From Wikipedia:

The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the differences in reactivity of the three classes of alcohols with hydrogen halides via an $\mathrm{S_N1}$ reaction:

$$\ce{ROH + HCl -> RCl + H2O}$$

The differing reactivity reflects the differing ease of formation of the corresponding carbocations. Tertiary carbocations are far more stable than secondary carbocations, and primary carbocations are the least stable.

So since alkyl halides are insoluble, they give a turbid solution upon formation, which easily happens for tertiary alcohols.

Consider a primary alcohol, like $\ce{(CH3)3-C-CH2OH}$, which forms a tertiary carbocation, $\ce{(CH3)2C+-CH2CH3}$ by an alkyl shift from primary carbocation during the substitution process. So will Lucas test give a false positive in this case, since a tertiary alkyl halide can be easily formed from a primary alcohol?


2 Answers 2


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

  1. 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)
  2. 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


Hydrochloric Acid ($\ce{HCl}$) reacts in the same way, although often Zinc (II) chloride (a Lewis acid) is added to help compensate for the lower nucleophilicity of chloride ion.

The mixture of $\ce{HCl}$ and $\ce{ZnCl2}$ is called the Lucas Reagent. Secondary and tertiary alcohols react via the $\mathrm{S_N1}$ mechanism with the Lucas reagent.

The $\ce{ZnCl2}$ coordinates to the hydroxyl oxygen and this generates a far superior leaving group.

Primary alcohols react in a similar fashion except the free cation is not generated, and the substitution is of $\mathrm{S_N2}$ type.

Source: Reaction of alcohols

  • $\begingroup$ But since a tertiary carbocation cam be formed by rearrangement, won't there be $\mathrm{S_N1}$ reaction for the alcohol mentioned in the question ? $\endgroup$ Jan 26, 2020 at 8:49
  • $\begingroup$ No @Gokul the carbocation does not form in the first place. Forget rearrangement. $\endgroup$ Jan 26, 2020 at 10:05
  • $\begingroup$ @Gokul carbocation is not formed in case of primary alcohol. Sn2 reaction takes place $\endgroup$
    – A Shafi
    Jan 26, 2020 at 11:38

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