Is \(\mathrm{C}_{22} \mathrm{H}_{46}\) an alkane, an alkene, or an alkyne? How can you tell?

Short Answer

Expert verified
\(\mathrm{C}_{22} \mathrm{H}_{46}\) is an alkane, as it matches the general formula for alkanes \(\mathrm{C}_n \mathrm{H}_{2n+2}\) with \(n=22\). It does not match the formulas for alkenes or alkynes.

Step by step solution

01

Recall General Formulas for Hydrocarbons

Alkanes have the general formula: \(\mathrm{C}_n \mathrm{H}_{2n+2}\) Alkenes have the general formula: \(\mathrm{C}_n \mathrm{H}_{2n}\) Alkynes have the general formula: \(\mathrm{C}_n \mathrm{H}_{2n-2}\)
02

Compare the given formula to general formulas

We have \(\mathrm{C}_{22} \mathrm{H}_{46}\), and we want to determine which of the hydrocarbon classes it belongs to. Let's test if it's an alkane: If it is an alkane, then \(2n+2= 46\), with \(n = 22\). Solving for \(n\), we get \(n = \dfrac{46-2}{2} = 22\). Since we get the correct value for n, we can conclude that \(\mathrm{C}_{22} \mathrm{H}_{46}\) is an alkane. However, let's check for the other classes of hydrocarbons: Alkene: If it's an alkene, then \(2n = 46\), with \(n = 22\). In this case, we get \(n = 46/2= 23\), which does not match the given \(n\) value (22). Alkyne: If it's an alkyne, then \(2n-2 = 46\), with \(n = 22\). Solving for \(n\), we get \(n = \dfrac{46+2}{2} = 24\). Again, it does not match the given \(n\) value (22).
03

Conclusion

Based on the comparison of the given formula to the general formulas for alkanes, alkenes, and alkynes, we can conclude that \(\mathrm{C}_{22} \mathrm{H}_{46}\) is an alkane.

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