
Which of the following compounds will give acetic acid with $KMn{O_4}/H/\Delta $.
A. $C{H_3} - CHO$
B. $C{H_3} - CH = CH - C{H_3}$
C. $C{H_3} - C \equiv C - C{H_3}$
D. $C{H_3}C{H_2}OH$
Answer
162k+ views
Hint: Acetic acid also known as ethanoic acid is the most important among the carboxylic acids. Moreover, $KMn{O_4}$ acts as an oxidizing agent in this reaction. It is considered as a strong oxidizing agent and is usually prepared from other minerals such as manganese.
Complete step by step answer:
When ethanol is oxidized with alkaline potassium permanganate, it gets oxidized to form ethanoic acid.
$C{H_3}C{H_2}OH + 2[O]\xrightarrow{{Alkaline\,KMn{O_4}}}C{H_3}COOH + {H_2}O$
$KMn{O_4}$is an oxidizing agent so it will convert the aldehyde group into acid.
$C{H_3}CHO\xrightarrow[{KMn{O_4}}]{\Delta }C{H_3} - COOH$
Ethanol is further oxidized to acetaldehyde on treatment with acidified potassium permanganate. Afterwards, acetaldehyde further gets oxidized to acetic acid.
The acidified potassium permanganate also breaks the carbon-carbon double and triple bonds. The two carbons attached to multiple bonds are oxidized to carboxylic acid.
Now, let’s consider the reaction with alkene,
$C{H_3} - CH = CH - C{H_3}\xrightarrow[{}]{{KMn{O_4}}}2C{H_3}COOH$
The product formed is acetic acid.
Now, in the last case also i.e. in case of alkynes, the product formed is acetic acid. The reaction is as shown:
$C{H_3} - C \equiv C - C{H_3}\xrightarrow{{KMn{O_4}}}2C{H_3}COOH$
Hence, all the four options on oxidation with acidified potassium permanganate gives acetic acid.
Moreover, when a solution of ethyl alcohol and alkaline $KMn{O_4}$ is heated, the pink color of the solution disappears. Since, $KMn{O_4}$ is a strong oxidizing agent, it oxidizes ethanol to ethanoic acid by donating nascent oxygen,
Hence, all the options are correct.
Note:
Potassium permanganate is used in qualitative analysis to determine the permanganate value. It is also used as a regeneration chemical in well water treatment for the removal of hydrogen sulphide and iron. This compound is also used as a disinfectant to cure certain skin conditions like foot fungal infections, dermatitis etc. This compound can even be used as a bleaching agent, as a pesticide and as an antiseptic.
Complete step by step answer:
When ethanol is oxidized with alkaline potassium permanganate, it gets oxidized to form ethanoic acid.
$C{H_3}C{H_2}OH + 2[O]\xrightarrow{{Alkaline\,KMn{O_4}}}C{H_3}COOH + {H_2}O$
$KMn{O_4}$is an oxidizing agent so it will convert the aldehyde group into acid.
$C{H_3}CHO\xrightarrow[{KMn{O_4}}]{\Delta }C{H_3} - COOH$
Ethanol is further oxidized to acetaldehyde on treatment with acidified potassium permanganate. Afterwards, acetaldehyde further gets oxidized to acetic acid.
The acidified potassium permanganate also breaks the carbon-carbon double and triple bonds. The two carbons attached to multiple bonds are oxidized to carboxylic acid.
Now, let’s consider the reaction with alkene,
$C{H_3} - CH = CH - C{H_3}\xrightarrow[{}]{{KMn{O_4}}}2C{H_3}COOH$
The product formed is acetic acid.
Now, in the last case also i.e. in case of alkynes, the product formed is acetic acid. The reaction is as shown:
$C{H_3} - C \equiv C - C{H_3}\xrightarrow{{KMn{O_4}}}2C{H_3}COOH$
Hence, all the four options on oxidation with acidified potassium permanganate gives acetic acid.
Moreover, when a solution of ethyl alcohol and alkaline $KMn{O_4}$ is heated, the pink color of the solution disappears. Since, $KMn{O_4}$ is a strong oxidizing agent, it oxidizes ethanol to ethanoic acid by donating nascent oxygen,
Hence, all the options are correct.
Note:
Potassium permanganate is used in qualitative analysis to determine the permanganate value. It is also used as a regeneration chemical in well water treatment for the removal of hydrogen sulphide and iron. This compound is also used as a disinfectant to cure certain skin conditions like foot fungal infections, dermatitis etc. This compound can even be used as a bleaching agent, as a pesticide and as an antiseptic.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
