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Question & Options

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

I get X as:

Product X

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.

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  • $\begingroup$ Friedel-Crafts acylation... $\endgroup$
    – orthocresol
    May 6, 2019 at 13:52
  • $\begingroup$ 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 $\endgroup$
    – McSuperbX1
    May 6, 2019 at 13:54
  • $\begingroup$ Ok, but you already (correctly) described what happens. What else do you really want? $\endgroup$
    – orthocresol
    May 6, 2019 at 13:57
  • $\begingroup$ @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. $\endgroup$
    – McSuperbX1
    May 6, 2019 at 13:58
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    $\begingroup$ 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. $\endgroup$
    – Waylander
    May 6, 2019 at 14:01

1 Answer 1

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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 $\ce{H+}$.

Something like this :-

Mechanism

It may look insignificant because the product is same, but this is the correct mechanism.

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    $\begingroup$ Welcome to Chemistry.SE! You might want to get acquainted with MathJax/mhchem formatting and drawing chemical structures, it helps make posts look more presentable. Have a great day. -From Review $\endgroup$
    – William R. Ebenezer
    May 7, 2019 at 5:29
  • $\begingroup$ And what is the reason why H3PO4 does this? Its dehydration properties or its much stronger acidity or both? What about acids such as H2SO4? $\endgroup$
    – McSuperbX1
    May 7, 2019 at 7:06

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