An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with Cl$$_2$$/h$$\nu$$, (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is __________.

Let the molecular formula of the alkane be (C_nH_{2n+2}).

The general combustion equation for an alkane is

$$C_nH_{2n+2}+\left(\frac{3n+1}{2}\right)O_2\rightarrow nCO_2+(n+1)H_2O.$$

Since 8 moles of (O_2) are required for complete combustion,

$$\frac{3n+1}{2}=8,$$

$$3n+1=16,$$

$$3n=15,$$

$$n=5.$$

Hence, the molecular formula of the alkane is

$$C_5H_{12}.$$

Among the three isomers of (C_5H_{12}):

• (n)-Pentane gives three monochlorinated products.
• 2-Methylbutane gives four monochlorinated products.
• 2,2-Dimethylpropane (neopentane) gives only one monochlorinated product because all twelve hydrogen atoms are equivalent.

Therefore, compound (Y) is neopentane with the structure

$$C(CH_3)_4.$$

A primary carbon is bonded to only one other carbon atom. In neopentane, each of the four methyl carbons is attached only to the central carbon and is therefore a primary carbon, while the central carbon is quaternary.

Hence, the number of primary carbon atoms in (Y) is 4.

Therefore, the correct answer is 4.