# Comparing relative negative inductive effect of substituents (dialkyl amine, amine, alkoxide, fluoride)

Which of the following is correct with respect to the negative inductive effect of the substituents? 1. $\ce{-NR2 < -OR < -F}$
2. $\ce{-NH2 < -OR < -F}$
3. $\ce{-NR2 > -OR > -F}$
4. $\ce{-NH2 > -OR > -F}$

Since it would be directly proportional to electronegativity, $\ce{-F}$ will be highest. That cancels out (3) and (4).

Now, to me both (1) and (2) seem correct. But the official answer (since this is from an entrance exam) is (2).

Are both (1) and (2) correct? Or is there something wrong with (1)?

The inductive effect is the polarization of σ (sigma) bonds within a chemical species caused usually by differences in electronegativity between the neighboring atoms. The higher the polarization it causes, the higher the inductive effect. In this case, the decreasing order of electronegativity is:

$\ce{-F > -OR > -NH2}$

So, for practical purposes, the correct answer should be option 2.

However, substituents far from the atom you are considering can also affect the inductive effect and an amino ($\ce{-NR2}$) group can be more withdrawing than an alkoxide ($\ce{-OR}$) group.

For example: $\ce{-N(-CF3)2 > -OCH3}$

This amino substituent is probably more electron-withdrawing than the alkoxide in the right, because the first is bound to two highly withdrawing groups ($\ce{-CF3}$) while the second is bound to an electron-donating group ($\ce{-CH3}$)

• It was more of a trick question because whenever we think of $\ce{-R}$, we think of alkyl groups intuitively – Abcd Jul 18 '18 at 10:16
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