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The least stable conformation of 2-methylbutane is :

  • A.Eclipsed Newman projection of 2-methylbutane viewed along C2-C3, with methyl groups at the top, left and lower right
  • B.Eclipsed Newman projection of 2-methylbutane viewed along C2-C3, with two methyl groups at the top and one at the lower right
  • C.Staggered Newman projection of 2-methylbutane, with methyl groups at the top, upper right and lower right and hydrogens at the upper left, lower left and bottom
  • D.Staggered Newman projection of 2-methylbutane, with methyl groups at the top, lower right and bottom and hydrogens at the upper left, upper right and lower left

Answer: B

The least stable conformation is (B).

Viewed along the C2-C3 bond, C2 carries two methyl groups and an H, and C3 carries one methyl (C4) and two H. Options C and D are staggered and so are more stable than either eclipsed form. Comparing the eclipsed forms with approximate torsional costs of CH3/CH3 about 11, CH3/H about 6 and H/H about 4 kJ mol-1:

\[ \begin{aligned} \text{(A)}&: 3(\text{CH}_3/\text{H}) \approx 18 \\ \text{(B)}&: 11 + 6 + 4 \approx 21\ \text{kJ mol}^{-1} \end{aligned} \]

(B) has a methyl eclipsing a methyl, the most costly interaction, so it has the highest energy. (A) spreads three methyls against hydrogens and is lower.

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