The treatment of $(C H_{3})_{2} C = C H C H_{3}$ with boiling $K M n O_{4}$ produces :

C H_{3} C O C H_{3} + C H_{3} C O O H

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{3} C O C H_{3} + C H_{3} C H O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} C H O + C O_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} C O C H_{3}

only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $C H_{3} C O C H_{3} + C H_{3} C O O H$

(Refer to Image)
Hot acidified $K M n O_{4}$ solution oxidizes the alkene by leaving carbon-carbon soluble bond and replacing it with two carbon-oxygen double bonds. But the aldehyde i.e $C H_{3} C H O$ is readily oxidized give carboxylic acid i.e. $C H_{3} C O O H$ .

Suggest Corrections

Similar questions

Q. The treatment of

(C H_{3})_{2} C = C H C H_{3}

with boiling

K M n O_{4}

produces :

The compound $C H_{3} | C H_{3} - C = C H - C H_{3}$ on reaction with $N a I O_{4}$ in the presence of $K M n O_{4}$ gives

Q. Products of reactions are

C H_{3} C = C C H_{2} C H_{3} \frac{K M n O_{4} . K O H}{373 - 383 K}

Q. Which of the following compounds contains a carboxylic acid group?

C H_{3} O H, C H_{3} C O O H, C H_{3} C H O, C H_{3} C O C H_{3}

X,Y and Z are three organic compounds. All the three undergo iodoform test .Only Y and Z can react with 2,4 -DNP. Only Z can reduce tollen's reagent. Here X,Y and Z respectively

Join BYJU'S Learning Program

The treatment of (CH3)2C=CHCH3 with boiling KMnO4 produces :

The correct option is A CH3COCH3+CH3COOH(Refer to Image)Hot acidified KMnO4 solution oxidizes the alkene by leaving carbon-carbon soluble bond and replacing it with two carbon-oxygen double bonds. But the aldehyde i.e CH3CHO is readily oxidized give carboxylic acid i.e. CH3COOH .

The treatment of $(C H_{3})_{2} C = C H C H_{3}$ with boiling $K M n O_{4}$ produces :