I'm assuming you're asking about the theory $ZF+\neg AC+CH$. The answer is yes - but you have to be clear about what you mean by "$CH$."

First, a trivial example: we can start with a model of $ZFC+CH$, and force a failure of $AC$ via a (symmetric submodel of) a $2^{\aleph_0}$-closed forcing extension. Since our forcing extension is sufficiently closed, it adds no new sets of reals, so $CH$ remains true; essentially, what's going on is that we're adding a failure of $AC$ to our model, but we're doing so at such a high level in the cumulative hierarchy that it doesn't affect the reals.

A more refined version of your question: can we have $ZF+\neg AC(\mathbb{R})+CH$? That is, a model of $ZF$ in which $CH$ holds but the reals are not well-ordered.

This is where we need to be precise about what $CH$ means. The useful version of $CH$ is "every set of reals is either at most countable, or can be bijected onto $\mathbb{R}$." Under this phrasing, we do indeed have the consistency of $ZF+\neg AC(\mathbb{R})+CH$!

To get $AC(\mathbb{R})$ to fail, we just need a model in which there is no injection from $\omega_1$ (which is defined to be the least uncountable ordinal) to $\mathbb{R}$: if $AC(\mathbb{R})$ held, we could send each countable ordinal $\alpha$ to the "least" real which codes a well-order of order type $\alpha$. To force the continuum hypothesis to hold requires a bit more subtlety. My favorite model of $ZF+\neg AC(\mathbb{R})+CH$ is $L(\mathbb{R})$ under the assumption of large cardinals: large cardinals imply that $L(\mathbb{R})\models AD$, the axiom of determinacy, which in turn implies that every set of reals has the perfect set property (so $CH$ holds) and also that there is no injection from $\omega_1$ to $\mathbb{R}$ (so $AC(\mathbb{R})$ fails). However, this model does involve a massive jump in consistency strength, past that of $ZF$, which is not necessary: Truss has a construction, which is a variation of a construction of Solovay (which *does* require large cardinals, albeit just one small one!), which does not require any more consistency strength than $ZF$ itself and satisfies $ZF+\neg AC(\mathbb{R})+CH$.