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Why does $\ce{-NH_2}$ shows more +M effect that $\ce{-NHR}$ and $\ce{-NR_2}$? My first conclusion was totally opposite of the reality. I thought, as the alkyl groups are attached to the Nitrogen the lone lone pair on nitrogen should be easily available. But it doesn't seem to be happening. What is the actual cause?

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It is because the bulky alkyl groups provide repulsion to the lone pair such that their orientations change. Due to the change in orientation the %s character changes and the lone pair becomes less available. This is not the case with hydrogen.

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The main reason for NH2 showing more +M than NHR and NHR more than NR2 is electronegativity As in NR2 two bulky groups are present so bond angle between the two groups will increase and we know that bond angle is directly proportional to electronegativity due to which the electronegativity of nitrogen atom of NR2 increases due to which NR2 becomes a better group to take electron density rather than donating its electrons which in turn reduces this groups tendency to donate it's lone pair of electron to other groups easily Same reason can be applied for NHR and NH2 Also because of the above reason the reverse order of -I effect is seen as NR2 most electronegative so more -I . Hence order of +M NH2 >NHR> NR2 Order of -I effect NH2< NHR< NR2 Thanks .

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