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In a recent exam, following question was asked:

Number of acidic hydrogen present in X are:

enter image description here

I've found 8 of them as shown:

enter image description here

7 of them (marked with numerals) were correct according to solution, but 8th one (pointing with red arrow) was not. Is it because of the fact that the carbon was bridge-headed or something else?

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  • $\begingroup$ I guess the trans conformation of the decalin-like ring might not permit the transition from the p orbital to the antibonding of π C=O for the marked hydrogen $\endgroup$ Commented Apr 4, 2020 at 11:37
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    $\begingroup$ @Yusuf Hasan: How do you know that the molecule has a trans conformation (sic)? Perhaps you meant configuration. I suspect that the inventors of the problem assumed that the order of deprotonation would be 6, 3(4), 1(2).... The enolate from 3(4) might render 8 less acidic. All in all, it is a terribly constructed question. $\endgroup$
    – user55119
    Commented Apr 4, 2020 at 15:01
  • $\begingroup$ It depends on the criteria used. In aqueous solution, only the H form COOH is acidic when heated by NaOH. The H7 is also acidic, but not in water. All H atoms are weak acids, sometimes extremely weak. $\endgroup$
    – Maurice
    Commented Apr 4, 2020 at 16:16

2 Answers 2

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An aldehyde or a ketone with an alpha hydrogen forms a carbanion that resonates to enolate form. This leads to two canonical structures that are in resonance. The example below is a polyphenol compound extracted from turmeric, called curcumin.

enter image description here

source: Biomaterials. 2010 May;31(14):4179-85. doi: 10.1016/j.biomaterials.2010.01.142. Epub 2010 Feb 23. (https://www.ncbi.nlm.nih.gov/pubmed/20181392)

The hydrogens between two keto groups are most acidic. This is due to formation of two equivalent enol structures stabilized by hydrogen bonding. The structural requirement would be planarity of atoms involved in hydrogen bonding.

Below are 3-D structures of the given compound in the question above.

  • Structure without hydrogens

Bridgehead carbon is 4. Keto groups are at 2 , 5 and 8 (shaded in grey). Keto groups on 5 and 8 are not in the same plane . Hence Hydrogen on 4 , 22 (see figure with hydrogens) does not form stable hydrogen bonded enol structures as discussed in the above example.

Therefore, hydrogen at 8th one (pointing with red arrow) may not be acidic.

Note: 3-D structures have been drawn on Chem3Dpro

enter image description here

  • Structure with hydrogens

enter image description here

  • Structure with hydrogens (2nd view)

enter image description here

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  • $\begingroup$ Thanks for your 3D structures, it was very helpful. But, is it true that those keto groups will always be non-coplanar (when both placed adjacent to bridgehead carbon), or is it specific to this one? (I'm asking this bcoz, I'd been provided only with 2D structure) $\endgroup$ Commented Apr 5, 2020 at 9:50
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    $\begingroup$ @Rahul verma if two keto groups are in two 6 membred rings seperated by a carbon bonded to these two rings , the two keto groups may not be in the same plane. $\endgroup$ Commented Apr 5, 2020 at 10:30
  • $\begingroup$ Why is hydrogen 5 acidic then? The two carbonyls are non co planar too. $\endgroup$
    – ba-13
    Commented Sep 9, 2020 at 3:10
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It is because of the fact that a double bond is not usually stable at the bridge head position in smaller fused rings. Therefore the hydrogen cannot be considered as acidic.

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    $\begingroup$ I don't understand. Where did the double bond come from? We're just asked to find the acidic hydrogens, right? $\endgroup$
    – Vishnu
    Commented Apr 4, 2020 at 15:44
  • $\begingroup$ You may relate this with hyperconjugation and look into what makes a hydrogen acidic $\endgroup$ Commented Apr 4, 2020 at 15:59
  • $\begingroup$ Are you referring to the Bredt's rule? $\endgroup$ Commented Apr 4, 2020 at 21:05
  • $\begingroup$ Yes I was indeed referring to Bredt's rule $\endgroup$ Commented Apr 5, 2020 at 13:09

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