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Are all the nitro-compounds having at least one alpha-$\ce{H}$ soluble in $\ce{NaOH}$? And what does solubility has to do with alpha-$\ce{H}$?

I was reading about tautomerism today and got stuck on this thing. Is it due to its conversion to the enol form? Does the keto form also show this? What could be the possible reasons?

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  • $\begingroup$ You should provide more details to your question , show some effort , and ask us where you got stuck and whats your doubt. Please note this is not a homework site ! $\endgroup$ – Tanuj Feb 26 '18 at 15:03
  • $\begingroup$ I was reading tautomerism today and got stuck on this thing. Is it due to its conversion to enol form? They also wrote that enol form is acidic. How is that possible when enol form contains one -OH group? $\endgroup$ – Kirti Agrawal Feb 26 '18 at 15:07
  • $\begingroup$ @kirtiagarwal a compound containing $-OH$ group isn't necessarily basic . See it this way. If some strong base in the solution is able to abstract $H^+$ from the $-OH$ of enol , the enol would be termed as an acid , not a base. $\endgroup$ – Tanuj Feb 26 '18 at 15:09
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    $\begingroup$ I think that it is only short chain nitro groups that are aq NaOH soluble $\endgroup$ – Waylander Feb 26 '18 at 15:52
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    $\begingroup$ Might help lie here? Google: Evans pKa $\endgroup$ – user55119 Feb 26 '18 at 16:11
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The compounds containing nitro groups when dissolved in basic medium, the following things can happen.
The base can take the $\ce{$\alpha$-H}$ of the nitro-group, which will create a negative charge on that $\ce{$\alpha$-C}$, which can further delocalise with the Nitro-group (similar to Nitro-Acinitro tautomerism).
So, the whole compound will be in the state of a Resonance Hybrid where there will be partial double bond character between $\ce{$\alpha$-C, N }$ and two $\ce O$ atoms. and also there will be a $\delta$- charge on the $\ce{$\alpha$-C }$ atom as well as on the $O$ atoms. , and $\delta$+ charge on the $N$ atoms.Due to such salt like compound formation, it becomes soluble.
This $\delta$- charge created on the Carbon atom is stable if the side chain is short, but starts to get destabilised as the side chain grows longer. So, I think the solubilities of the Nitro-compounds decrease as the side chain grows longer, and the compound with $10-12$ C atoms may not be at all soluble in base because of such large Hydrophobic part and destabilisation of the $\delta$- charge.

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  • $\begingroup$ Didn't I put you a link to a page about proper formatting of posts on Chem.SE? Even when edit is started you automatically can see some tips, like using enters instead of breaklines... $\endgroup$ – Mithoron Feb 26 '18 at 18:44

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