4. In a cube $A C^{\prime}$ with an edge length of 1, $E$ and $F$ are the midpoints of edges $B C$ and $D D^{\prime}$, respectively. The volume of the tetrahedron $A B^{\prime} E F$ is $\qquad$ .
4. $\frac{5}{24}$.

Although the areas of the tetrahedron's faces can be calculated, it is difficult to find any height, so it is not advisable to directly calculate the volume. Instead, consider an equivalent volume transformation, making one face of the tetrahedron $A B^{\prime} E F$ parallel to the base of the cube. Note that the tetrahedron $A B^{\prime} E F$ has two vertices $A$ and $E$ within the base $A B C D$ of the cube. Therefore, consider transforming another vertex to the plane $A B C D$. Suppose the base $A E F$ remains fixed, while the vertex $B^{\prime}$ moves along a line $l$ parallel to the plane $A E F$. Draw a line $l \parallel A F$ through $B^{\prime}$, which must intersect the extension of $C B$ at a point $G$. Let the midpoint of $B B^{\prime}$ be $F^{\prime}$. Take a point $K$ on $B G$ such that $B K=1$, then $F^{\prime} K \parallel F A$. Therefore, $F^{\prime} K \parallel B^{\prime} G$.
Since $B F^{\prime}=F^{\prime} B^{\prime}$, it follows that $G K=K B=1, E G=\frac{5}{2}$.
Thus, $V=V_{C-A E F}=V_{F-A C E}=\frac{1}{6} A B \cdot E G \cdot F D=\frac{5}{24}$.