Identify the reagents X, Y, Z for the followi | vedclass

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

Question

(A) The dehydration of alcohols depends on the degree of substitution of the alcohol:$1$. Primary $(1^{\circ})$ alcohols like ethanol undergo dehydrat

Vedclass · Accepted Answer

$X$$Y$$Z$$H_2SO_4, 443 \ K$$85 \% \ H_3PO_4, 440 \ K$$20 \% \ H_3PO_4, 358 \ K$

Vedclass · Answer

(A) The dehydration of alcohols depends on the degree of substitution of the alcohol:$1$. Primary $(1^{\circ})$ alcohols like ethanol undergo dehydration using concentrated $H_2SO_4$ at $443 \ K$.$2$. Secondary $(2^{\circ})$ alcohols like propan$-2-$ol undergo dehydration using $85 \% \ H_3PO_4$ at $440 \ K$.$3$. Tertiary $(3^{\circ})$ alcohols like tert-butyl alcohol undergo dehydration using $20 \% \ H_3PO_4$ at $358 \ K$.Therefore,$X = H_2SO_4, 443 \ K$; $Y = 85 \% \ H_3PO_4, 440 \ K$; $Z = 20 \% \ H_3PO_4, 358 \ K$.

Identify the reagents $X, Y, Z$ for the following dehydration reactions of alcohols:
$CH_3-CH_2-OH \xrightarrow{X} CH_2=CH_2 + H_2O$
$CH_3-CH(OH)-CH_3 \xrightarrow{Y} CH_3-CH=CH_2 + H_2O$
$(CH_3)_3C-OH \xrightarrow{Z} (CH_3)_2C=CH_2 + H_2O$

Similar Questions

Primary alcohols on dehydration give

Product $(C)$ is

Ethyl alcohol exhibits acidic character on reacting with

Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

$X + HCl$ $\xrightarrow[\text{Addition}]{\text{Anhy. } AlCl_3} C_2H_5Cl \xleftarrow[\text{Substitution}]{\text{Anhy. } ZnCl_2/HCl} Y$. The conversion of $Y$ to $X$ can be achieved by heating with .......... at .......... temperature.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$X$	$Y$	$Z$
$H_2SO_4, 443 \ K$	$85 \% \ H_3PO_4, 440 \ K$	$20 \% \ H_3PO_4, 358 \ K$

Identify the reagents $X, Y, Z$ for the following dehydration reactions of alcohols:$CH_3-CH_2-OH \xrightarrow{X} CH_2=CH_2 + H_2O$$CH_3-CH(OH)-CH_3 \xrightarrow{Y} CH_3-CH=CH_2 + H_2O$$(CH_3)_3C-OH \xrightarrow{Z} (CH_3)_2C=CH_2 + H_2O$