2 orgo questions

[issa]

---

Members don't see this ad.

lets see how you answer these.

 

[asckwan]

#1 C #2 A

 

[djeffreyt]

I don't know if these answers are right...let me know cause I started Organic Chem studying first and haven't looked at it in a while.

For the first one, (D), they are diastereomers because they are obviously isomers cause they have the same connectivity and atoms. They are chiral, but aren't mirror images, so they aren't enantiomers. and I think diastereomers were chiral isomers that aren't mirror images.

For the second one, I have to guess because they all look chiral to me. I think (A). (A) looks like a meso compound cause it has an internal line of symmetry. (B) is similar but the connections to the chirl centers seem like they might not line up in a mirror image, so I don't think it is a meso compund, and all the other answers look chiral on both the C2 and C3 carbons. The only one that stands out is (A) as a meso compound, which is still chiral I thought, but it's the only answer that seems reasonably different.

 

[issa]

#1 C #2 A

for the second one i thought the answer is E because chiral compounds must bond to 4 different groups. E is the only choice that is not bonded to 4 different groups. any one have an explanation?

 

[beastly115]

12. D


31. E

Edit: Chirality = handedness. A chiral center is a carbon bonded to four _different_ groups. An achiral carbon is one _not_ bonded to four different groups.

 

[RozhonDDS]

for the second one i thought the answer is E because chiral compounds must bond to 4 different groups. E is the only choice that is not bonded to 4 different groups. any one have an explanation?

A for number 2 isn't bonded to 4 groups either

 

[RozhonDDS]

1)=Def D, 2) either A or B, all the other configurations wouldn't allow it be superimpossible on itself

 

[DDS:)]

lets see how you answer these.

c.e

 

[djeffreyt]

okay....ISSA what is the answer that is given?

We have a fair consensus that #1 is (D)

For #2, I still think it is (A) for the reasons I wrote earlier. But (E) is chiral because one of its carbons is chiral. The two carbons on the ends aren't chiral, but the middle carbon is; it is bound to four different groups (a bromine, a hydrogen, a methyl group, and a dibromomethyl group)

Still, this question sucks and I would like to know the answers

 

[djeffreyt]

sorry to keep replying...question 2 was bugging me so i broke out my o-chem model kit and actually made models and tried to superimpose them.

So chiral is defines as a molecule that cannot be superimposed on its mirror image. Achiral is a molecule that can be superimposed on its mirrow image. If you test out all five answers, only answer (A) is superimposable on its mirror image. I actually did the models for all five answers and (A) was the only one where all the atoms line up perfectly.

 

[RozhonDDS]

sorry to keep replying...question 2 was bugging me so i broke out my o-chem model kit and actually made models and tried to superimpose them.

So chiral is defines as a molecule that cannot be superimposed on its mirror image. Achiral is a molecule that can be superimposed on its mirrow image. If you test out all five answers, only answer (A) is superimposable on its mirror image. I actually did the models for all five answers and (A) was the only one where all the atoms line up perfectly.

Thats what i said, because the question is asking for achiral, not chiral, and everyone is talking about it being E being chiral because it has 4 different components, but the question didn't ask for that

 

[hoabangkhuang]

1. C Enantiomers 2. E the R group on the right is achiral.

 

[issa]

the answer for the first one is C. for the 2nd one the answer is A.

 

[SgtSadhu]

the answer for the first one is C. for the 2nd one the answer is A.

yep C and A

for the first one all that was done was two of the groups were switched so they are enantiomers

and for the second its A cuz its a meso compound

 

[issa]

these two questions from the company that wrote the actual questions for the dat. their predict the product questions are a lot easier.

 

[brad219]

The first one is C NOT D! If you have to, please build a model of the structures to convince yourself...it's also good practice for the PAT if you can manipulate the images in your mind to see that they are in fact mirror images of each other.

 

[Futureortho24]

why is the 2nd one not E? Chiral means 4 DIFF substituents attached to the carbon; in choice E, Br appears 2 times making it achiral! I may very well be wrong, but why?

 

[beastly115]

why is the 2nd one not E? Chiral means 4 DIFF substituents attached to the carbon; in choice E, Br appears 2 times making it achiral! I may very well be wrong, but why?

Earlier I said it was E, but the other guys are right that it is A. A meso compound is considered achiral. The mistake I made was that 1 assymetric carbon + 1 achiral carbon = an achiral molecule, however it does not.

 

[mlle]

Earlier I said it was E, but the other guys are right that it is A. A meso compound is considered achiral. The mistake I made was that 1 assymetric carbon + 1 achiral carbon = an achiral molecule, however it does not.

yea, but a and b are the same if you look closely. they are identical!
so if it's a (which I agree with) then it should be b as well...

 

[Clover]

yea, but a and b are the same if you look closely. they are identical!
so if it's a (which I agree with) then it should be b as well...

Regarding problem 2:
choice a and b are not the same.

For b: Ignoring the wedges and dashes for a moment, if you rotate carbon #2 so Br and Br are syn like in choice a, you will find that the hydrogen sticks of the plane of the screen and the methyl (CH3) is in the rear. This differs from choice a, where both carbon's methyl are in front.

Regarding problem 1:

This is a 1 chiral carbon structure, there are no diasteriomers.

Even if you didn't know that, just do R & S.
structure 1 = R
structure 2 = S
Simply, R and S = enantiomers

I hope this helps. Don't let the wedges and dashes fool you. Also I've said it before but I'll say it here too - DAT Destroyer has some great problems and tips/tricks on R & S config. It's a simple concept that will screw you up if you don't see the problem for what it is.

 

[mlle]

Regarding problem 2:
choice a and b are not the same.

For b: Ignoring the wedges and dashes for a moment, if you rotate carbon #2 so Br and Br are syn like in choice a, you will find that the hydrogen sticks of the plane of the screen and the methyl (CH3) is in the rear. This differs from choice a, where both carbon's methyl are in front.

thanks for trying to explain it to me. but here's what I don't get: the h's are both behind, the ch3's are both in front, and the br's are both in the plane according to the dashes/wedges, aren't they? Isn't that what matters? I must be missing something basic here.

 

[Clover]

in the problem, the wedges and dashes are just there to confuse you.

For choice b:

The Br groups are anti. In order to make them syn (to compare with choice a), you must rotate carbon 2 while holding carbon 1.

When you rotate, the methyl group will go to the back and hydrogen comes forward.

** the carbon must be rotated, you can't just push the methl and hydrogen groups down while pushing Br up.

you will need build a structure if you can't picture it

 

[Futureortho24]

I understand what everyone is saying, however, I still do not understand how choice E for problem #2 is not achiral - 2 of the same substituent appear on the molecule - since when does that not matter! I may be missing the obvious 🙁

Ok i looked at the figure again - is A the right answer b/c it had a plane of symmetry and meso compunds are achiral and have a plane of sym??? maybe?!!!

 

[asckwan]

You are right in that the right hand side carbon is achiral, but because the left hand side carbon is chiral it will make the molecule chiral.

 

[DonExodus]

Okay, for #1, the compounds are non-superimposable, but they are NOT mirror images, hence, they are NOT enantiomers.

1= D

For 2, E has only 1 chiral center. The right side is Achiral, but the left IS chiral, hence the molecule itself being chiral.

A has 2 chiral centers, but it IS superimposable on its mirror image, making it a meso compound.

Meso compounds are achiral, hence....

2= A

-=DonExodus=-

 

[Orthodox Bear]

1)c 2)a

 

[mlle]

in the problem, the wedges and dashes are just there to confuse you.

For choice b:

The Br groups are anti. In order to make them syn (to compare with choice a), you must rotate carbon 2 while holding carbon 1.

When you rotate, the methyl group will go to the back and hydrogen comes forward.

** the carbon must be rotated, you can't just push the methl and hydrogen groups down while pushing Br up.

you will need build a structure if you can't picture it

i get how to rotate it, and I can picture that, but don't the dashes/wedges mean anything in relation to the other carbon's dashes/wedges? Are they just there for consideration within the one carbon as opposed to position comparison with the other? does what I'm saying make sense....

thanks guys.