Question

A hydrocarbon $\left(A\right)$ $C_{n}H_{2 n - 4}$ on ozonolysis gives
$\left(C H_{3}\right)_{2}CHCH_{2}CHO,OHCCH_{2}CH_{2}CHO$ and $CH_{3}COCH_{3}$ .
The value of n is

Answer 1

Based on the formula of hydrocarbon $C_{n}H_{2 n - 4}$ the degree of unsaturation is 3. So, there is two case:
(I) 3 double bonds
(II) 1 double bond and 1 triple bond
Second case can be easily rejected because products of ozonolysis are aldehyde or ketone rather than carboxylic acid.
So the alkene would be:
$\left( CH _{3}\right)_{2} CHCH _{2} CH = CHCH _{2} CH _{2} CH = CHCH _{2} CH _{2} CH = C \left( CH _{3}\right)_{2}$
Which on ozonolysis will give products as:
$ \left( CH _{3}\right)_{2} CHCH _{2} CH = CHCH _{2} CH _{2} CH = CHCH _{2} CH _{2} CH = C \left( CH _{3}\right)_{2} $
$\underset{\substack{\text { Ozonolysis }}}{\rightarrow}\left( CH _{3}\right)_{2} CHCH _{2} CHO +2 OHCCH _{2} CH _{2} CHO + CH _{3} COCH _{3}$