Tertiary alcohols (3^{circ}) having at least | vedclass

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

Question

(D) Tertiary alcohols $(3^{\circ})$ are resistant to oxidation under mild conditions. However,under drastic conditions (e.g.,using concentrated $HNO_3

Vedclass · Accepted Answer

All the above three options are correct

Vedclass · Answer

(D) Tertiary alcohols $(3^{\circ})$ are resistant to oxidation under mild conditions. However,under drastic conditions (e.g.,using concentrated $HNO_3$ or $KMnO_4$ at high temperatures),they undergo carbon-carbon bond cleavage.First,the tertiary alcohol is oxidized to a ketone by losing one carbon atom.Then,the resulting ketone undergoes further oxidation,which involves the cleavage of the carbon-carbon bond,resulting in the loss of more carbon atoms to form carboxylic acids.Specifically,for a tertiary alcohol with $n$ carbon atoms,the oxidation process involves the loss of carbon atoms at each step of the cleavage,leading to products with fewer carbon atoms than the parent alcohol. The reaction follows the general pathway: $R_3C-OH$ $\xrightarrow{[O]} R_2C=O$ $\xrightarrow{[O]} R-COOH$.Thus,the resulting carboxylic acid has fewer carbon atoms than the original tertiary alcohol.

Tertiary alcohols $(3^{\circ})$ having at least four carbon atoms upon drastic oxidation yield carboxylic acid with:

Similar Questions

$CH_3 CH_2 OH \underset{\text{step-}1}{\stackrel{Cl_2}{\longrightarrow}} CH_3 CHO$ $\underset{\text{step-}2}{\stackrel{3 Cl_2}{\longrightarrow}} Cl_3 CCHO$
In the above reactions,the role of $Cl_2$ in step-$1$ and step-$2$ respectively is:

Dehydration of $2-$butanol yields:

Methanol can be distinguished from ethanol by

Glycerol reacts with potassium bisulphate to produce

What is the major product of the following reaction?
$HO-CH_2-CH_2-CH_2-CH_2-Cl \xrightarrow{Alc. KOH} ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module