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This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it.

Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this questionthis question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you).

Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.

This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it.

Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you).

Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.

This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it.

Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you).

Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.

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Hailong Dao
Hailong Dao
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This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it.

Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you).

Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.