[Roddy]

I can't figure out where to start here.  I am given 4-chlorobutyric acid (ClCH2CH2CH2COOH) + OH- and asked to show the product of the reaction.  I understand the acid-base reaction will be faster than the nucleophilic substitution, but does the OH- remove the H from the C-OH?  I am assuming the final product will be C=C-C-COOH  ... is that right, or am I not even on the right track?
               

[eunChae]

In my humble opinion,
the hydrogen atom in the OH group has the most acidic character in this molecule. So the molecule wants to give this hydrogen in the presence of a base. OH- is a strong base and takes this hydrogen, forms water and 4-chlorobutyrate ion.

But I couldnt understand how you got CH2=CH-CH2-COOH? Could you plz explain it?

[Roddy]

Yes, that's what I thought it would start at.  My thought was the OH- would take the H from the OH group attached and leave the ion, but I am not sure what happens next.  I think I am suposed to eliminate the Cl somehow (through SN2 reaction).  Any ideas?

[Roddy]

OK, so I tried something else a little different and here is what I got... let me know if this is right!  The OH- takes the H off the OH of the carboxylic acid and the newly formed O- forms a ring to the primary carbon with the Cl attached.  This kicks off the Cl and then the H20 formed from the OH+H comes and attaches to the primary carbon and breaks the ring, giving HOCH2CH2CH2COO-.  (starting from ClCH2CH2CH2COOH + OH-)  Does that look right?

[orgopete]

That doesn't look right. Let's do a quick run down. HO^- reacts with a RCOOH to give H₂O and RCOO^-. RCOO^- reacts with RCH2Cl to give RCOOCH2R and Cl^-. In each reaction, the product is a weaker base than the starting material. Why should your suggested last step, lactone + H₂O -> HOCH₂CH₂CH₂COO^- + H₃O⁺, reverse this trend?