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I've seen the related questions 1 and 2, they're not duplicates

So. when introducing Red P+HI reduction, our teacher said that it is the strongest reducing agent, and it can reduce any functional group i.e. all of these: carboxylic acid, acid anhydride, ether, amide, acid halide, carbonyl, and nitrile

However, today I stumbled upon a question in my book involving a step in the synthesis of ibuprofen, where they did not reduce the carboxylic acid group even after using RedP+HI. I searched a lot on the internet, and finally came across this paper which confirms this:

enter image description here

paper - notice they've only reduced the alcohol in step (ii)

This now makes me wonder if what I was taught is actually correct. Which brings me to the question:

Does RedP+HI really reduce all carbon functional groups to alkane?

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Nope, in general it doesn't. You are correct. See also for example:

  • A. Aramini, M. R. Sablone, G. Bianchini, A. Amore, M. Fanì, P. Perrone, A. Dolce, M. Allegretti, Tetrahedron 2009, 65 2015–2021. DOI: 10.1016/j.tet.2009.01.005:

enter image description here enter image description here

  • M. Dobmeier, J. M. Herrmann, D. Lenoir, B. König, Beilstein J. Org. Chem. 2012, 8, 330–336. DOI: 10.3762/bjoc.8.36 Open access PDF. (This also includes your one as a reference, too.):

enter image description here enter image description here

If it did, it would be a pretty poor reagent since it would have no functional group tolerance, so not very useful. However, it is not unlikely that under special conditions it might be able to reduce carboxylic groups as well.

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    $\begingroup$ Instead of just linking the entire pdf, if you could cite it as a reference, and pick out the specific extracts, or page numbers from it, that'd be better. In fact, if you could also list the carbon functional groups that are actually reduced by red P +HI, that would make your answer much more complete. Thank you! ^_^ $\endgroup$
    – Gaurang Tandon
    Feb 17 '18 at 5:36
  • $\begingroup$ @GaurangTandon From the looks of it, HI + P is mainly aimed at reducing the OH to H. If you read the second paper they also have a table where they report decomposition and lack of selectivity using the reagent for some compounds. $\endgroup$
    – AMM
    Feb 18 '18 at 1:34
  • $\begingroup$ Thanks for the details! Yes, you're correct. This does seem to be primarily aimed at alcohols. From the table, it's clear that they did not (1) reduce either at step 5 (2) ketone at step 4 (3) amide at step 6 (4) ester at step 12. Though, I'll wait for some conclusive evidence before the green checkmark. Thanks for answering though! :) $\endgroup$
    – Gaurang Tandon
    Feb 18 '18 at 2:09

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