[einstein99]

Can't figure out how to go from CH3CH2CH2C=OH to CH3CH2CH2CH2OCH2CH2CH2CH2CH3.  An aldehyde to an ether with the addition of five carbons.  I think I can use a Grignard to add the carbons but how does the ether form?  A Grignard plus an aldehyde yields a secondary alcohol.  What am I missing?

[AWK]

Aftre reducing an aldehyde to an alcohol use Williamson method for ether synthesis.

http://members.aol.com/logan20/eth_syn.html

[Albert]

Forget about Grignards. If I were you, I'd reduce the aldehyde to primary alcohol with NaBH4 (solvent is ethanol).

Then you can follow two different pathways: the former is a Williamson reaction with 1-bromopentane, the latter is the halogenation of your primary alcohol with concentrated HBr (very easy), followed by the williamson with pentanol.
If asked, I'd opt for the latter.

[einstein99]

Thanks guys

[einstein99]

one more problem...i can't figure out how to get CH3CH2CH2C=OH (butanaldehyde) from ethanol.  PCC would oxidize the ethanol to acetaldehyde but how can i add the two extra carbons?

[Albert]

Why should you do such a thing? I mean, you're taking about an exercise, aren't you?

[einstein99]

yes, it's a synthesis problem using only ethanol and going to CH3CH2CH2C=OCH2CH3 (KETONE).  i started by making a Grignard EtMgBr + CH3CH2CH2CHO (aldehyde) in acid to make a secondary alcohol.  then i added PCC to oxidize this alcohol to get the final ketone product.

BUT, I cannot figure out how to get that necessary aldehyde from ethanol in the beginning.

[Albert]

If we talk about exercises and NOT about REAL experiments, where yields are paramount, you can consider this. However, it has to be said that the last step yield a mixture of products, where the one you're looking for should represent a mere 25%.

[HP]

Hi Albert! Would you explain the last reaction of CH2 addition to the butanone with ethyliodide?

[Albert]

Well, I used my personal experience in organic lab: I did a reaction using similar conditions in order to alkylate an asymmetrical ketone.

In a nutshell, a carbanion is formed and an SN2 reaction takes place with CH3CH2I.

I forgot to say that at low temperatures you get more products coming from the alkylation of the most substituted C, while, at higher temperatures, the reaction particularly involves the less one.

[HP]

Aha, i think close to imagine and the thermokinetic control  Do you first ading NaH to the cetone and last alkyliodide or at once all? i think its posible and reaction with diazomethane(CH2N2) as in lactones CH2 addition. but may be this is some dangerous process...

[Albert]

I can't remember exactly. But I think it's the former: first we added NaH and then the other compound.
Generally speaking, because we are considering an exercise, you could use LDA instead of NaH as well.

[einstein99]

So to clarify the mechanism of the last step, the NaH (a very strong base) deprotonates one of the C in an ether solution.  then this carbanion pushes out the I in the SN2?  BTW, could just plain Na+ metal be used to deprotonate or is sodium hydride needed?

[Albert]

So to clarify the mechanism of the last step, the NaH (a very strong base) deprotonates one of the C in an ether solution.  then this carbanion pushes out the I in the SN2?

Yes, exactly.

could just plain Na+ metal be used to deprotonate or is sodium hydride needed?

I can't get the meaning of plain Na+ metal. You meant sodium metal, didn't you? Well, if that's how things are, the answer is absolutely NO.
I said I think you could also use LDA.

By the way, remember what I added:

I forgot to say that at low temperatures you get more products coming from the alkylation of the most substituted C, while, at higher temperatures, the reaction particularly involves the less one.

You know, just in case your professor would ask you to discuss your answer.