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Chapter 24: Problem 26

Give the general formula for any cyclic alkene, that is, a cyclic hydrocarbon with one double bond.

Short Answer

Expert verified
The general formula for any cyclic alkene, a cyclic hydrocarbon with one double bond, is \(C_nH_{2n-2}\).

Step by step solution

01

Identify the type of hydrocarbon

A cyclic alkene is a type of hydrocarbon with a cyclic structure and only one double bond in the molecule. The presence of a double bond means that alkenes are unsaturated hydrocarbons.
02

Determine the basic formula for unsaturated hydrocarbons

The general formula for alkenes (unsaturated hydrocarbons) is \(C_nH_{2n}\), where n represents the number of carbon atoms in the molecule.
03

Adjust the formula for a cyclic structure

In a non-cyclic alkene, there are two additional hydrogen atoms due to the free ends of the chain. However, in a cyclic structure, these free ends are closed to form a ring, meaning that there will be two fewer hydrogen atoms in the molecule. So, we must subtract 2 from the number of hydrogen atoms in the general alkene formula. The adjusted formula becomes \(C_nH_{2n-2}\), where n represents the number of carbon atoms in the cyclic alkene molecule.
04

Final Answer

The general formula for any cyclic alkene, a cyclic hydrocarbon with one double bond, is \(C_nH_{2n-2}\).

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Most popular questions from this chapter

Write the condensed structural formulas for as many alkenes and alkynes as you can think of that have the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{10}\)

Acetic anhydride is formed from acetic acid in a condensation reaction that involves the removal of a molecule of water from between two acetic acid molecules. Write the chemical equation for this process, and show the structure of acetic anhydride.

Explain why the boiling point of ethanol \(\left(78^{\circ} \mathrm{C}\right)\) is much higher than that of its isomer, dimethyl ether \(\left(-25^{\circ} \mathrm{C}\right),\) and why the boiling point of \(\mathrm{CH}_{2} \mathrm{~F}_{2}\left(-52{ }^{\circ} \mathrm{C}\right)\) is far above that of \(\mathrm{CH}_{4}\left(-128^{\circ} \mathrm{C}\right)\)

Why is geometric isomerism possible for alkenes but not for alkanes and alkynes?

What is the empirical formula of glycogen? What is the unit that forms the basis of the glycogen polymer? What form of linkage joins these monomeric units?
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