What is the correct order of reactivity of alcohols in the following reaction ?
$R - O H + H C l Z n C l_{2} - --- \to R - C l + H_{2} O$

1^{0} > 2^{0} > 3^{0}

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1^{0} < 2^{0} < 3^{0}

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3^{0} > 2^{0} > 1^{0}

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3^{0} > 1^{0} > 2^{0}

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Solution

The correct option is C $3^{0} > 2^{0} > 1^{0}$
Reagent mixture of conc $H C l$ and dry anhydrous $Z n C l_{2}$ . is called Lucas reagent and the reaction is called the Lucas test. It is used to recognize whether the given alcohol is primary, secondary or tertiary in nature. In this reaction, the carbocation is formed as an intermediate and thus the stability and ease of carbocation formation determine the rate of the reaction as that's the slow step of the mechanism. Since the stability of tertiary carbocation is the most due to hyperconjugation effect. Hence the order of reactivity is tertiary alcohol>secondary alcohol by $S_{N} 1$ mechanism.
Thus with Lucas reagent, tertiary alcohol reacts fastest and by $S_{N} 1$ mechanism, followed by secondary and then primary.
The correct order of reactivity of alcohols in the following reaction:
$R - O H + H C l Z n C l_{2} \to R - C l + H_{2} O$ is
$3^{\circ} > 2^{\circ} > 1^{\circ}$

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What is the correct order of reactivity of alcohols in the following reaction ?R−OH+HClZnCl2−−−−→R−Cl+H2O

What is the correct order of reactivity of alcohols in the following reaction ?
$R - O H + H C l Z n C l_{2} - --- \to R - C l + H_{2} O$