Let $A$ be a finite type algebra over $\mathbb{C}$. Does there exist a finite type $\mathbb{C}$-algebra $B$ and a nonzero divisor $b \in B$ such that $B/b \cong A$ and $B[1/b]$ is Cohen-Macaulay (or, even better, regular)?

Let $A$ be a finite type algebra over $\mathbb{C}$. Does there exist a finite type $\mathbb{C}$-algebra $B$ and a nonzero divisor $b \in B$ such that $B/b \cong A$ and $B[1/b]$ is Cohen-Macaulay (or, even better, regular)?

EDIT: In the view of the very helpful counterexample of Jason Starr given in the comments below, let us assume in addition that $A$ is (S$_2$).