[Lo.Lee.Ta.]

1. "Students were instructed to mix the benzaldehyde and acetone starting materials in their conical vials before adding the ethanolic sodium hydroxide solution.
Why was this essential to the success of the reaction?
What would have been the most likely product formed if the sodium hydroxide solution were added to the vial first, followed by acetone and then waiting a few minutes to add the benzaldehyde?
Provide a balanced equation showing this reaction:"


2. This is a post-lab question from my lab manual...
I wrote out what I thought the beginning of the mechanism might be:



So I thought things would start off by NaOH deprotonating acetone's alpha carbon, creating Na+ and water.
Then acetone's negative alpha carbon would attack the carbonyl carbon of a normal acetone, breaking the its double bond with oxygen.
The negative O then takes the previously-formed water molecule's H, reforming the NaOH catalyst.
The new hydroxide then takes the alpha carbon's H, becoming H₂0+.
H₂0+ leaves, becoming a free H₂O, and the negative alpha carbon creates a double bond in place of the H₂O+.

From this mechanism, I'm now left with 4-methyl-3-penten-3-one + H₂O + NaOH.

I guess this reaction could happen all over again with the alpha carbon on the other side, resulting in 2,6-dimethyl-2,5-heptadien-4-one + 2H₂O + NaOH...

Now I'm supposed to add benzaldehyde... hm.

I guess the two alpha carbons of 2,6-dimethyl-2,5-heptadien-4-one still do have 1 H each, so would there be deprotonation again by NaOH?

If so, then I guess the negative alpha carbon would attack benzaldehyde's carbonyl carbon...
The resulting negative O would take the H from water.
This would happen again on the other side, but the steric strain might force the 2 hydroxyls to undergo a dehydration reaction, forming an ether bond between them...?

So from this idea, the final product would be:



...This seems like such a weird product, though.
Is this right? ...Or what am I doing wrong here?
Thanks!

[Archer]

I can't see your images, however from the text I think I can work out what you have said.

I think that the only product from this reaction would be the self condensation product of acetone.

[orgopete]

While that pryone could form, I think thus is going too far. If acetone self condenses, it can continue to do so. Rather than benzaldehyde, this could have been acetone, however the products will more likely show the acetone condensation and no reaction with the benzaldehyde. That is, if mesityl oxide should react further, it should do so with acetone or itself until those reaction are slowed or depleted. I think the principle they are getting at is that if the enolization occurs in the presence of benzaldehyde, then a (relatively) faster reaction occurs with the benzaldehyde and in its absence, it reacts with itself. I don't think they were going for an illustration of all of the possible products that acetone can provide.

[Lo.Lee.Ta.]

Oh, okay! Thanks for the help, you guys!