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  • What is the correct order of reactivity of alcohols in the following reaction? {R-OH+HCl}rightarrow (ZnCl_{2}){R-Cl}+{H}_{2} {O}

What is the correct order of reactivity of alcohols in the following reaction?

\text {R-OH+HCl} \overset{ZnCl_2}{\rightarrow} \text {R-Cl}+\text {H}_{2}\text {O}

(i) 1^{o}>2^{o}>3^{o}

(ii) 1^{o}<2^{o}>3^{o}

(iii) 3^{o}>2^{o}>1^{o}

(iv) 3^{o}>1^{o}>2^{o}

Answers (1)

The answer is option (iii). When dry anhydrous \text {ZnCl}_{2} is mixed with concentrated \text {HCl}, it forms Lucas reagent and the reaction involved is knows as the Lucas test. Formation of an intermediate carbocation takes place during the reaction, this reaction being a slow step of the mechanism, this carbocation dictates the rates of reaction.

According to the \text {S}_N{1} mechanism, the alcohol reactivity order is 3^{o}>2^{o}. But tertiary alcohol reacts very fast with Lucas reagent, so the reactivity order of alcohols in this particular reaction is –

\text {R-OH+HCl} \overset{ZnCl_2}{\rightarrow} \text {R-Cl}+\text {H}_{2}\text {O}

3^{o}>2^{o}>1^{o}

So the correct option is (iii) 3^{o}>2^{o}>1^{o}

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infoexpert23

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