Question

An unsaturated hydrocarbon 'A' adds two molecules of H2 ​and on reductive ozonolysis gives butane-1, 4-dial, ethanal and propanone. What will be the IUPAC name of ‘A.’

Answer 1

Ozonolysis:

• Ozonolysis is an organic process in which ozone cleaves the unsaturated bonds of alkenes, alkynes, or azo compounds.
• Alkenes and alkynes form organic compounds with a carbonyl group replacing the numerous carbon-carbon bonds, whereas azo compounds form nitrosamines.

Finding unsaturated carbon ‘A’:

• The chemical reaction involved is:

${\mathrm{CH}}_3\mathrm{CH}=\mathrm{O}+\mathrm{O}=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{O}+\mathrm{O}=\mathrm{C}\left({\mathrm{CH}}_3\right){\mathrm{CH}}_3⟶{\mathrm{CH}}_3\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_2{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{C}\left({\mathrm{CH}}_3\right)-{\mathrm{CH}}_3$

• Since there are eight carbon atoms in the main chain so octa will be used
• suffix $2,6-$diene will be used as there are $2$double bonds present in carbon no. six and two
• The methyl group is present in $2$nd carbon atom so $2$- methyl prefix will be used

Hence the IUPAC name of the compound ‘A’ is $2-\mathrm{methylocta}-2,6-\mathrm{diene}$