Question

An unsaturated hydrocarbon 'A' adds two molecules of $H_2$ and on reductive ozonolysis gives butane-$1$, $4$-dial, ethanal and propanone. Give the IUPAC name of A.

• A. $3$-methylocta-$2$, $6$-diene
• B. $2$-methylocta-$2$, $5$-diene
• C. $2$-methylocta-$2$, $6$-diene
• D. $2$-methylocta-$3$, $5$-diene

Answer 1

Answer: (C)

$2$-methylocta-$2$, $6$-diene

$\underset{\left(Ethanal\right)}{C{H}_3\stackrel{H}{\stackrel{|}{C}}}=O+O=\stackrel{H}{\stackrel{|}{C}}-C{H}_2-\underset{(Butane-1,4-dial)}{C{H}_2-\stackrel{H}{\stackrel{|}{C}}=O}+\underset{\left(Propanone\right)}{O=\stackrel{C{H}_3}{\stackrel{|}{C}-}C{H}_3}⟶$

$C\stackrel{8}{H_3}\stackrel{7}{CH}=\stackrel{6}{CH}-\stackrel{5}{C{H}_2}-\stackrel{4}{C{H}_2}-\stackrel{3}{CH}=\stackrel{C{H}_3}{\stackrel{2|}{C}-}C{H}_3$

(A) 2-methylocta-2, 6-diene

The structure of compound, A is.