[pecan66]

Compound A was prepared from butanol, benzaldehyde and dimethylamine in the presence of an acid catalyst.

Compound A + + MgBr

Compound A (compound with the phenyl ring) (shown above) reacts with unsaturated Grignard reagents to give (after
working up) different unsaturated amines; B and C are isomers. Where B has a 70 percent yield and C a 30 percent yield.
Identify B and C.

Now I can only identify two enantiomers in the product but I cannot see how these are formed in different proportions?

The synthesis of A should produce a racemic mixture of A so how are different isomers formed when using A for the next part?

Thanks

[Dan]

+ MgBr

Compound A (compound with the phenyl ring) (shown above) reacts with unsaturated Grignard reagents

Can you clarify the structure of the Grignard reagent? Do you mean:

to give (after working up) different unsaturated amines; B and C are isomers. Where B has a 70 percent yield and C a 30 percent yield. Identify B and C.

Now I can only identify two enantiomers in the product but I cannot see how these are formed in different proportions?

I don't think B and C refer to two enantiomers. As you point out, a racemate will form, so B and C cannot be enantiomers.

I think the second product may be a consequence of the chemistry of alkynes under basic conditions (but I'm not sure, I might be missing something here). What do you know about the stability of internal alkynes in strongly basic media?

[pecan66]

Thanks for the reply

Yes that is the structure of the Grignard reagent.

I'm thinking that the alkyne is itself susceptible to nucleophilic attack from the Grignard reagent perhaps after an initial reaction with A to form  a compound containing both an alkyne and an alkene. Although this does not agree with the nmr.

However, there is nothing there to stabilize  the negative charge that would result.

I should note that I was given some nmr data of B(70 percent yield)

1.53 (3H, dd)
2.13(6H, s)
3.50(1H, s)
4.43-4.80(2H, m)
7.19(5H , m)

Are you perhaps suggesting that the alkyne can be deprotonated by the Grignard reagent?

[orgopete]

The NMR is helpful. What do you think B is? (Dan's suggestion is helpful.)

[pecan66]

Ok so I think that this compound is B

Now I have drawn a mechanism to produce the terminal alkyne which involved a series of deprotonations and what not. The resulting terminal alkyne then acts as a nucleophile to give:



This is what I think may be compound C. It is formed in a lower proportion due to the initial steps required for the terminal alkyne to form. As well as the initial alkyne being more stable(hyperconjugation)

Hopefully this is correct!

Thanks