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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?

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  • $\begingroup$ And where will ester ( O-CO-CH3 ) lie in this order? $\endgroup$ – the.eleventh.letter Aug 24 at 4:09
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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.

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$\ce{COOH}$ 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.

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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.

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