In the following sequence of reactions, the alkene affords the compound 'B'
$C H_{3} C H = C H C H_{3} O_{3} - \to A H_{2} O - - \to Z n B$ The compound B is:

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

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

C H_{3} C H_{2} C O C H_{3}

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

C H_{3} C H O

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

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

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

Open in App

Solution

The correct option is C $C H_{3} C H O$
The compound A is intermediate ozonide or zwitter ion formation by 1,3 cycloaddition to a symmetrical alkene. It is reduced by reducing agents like Zn dust, thiourea or dimethyl sulphide producing aldehyde like acetaldehyde as compound B.

Suggest Corrections

Similar questions

B

C H_{3} C H = C H C H_{3} O_{3} - \to A Z n / H_{2} O - ---- \to B

C H_{3} C H = C H C H_{3} O_{3} - \to A H_{2} O - - \to Z n B

B

C H_{3} C H = C H C H_{3} + O_{3} Z n - - \to H_{2} O B

C H_{3} C H O

C H_{3} C O C_{2} H_{5}

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

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

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

Join BYJU'S Learning Program