# Question 255f6

Jan 9, 2017

WARNING! Long answer! Here's how I do it.

#### Explanation:

The name of your compound is 1,10-dimethylbicyclo[4.4.0]decan-9-ol.

For easy reference, I have inserted the numbering for you.

The chiral centres appear to be carbons 1, 9 and 10 (all the others have at least two $\text{H}$ atoms, so they can't be chiral).

Configuration at $\text{C1}$:

$\text{C1}$ is attached to $\text{C2, C9, C11}$, and $\text{H}$.

$\text{H}$ is obviously Priority 4, but the carbon atoms are all tied. We must go one atom further out from each atom to break the tie.

$\text{C2}$ is attached to $\text{C3, H}$, and $\text{H (C,H,H)}$.

$\text{C9}$ is attached to $\text{O, C8}$, and $\text{C10 (O,C,C)}$.

$\text{C11}$ is attached to $\text{H, H}$, and "H (H,H,H").

We compare these atoms at the first point of difference.

Since $\text{O > C >H}$, $\text{C9}$ is Priority 1, $\text{C2}$ is Priority 2, and $\text{C11}$ is Priority 3.

$\text{H}$ is in the rear, and the 1 → 2 → 3 direction is clockwise, so $\text{C1}$ has the ($R$) configuration.

Configuration at $\text{C9}$:

$\text{C9}$ is attached to $\text{O, C1, C8}$, and $\text{C10}$.

$\text{O}$ is obviously Priority 1, but the carbon atoms are all tied.

$\text{C1}$ is attached to $\text{C2, C11}$, and $\text{H (C,C,H)}$.

$\text{C8}$ is attached to $\text{C7, H}$, and $\text{H (C,H,H)}$.

$\text{C10}$ is attached to $\text{C4, C5}$, and $\text{C12 (C,C,C)}$.

$\text{C10 (C,C,C) > C1 (C,C,H) > C7 (C,H,H)}$, so $\text{C10}$ is Priority 2, $\text{C1}$ is Priority 3, and $\text{C8}$ is Priority 4.

If you view along the $\text{C9-C8}$ bond with the $\text{C9-C10}$ bond in the plane of the paper, $\text{C1}$ comes in front of the paper.

The 1 → 2 → 3 direction is counterclockwise, so the configuration at $\text{C9}$ is ($S$).

Configuration at $\text{C10}$:

$\text{C10}$ is attached to $\text{C4, C5, C9}$, and $\text{C12}$.

$\text{C4}$ is attached to $\text{C3, H}$, and $\text{H (C,H,H)}$.

$\text{C5}$ is attached to $\text{C6, H}$, and $\text{H (C,H,H)}$.

$\text{C9}$ is attached to $\text{O, C3}$, and $\text{C8 (O,C,C)}$.

$\text{C12}$ is attached to $\text{H, H}$, and $\text{H (H,H,H)}$.

$\text{C9 (O,C,C) > C12 (H,H,H)}$, so $\text{C9}$ is Priority 1 and $\text{C12}$ is priority 4.

To break the tie between $\text{C4}$ and $\text{C5}$ for Priorities 2 and 3, we move out another three carbons.

That brings $\text{C4}$ to $\text{C1 (C,C,H)}$ and $\text{C5 to C8 (C,H,H)}$.

Thus, $\text{C4}$ is Priority 2, and $\text{C5}$ is Priority 3.

The 1 → 2 → 3# direction is clockwise, so we would assign the configuration ($R$).

However, the lowest priority group is in front, so we reverse the assignment to ($S$).

The compound is [$1 R , 9 S , 10 S$]-1,10-dimethylbicyclo[4.4.0]decan-9-ol.