Question 1: Which of the following Newman projections represent the most stable confirmation of the diastereomer of the molecule given below?

most stable confirmation of the diastereomer

I guess my main question is looking only at the Newman projection, how does one determine the stereochemistry?

In the molecule to the right, I get for the first carbon, ($\ce{Et, H, Me}$) R (I flipped it after moving the $\ce{H}$ back) and for $\ce{C}$2 ($\ce{Et, Br, Me}$) I get S.

Question 2: Which of the following Newman projections represents a confirmation of the enantiomer of the molecule given below?

a confirmation of the enantiomer

In the same vein, for the second question, I thought I should flip all the stereocenters to get the enantiomer since it is a mirror image. But I am not getting the answers the professor provided to us.

  • $\begingroup$ A side note: For the question in title, you could just move on with the Newman projections to determine the relative stabilities of various conformers. Large groups which are anti to each other have the least energy possible, and hence are more stable. I don't see why you want to convert the Newman structures to 3d bond line structures for the question and I consider it to be unnecessary for this question. $\endgroup$
    – Vishnu
    Mar 2, 2020 at 13:26

2 Answers 2


To answer your professor's questions, there is no need to determine the R/S configuration of the two carbon atoms. We have to check the clockwise/anticlockwise arrangement of the substituents around the carbon atom, but checking specifically whether it is R or S is unnecessary.

Before moving further, let's have a look at the following image so that we are clear on how to visualise Newman Projections:

Newman Projection

Since in the options provided in the question the carbon atom having bromine is at the back, we'll have to look at the molecule in the following manner:

enter image description here

Hence the Newman Projection obtained would be this: enter image description here

Now, in question 1, option A can be rejected as it's an entirely different molecule. Option 4 has the atoms in the same order as above, so it is the exactly same molecule. In option 2 and 3, the arrangement of atoms is same for the back carbon but different for the front carbon. Therefore, without finding R or S configuration, we can say that options 2 and 3 are the diastereomers!

Out of these two structures, option 3 is more stable as the bulky ethyl groups are at the farthest position from each other.

Moving on to question 2:

You are right in flipping the arrangement about both the carbon atoms and you should get the answer that way. Maybe you are doing some mistake in getting the right Newman projection. Look again at the Newman projection I've provided above and then compare the options with it to see which option has different arrangement of substituents on both the carbons.


A table of A-values indicate that $\ce{Br}$ substituent on a long chain organic molecule has relatively less steric effect than that of $\ce{Me}$- or $\ce{Et}$-group ($1.7$ and $\pu{1.75 kcal mol-1}$, respectively compared to $\pu{0.38 kcal mol-1}$ for $\ce{Br}$), and even smaller than that of $\ce{Cl}$ ($\pu{0.43 kcal mol-1}$), probably due to longer $\ce{C-Br}$ bond. Thus, it is safe to assume that both $\ce{Br}$-$\ce{Me}$ and $\ce{Br}$-$\ce{Et}$ gauche interactions are smaller than ether of $\ce{Me}$-$\ce{Me}$ or $\ce{Me}$-$\ce{Et}$ or $\ce{Et}$-$\ce{Et}$ gauche interactions.

Now let's look at the questions in hand:

Question 1 & 2

OP's main question: Looking only at the Newman projection, how does one determine the stereochemistry?

I already gave the comparative gauche interactions by $\ce{Me}$, $\ce{Et}$, and $\ce{Br}$ groups. First, I want to point out that the compound $\bf{1}$ is either written wrong or intentionally put a $\ce{Me}$ group on the back carbon (see the group marked by green circle). I think it should be an $\ce{Et}$ group so that it would relate to the given reference compound, $\mathrm{\bf{I}}$. So, I made the adjustment as of it is an $\ce{Et}$ group.

Since the given compound is (3S,4R), its diastereomers should be (3R,4R) and (3S,4S) as depicted. Therefore, it is necessary to find out that all given compounds are diastereomers before you work on the most stable conformer. As indicated, compounds $\bf{1}$, $\bf{2}$, $\bf{3}$, and $\bf{4}$ are either diastereomer at different conformation. I have put how many gauche interactions in each compound. When roughly calculated, $\bf{1}$ and $\bf{2}$ have almost similar energy. However, $\bf{3}$ has less energy than that of $\bf{1}$ and $\bf{2}$ (e.g., $\bf{3}$ has an extra $\ce{Me}$-$\ce{Br}$ interaction in place of $\ce{Me}$-$\ce{Me}$ interaction in $\bf{1}$). However, if you look at carefully, $\bf{3}$ is exact mirror image of $\bf{4}$, thus have the identical energy (See the diagram above). Therefore, actual answer is both $\bf{3}$ and $\bf{4}$ ($\mathrm{C}$ and $\mathrm{D}$ in the grid).

The easiest method to answer OP's second question is to identify the (R,S) configurations of given answers. Since that of the given compound is (3S,4R), its enantiomer is (3R,4S), which is $\bf{6}$ ($\mathrm{B}$ in the grid; I have changed the original numbering for clarity). Other three options: $\bf{5}$ and $\bf{7}$ are diastereomers of $\mathrm{\bf{I}}$ while $\bf{8}$ is identical to $\mathrm{\bf{I}}$.

Note: I have also put relevant gauche interactions in each compound of second question for the readers' benefit.


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