Derivation for drug half life
Well, $Cl = R/Cp$ is correct, but $R$ is not $Cp \cdot k$.

Given $X(t)$, the actual quantity (ie weight or moles) of drug in the system, $R$ should have the dimension of $\frac{dX(t)}{dt}$, that is $weight \cdot time^{-1}$ or $mole \cdot time^{-1}$ because the base definition of clearance is given by $$- \frac{dX(t)}{dt} = Cl \cdot C(t)$$. 

The volume of distribution $Vd$ comes into play when converting from $X(t)$ to $C(t)$. Following the same steps as you did :

$$ R = \frac{dX(t)}{dt} = \frac{dC(t)}{dt} * Vd = C_p  \cdot k \cdot V_d$$
because (due to 1st order kinetics)
$$\frac{dC(t)}{dt} = Cp \cdot k$$


  Substituting, 
  $$ t_{\frac{1}{2}} = \frac{\ln(2)}{k}$$
  $$ t_{\frac{1}{2}} = \frac{\ln(2) \cdot C_p \cdot V_d}{R}$$
  Since $ Cl = \frac{R}{Cp}$, substituting would give,
  $$ t_{\frac{1}{2}} = \ln(2) \cdot \frac{1}{Cl}  \cdot V_d$$
  $$ Cl = \ln(2) \cdot \frac{1}{t_{\frac{1}{2}}}  \cdot V_d$$
  $$ Cl = k  \cdot V_d$$
  $$ k = \frac{Cl}{V_d}$$