In the Newman projection of butane viewed along C2-C3, the front carbon has CH3 at 12 o'clock, H at 4 and H at 8 o'clock. The rear carbon, eclipsed with the front one, has H at 12, CH3 at 4 and H at 8 o'clock. Use: H/H eclipsing 4.0, H/CH3 eclipsing 6.0, CH3/CH3 eclipsing 11.0 kJ/mol, and take the anti conformer as zero. The energy of this conformer is

JEE Main levelSingle correct optionDifficulty 2 of 3st133

In the Newman projection of butane viewed along C2-C3, the front carbon has $CH X_{3}$ at 12 o'clock, H at 4 and H at 8 o'clock. The rear carbon, eclipsed with the front one, has H at 12, $CH X_{3}$ at 4 and H at 8 o'clock. Use: H/H eclipsing 4.0, H/ $CH X_{3}$ eclipsing 6.0, $CH X_{3}$ / $CH X_{3}$ eclipsing 11.0 kJ/mol, and take the anti conformer as zero. The energy of this conformer is

The eclipsed pairs are: front

CH X_{3}

(12) with rear H (12) = 6.0; front H (4) with rear

CH X_{3}

(4) = 6.0; front H (8) with rear H (8) = 4.0. Total = 16.0 kJ/mol. The two methyl groups are

12 0^{\circ}

apart, so this is not the fully eclipsed (19.0) form.