Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. If $A$ is semisimple, then $A$ is Morita equivalence with a commutative algebra, that is $A \backsim K^n$ where $n$ is the number of isomorphism classes of simple representations of $A$. For a general, non-semisimple algebra, we cannot expect anymore that $A$ will be Morita equivalent with a commutative algebra: indeed, if $A$ is Morita equivalent with a commutative algebra $B$ then $Z(A)\cong Z(B) = B$, and there are examples of finite dimensional algebras which are not Morita equivalent with their centers. My question is the following:

For a finite dimensional $K$-algebra $A$, can we always find a Morita equivalent finite dimensional $K$-algebra $B$ such that $B/J(B)$ is commutative? Moreover, in case such an algebra exists, will it necessarily be unique?

Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. If $A$ is semisimple, then $A$ is Morita equivalence with a commutative algebra, that is $A \backsim K^n$ where $n$ is the number of isomorphism classes of simple representations of $A$. For a general, non-semisimple algebra, we cannot expect anymore that $A$ will be Morita equivalent with a commutative algebra: indeed, if $A$ is Morita equivalent with a commutative algebra $B$ then $Z(A)\cong Z(B) = B$, and there are examples of finite dimensional algebras which are not Morita equivalent with their centers. My question is the following:

For a finite dimensional $K$-algebra $A$, can we always find a Morita equivalent finite dimensional $K$-algebra $B$ such that $B/J(B)$ is commutative? ($J(B)$ stands here for the Jacobson radical of $B$). Moreover, in case such an algebra exists, will it necessarily be unique?