An alkene has moleuclar mass in the range of 80-85. A 10.02mg of its sample took up 8.4mL of `H_(2)` gas measured at `0^(@)C` and 760 mn pressure. Upon ozonolysis it yielded glyoxal and methanol. The alkene is

Options

A. Hexane-1
B. Hexyne-1
C. 1,3-Hexadien
D. 1,3,5-Hexatriene

Answer

D

Explanation

Vol. of `H_(2)` at NTP=8.4mL
Moles of `H_(2)=(8.4mL)/ (22400mL"mol") =3.7xx10^(-4)`
Mol of alkene = 10.02mg=`10.02xx10^(-3)g`
mol. Mass of alkene =80g`mol^(-1)`
Mol pf alkene=`(10.2xx10^(-3)g)/(80mgmol^(-))` =`1.25xx10^(4)mol`
`thererfore` 1 mol of alkene `-=3 "mol of" H_(2)`
Therefore, products of ozonolysis should be 2 moles each of glyoxal and methanl.