Question

question 9 (1 point)
which compound has one triple carbon bond?
a) c₄h₆
b) c₄h₈
c) c₄h₁₀
d) c₆h₆
e) all of the above

Explanation:

Step1: Calculate degree of unsaturation

The formula for degree of unsaturation \(DU=\frac{2C + 2 - H}{2}\), where \(C\) is the number of carbon atoms and \(H\) is the number of hydrogen atoms.
For option a) \(C = 4\), \(H=6\). Then \(DU=\frac{2\times4 + 2-6}{2}=\frac{8 + 2-6}{2}=\frac{4}{2}=2\). One triple bond has a degree of unsaturation of \(2\).
For option b) \(C = 4\), \(H = 8\). Then \(DU=\frac{2\times4+2 - 8}{2}=\frac{8 + 2-8}{2}=1\) (one double bond or one ring).
For option c) \(C = 4\), \(H = 10\). Then \(DU=\frac{2\times4+2-10}{2}=\frac{8 + 2 - 10}{2}=0\) (saturated, only single bonds).
For option d) \(C = 6\), \(H = 6\). Then \(DU=\frac{2\times6+2 - 6}{2}=\frac{12 + 2-6}{2}=4\) (in benzene, there are three double bonds and one ring).

Answer:

A. \(C_4H_6\)