Solution 20

Last updated

Jun 3, 2019
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- Solution 2
- Solution 21

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( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

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$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

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$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

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$\newcommand{\norm}[1]{\| #1 \|}$

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$\definecolor{fillinmathshade}{gray}{0.9}$

H due to resonance (see conjugate bases below). The charge due to removal of the red proton on the right is localized on the Nitrogen. However, the charge due to removal of the blue proton is stabalized by resonance. Since it is more stable, it is therefore more acidic.

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