# Dehydration of carboxylic acids with phosphoric acid

To find Z one must find X. Finding Y is of no use to find Z.

I get X as:

After addition of $$\ce{H2/Pd-C}$$, the unsaturated portion becomes saturated and the carboxylic acid remains unchanged.

However I am unsure what does $$\ce{H3PO4}$$ do to this (new) carboxylic acid? I was unable to find any information. I did find information regarding dehydration with $$\ce{P2O5}$$, but unsure what to make use out of it.

After looking at (unofficial/3rd party) solutions, I saw that $$\ce{H3PO4}$$ protonates the $$\ce{-OH}$$ part of the carboxylic acid (since it is a much stronger acid than the organic acid?) then $$\ce{H2O}$$ leaves, rendering a carbo-cation at the carbonyl carbon, followed by EAS at the ring.

However, I do not see how $$\ce{H3PO4}$$ exactly works with the carboxylic acid. I have merely guessed that the $$\ce{-OH}$$ portion gets protonated (since, this was the only thing I predicted would happen). I really do not know why this happens and how it happens.

Hints and reference material or examples would be appreciated.

The original question is taken from the JEE (Advanced) [2018] Paper-1.

• Friedel-Crafts acylation... – orthocresol May 6 at 13:52
• Sure I do know that happens, however for that to happen, a carbonyl cation has to form. I do not know how exactly the carbonyl cation forms here by H3PO4. @orthocresol – McSuperbX1 May 6 at 13:54
• Ok, but you already (correctly) described what happens. What else do you really want? – orthocresol May 6 at 13:57
• @orrthocresol Please see that I said the sources are third party and unofficial. I would like to know H3PO4 exactly causes dehydration of the carboxylic acid - because I have never seen this before. I will reframe my question to make it more clearer. – McSuperbX1 May 6 at 13:58
• phosphoric acid, usually in the form of polyphosphoric acid, is a powerful dehydrating agent. It can be regarded as the anhydride of phosphoric acid capturing OH by formation of an HO-P bond. A brief review is here: reag.paperplane.io/00002328.htm Its major advantage over sulfuric acid for cyclisations is that it is not an oxidising agent. – Waylander May 6 at 14:01

The $$\ce{-OH}$$ part will not get protonated, since its lone pair is in resonance with the carbonyl part of the $$\ce{-COOH}$$ group. Instead, the lone pair of the carbonyl part will give its lone pair to $$\ce{H+}$$.