An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with Cl2/h , (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is .
Answer
4
Solution
CnH2n+2 + 3n+12O2 nCO2 + (n+1)H2O
3n+12 = 8 ⇒ n = 5 ⇒ Y is C5H12
Only one monochlorinated product ⇒ all 12 H atoms equivalent ⇒ Y = neopentane (2,2-dimethylpropane), C(CH3)4
Z = (CH3)3CCH2Cl
Primary carbons: the four CH3 carbons (each bonded to one C); the central C is quaternary.
∴ Number of primary carbon atoms = 4
Answer: 4