[THC]

Q: Reaction of butanone with HCN yields a cyanohydrin product, R₂C(OH)CN, having a new stereocenter. Explain why the product is not optically active.

A: The only reason I can think of is that the product is a racemic mixture. But that seems wrong. HCN is added to ketones by a SN2 reaction where HCN attacks the C from the opposte site of =O (acidic solution, I assume). Help?

[macman104]

If you were to draw butanone in 3d space, what angles could the HCN attack the carbonyl group from?  Is there more than one way HCN could come at the carbonyl?

[THC]

The carbonyl is protonated:
R-CO-R -> R-COH⁺-R

The HCN is a nucleophile, so it attacks the carbonyl C:
HCN-COH⁺ -> CN-C(OH)

Then HCN must attack the carbonyl C from the opposite site of the carbonyl O, right? Isn't nucleophilic addition reaction to carbonyls generally SN2?

[sjb]

OK, let's try a different idea.

What's the hybridisation state, and hence the local shape of butanone at C₂?

S

[Rico]

Hey THC

Your answer, that the reason why the product is not optically active is because its a racemic mixture, is  correct! The two faces of the carbonyl group in butanone is enantiotopic, giving upon addition of a nucleophile like cyanide, two enantiomeric products as a racemic mixture (which is not optically active).

With regards to the mechanism, it is a nucleophilic addition to the carbonyl-group and not an SN2 reaction. Remember that SN2 stands for substitution nucleophilic bimolecular, and therefore you need to do a substitution, in other words you need to have a leaving group, for the reaction to have an SN2-mechanism. In the nucleophilic addition to a carbonyl group you don't have a leaving group, you just add a nucleophile and rearranges some electrons in the molecule. I have sketched the two mechanisms in a jpeg attached.

Hopes this answers your question

Rico

[THC]

OK, let's try a different idea.

What's the hybridisation state, and hence the local shape of butanone at C₂?

S

Ah, now I get it
The C atom in the carbonyl group is sp2 hybridised, therefore planar, so HCN can attack from both sides - resulting in a racemic mixture!
Thanks!

Hey THC

Your answer, that the reason why the product is not optically active is because its a racemic mixture, is  correct! The two faces of the carbonyl group in butanone is enantiotopic, giving upon addition of a nucleophile like cyanide, two enantiomeric products as a racemic mixture (which is not optically active).

With regards to the mechanism, it is a nucleophilic addition to the carbonyl-group and not an SN2 reaction. Remember that SN2 stands for substitution nucleophilic bimolecular, and therefore you need to do a substitution, in other words you need to have a leaving group, for the reaction to have an SN2-mechanism. In the nucleophilic addition to a carbonyl group you don't have a leaving group, you just add a nucleophile and rearranges some electrons in the molecule. I have sketched the two mechanisms in a jpeg attached.

Hopes this answers your question

Rico

Yes, you're absolutely right. I just compared the reaction mechanism to SN2, but it's a nucleophile addition. And addition reactions are always bimolecular
Thank you for the correction.

[THC]

The product is always a racemic mixture, no matter what reagents are added to the carbonyl, right? What about Ph-MgBr?

[sjb]

The product is always a racemic mixture, no matter what reagents are added to the carbonyl, right?

For some level of "no matter what reagents", yes. However, moving along, consider adding (R)-1-phenylethylamine to 3-methylbutanone. Is this the same?

What about Ph-MgBr?

As far as I know this should result in a racemic mixture.

[pacman12]

Hey THC

Your answer, that the reason why the product is not optically active is because its a racemic mixture, is  correct! The two faces of the carbonyl group in butanone is enantiotopic, giving upon addition of a nucleophile like cyanide, two enantiomeric products as a racemic mixture (which is not optically active).

With regards to the mechanism, it is a nucleophilic addition to the carbonyl-group and not an SN2 reaction. Remember that SN2 stands for substitution nucleophilic bimolecular, and therefore you need to do a substitution, in other words you need to have a leaving group, for the reaction to have an SN2-mechanism. In the nucleophilic addition to a carbonyl group you don't have a leaving group, you just add a nucleophile and rearranges some electrons in the molecule. I have sketched the two mechanisms in a jpeg attached.

Hopes this answers your question

Rico

Hello

Could you please reattach those images? I have the products but I'm not sure if they are right. The CN is wedged and the OH is Dashed for the S product and vice versa for the R product?

[Honclbrif]

The product is always a racemic mixture, no matter what reagents are added to the carbonyl, right? What about Ph-MgBr?

If your starting materials are achrial, your products will be achiral.

[Honclbrif]

Perhaps I should clarify. If your starting materials are achiral, your products will be achiral or a racemic mixture (depending on the <a target="_blank" href="http://en.wikipedia.org/wiki/Topicity">topicity</a> of the reaction site). Think about the transition states (TS*) when forming the R-product, or the S-product from the attack of an achiral nucleophile on an achiral electrophile

N + E  TS_R^*  P_R

N + E  TS_S^*  P_S

TS_R and TS_S are enantiomers, and therefore have the same energy. Because of this, no one pathway is favored, and no one product is formed in excess.

 If one or more of the starting materials are chiral, then the transition states for the various pathways are themselves diasteromers with different energies, resulting in a product enriched in one of the products.