(Most likely) no. Your question is a bit hard to answer as you do not say what your hypothesis and your test statistic are, nor does it define $\chi_{2n,\alpha}$, so we will make assumptions.
First, typically, statistics $T(X)$ that have a chi-square null distribution (that is how I read the quantile $\chi_{2n,\alpha}$) are (as the square indicates) squared expressions, i.e., any negative signs get removed by squaring so that, generally, large values of the test statistic provide evidence against the null. A leading example is tesing if all slope coefficients of a regression model are zero, see e.g. here. Another is that no study in a meta-analysis provides evidence against the null, see e.g. here.
Hence, you reject if the statistic is larger than an extreme upper quantile of the null distribution, which, in your notation, should then be $\chi_{2n,\alpha}$. At the 5% level, this notation could indicate the 95% quantile. Then, $\chi_{2n,1-\alpha}$ would be the 5% quantile and rejecting if $T(X) < \chi_{2n,1-\alpha}$ would mean to reject if the test statistic is very small.
Now, it is not impossible to think of applications of test statistics with a chi-square null distribution that reject either if $T$ is small or if $T$ is large (see e.g. here), but, by my argument above, that is not common.