# Which conformations of 2,3-dimethylbutane are more stable?

I am not sure what the proper arrangement of these conformations are from least to most stable. I got the wrong answer, but my thinking was this: First, I thought that because 2 and 4 are in eclipsed conformations they are least stable. From there I thought that 4 was less stable than 2 because of the steric interactions of the methyl groups. Then of the two staggered conformations 1 and 3, it appears to me that 1 has more steric interaction of the methyl groups than 3. So my ordering from least to most stable was 4,2,1,3. Is there an error in my line of reasoning?

• I followed your reasoning and actually, I would have reached the same answer as you did. I suppose the only possible area up for debate is between 1 and 2 - it basically comes down to whether 3x Me-Me gauche interactions are worse than 1x Me-Me syn - without running some calculations it is difficult to answer that question so I would not fault you (and myself) for getting the wrong answer. – orthocresol Sep 7 '16 at 14:10
• ..so I just checked the assignment for the millionth time.. because for the past 2 days I've been sure I'm right about this....just discovered the mc answer option I selected was 4,2,3,1 (swapped 1 and 3) and 4,2,1,3 was not listed..the correct answer was none of the above. So I'm pretty sure that my reasoning is correct here. Unless anyone can point me to an error still? – MegaboofMD Sep 7 '16 at 18:10
• Nah, I figure you and I are correct. Sucks to have questions that are like that, sometimes feels more like a test of reading carefulness than chemistry. – orthocresol Sep 7 '16 at 18:11

The correct arrangement (from least stable to most stable) is probably: $4 > 2 > 3 \gtrapprox 1$ based on my calculations for the four gas phase rotamers you can see in the figure below.

Your reasoning for 4 over 2 sounds good, that would also be my point. For the two minima 1 and 3 (with Boltzmann weights of 46%:54% at the highest calc. niveau (red), 33%:66% at the lowest (blue)) you might also be right. But from the really small energy difference I don’t want to argue too much about it.

My calculation methods seem a bit random, but at first I only wanted to make that fast wB97X-D3/def2-TZVP//HF-3c scan and then decided to calculate the bigger one.

I found a question in Chem 220 - Organic Chemistry Problem Set 2.

The question is such:

Draw Newman projections for the eclipsed and staggered conformations of 2,3-dimethylbutane viewed along the $$C_2-C_3$$ axis. Calculate the energy of each conformation, both staggered and eclipsed.

Eclipsed (kcal/mol): H/H, 1.0; CH3/H, 1.3; CH3/CH3, 3.0 kcal/mol. Staggered (kcal/mol): H/H, 0; CH3/H, 0; CH3/CH3, 0.9.

Solution given:

This solution seems reasonable,your book might be wrong.

The solution given by user16347 seems logical but is not really the correct solution. According to the answer the anti form is more stable since it has the least energy, this is actually false. Experimentally, the gauche form with 3 gauche interactions ( that is structure with energy 2.7 ) is more stable than the anti form. The reason being that in the anti form the geminal repulsions due to the methyl on adjacent carbon hakes the system more unstable than the gauche interactions. This for this particular molecule the gauche conformer is more stable.

For details into this concept of geminal repulsion, refered Chapter 2 - Strain and Staibilty from the book Physical Organic Chemistry by anslyn doherthy