An alcohol X left( C_5 H_{12} O right) produc | vedclass

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(D) $1$. The Lucas test (conc. $HCl / ZnCl_2$) is used to distinguish between primary,secondary,and tertiary alcohols. Tertiary alcohols react instant

Vedclass · Accepted Answer

$3-$methylbutan$-1-$ol,$2$-methylbutan$-2-$ol

Vedclass · Answer

(D) $1$. The Lucas test (conc. $HCl / ZnCl_2$) is used to distinguish between primary,secondary,and tertiary alcohols. Tertiary alcohols react instantly to produce turbidity.$2$. Alcohol $X$ produces turbidity instantly,meaning it must be a tertiary alcohol. Among the isomers of $C_5 H_{12} O$,$2$-methylbutan-$2$-ol is a tertiary alcohol.$3$. Alcohol $Y$ undergoes dehydration with conc. $H_2 SO_4$ at $443 \ K$. This is a characteristic reaction for alcohols to form alkenes.$4$. Comparing the options,$2$-methylbutan-$2$-ol is the tertiary alcohol $(X)$ and $3$-methylbutan-$1$-ol is a primary isomer $(Y)$.$5$. Therefore,the correct pair is $X = 2$-methylbutan-$2$-ol and $Y = 3$-methylbutan-$1$-ol,which corresponds to option $D$.

Column-$I$	Column-$II$ (Common Name)	Column-$III$ ($\text{IUPAC}$ Name)
$(A)$ Phenol	$(1)$ Resorcinol	$(i)$ Benzene$-1,2-$diol
$(B)$ Benzene$-1,3-$diol	$(2)$ Catechol	(ii) Phenol
$(C)$ Benzene$-1,2-$diol	$(3)$ Phenol	(iii) Benzene$-1,4-$diol
$(D)$ Benzene$-1,4-$diol	$(4)$ Hydroquinone	(iv) Benzene$-1,3-$diol

An alcohol $X \left( C_5 H_{12} O \right)$ produces turbidity instantly with conc. $HCl / ZnCl_2$. Isomer $(Y)$ of $X$ undergoes dehydration with conc. $H_2 SO_4$ at $443 \ K$. $X$ and $Y$ respectively are

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