Question #255f6

Question

Question #255f6

1 Answer

Impact of this question

2265 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-09T23:14:08

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 $H$ $"H"$ atoms, so they can't be chiral).

Configuration at $C1$ $"C1"$ :

$C1$ $"C1"$ is attached to $C2, C9, C11$ $"C2, C9, C11"$ , and $H$ $"H"$ .

$H$ $"H"$ 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.

$C2$ $"C2"$ is attached to $C3, H$ $"C3, H"$ , and $H (C,H,H)$ $"H (C,H,H)"$ .

$C9$ $"C9"$ is attached to $O, C8$ $"O, C8"$ , and $C10 (O,C,C)$ $"C10 (O,C,C)"$ .

$C11$ $"C11"$ is attached to $H, H$ $"H, H"$ , and $H (H,H,H)$ $"H (H,H,H")$ .

We compare these atoms at the first point of difference.

Since $O > C >H$ $"O > C >H"$ , $C9$ $"C9"$ is Priority 1, $C2$ $"C2"$ is Priority 2, and $C11$ $"C11"$ is Priority 3.

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

Configuration at $C9$ $"C9"$ :

$C9$ $"C9"$ is attached to $O, C1, C8$ $"O, C1, C8"$ , and $C10$ $"C10"$ .

$O$ $"O"$ is obviously Priority 1, but the carbon atoms are all tied.

$C1$ $"C1"$ is attached to $C2, C11$ $"C2, C11"$ , and $H (C,C,H)$ $"H (C,C,H)"$ .

$C8$ $"C8"$ is attached to $C7, H$ $"C7, H"$ , and $H (C,H,H)$ $"H (C,H,H)"$ .

$C10$ $"C10"$ is attached to $C4, C5$ $"C4, C5"$ , and $C12 (C,C,C)$ $"C12 (C,C,C)"$ .

$C10 (C,C,C) > C1 (C,C,H) > C7 (C,H,H)$ $"C10 (C,C,C) > C1 (C,C,H) > C7 (C,H,H)"$ , so $C10$ $"C10"$ is Priority 2, $C1$ $"C1"$ is Priority 3, and $C8$ $"C8"$ is Priority 4.

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

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

Configuration at $C10$ $"C10"$ :

$C10$ $"C10"$ is attached to $C4, C5, C9$ $"C4, C5, C9"$ , and $C12$ $"C12"$ .

$C4$ $"C4"$ is attached to $C3, H$ $"C3, H"$ , and $H (C,H,H)$ $"H (C,H,H)"$ .

$C5$ $"C5"$ is attached to $C6, H$ $"C6, H"$ , and $H (C,H,H)$ $"H (C,H,H)"$ .

$C9$ $"C9"$ is attached to $O, C3$ $"O, C3"$ , and $C8 (O,C,C)$ $"C8 (O,C,C)"$ .

$C12$ $"C12"$ is attached to $H, H$ $"H, H"$ , and $H (H,H,H)$ $"H (H,H,H)"$ .

$C9 (O,C,C) > C12 (H,H,H)$ $"C9 (O,C,C) > C12 (H,H,H)"$ , so $C9$ $"C9"$ is Priority 1 and $C12$ $"C12"$ is priority 4.

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

That brings $C4$ $"C4"$ to $C1 (C,C,H)$ $"C1 (C,C,H)"$ and $C5 to C8 (C,H,H)$ $"C5 to C8 (C,H,H)"$ .

Thus, $C4$ $"C4"$ is Priority 2, and $C5$ $"C5"$ is Priority 3.

C10

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

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

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

Science

Math

Humanities

... and beyond

Question #255f6

1 Answer

Explanation:

Related questions

Impact of this question