An organic compound 'A' (C_4H_8) on treatment | vedclass

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Question

(B) The reaction of an alkene with $KMnO_4 / H^{+}$ (oxidative cleavage) or ozonolysis ($O_3$ followed by $Zn/H_2O$) breaks the double bond.Compound $

Vedclass · Accepted Answer

$2-$Methylpropene

Vedclass · Answer

(B) The reaction of an alkene with $KMnO_4 / H^{+}$ (oxidative cleavage) or ozonolysis ($O_3$ followed by $Zn/H_2O$) breaks the double bond.Compound $'A'$ $(C_4H_8)$ yields $C_3H_6O$ (acetone) as one of the products.$2-$Methylpropene $(CH_3-C(CH_3)=CH_2)$ undergoes oxidative cleavage with $KMnO_4 / H^{+}$ to form acetone $(CH_3COCH_3)$ and $CO_2$.Similarly,ozonolysis of $2-$Methylpropene gives acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$.Since the question specifies that $'A'$ yields $'B'$ $(C_3H_6O)$,$2-$Methylpropene is the correct structure for $'A'$.

An organic compound $'A'$ $(C_4H_8)$ on treatment with $KMnO_4 / H^{+}$ yields compound $'B'$ $(C_3H_6O)$. Compound $'A'$ also yields compound $'B'$ on ozonolysis. Compound $'A'$ is:

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