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Oxidation of Alkenes

448 mL of a h...

Question

$448$ mL of a hydrocarbon (A) having C ( $87.80$ %), H ( $12.19$ %) weigh $1.64$ g at NTP. On hydrogenation it gives 2 methyl pentane. Treatment of (A) with acidic $H g S O_{4}$ gives a new compound (B) of molecular weight $C_{6} H_{12} O$ . Compound (A) does not react with ammoniacal $A g N O_{3}$ . What is the structure of (A)?

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Solution

For compounds
No. of moles $= \frac{v o l u m e}{M o l a r v o l .} = \frac{w t}{M o l . w t} \Rightarrow \frac{448}{22400} = \frac{1.64}{M o l . w t} \Rightarrow M o l . w t = \frac{164}{2} = 82 g$

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