[nameless]

im having some difficulty with this particualar question. Im sure that the IR ranges are for C=C and alcohol respectively. Can someone have a look at my work/answer and help me figure this question out? Thanks.

[orgopete]

The key information is that B is formed from A by reductions that give a cis-alkene (Lindlar's cat.) and a trans-alkene (Na/NH3). That is necessary so a mixture does not result from the reaction of B with acid.

[nameless]

yea im really stuck on this question, im not even sure if my A is correct. Can you explain your answer, not quite sure what u mean....

[tmartin]

It says that compound C has a racemic center.. but your product C has no stereocenters.  What is the product of treating any generic alkene you can think of with acid?  Perhaps thinking about that mechanism will give a hint as to compound C.

[Dan]

im not even sure if my A is correct.

I don't think it is. Your structure would give a different product on treatment with hydrogen/Lindlar's catalyst than metal/ammonia reduction. Think about Orgopete's clue - it is indeed the key to the problem. The fact that the two reduction methods give the same product gives you vital information about the triple bond.

[nameless]

well i know that adding an acid to an alkene gives an alcohol but im not sure how to get rid of the alcohol group cause the IR range doesnt show an alcohol (i think). i was thinking that from B to C it forms an enol which tautamerizes to a ketone.

[Dan]

well i know that adding an acid to an alkene gives an alcohol

Slow down, you're on the right track but what is the reactive intermediate you get before the alcohol forms?

the IR range doesnt show an alcohol (i think).

You are correct, the product is not an alcohol...

[nameless]

do you mean the step when the whole water molecule attaches and has the partial positive charge on the oxygen?

[nameless]

well heres another guess at the answer, am i close? any additional help would be nice!

[Dan]

Yup, that's the product I had too.

What I was getting at before was that when you protonate the double bond to get the secondary carbocation, instead of adding water to it, you add the internal alcohol. Intramolecular reactions are typically much faster than intermolecular reactions, so I'm not sure the reaction will proceed via the diol you have drawn.

[nameless]

thanks alot! these forums are really helpful.