An alcohol (x) gives blue colour in victormayer’s test and 3.7g of X

1.	An alcohol (x) gives blue colour in victormayer’s test and 3.7g of X when treated with metallic sodium liberates 560 mL of hydrogen at 273 K and 1 atm pressure what will be the possible structure of X?(a) CH3 CH (OH) CH2CH3 (b) CH3 – CH(OH) – CH3(c) CH3 – C (OH) (CH3)2(d) CH3 – CH2 – CH (OH) – CH2 – CH3
Answer» (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 C_n H_2n+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 CH_n CH(OH)CH₂CH₃

An alcohol (x) gives blue colour in victormayer’s test and 3.7g of X when treated with metallic sodium liberates 560 mL of hydrogen at 273 K and 1 atm pressure what will be the possible structure of X?(a) CH3 CH (OH) CH2CH3 (b) CH3 – CH(OH) – CH3(c) CH3 – C (OH) (CH3)2(d) CH3 – CH2 – CH (OH) – CH2 – CH3

Answer»

(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 C_n H_2n+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 CH_n CH(OH)CH₂CH₃