Question

I have to deduce a couple of compounds based on NMR spectra, formulas given, and the reactions they undergo. I'm having trouble with figuring out the product of a couple of reactions- as in, whether they are alcohols, ketones,carbonyls (of that sort- no halides or anything, all of the general formula CxHyOz).

Organic Chem

Compound A, C9H18O slowly reacted with metallic sodium to give off hydrogen gas- after all the sodium was reacted, it was treated with iodomethane to give compound B, C10H20O.

Compound B did not react with sodium, and is converted into compound C, C9H18O2, if heated with a solution of potassium dichromate in sulfuric acid.

My question is what compound reacts with metallic sodium and then an alkyl halide, and then the product of that undergoes the oxidation of potassium dichromate? At first I thought B was an ether, but ethers cant be oxidized to COOH. Thus, what type of organic molecule is A and B? I can draw out the structures myself, I just want to know whether A and B are alcohols, ketones, aldehydes, etc. Also, what kind of molecule slowly reacts with metallic sodium to give off hydrogen gas?

Answer 1

This is a really nice question! Enjoyed working it out!

You see, first of all, you compare a few things between compounds A, B and C. Compounds A and B differ by a CH2 unit while compounds A and C differ by one O atom. Compounds B and C however differ by both one CH2 and one O atom.

Now take into consideration the last comparison b/w B and C. It seems like one CH2 group as been replaced by an O atom (That is how we end up with a CH2 being less and an O being extra in compound C) while converting from B to C. So definitely, it must be an oxidation process where a CH2 is being ripped off and replaced by O. And we can see such cases in terminal double bond oxidations.

So our B must have a double bond to be chopped off, and that too a terminal one 'cause that's how you're gonna get one CH2 less.

Now coming to A. A is seen to react slowly with Na right? Remember acidic hydrogens react with Na metal to liberate hydrogen gas. Also remember: alpha hydrogens of aldehydes and ketones are weakly acidic. Correlate these two facts and you have some idea about the structure of A - probably a methyl ketone.

Now in the enolate we have the required terminal double bond, but enolates are not isolable intermediates! We need to trap 'em with an electrophile. There comes your halide. So,

And now we have a vinyl ether which is your compound B, your compound A being a methyl ketone.

This when reacted with an oxidizing agent readily chopps off the double bond to give you an ester which is your compound C.

You can verify the result by matching the molecular formula for each of the species I mentioned.