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What I know:

  1. All alkanes are saturated hydrocarbons with the general formula, $C_nH_{2n+2}$
  2. All alkenes are unsaturated hydrocarbons with the general formula, $C_nH_{2n}$

What's Confusing Me:

enter image description here

This is, if I am right, is called a "cyclobutane"; when I look back to the definitions I have, this compound seems to be bending them slightly. It's an alkane, which means that it is supposed to be saturated and be in the form $C_nH_{2n+2}$.

  • Saturated? Yes
  • Same general formula? No, this is $C_4H_8$, which means that it's an alkene?

Which homologous series does this belong to? Alkanes or alkenes? What is wrong with the "definitions" that I have?

If a were to draw its isomers, which ones would I draw?

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    $\begingroup$ Cycloalkanes do not follow the general formula for alkanes. They follow the general formula of alkenes because they have a single hydrogen deficiency like alkenes but it is not on the form of a double bond. $\endgroup$ – Nisarg Bhavsar Jun 16 at 13:31
  • $\begingroup$ Why exactly would you think all hydrocarbons are either alkanes or alkenes? How about cycloalkanes, alkynes, dienes, arenes, etc. etc. $\endgroup$ – Mithoron Jun 16 at 17:33
  • $\begingroup$ So cycloalkanes are in an entirely different group/homologous series? $\endgroup$ – Shane Jun 17 at 3:32
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Let me bring clarity to the matter.

All hydrocarbons are divided into saturated and unsaturated. In turn, each of them is divided into aliphatic and cyclic hydrocarbons.

So, alkanes and cycloalkanes refer to saturated hydrocarbons, BUT: alkanes are aliphatic and cycloalkanes are cyclic. Alkenes refer to unsaturated aliphatic hydrocarbons. They are unsaturated because they have double bond in their structure.

It is very important to understand that cycloalkanes and alkenes are just isomers due to having same general formula $\ce{C_{n}H_{2n}}$. But it doesn't mean that they are both unsaturated!

The general formula doesn't determine structure of compounds, but only the quantitative and qualitative compositions of their molecules.

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    $\begingroup$ I'm not so sure your definition of "unsaturation" is widely used. See for examplle en.wikipedia.org/wiki/Degree_of_unsaturation , where it is clear that cyclohexane and 1-hexene would both have the same degree of "unsaturation", that is, that cyclohexane would be regarded as unsaturated. I'd recommend revising this answer to make that clear. $\endgroup$ – Curt F. Jun 16 at 16:47
  • $\begingroup$ According to Wikipedia, cycloalkanes are saturated hydrocarbons. For explaining this particular matter, they can be considered as saturated. Link: en.wikipedia.org/wiki/Cycloalkane $\endgroup$ – Exeplone Jun 16 at 18:49
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What is wrong with the "definitions" that I have? Yes, you are right. Employing common sense, All alkenes are

  1. unsaturated
  2. hydrocarbons
  3. the general formula, $\ce{C_nH_{2n}}$

Cycloalkane

  1. is a hydrocarbon
  2. is saturated
  3. has the general formula, $\ce{C_{n}H_{2n}}$

All alkanes are

  1. saturated
  2. hydrocarbons
  3. general formula, $\ce{C_nH_{2n+2}}$

There are three entities. If you will, the suffix -ane means saturation.

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The formula $\ce{C_{n}H_{2n+2}}$ is useful only for saturated, aliphatic alkanes. Not for cycloalkanes.

Cycloalkanes belong to the homologous series with general formula $\ce{C_{n}H_{2n}}$.

For example, the following is a straight chain alkane:

Straight chain alkane

Here, the two terminal alkanes have 3 bonded hydrogens and 1 bond with carbon, while the other intermediary carbons have 2 bonded hydrogens and 2 bonds with adjacent carbons.

On the other hand, here is cyclobutane:

enter image description here]1

In this compound, each carbon has 2 bonded hydrogens and 2 bonds with adjacent carbons.

Thus, overall, any straight chain alkane (and even branched alkanes) has two hydrogens more than the corresponding cycloalkane with the same number of carbons. Both of them are saturated.

Hence they belong to different homologous series with different general formulae.

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    $\begingroup$ Hi, thank you for your answer. If you could please provide a more fuller response to my question, I think I would have a better grounding on the matter! $\endgroup$ – Shane Jun 16 at 13:52
  • $\begingroup$ @Shane I have edited the answer. $\endgroup$ – Nilabja Jun 16 at 14:28

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