Optically active 2-octanol rapidly loses its | vedclass

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

Question

(A) The optical activity of $2-$octanol is due to the presence of a chiral carbon atom at the $C-2$ position.When treated with dilute acid,$2-$octanol

Vedclass · Accepted Answer

dilute acid

Vedclass · Answer

(A) The optical activity of $2-$octanol is due to the presence of a chiral carbon atom at the $C-2$ position.When treated with dilute acid,$2-$octanol undergoes an acid-catalyzed dehydration reaction.This reaction leads to the formation of an alkene (octene),which is an achiral molecule.Since the chiral center is destroyed during the formation of the double bond,the molecule loses its optical activity.$CH_3-CH(OH)-(CH_2)_5-CH_3 \xrightarrow{dil. \, acid} CH_3-CH=CH-(CH_2)_4-CH_3 + H_2O$

Optically active $2-$octanol rapidly loses its optical activity when exposed to

Similar Questions

Of the following alcohols,the one that would react fastest with conc. $HCl$ and anhydrous $ZnCl_{2}$ is,

By means of calcium chloride,which of the following can be dried?

Consider the following reaction sequence:
$\text{Cyclohexanol}$ $\xrightarrow{CrO_3} A$ $\xrightarrow{CH_3MgBr / H^+, H_2O} B$ $\xrightarrow{H_2SO_4, \Delta} C$ $\xrightarrow{B_2H_6, H_2O_2, OH^-} D$
Identify the final product $D$.

When $3, 3-$dimethyl$-2-$butanol is heated with ${H_2SO_4}$,the major product obtained is:

Heating $I_2$ and $NaOH$ with which of the following produces iodoform?

Vedclass Test Series

Exam Paper Generator

Online Exam Module