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Why $\ce{-NR3+}$ has more -I effect than $\ce{-NH3+}?$

And what about order of $\ce{-NHR2+}$ and $\ce{-NH2R+}?$

My attempts

I got idea about electronegativity of carbon atom in $\ce{R}$ is more than hydrogen atom, hence positive charge gets more concentrated on nitrogen atom and it exerts more electronic attraction. But this didn't worked out in latter case when -I effect of $\ce{-NHR2+}$ is less than $\ce{-NH2R+}.$

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The inductive effect is both distance-dependent and χ-dependent. N-H bonds are 1.01 Å, yet the χ of H is only 2.1. N-C bonds are 1.42 Å, yet the χ of C is 2.5. At the extremes, where there are only N-C or N-H bonds, the ordering of -I is easy to rationalize.

However, the two, short N-H bonds in -NH2R+ outweigh the higher χ of N-C bonds, giving -NH2R+ a higher -I than -NHR2+.

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  • $\begingroup$ What that X means dipole moment $\endgroup$
    – Dracula
    Commented Jul 12, 2019 at 16:44
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    $\begingroup$ @user49564 It's not "X", it's a Greek letter χ ("chi") . Guessing from its numerical values and the fact it's dimensionless, my bet is it's electronegativity. Mark, it would be nice to add clarification to the answer. $\endgroup$
    – andselisk
    Commented Jul 12, 2019 at 18:41
  • $\begingroup$ Oh yeah, it's electronegativity. Thanks :) $\endgroup$
    – Dracula
    Commented Jul 12, 2019 at 18:53
  • $\begingroup$ Can anyone tell me why are not we considering the +I effect shown by alkyl group and hydrogen atom to determine the order of -I effect in $NR_3+$ and $NH_3+$? $\endgroup$ Commented Feb 28, 2020 at 9:03

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