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Textbooks I've read state that R-NH3+ has exhibits a greater -I inductive effect than R-NH2 does.

However, I've also read in places that the + charge in R-NH3+ only represents a formal charge and has no physical significance.

If this is the case what explains the greater inductive effect shown by R-NH3+?

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    $\begingroup$ Does a proton have a formal charge without physical significance? As the extra proton is the difference between R-NH2 and R-NH3+. $\endgroup$
    – Poutnik
    Commented Jan 13 at 16:39
  • $\begingroup$ I'm actually unsure. I've read in textbooks that compounds like NH4+ and R-NH3+ are formed as a result of coordinate covalent bonds, in which we only represent the + and - "formal" charge to denote exclusively, the donor and acceptor atoms $\endgroup$
    – Aarchaeus
    Commented Jan 14 at 16:22
  • $\begingroup$ All N-H bonds are equivalent there, partial delocalized positive charge distributed on all of them. $\endgroup$
    – Poutnik
    Commented Jan 14 at 21:05

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Thanks for your interesting question, To my knowledge, depending on the solvent you use, you can have a protontaed or a non protonated species. if you are using an aliphatic amine in water, for example, the compound is mostly protonated. For sure the protonated form has a stronger electron-withdrawing effect than its non-protonated form. To answer your question, sure! the charge on the RNH3 group is important and it has a physical significance. It makes the groups more able to attract negative charge through single bond thus increasing the inductive effect on the carbon skeleton of your compound.

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