(a) CH3 CH (OH) CH2CH3
2R – OH + Na → 2RONa + 2H2 ↑ 2 moles of alcohol gives 1 mole of H which occupies 22.4L at 273K and 1 atm
number of moles of alcohol = \(\frac{2\,moles\,of\,R-OH}{22.4L\,of\,H_2}\times560\) mL
= 0.05 moles
number of moles = \(\frac{mass}{molar\,mass}\)
= molar mass = \(\frac{3.7}{0.05}\) = 74 g mol-1
General formula for
R – OH Cn H2n+1 – OH
n(12) + (2n+1) (1) + 16 +1 = 74
14n = 74 – 18
14n = 56
n = \(\frac{56}{14}\) = 4
The 2° alcohol which contains 4 carbon is CHn CH(OH)CH2CH3