Q.72. An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with , (Y

Q: What is the correct answer for this exam Chemistry Q72?

The correct answer is 4. Sol. The general combustion equation for an alkane $C_nH_{2n+2}$ is: $C_nH_{2n+2} + \left( n + \frac{2n+2}{4} \right) O_2 \rightarrow nCO_2 + (n+1)H_2O$.This simplifies to $C_nH_{2n+2} + \left( \frac{3n+1}{2} \right) O_2 \rightarrow nCO_2 + (n+1)H_2O$.Given that 8 moles of oxygen are required, so $\frac{3n+1}{2} = 8$.$3n+1 = 16 \Rightarrow 3n = 15 \Rightarrow n = 5$.So, the alkane (Y) is $C_5H_{12

Q: Q.72. An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with , (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is…

The correct answer is 4. Sol. The general combustion equation for an alkane $C_nH_{2n+2}$ is: $C_nH_{2n+2} + \left( n + \frac{2n+2}{4} \right) O_2 \rightarrow nCO_2 + (n+1)H_2O$.This simplifies to $C_nH_{2n+2} + \left( \frac{3n+1}{2} \right) O_2 \rightarrow nCO_2 + (n+1)H_2O$.Given that 8 moles of oxygen are required, so $\frac{3n+1}{2} = 8$.$3n+1 = 16 \Rightarrow 3n = 15 \Rightarrow n = 5$.So, the alkane (Y) is $C_5H_{12

Question

Q.72. An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with , (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is ____.

FirstInTest · Accepted Answer

Sol. The general combustion equation for an alkane $C_nH_{2n+2}$ is: $C_nH_{2n+2} + \left( n + \frac{2n+2}{4} \right) O_2 \rightarrow nCO_2 + (n+1)H_2O$.This simplifies to $C_nH_{2n+2} + \left( \frac{3n+1}{2} \right) O_2 \rightarrow nCO_2 + (n+1)H_2O$.Given that 8 moles of oxygen are required, so $\frac{3n+1}{2} = 8$.$3n+1 = 16 \Rightarrow 3n = 15 \Rightarrow n = 5$.So, the alkane (Y) is $C_5H_{12}$.The alkane $C_5H_{12}$ gives only one monochlorinated product (Z) upon chlorination with $Cl_2/hv$. This implies that all hydrogen atoms in the alkane are equivalent. The only isomer of $C_5H_{12}$ that fits this description is neopentane (2,2-dimethylpropane).Structure of neopentane: $C(CH_3)_4$.All 12 hydrogen atoms in neopentane are equivalent, so chlorination yields only one monochlorinated product, 1-chloro-2,2-dimethylpropane.The question asks for the total number of primary carbon atoms in (Y), which is neopentane.Neopentane has a central quaternary carbon atom bonded to four methyl groups. Each methyl group contains a primary carbon atom.Therefore, there are 4 primary carbon atoms in neopentane.

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