No. It does not.
In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions
In chemical kinetics, there are two ways to deal with multiple-step mechanisms
- Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
- Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)
So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.
Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test