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

$CH_3-CH=CH_2 + NOCl \to$ Product (Major). Identify the product.

With anhydrous zinc chloride,ethylene glycol gives:

$CH_3-CH=CH_2 \xrightarrow[{(low \, conc.)}]{Br_2/h\nu} (A)$; Product $(A)$ of the reaction is

Aqueous sulphuric acid reacts with $2-$methyl$-1-$butene to give predominantly

The reaction of $HBr$ with the following compound would produce:

Vedclass Test Series

Exam Paper Generator

Online Exam Module