This is probably a trivial question, but I don't see the answer, and I haven't found it on Wikipedia, nLab, nor MathOverflow.

Let $\text{ComAlg}$ denote the category whose objects are commutative algebras over a fixed field $\mathbb K$ and whose morphisms are homomorphisms of algebras, and let $\text{ComAlg}^{\rm op}$ denote its opposite category. Given commutative algebras $A,B$, let $\operatorname{hom}(A,B)$ denote the set of algebra homomorphisms $A\to B$, so that $\operatorname{hom}$ is the usual functor $\text{ComAlg}^{\rm op} \times \text{ComAlg} \to \text{Set}$. The short version of my question:

Is $\text{ComAlg}^{\rm op}$ Cartesian closed?

The long version of my question (if I've gotten all the signs right):

Is there a functor $[,] : \text{ComAlg} \times \text{ComAlg}^{\rm op} \to \text{ComAlg}$ such that there is an adjunction (natural in $A,B,C$, i.e. an isomorphism of functors $\text{ComAlg}^{\rm op} \times \text{ComAlg} \times \text{ComAlg} \to \text{Set}$) of the form: $$ \operatorname{hom}([A,B],C) \cong \operatorname{hom}(A,B\otimes C) ?$$

Recall: $\otimes$ is the coproduct in $\text{ComAlg}$, hence the product in $\text{ComAlg}^{\rm op}$.

Motivation: $\text{ComAlg}^{\rm op}$ is complete and cocomplete, and so many constructions that make sense in $\text{Set}$ and $\text{Top}$ transfer verbatim to the algebraic setting. I would like to know how many.