Identify the product B in the following serie | vedclass

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

Question

(B) Step $1$: Dehydration of $CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol) with $Al_2 O_3$ at $623 \ K$ gives $A$,which is Propene $(CH_3-CH=CH_2)$.Step $2$: Pr

Vedclass · Accepted Answer

Propan$-2-$ol

Vedclass · Answer

(B) Step $1$: Dehydration of $CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol) with $Al_2 O_3$ at $623 \ K$ gives $A$,which is Propene $(CH_3-CH=CH_2)$.Step $2$: Propene reacts with conc. $H_2 SO_4$ followed by hydrolysis $(H_2 O)$ to undergo hydration according to Markovnikov's rule.Step $3$: The reaction follows the mechanism: $CH_3-CH=CH_2 + H_2 SO_4$ $\rightarrow CH_3-CH(OSO_3 H)-CH_3$ $\xrightarrow{H_2 O} CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).Thus,the product $B$ is Propan$-2-$ol.

Identify the product $B$ in the following series of reactions.
$CH_3-CH_2-CH_2-OH$ $\xrightarrow{Al_2 O_3, 623 \ K} A$ $\xrightarrow[(ii) H_2 O]{(i) \text{conc. } H_2 SO_4} B$

Similar Questions

$A$ compound with formula $C_4H_{10}O$ on oxidation gives $C_4H_8O$,which forms an oxime and gives a positive iodoform test. The original compound on reaction with concentrated $H_2SO_4$ gives $C_4H_8$. What is the structure of the compound?

The product obtained by heating ethanol with concentrated $H_2SO_4$ at $165^{\circ}C - 170^{\circ}C$ is:

Explain: Ethanol has a higher boiling point than methoxymethane.

Which of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?

An alcohol of formula $C_9H_{12}O$ reacts with $Na_2Cr_2O_7$ to form a compound having formula $C_9H_{10}O$. The original alcohol might be

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify the product $B$ in the following series of reactions.$CH_3-CH_2-CH_2-OH$ $\xrightarrow{Al_2 O_3, 623 \ K} A$ $\xrightarrow[(ii) H_2 O]{(i) \text{conc. } H_2 SO_4} B$