Given $\cos \alpha \cdot \tan \alpha\ \ \lt 0$, we are dealing with the product of two trigonometric functions that is less than zero. This means one of the functions is positive and the other is negative. Let's analyze this condition step by step:

1. Recall that $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$. Thus, the expression $\cos \alpha \cdot \tan \alpha$ simplifies to $\sin \alpha$.

2. The sign of $\sin \alpha$ is determined by the y-coordinate in the unit circle, which is positive in the first and second quadrants and negative in the third and fourth quadrants.

3. However, for $\cos \alpha \cdot \tan \alpha\ \ \lt 0$ to hold, considering $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$, we must have opposite signs for $\cos \alpha$ and $\sin \alpha$ since $\tan \alpha$ involves division by $\cos \alpha$ and multiplication by itself doesn’t change the sign of $\sin \alpha$.

4. $\cos \alpha$ is positive in the first and fourth quadrants and negative in the second and third quadrants. Given the product is negative, and considering the sign patterns:
- In the first quadrant, both $\sin \alpha$ and $\cos \alpha$ are positive, so $\cos \alpha \cdot \tan \alpha > 0$.
- In the second quadrant, $\sin \alpha > 0$ and $\cos \alpha 0$, making $\cos \alpha \cdot \tan \alpha$ negative as required since $\tan \alpha$ would be negative (negative $\sin \alpha$ over positive $\cos \alpha$).

Thus, for $\cos \alpha \cdot \tan \alpha\ \ \lt 0$, $\alpha$ must lie in the quadrants where the product of $\cos \alpha$ and $\tan \alpha$ is negative, which are the third and fourth quadrants.

$\boxed{\text{C: Quadrants III and IV}}$ is the correct answer.