Let $X_1,\ldots,X_n$ be a random sample of size $n$ from a Beta distribution with parameters $α$ and $β=1,$ that the pdf is given by $f(x) = αx^{α-1}$
Find the distribution of $-2\alpha\sum\log x_i$ and construct a two-tailed CI for α, with coverage $100(1-r)\%$ with $0 < r < 1,$ given $\Pr\{\chi^2 > \chi^2_{r/2}\} = r/2$
Using mgf's I derived the distribution of $-2\alpha\sum\log x_i$ as a Chi-squared r.v. with $2n$ degrees of freedom
Then I constructed a confidence interval using $\Pr\left\{ \chi^2_{1-r/2} < -2\alpha\sum\log x_i < \chi^2_{r/2} \right\} = 1 - r,$
From this I obtained the CI for $α$ as $\Pr\left\{ χ^2_{r/2}/2\sum\log(x_i) < α < χ^2_{1-r/2}/2\sum\log (x_i)\right\} = 1 - r$
So overall the CI is given by the interval $\left[χ^2_{r/2}/2\sum\log(x_i), χ^2_{1-r/2}/2\sum\log(x_i)\right],$ but surely this is incorrect as the upper bound of this interval is smaller than the lower bound
$$ -2\alpha\sum\log x_i \sim\chi^2_{2n} $$ I haven't checked the above, but for now I'll assume it is correct. You have $$ \Pr\left\{ \chi^2_{1-r/2} < -2\alpha\sum\log x_i < \chi^2_{r/2} \right\} = 1 - r. $$ Note that $-2\sum\log x_i$ is positive since $\log x_i$ is negative. Your confidence interval is then to be derived from the two inequalities $$ \chi^2_{1-r/2} < -2\alpha\sum\log x_i < \chi^2_{r/2}. $$ Dividing all three members by the positive number $-2\sum\log x_i$ yields $$ \frac{\chi^2_{1-r/2}}{-2\sum\log x_i} <\alpha < \frac{\chi^2_{r/2}}{-2\sum\log x_i}. $$ Since the number by which we divided is positive, the inequalities still go in the same direction.
It looks as if your only mistake may have been losing the minus sign.