formation of 4-oxo-2-hexenal

The pyridine is there to scavenge the HBr produced from the formation the radical.

THF and acetone are solvents. 

The HBr and water can also complete the acid catalyzed hydrolysis of an enol ether.  This would help with the formation of the compound.

Slowly but surely....

So I think where I went wrong is immediately assuming since it was NBS that it had to be allylic bromination through a radical mechanism.  But I started thinking (with a nudge from Dan) that NBS is just a constant, low concentration source of bromine.  So I thought about it and after some google help, a article said the bromination of furan happens through a 1,4-addition type mechanism.  Attached is the second half of the mechanis, then the bromination mechanism.
Does this seem more logical? Does to me.

Thanks Dan

I am writing the mechanism out and it hits me.  I do not exactly know which position to brominate. What makes sense is that when the double bond is formed thourgh E2 elimination at the 2 position, the proton closest to the carbonyl (aldehyde) would be the most labile. 

Also, this is enol ether chemistry.  Which I think there is the hydrolysis of enol ether going on.  I have to look some more.  Ill post what I have and we can go from there.

You've skipped too far ahead with your bromination.

What is the full mechanism for the bromination and why have you brominated at that position?

Since light or a radical initiator are not given in the question, I would not assume a radical pathway. Google furan bromination.

Initially it might be better to invoke the oxygen lone pair to move the electrons around.

I think that is even better.  Thanks Disco!!