Which could exist as an enantiomer? (EK)

[stester77s]

Which of the following properly named compounds could exist in enantiomeric form?

A) 3-chloro-1-propene
B) 3-chloro-1,4-dichlorocyclohexane
C) trans-1,4-dichlorocyclohexane
D) 4-chloro-1-cyclohexene


D is listed as the correct answer. ExamKrackers says C could not be enantiomeric because it is meso. Is that correct? I am failing to see how it is meso given that the chloro groups are trans. Where is the line of symmetry?

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[BrawnsNBrain]

Is there an image with this question? You could just look at the image, find chiral centers, determine the chirality and prove it's meso

 

[stester77s]

No image is included in the question. Please try to draw it yourself and see that it's not meso, unless I'm mistaken, which is why I posted this question.

 

[dudewheresmymd]

Which of the following properly named compounds could exist in enantiomeric form?

A) 3-chloro-1-propene
B) 3-chloro-1,4-dichlorocyclohexane
C) trans-1,4-dichlorocyclohexane
D) 4-chloro-1-cyclohexene


D is listed as the correct answer. ExamKrackers says C could not be enantiomeric because it is meso. Is that correct? I am failing to see how it is meso given that the chloro groups are trans. Where is the line of symmetry?

Ek is right for the wrong reason. Choice C could not be enatiomeric, but not b/c it's a meso compound. You are right in that the compound is not meso b/c it's not even a chiral compound to begin with as it doesn't have 2 or more stereocenters. To be meso you need 2 or more stereocenters by definition. The two carbons do not have 4 different things attached to them if you trace them back from left to right they end up at the same carbon, thus they are not stereocenters. EK 10001 has tons of mistakes, don't use it if you wish to keep your sanity.

Choice D definitely has a stereocenter and is the right choice.

References:

Wiki:
A meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active.[1][2] This means that despite containing two or more stereocenters (chiral centers) it is not chiral.


https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/sterism3.htm

MSU CHEM

 

[stester77s]

Ek is right for the wrong reason. Choice C could not be enatiomeric, but not b/c it's a meso compound. You are right in that the compound is not meso b/c it's not even a chiral compound to begin with as it doesn't have 2 or more stereocenters. To be meso you need 2 or more stereocenters by definition. The two carbons do not have 4 different things attached to them if you trace them back from left to right they end up at the same carbon, thus they are not stereocenters. EK 10001 has tons of mistakes, don't use it if you wish to keep your sanity.

View attachment 180360

Choice D definitely has a stereocenter and is the right choice.

References:

Wiki:
A meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active.[1][2] This means that despite containing two or more stereocenters (chiral centers) it is not chiral.


https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/sterism3.htm

MSU CHEM

View attachment 180364

Thanks for the excellent reply. You're right, it's not even a chiral carbon. Thanks!