<*< W(U[ V M(1, W Or MhdepookMTC...
5 answers
Question 3 by induction or otherwise to show that (a) Prove De Moivrets Theorem by-"Ccos(8) isin(n0)) [r(cose isin8)]"
Question 3 by induction or otherwise to show that (a) Prove De Moivrets Theorem by-"Ccos(8) isin(n0)) [r(cose isin8)]"...
5 answers
Points)Write tne detinitionfunction named which the size of the population that begins yearwth 2490 members and decreases ata 3.5 % annual rate. Assume that time measured years.Tne annual inflajion rate 3.5% per year with inflalion:movie ticket costs 58.50 today; write the deiiniiion ofa funciion named the pricemovie ticket years from today; assuming that movie tickets keep UpAccording your formula; how much will movie Iicket cost in 15 years?help (numbers)
points) Write tne detinition function named which the size of the population that begins year wth 2490 members and decreases ata 3.5 % annual rate. Assume that time measured years. Tne annual inflajion rate 3.5% per year with inflalion: movie ticket costs 58.50 today; write the deiiniiion ofa funcii...
2 answers
(b) Compute the integralf(z) COS Z dzwhere C is the unit circle |z/ = 1/4 traversed once clockwise
(b) Compute the integral f(z) COS Z dz where C is the unit circle |z/ = 1/4 traversed once clockwise...
5 answers
5 , Show thel 75 204 (X;-XJ2 24 LX; ~Xjj 2 n(n-) j2(
5 , Show thel 75 204 (X;-XJ2 24 LX; ~Xjj 2 n(n-) j2(...
5 answers
222 F } 1 HH 1 L 1 L D [ 1 72 0 : F M M 4 m 1 [ 1 JH [ | 8 1 1 1 7 1 1 J
222 F } 1 HH 1 L 1 L D [ 1 72 0 : F M M 4 m 1 [ 1 JH [ | 8 1 1 1 7 1 1 J...
5 answers
Graph of gFor x 2 ], the continuous function g is decreasing and positive A portion ofthe graph ofgis shown above. For n 2 1, the nth term of the series 4n isdefined by & = g(n). If6~ x() dx converges to 8, which ofthe following could be true? Ea,=6(B)2 0,= 8 (C) 2 an =10(D) 2 Qn diverges
Graph of g For x 2 ], the continuous function g is decreasing and positive A portion ofthe graph ofgis shown above. For n 2 1, the nth term of the series 4n isdefined by & = g(n). If 6~ x() dx converges to 8, which ofthe following could be true? Ea,=6 (B) 2 0,= 8 (C) 2 an =10 (D) 2 Qn diverges...
5 answers
A positive charge is moving to the right and experiences an upward magnetic force; as shown in this figure:+Q In which direction must the magnetic field have a component? A to the right B. to the left C upward D. out of the page E into the pageHow would you convince someone that your answer is correct? Please describe in detail the steps of the Right-Hand Rule used:
A positive charge is moving to the right and experiences an upward magnetic force; as shown in this figure: +Q In which direction must the magnetic field have a component? A to the right B. to the left C upward D. out of the page E into the page How would you convince someone that your answer is cor...
5 answers
Use Algorithm 5 to find $123^{1001}$ mod 101 .
Use Algorithm 5 to find $123^{1001}$ mod 101 ....
5 answers
Plot the function $f(x)$ over the interval [1.5,2.5] Zoom in on the graph of each function to determine how close $x$ must be to 2 in order that $f(x)$ is within 0.002 of $4 .$ Your answer should be of the form "If $x$ is within __________ of 2, then $f(x)$ is within 0.002 of $4 . "$$$f(x)=x^{2}$$
Plot the function $f(x)$ over the interval [1.5,2.5] Zoom in on the graph of each function to determine how close $x$ must be to 2 in order that $f(x)$ is within 0.002 of $4 .$ Your answer should be of the form "If $x$ is within __________ of 2, then $f(x)$ is within 0.002 of $4 . "$ $$f(x...
5 answers
H2.1 Level 2 Homework UnansweredIdentify the alkyl substituents' systematic name and common nameSelect an answer and submit: For keyboard navigation, use the upldown arrow keys select an answer3-methylbutyl (isopentyl)1-methylpropyl (sec-butyl)2,2-dimethylpropyl (neopentyl)1-methylethyl (isopropyl)
H2.1 Level 2 Homework Unanswered Identify the alkyl substituents' systematic name and common name Select an answer and submit: For keyboard navigation, use the upldown arrow keys select an answer 3-methylbutyl (isopentyl) 1-methylpropyl (sec-butyl) 2,2-dimethylpropyl (neopentyl) 1-methylethyl (...
5 answers
In Exercises $59-70,$ solve the equation and express each solution in the form $a+b i$.$$x^{2}+1770.25=-84 x$$
In Exercises $59-70,$ solve the equation and express each solution in the form $a+b i$. $$x^{2}+1770.25=-84 x$$...
5 answers
What is the mean number of kilowatt-hours used each month by the four families for hot water?
What is the mean number of kilowatt-hours used each month by the four families for hot water?...
5 answers
Solve the followings equations:i) ((3𝑥 + 2)^2)𝑦′′ + 5(3𝑥 + 2)𝑦′ − 3𝑦 = 𝑥^2 + 𝑥 + 1ii) (𝑥^2)𝑦′′ + 𝑥𝑦′ + 𝑦 = 2 cos^2( log 𝑥)
Solve the followings equations: i) ((3𝑥 + 2)^2)𝑦′′ + 5(3𝑥 + 2)𝑦′ − 3𝑦 = 𝑥^2 + 𝑥 + 1 ii) (𝑥^2)𝑦′′ + 𝑥𝑦′ + 𝑦 = 2 cos^2( log 𝑥)...
5 answers
Complete and balance each of the following equations. If noreaction occurs, enter NOREACTION.Express your answer as a chemical equation.Enter NOREACTION if no reaction occurs. Identify all ofthe phases in your answer. Please answer A-DA- Ca(NO3)2(aq)+KCl(aq)→B- NaCl(aq)+Hg2(C2H3O2)2(aq)→C- (NH4)2SO4(aq)+CuCl2(aq)→(NH4)2SO4(aq)+CuCl2(aq)→D- NH4Cl(aq)+AgNO3(aq)→
Complete and balance each of the following equations. If no reaction occurs, enter NOREACTION. Express your answer as a chemical equation. Enter NOREACTION if no reaction occurs. Identify all of the phases in your answer. Please answer A-D A- Ca(NO3)2(aq)+KCl(aq)→ B- NaCl(aq)+Hg2(C2H3O2)2(aq)...
5 answers
The time (iniseconds)ltiakes to get a ball to reach 30 feet is modeled by the following quadratic: 30 = 11647 12644271 Solve the quadratic for the variable Kwhich represents seconds) then solutions for t at which the ball wlkelBo feet give the two possible high. t#3/2 seconds, 4/8 seconds[72 seconds, ( 8 seconds[52/3 seconds; t-8 secondst=l secondit-B seconds
The time (iniseconds)ltiakes to get a ball to reach 30 feet is modeled by the following quadratic: 30 = 11647 12644271 Solve the quadratic for the variable Kwhich represents seconds) then solutions for t at which the ball wlkelBo feet give the two possible high. t#3/2 seconds, 4/8 seconds [72 second...
2 answers
HospitalTatulTot4292699 |432922,252IMRS[95270246242953Admittcd12771558665:9844485Not admiuted382051634728 310316.8145640
Hospital Tatul Tot 4292 699 | 4329 22,252 IMRS [95 270 246 242 953 Admittcd 1277 1558 665: 984 4485 Not admiuted 3820 5163 4728 3103 16.814 5640...

-- 0.022858--