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(D) The barrier to inversion for $N(i-Pr)_2(CH_3)$ is less than for $N(CH_3)_3$ because going from $sp^3$ (tetrahedral) to $sp^2$ (trigonal planar) sp

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All of these

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(D) The barrier to inversion for $N(i-Pr)_2(CH_3)$ is less than for $N(CH_3)_3$ because going from $sp^3$ (tetrahedral) to $sp^2$ (trigonal planar) spreads the bulky isopropyl groups further apart and relieves steric crowding. In other words,the isopropyl groups destabilize the pyramidal amine more than the planar transition structure.$(b)$ The $sp^2$ transition state has an ideal bond angle of $\sim 120^{\circ}$. The three-membered ring transition state is highly strained because it cannot achieve $120^{\circ}$ due to the small ring locking the angles at $\sim 60^{\circ}$. The $5$-membered ring can easily adopt the $120^{\circ}$ angle.$(c)$ As the nitrogen goes from $sp^3 \rightarrow sp^2$,it becomes effectively more electronegative. (Remember: more $s$-character means more electronegative). The electronegative chlorine atoms pull electron density away from $N$ more than the methyl groups,creating an electron-deficient $N$ and destabilizing the transition state $(T.S.)$.

Several factors (steric,electronic,orbital interactions,etc.) can affect the inversion barrier of an amine. In the given pairs,which data is correctly placed?

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