-1
$\begingroup$

I inserted a picture to help show where my misunderstanding lies. I know that in the nucleophilic addition of ketones and aldehydes , a racemix mixture is produced . And I know that the reason is that the molecule can be attacked by the nucleophile from “either side of the molecule” or “above and below” But I can’t seem to picture that . In the picture I added , I show the one place/direction of attack that makes sense to me , could someone please show me the other “side” ? enter image description here

$\endgroup$

closed as unclear what you're asking by Mithoron, Jon Custer, Todd Minehardt, andselisk, A.K. Dec 15 '18 at 19:13

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Draw in 3D or build a model of the enantiomers of the cyanohydrin to get you started. $\endgroup$ – user55119 Dec 14 '18 at 20:39
1
$\begingroup$

Here's a link to an animation of cyanohydrin formation. You have to click on the reaction arrow and the individual step will play. http://www.chemtube3d.com/Nucleophilic%20substitution%20at%20the%20carbonyl%20group%20-%20Cyanohydrin%20formation.html

Since the carbonyl is completely planar, the nucleophile could either approach from the front or the back. While your diagram seems to indicate the nucleophile attacks in the same plane as the carbonyl (180° angle from O–C–nuc), in reality the angle of approach is roughly 107° (Bürgi–Dunitz angle).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.