Why in this problem there isn't a diastereomer with the $\ce{-CH3}$ in the back (dash) and the $\ce{-CH2CH3}$ in front (wedge)?
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4$\begingroup$ The structure that you are asking about is essentially the first structure (the left-most one) given. $\endgroup$– AniruddhaCommented May 12, 2021 at 15:47
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$\begingroup$ Try assigning (R) and (S) to the two stereocenters. $\endgroup$– ZheCommented May 12, 2021 at 22:28
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2 Answers
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To answer this question I would first assign the R-S configuration to the original molecule and the molecule proposed by you and show you that both are identical. At the end I will also show you a simpler way to tell that both are identical.
Original molecule:
Proposed molecule:
As you can see, both the molecules have the same R-S configuration and thus are identical.
Now to see this easily and intuitively just assume you are looking at the original molecule from inside the screen. What do you see? You will observe that the original molecule will look just as same as the proposed molecule which means that both are identical. This method works because our frame of reference doesn't change the optical properties of a molecule until and unless we change the configuration of the molecule by doing some reflections.
answered May 13, 2021 at 5:08
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Actually for this question if we draw the isomer with methyl group in the back and ethyl group in the front as in:
we would end up getting the enantiomer of the given compound in the question (you could visualise it by flipping the above compound (plane of compound) by 180 degree and then rotate the compound through an angle of 180 degree in the same plane)
And if we draw the isomer like this:
we would end up getting the same initial compound!!( you visualise it in the same way as mentioned before)
So in either case the compound with methyl in front and ethyl at back would not yield a diastereoisomer.
Hope it helps!!!
