We can obtain a representation of the Pauli matrices through the exponential map: $$ \exp(-i \vec{\sigma} \cdot \hat{n} \phi)= \textbf{1} \cos(\phi) - i \vec{\sigma}\cdot \hat{n} \sin(\phi) .$$
Why is it that the exponential of the $\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$ Pauli matrix specifically, which is: $$ \exp(-i \sigma_2 \phi) = \begin{pmatrix} \cos(\phi) & - \sin(\phi) \\ \sin(\phi) & \cos(\phi) \end{pmatrix} $$ correspond to a rotation in 2 dimensions? It seems odd that the generator along $\hat{n} = \hat{y}$ corresponds to a 2d real rotation matrix (and not along any other direction).
Not sure if this is the answer you are looking for, but the matrix exponential of any anti-symmetric matrix $A^\top=-A$ is an orthogonal matrix (hence a rotation matrix). since for $R=e^{A}$ we have $$R^{\top} R=e^{A^\top}e^A = e^{A^\top+A}=e^0 = I.$$ This means that you can generate rotations in 3 and 4 dimensions as well, simply by exponentiating 3x3 and 4x4 anti-symmetric matrices.