A hydrocarbon ‘X’ is found to have molar mass of $80$ . A $10.0 m g$ of compound ‘X’ on hydrogenation consumed $8.40 m L$ of $H_{2}$ gas (measured at STP). Ozonolysis compound ‘X’ yields only formaldehyde and dialdehyde. The total number of fragments/molecules produced from the ozonolysis of compound ‘X’ is _

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Detailed Solution

moles of $X = \frac{10 m g}{80} = 0.125 m m o l .$
$\begin{array}{l} M o l e s c o n s u m e d o f H_{2} = \frac{8.4}{22.4} = 0.375 m m o l . \\ \frac{^{n} H_{2}}{n_{X}} = \frac{0.375}{0.125} = 3 \end{array}$
So, the compound X have 3 double bonds. Ozonolysis of the compound yields only formaldehyde and dialdehyde, so the compound should be

$\begin{array}{l} H_{2} C = C H - C H = C H - \overset{\overset{H}{|}}{C} = C H_{2} \to \underset{4 m o l e c u l e s}{\underset{⎵}{2 H C H O + 2 C_{2} O_{2} H_{2}}} \\ M o l e c u l a r m a s s = (12 \times 6) + 1 \times 8 = 72 + 8 = 80 a m u \\ O z o n o l y s i s f o r m : \end{array}$