A rotation $R(\hat{\textbf{n}},\phi)$ in the three-dimensional physical space, about an arbitrary axis $\hat{\textbf{n}}$ through an angle $\phi$ is given by $$R(\hat{\textbf{n}},\phi)=e^{-i(\textbf{j}\cdot\hat{n})\phi}\tag{1}$$ where $\textbf{j}=(j_1,j_2,j_3)$ is dimensionless and $j_i=-j_i^T$ ($i=1,2,3$) owing to the condition $R^TR=\mathbb{1}$.

In quantum mechanics, by Wigner's theorem, the rotation will be represented by an unitary operator $U(R)$. The operator $U(R)$ is usually quoted to have the form (see Sakurai's Modern Quantum Mechanics, for example) $$U(R(\hat{\textbf{n}},\phi))=e^{-i(\textbf{J}\cdot\hat{\textbf{n}})\phi/\hbar}\tag{2}$$ where $\textbf{J}^\dagger=\textbf{J}$ (follows from $U^\dagger U=\mathbb{1}.$)

Is it possible to derive equation (2) from Eq. (1)? In other words, given $R$, how do we construct the map U(R) that acts on the Hilbert space?

I used two different symbols $\textbf{j}$ and $\textbf{J}$ in (1) and (2) respectively because unless we actually construct that map $U: R\to U(R)$ it's not clear what $\textbf{j}$ in relation (1) has to do with $\textbf{J}$ in relation (2).

Attempt 1 Since rotations form a Lie group all representations have the exponential form. Therefore, without any loss of generality, the unitary representation in the Hilbert space must have the form $$U(R)=e^{-i\textbf{J}(\textbf{j})\cdot\hat{\textbf{n}}\phi}$$ where $\textbf{J}(\textbf{j})$ is a function of $\textbf{j}$, and unitarity implies $\textbf{J}(\textbf{j})^\dagger=\textbf{J}(\textbf{j}).$ Now it remains to find what $\textbf{J}(\textbf{j})$ is.

Attempt 2 Another option might be to expand $U(R(\hat{\textbf{n}},\delta\phi))=U\Big(\mathbb{1}-i(\textbf{j}\cdot\hat{\textbf{n}})\delta\phi\Big)$ in a Taylor series? But I'm not sure how to carry it out.