The problem is that the \begin{...} and \end{...} pair commands automatically create a "group" so that, in effect, the & and \\ are "out of scope" for the tabular, while inside the "production" environment they just show up at a place where the compiler is not expecting them.
A second problem with your definitions is that, even if they would work, they would be adding an extra \\ at the end of the tabular, adding an unwanted space at the end. Perhaps some more appropriate definitions would be
\documentclass[a4paper,10pt]{article}
\newcommand{\production}[1]{#1 ::= &}
\newenvironment{grammar}{\tabular{p{3cm}l}}{\endtabular}
\begin{document}
\begin{grammar}
\production{XmlStartTag} ... \\
\production{XmlOtherTag} ... \\
\production{XmlEndTag} ...
\end{grammar}
\end{document}
Note no \\ at the end of the last production. Also in the definition of the grammar you don't need to repeat the work of \begin/\end, and you can instead directly use \tabular and \endtabular.
The \ifthenelse condition ends prematurely and leaves an open environment hanging around in the middle of nowhere.
In conjunction with tcolorbox environment, the end - delimiter is \endtcolorbox and I suggest to use two \ifthenelse statements, one for the start code of the environment and another one for the end code.
A better approach would use \DeclareTColorbox, in my opinion or a weird \scantokens construct.
Also possible: Use \tcolorboxenvironment to wrap around an existing solution environment.
\documentclass{article}
\usepackage{ifthen}
\usepackage[most]{tcolorbox}
\newboolean{solution}
\newenvironment{solution}{%
\ifthenelse{\boolean{solution}}{%
\tcolorbox[breakable, width=\textwidth, colframe=red, colback=white]
}{%
}%
}{\ifthenelse{\boolean{solution}}{\endtcolorbox}{}}
\begin{document}
\setboolean{solution}{true}
\begin{solution}
\begin{align*}
x^2 + y^2 &= z^2\\
\Rightarrow x &= \sqrt{z^2 - y^2}\\
&= ...
\end{align*}
\end{solution}
\setboolean{solution}{false}
\begin{solution}
\begin{align*}
x^2 + y^2 &= z^2\\
\Rightarrow x &= \sqrt{z^2 - y^2}\\
&= ...
\end{align*}
\end{solution}
\end{document}
Cleaner solution with two different environments
\documentclass{article}
\usepackage[most]{tcolorbox}
\tcbset{
commonboxes/.style={nobeforeafter},
nobox/.style={commonboxes,blank,breakable},
solutionbox/.style={commonboxes,breakable, colframe=red, colback=white}
}
\newtcolorbox{solutionbox}[1][]{
solutionbox,#1
}
\newtcolorbox{solutionbox*}[1][]{%
nobox,#1
}
\begin{document}
\begin{solutionbox*}
\begin{align*}
x^2 + y^2 &= z^2\\
\Rightarrow x &= \sqrt{z^2 - y^2}\\
&= ...
\end{align*}
\end{solutionbox*}
\begin{solutionbox}
\begin{align*}
x^2 + y^2 &= z^2\\
\Rightarrow x &= \sqrt{z^2 - y^2}\\
&= ...
\end{align*}
\end{solutionbox}
\end{document}
Third installment of a solution with \NewEnviron and the \BODY command.
\documentclass{article}
\usepackage{environ}
\usepackage{ifthen}
\usepackage[shortlabels]{enumitem}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage[most]{tcolorbox}
\newboolean{solution}
\tcbset{
commonboxes/.style={nobeforeafter,breakable},
nobox/.style={commonboxes,blank,breakable},
solutionbox/.style={commonboxes,breakable, colframe=red, colback=white}
}
\NewEnviron{solution}[1][]{%
\ifthenelse{\boolean{solution}}{%
\tcolorbox[solutionbox, width=\textwidth,#1]
\BODY
}{%
}%
}[\ifthenelse{\boolean{solution}}{\endtcolorbox}{}]
\begin{document}
\begin{enumerate}[label={\alph*)}]
\item Compute the Fourier transform of $e^{-|x|}$ for $x\in \mathbb{R}$.
\begin{solution}[colframe=blue]
\begin{align*}
\hat{f}(\xi)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ix\xi}dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x-ix\xi}dx+\int_{-\infty}^0e^{x-ix\xi}dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}(e^{-x-ix\xi}-e^{-x+ix\xi})dx\\
&=\frac{1}{\sqrt{2\pi}}[\frac{1}{-(1+i\xi)}(-1)-\frac{1}{-1+i\xi}(-1)]\\
&=\frac{1}{\sqrt{2\pi}}[\frac{1-i\xi}{1+\xi^2}+\frac{-(1+i\xi)}{1+\xi^2}]\\
&=\frac{1}{\sqrt{2\pi}}\frac{-2i\xi}{1+\xi^2}\\
&=-\sqrt{\frac{2}{\pi}}\frac{i\xi}{1+\xi^2}
\end{align*}
\end{solution}
\item Compute the Fourier transform of $e^{-a|x|^2},~a>0$, directly, where $x\in \mathbb{R}$.\\
\begin{solution}
\begin{align*}
\hat{f}(\xi)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a|x|^2}e^{-ix\xi}dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a(x+\frac{i\xi}{2a})^2+\frac{-\xi^2}{4a}}dx~~~~~~~~x'\doteq x+\frac{i\xi}{2a}\\
&=\frac{1}{\sqrt{2\pi}}e^{-\frac{\xi^2}{4a}}\int_{-\infty}^{\infty}e^{-ax^2}dx\\
&=\frac{e^{-\frac{\xi^2}{4a}}}{2a}
\end{align*}
\end{solution}
\end{enumerate}
\setboolean{solution}{true}
\begin{enumerate}[label={\alph*)}]
\item Compute the Fourier transform of $e^{-|x|}$ for $x\in \mathbb{R}$.
\begin{solution}[colframe=blue]
\begin{align*}
\hat{f}(\xi)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ix\xi}dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x-ix\xi}dx+\int_{-\infty}^0e^{x-ix\xi}dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}(e^{-x-ix\xi}-e^{-x+ix\xi})dx\\
&=\frac{1}{\sqrt{2\pi}}[\frac{1}{-(1+i\xi)}(-1)-\frac{1}{-1+i\xi}(-1)]\\
&=\frac{1}{\sqrt{2\pi}}[\frac{1-i\xi}{1+\xi^2}+\frac{-(1+i\xi)}{1+\xi^2}]\\
&=\frac{1}{\sqrt{2\pi}}\frac{-2i\xi}{1+\xi^2}\\
&=-\sqrt{\frac{2}{\pi}}\frac{i\xi}{1+\xi^2}
\end{align*}
\end{solution}
\item Compute the Fourier transform of $e^{-a|x|^2},~a>0$, directly, where $x\in \mathbb{R}$.\\
\begin{solution}
\begin{align*}
\hat{f}(\xi)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a|x|^2}e^{-ix\xi}dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a(x+\frac{i\xi}{2a})^2+\frac{-\xi^2}{4a}}dx~~~~~~~~x'\doteq x+\frac{i\xi}{2a}\\
&=\frac{1}{\sqrt{2\pi}}e^{-\frac{\xi^2}{4a}}\int_{-\infty}^{\infty}e^{-ax^2}dx\\
&=\frac{e^{-\frac{\xi^2}{4a}}}{2a}
\end{align*}
\end{solution}
\end{enumerate}
\end{document}
The \BODY command contains the environment 'text' and is printed only in the case solution is true.
Best Answer
I'm not sure why but you can't use
\begin{align}and\end{align}in the definition of a new environment; you have to use the "lower-level" macros\alignand\endaligninstead. Edit: as pointed by alexwlchan in his comment, you can find more details about that in section 6 of Technical notes on theamsmathpackage.Here I've used the equivalent of an
align*environment (see Herbert's answer to Define a custom align, and align* environment).Note that you will get an error if you try to reset your
grammarcounterat each chapter in thearticleclass, because the latter doesn't have chapters;\sectionis the most high-level sectioning command in thearticleclass. Did you meaninstead?