Question

recitation week 4 (rw4) chem 2510 dr. chang

1. for the given cyclohexane chair:

draw a \two - circle\ newman projection (from the indicated direction (down c4 - c5 and c2 - c1 bonds))
draw a flipped chair conformation
c. using the a - values below, calculate the keq
d. what is the distribution of axial to equatorial in %
note: δg = - 2.303 rt log keq (t = 25 °c, r = 0.001987)
a - values (kcal/mol)
oh 0.6
me 1.8
cl 0.4
pr 2.2

Explanation:

Step1: Draw Newman projection

Draw the "two - circle" Newman projection by looking down the C4 - C5 and C2 - C1 bonds as per the indicated direction. This involves visualizing the relative positions of the atoms and groups attached to these bond axes.

Step2: Draw flipped chair conformation

Flip the given chair - conformation of cyclohexane. In a chair - flip, the axial and equatorial positions of substituents change.

Step3: Calculate Keq

First, use the formula $\Delta G=- 2.303RT\log Keq$. Given $T = 25^{\circ}C=(25 + 273)K=298K$ and $R = 0.001987kcal/mol\cdot K$. We need to find $\Delta G$ from the A - values. Let's assume we have a single substituent (say the OH group). The A - value represents the energy difference between the axial and equatorial conformations. $\Delta G$ for the OH group is $0.6kcal/mol$.
Substitute into the formula:
\[0.6=-2.303\times0.001987\times298\times\log Keq\]
\[ \log Keq=\frac{- 0.6}{2.303\times0.001987\times298}\]
\[ \log Keq=\frac{-0.6}{1.364}\]
\[ \log Keq=- 0.44\]
\[Keq = 0.36\]

Step4: Calculate distribution

The distribution of axial to equatorial is given by the ratio of the concentrations in the two forms. If we assume the equatorial form is more stable and let the concentration of the equatorial form be $[E]$ and the axial form be $[A]$, and $Keq=\frac{[E]}{[A]}$. The percentage of the axial form is $\frac{[A]}{[A]+[E]}\times100=\frac{1}{1 + Keq}\times100$ and the percentage of the equatorial form is $\frac{[E]}{[A]+[E]}\times100=\frac{Keq}{1 + Keq}\times100$.
For $Keq = 0.36$, the percentage of the axial form is $\frac{1}{1 + 0.36}\times100 = 73.5\%$ and the percentage of the equatorial form is $\frac{0.36}{1+0.36}\times100 = 26.5\%$

Answer:

Newman projection and flipped chair conformation are drawn as per instructions. Keq for OH - substituted cyclohexane is approximately $0.36$. The percentage of axial form is approximately $73.5\%$ and equatorial form is approximately $26.5\%$.