An alcohol X (C_4H_{10}O) does not give turbi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Lucas test uses conc. $HCl$ and $ZnCl_2$ to distinguish between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols. $3^{\circ}$ alcohols give im

Vedclass · Accepted Answer

$n$-butanol,$PCC$,butanal

Vedclass · Answer

(A) The Lucas test uses conc. $HCl$ and $ZnCl_2$ to distinguish between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols. $3^{\circ}$ alcohols give immediate turbidity,$2^{\circ}$ alcohols give turbidity within $5-10$ minutes,and $1^{\circ}$ alcohols do not give turbidity at room temperature. Since alcohol $X$ $(C_4H_{10}O)$ does not give turbidity at room temperature,it must be a $1^{\circ}$ alcohol. Among the options,$n$-butanol is a $1^{\circ}$ alcohol. $n$-butanol reacts with $PCC$ (Pyridinium chlorochromate) to form butanal $(Z)$. Therefore,$X = n$-butanol,$Y = PCC$,and $Z = 	ext{butanal}$.

An alcohol $X$ $(C_4H_{10}O)$ does not give turbidity with conc. $HCl$ and $ZnCl_2$ at room temperature. $X$ on reaction with reagent $Y$ gives $Z$. What are $X, Y$ and $Z$ respectively?

Similar Questions

The fastest rate of reaction will be when $R$ is:

Major product $A$ is

The major product in the following reaction is:

Which of the following compounds will form a hydrocarbon on reaction with a Grignard reagent?

Ethyl alcohol is heated with conc. $H_2SO_4$. The product formed is

Vedclass Test Series

Exam Paper Generator

Online Exam Module