An optically active alkene having molecular f | vedclass

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Question

(B) $1$. Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.$2$. The formation of acetone $(CH_3)_2C=O$

Vedclass · Accepted Answer

$3,4-$dimethylhex$-2-$ene

Vedclass · Answer

(B) $1$. Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.$2$. The formation of acetone $(CH_3)_2C=O$ indicates that the alkene must have a $(CH_3)_2C=$ group.$3$. The molecular formula is $C_8H_{16}$. If one part is acetone $(C_3H_6O)$, the other part must be a $C_5$ aldehyde or ketone.$4$. For the alkene to be optically active, it must contain a chiral center. In the structure $2,3$-dimethylhex-$2$-ene, the molecule is achiral. In $3,4$-dimethylhex-$2$-ene, the carbon at position $3$ is chiral because it is bonded to four different groups: $-H$, $-CH_3$, $-CH_2CH_3$, and $-C(CH_3)=C(CH_3)_2$.$5$. The structure shown in option $B$ corresponds to $3,4$-dimethylhex-$2$-ene, which contains a chiral center (marked with an asterisk) and produces acetone upon ozonolysis.

An optically active alkene having molecular formula $C_8H_{16}$ gives acetone as one of the products on ozonolysis. The structure of the alkene is:

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