Ozonolysis of one mole of a symmetrical alken | vedclass

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

Question

(D) The general reaction for the ozonolysis of a symmetrical alkene is: $R-CH=CH-R + O_3 \rightarrow 2R-CHO$. Given that the product is an aldehyde wi

Vedclass · Accepted Answer

$2$-Butene

Vedclass · Answer

(D) The general reaction for the ozonolysis of a symmetrical alkene is: $R-CH=CH-R + O_3 \rightarrow 2R-CHO$. Given that the product is an aldehyde with a molar mass of $44 \ u$. The general formula for an aldehyde is $R-CHO$. The molar mass of $CHO$ is $12 + 1 + 16 = 29 \ u$. Therefore,the mass of the alkyl group $R$ is $44 - 29 = 15 \ u$. An alkyl group with a mass of $15 \ u$ is a methyl group $(-CH_3)$. Thus,the aldehyde is acetaldehyde $(CH_3CHO)$. Substituting this back into the reaction: $CH_3-CH=CH-CH_3 + O_3 \rightarrow 2CH_3CHO$. The reactant is $2$-butene.

Ozonolysis of one mole of a symmetrical alkene yields two moles of an aldehyde with a molar mass of $44 \ u$. Identify the alkene.

Similar Questions

Ozonolysis of which compound gives $OHC-CH_2-CH(CH_3)-CH_2-CHO$?

The reaction $CH_2 = CH_2 + H_2 \xrightarrow[{250 - 300 \, ^oC}]{{Ni}} CH_3 - CH_3$ is known as:

The major product formed in the following reaction is :

The compounds that will give an isomer of $2,2$-dimethylpropane on catalytic hydrogenation are:
$(1)$ $CH_3-CH=C(CH_3)_2$
$(2)$ $CH_3-CH=CH-CH_3$
$(3)$ $CH_3-CH=CH-CH_2-CH_3$
$(4)$ $CH_3-C(CH_3)=C(CH_3)-CH_3$

$1-$Phenylbut$-1-$ene on ozonolysis gives

Vedclass Test Series

Exam Paper Generator

Online Exam Module