If I add $aq. \ce{KOH}$ to both of the compounds substitution takes place in the case of benzyl chloride and then if I oxidize with $\ce{KMnO4}$ I will get benzene carboxylic acid which will give effervescence on reaction with $\ce{NaHCO3}$. But, does $\ce{KMnO4}$ have any effect on chloro benzene?


When benzyl chloride is treated with $\ce{NaOH}$ and $\ce{AgNO3}$ , a white precipitate is formed. Chlorobenzene does not form white precipitate reacting with $\ce{NaOH}$ and $\ce{AgNO3}$.

  • $\begingroup$ Can you expand on why this is so and the reactions involved? $\endgroup$ – Satwik Pasani Mar 8 '14 at 7:29
  • 1
    $\begingroup$ Nice! However, you can force the transformation of chlorobenzene to phenol at ~ 450 °C when streaming mixtures of water and chlorobenzene (in the vapour phase) over oxidic catalysts. That's the old (and long replaced) phenol synthesis by Raschig. $\endgroup$ – Klaus-Dieter Warzecha Mar 8 '14 at 10:01

Your observations are correct and on the short term, i.e. in the time scale of typical reactions in the lab, you'll hardly see an effect of the permanganate on chlorobenzene.

However, there seemingly have been attempts to decontaminate soil, e.g. on production sites, using permanganate for the complete oxidation of pollutants to carbon dioxide. But these processes are supposed to take much longer ( > 1 year) and usually don't play a role in considerations on what will happen under typical lab conditions.

But referring to the title, I'd rather suggest to simply record $^1H$ NMR spectra of the compounds. Even on an ancient 90 MHz spectrometer, you'll be able to spot the signals of the benzylic protons at $\delta$ = 4.5 ppm.


protected by orthocresol Jul 10 '17 at 13:05

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.