A chemistry reference book that I have been reading gives the order of inductive effect ($-I$) as follows:

inductive effect ordering as per the reference book

While the Wikipedia page on inductive effect gives the following order:

$\ce{-NH3+} > \ce{-NO2} > \ce{-SO2R} > \ce{-CN} > \ce{-SO3H} > \ce{-CHO} > \ce{-CO} > \ce{-COOH} > \ce{-COCl}> \ce{-CONH2} > \ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I} > \ce{-OR} > \ce{-OH} > \ce{-NH2} > \ce{-C6H5} >\ce{-CH=CH2} >\ce{-H}$

The reference book has $\ce{COOH}$ ahead of $\ce{CHO}$ but Wikipedia has it the other way around. Which source is correct?

  • 1
    $\begingroup$ And where will ester ( O-CO-CH3 ) lie in this order? $\endgroup$ Aug 24 '20 at 4:09

I believe the Wiki page is more accurate, because the Hammett Equation substituent constants (para) indicate -CHO should be more "electron withdrawing" than -COOH, but there is little reason to use such lists as "absolute" - I think they are simply good guidelines for reactivity.

I looked at the data in this link: https://www.bluffton.edu/homepages/facstaff/bergerd/classes/CEM311/handouts/sigmas.pdf

However, substituent constants are not absolute proof of "inductive effect" and it is difficult to conclude from such data that a particular substituent is more "inductively" electron withdrawing than another. To do that, we would need a model (or a set of compounds, equivalently) where we can achieve separability of sigma and pi electrons.


$\ce{COOH}$ will show more -i effect than $\ce{CHO}$ because $\ce{COOH}$ hybridisation considering free radical as a sigma bond will be $\ce{sp^2}$ and both oxygen will also attract electron in angle 120° so that re will be more positive than aldehyde therefore it will attract electron more and show more -i effect.


CHO group will have more -I effect than COOH because OH group in COOH group will decrease the charge on the carbonyl C which reduces it's withdrawing effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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