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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? $\endgroup$ Aug 24, 2020 at 4:09
  • $\begingroup$ I also find it strange that the acyl chloride COCl is considered less electron-withdrawing than aldehydes (CHO) and ketones (CO) in the Wikipedia list. $\endgroup$ Nov 24, 2021 at 13:44

4 Answers 4

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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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  • $\begingroup$ Shouldn't decreasing the charge on C increase the electron-withdrawing effect (on nearby substituents)? $\endgroup$ Nov 24, 2021 at 13:42
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Note that electron density is enhanced at the carbonyl carbon in carboxylic acid by resonance effect ( lone pair on oxygen is in conjugation with pi bond on C-O bond ) . Hence, the R effect increases electron density at carbon in carboxylic acid while there is no R effect in aldehyde .

And , we know that ;

The more electron rich is carbon , the less electronegative it is and the less electronegative it is, the less negative inductive effect it shows .

Hence , aldehyde group would show more I effect than carboxylic acid.

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  • $\begingroup$ Can anyone do a constructive criticism that what is wrong in my answer ? $\endgroup$
    – Get_ Maths
    Apr 13 at 15:15

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