Ozonolysis of an alkene with molecular formul | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Ozonolysis involves the cleavage of the $C=C$ double bond. Acetone is $CH_3-CO-CH_3$ and $butan-2-one$ is $CH_3-CO-CH_2-CH_3$. To find the origina

Vedclass · Accepted Answer

$2, 3-dimethylpent-2-ene$

Vedclass · Answer

(D) Ozonolysis involves the cleavage of the $C=C$ double bond. Acetone is $CH_3-CO-CH_3$ and $butan-2-one$ is $CH_3-CO-CH_2-CH_3$. To find the original alkene,remove the oxygen atoms from the carbonyl groups and join the carbon atoms with a double bond: $(CH_3)_2C=O + O=C(CH_3)CH_2CH_3 \rightarrow (CH_3)_2C=C(CH_3)CH_2CH_3$. The structure is $CH_3-C(CH_3)=C(CH_3)-CH_2-CH_3$. The longest chain has $5$ carbons (pentene). Numbering from the right gives the double bond at position $2$ and methyl groups at positions $2$ and $3$. Thus,the $IUPAC$ name is $2, 3-dimethylpent-2-ene$.

Ozonolysis of an alkene with molecular formula $C_{7}H_{14}$ in the presence of $Zn$ dust gives a mixture of acetone and $butan-2-one$. The $IUPAC$ name of the alkene is ..........

Similar Questions

In the following reaction,$A$ and $B$ respectively are
$A \xrightarrow{HBr} C_2H_5Br \xrightarrow{B} A$

What is the product formed when ethene reacts with acidic $KMnO_4$ solution?

The reaction conditions used for converting $1, 2-$dibromoethane to ethene are

What is $A$ in the following reaction?
$C_6H_5-CH_2-CH=CH_2 + HCl \rightarrow A$

The product $(A)$ of the given oxymercuration-demercuration reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module