Newman projection que.

[predentgal]

Hey everyone.. I'm having trouble with this question. Answer choice B looks more like cis to me than does C. How do you approach this problem?

http://img38.imageshack.us/img38/8246/18280336.png

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[UndergradGuy7]

Well I would first start by posting the question (jk)?

 

[azar87]

that would be a good starting point let us know what question. What book are you using kaplan or destroyer? or whatever else

 

[predentgal]

Oopsie. Im an idiot. Forgot to post the link. Im in class now ill post it n a bit

 

[UndergradGuy7]

Well, I haven't started reviewing Organic Chem, but I know that a metal catalyzed hydrogenation is a syn-addition. Since it adds to the same side of the double bond I thought B would be the answer.

C. is more stable since the CH3 are trans to each other making it more stable and the D are also trans. While in B only the CH3 is trans, but not the D.

So, I dno?

 

[UCB05]

since you're now differentiating between H and D, the orientation of the bonds matter. You initially start out with cis-2-butene. Addition of D2 over metal catalyst adds these D's in a syn fashion. This leads to a specific orientation where, looking down the C2-C3 bond, the -C(1)H3 and -C(4)H3 are eclipsed, and the two new D's are eclipsed. This is not one of the answer choices, so you also have to take into account the fact that the C2-C3 bond can now freely rotate. You have to keep the same bond orientation, but if you rotate about C2-C3, you can get the answer shown in C. The incorrect answer given in B can be arrived at if the D's added in anti fashion.

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[predentgal]

How do you know that the d's and the h's are eclipsed?

 

[JBarr29]

Your starting material was cis meaning the C1 and C4 are in the same plane. because the orientation of the entering Ds are syn and enter at the same face looking down the carbon bond from C2-C3 they are eclipsed.

 

[kpanesar]

i just drew it out... so you have your cis-2-butene and after reduction i had the dueterium coming out at my on C2 and C3 and the C1 and C4 going away and then the hydrogen pointing down.

so for newman's i just start by drawing C2 so if you're looking at the propellor then the methyl is at 10 o clock, the dueterium at 2 o clock and the hydrogen at 6 o clock. then you just rotate bond btw C2-C3 so that the substituents are on opposite sides (which is the case because rotating 180 ends up with the deuterium now going away from you and the methyl coming towards you.

 

[RSD2014]

Hey everyone.. I'm having trouble with this question. Answer choice B looks more like cis to me than does C. How do you approach this problem?

http://img38.imageshack.us/img38/8246/18280336.png

First draw Cis-2-butene and reduce it with D2 (Syn-Addition).
In converting a perspective drawing of a molecule into Newman projection, we keep the top section, and rotate the bottom section. Looking at C2, we have CH3 at Top, D at Left, H at Right. We rotate C3 180 degree, CH3 at Bottom, D at Right, H at Left. Which is Answer C.