Reaction of Grignard reagent,C_{2}H_{5}MgBr w | vedclass

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(C) $1$. The molecular formula of the reactant is $C_{8}H_{8}O$. Given the reaction with $C_{2}H_{5}MgBr$ (Grignard reagent) and subsequent hydrolysis

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$2-$chloro$-2-$phenylbutane

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(C) $1$. The molecular formula of the reactant is $C_{8}H_{8}O$. Given the reaction with $C_{2}H_{5}MgBr$ (Grignard reagent) and subsequent hydrolysis,it is likely an aldehyde or ketone. Acetophenone $(C_{6}H_{5}COCH_{3})$ fits the formula $C_{8}H_{8}O$.$2$. Reaction: $C_{6}H_{5}COCH_{3} + C_{2}H_{5}MgBr \rightarrow C_{6}H_{5}C(OH)(CH_{3})(C_{2}H_{5})$. This product $A$ is $2$-phenylbutan-$2$-ol.$3$. Compound $A$ is a tertiary alcohol. Tertiary alcohols react instantly with Lucas reagent $(conc. \ HCl + ZnCl_{2})$ to form alkyl chlorides.$4$. The reaction of $2$-phenylbutan-$2$-ol with $HCl$ replaces the $-OH$ group with $-Cl$ to form $2$-chloro-$2$-phenylbutane $(C_{10}H_{13}Cl)$.$5$. Comparing this with the given options,the structure corresponding to $2$-chloro-$2$-phenylbutane is option $C$.

Reaction of Grignard reagent,$C_{2}H_{5}MgBr$ with $C_{8}H_{8}O$ followed by hydrolysis gives compound $A$ which reacts instantly with Lucas reagent to give compound $B$,$C_{10}H_{13}Cl$. The compound $B$ is

Similar Questions

Lucas test is used for:

An alcohol $X$ $(C_4H_{10}O)$ on dehydration gives an alkene $(C_4H_8)$ as the major product,which on bromination followed by treatment with $Y$ gives an alkyne $C_4H_6$. The alkyne $C_4H_6$ does not react with sodium metal. What are $X$ and $Y$?

Statement-$1$ : Secondary $\&$ Tertiary alcohols can be distinguished by Lucas reagent.
Statement-$2$ : Secondary $\&$ Tertiary alcohols can be distinguished by dichromate test.

Which of the following alcohols is least soluble in water?

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