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3° carbocations are the most stable because of the +I effect of -CH3 group. This group gives a slight negative charge to the central carbon atom, which makes it less positively charged and thus, more stable.

But, in the case of -N(CH3)3 and -NH3, the central N atom has a positive charge.

-CH3 group has +I effect on the central N atom, making N atom slightly less positively charged. So, -N(CH3)3 group should have lesser -I effect. Isn't this statement correct?

But, -N(CH3)3 group has greater -I effect than -NH3. Why does this happen?

Why does -N(CH3)3 have greater -I effect than -NH3?

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    $\begingroup$ The R3N- group has a full positive charge that you omitted from the structure. $\endgroup$
    – jimchmst
    Nov 30, 2021 at 23:25
  • $\begingroup$ Yeah, but still the +I effect of R will make the N atom less positively charged, similar to the case of carbocations. So, in R3N- group, the central N atom, being slightly less positively charged than that of H3N- group, should give lesser -I effect than H3N- , right? $\endgroup$
    – user119416
    Dec 1, 2021 at 9:12
  • $\begingroup$ Sorry if I have made any mistake. I am in high school now. I cannot understand this phenomenon. Why does -N(CH3)3 gives more -I effect than -NH3? $\endgroup$
    – user119416
    Dec 1, 2021 at 9:13
  • $\begingroup$ Are you speaking of the positive $\ce{NH3^+}$ ion, or of the negative (-)$\ce{NH3^-}$ ion, or of the neutral molecule $\ce{NH3}$ ? Or even of the $\ce{R-NH3^+}$ ion ? $\endgroup$
    – Maurice
    Dec 1, 2021 at 11:14
  • $\begingroup$ NH3+ ion and NR3+ ion. "-NH3" and "-NR3" are common ways of representation of those groups. $\endgroup$
    – user119416
    Dec 1, 2021 at 15:07

1 Answer 1

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The methyl group carbon has a partial negative charge due to the polarity present in the C-H bonds. This partial negative charge is preferentially donated, leading the methyl group to have the +I effect.

An important way to consider the stability of a tertiary carbocation is through the process of hyperconjugation. Neighboring C-H or C-C bonds, parallel to the vacant p-orbital of a carbocation, can donate some of their electron density to the empty p-orbital, thus stabilizing the positive charge. Tertiary carbocations have a greater capability of providing this electron density, and thus are more stable.

The positive charge present on the nitrogen in either the -NR3+ or -NH3+ groups induces the extent of the -I effect exhibited by both groups. Although it may seem that the presence of alkyl groups would lend a greater +I effect to the -NR3+ group, there are other factors that result in this "anomaly" of sorts.

The factor that causes this is the issue of steric hindrance. The alkyl groups surrounding the nitrogen are relatively large and do not want to be in the vicinity of each other. Although the structure of the -NR3+ group would appear to be tetrahedral, as the alkyl groups push away from each other, they move towards what seems to be more of a planar structure. This is an increase in bond angle from the tetrahedral structure.

It is important to note that the greater the bond angle, the greater %s character. S orbitals are typically closer to the nucleus as compared to the p orbitals, and thus, the electrons are held closer -- the electronegativity of the nitrogen atom will be greater than if it was in the -NH3+ group. As such, it exhibits a greater -I effect.

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