An alkene of molecular formula C_9H_{18} on o | vedclass

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

Question

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon atom to form carbonyl compo

Vedclass · Accepted Answer

$2,2,4$-trimethylhex$-3$-ene

Vedclass · Answer

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon atom to form carbonyl compounds.Given products are $2,2$-dimethylpropanal $(CH_3)_3C-CHO$ and butan$-2$-one $(CH_3-CO-CH_2CH_3)$.To find the original alkene,we join the carbonyl carbons by removing the oxygen atoms and forming a double bond between them:$(CH_3)_3C-CH=O + O=C(CH_3)CH_2CH_3 \rightarrow (CH_3)_3C-CH=C(CH_3)CH_2CH_3$.The resulting structure is $2,2,4$-trimethylhex$-3$-ene.

An alkene of molecular formula $C_9H_{18}$ on ozonolysis gives $2,2$-dimethylpropanal and butan$-2$-one. The alkene is:

Similar Questions

Two isomers $A$ and $B$ with molecular formula $C_4H_8$ give different products on oxidation with $KMnO_4$ in acidic medium. Isomer $A$ on reaction with $KMnO_4 / H^{+}$ results in effervescence of a gas and gives a ketone. The compound $A$ is ...... .

Pent-$1$-ene reacts with diborane to form $X$. $X$ on oxidation using $H_2O_2$ in the presence of aqueous $NaOH$ gives $Y$. Compound $Y$ is

Which reagent converts propene into $1$-propanol?

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:

Reaction of $Br_2$ on ethylene in the presence of $NaCl$ gives:

Vedclass Test Series

Exam Paper Generator

Online Exam Module