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I understand that a homologous series is a group of compounds with the same functional group attached to it.

I also know of an alternate definition of it as being a series of compounds with the same general formula.

But if the first definition is to be considered then how can alkanes, alkenes or alkynes be considered homologous series since there is no functional group attached to any of these, as opposed to, e.g., the homologous series of alcohols?

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For the linear alkanes, the internal methylene groups $\left(-\ce{CH2}-\right)$ are considered to be the functional group defining the homologous series:

$$ \ce{H3C-\underline{(\ce{CH2})_n}-CH3}~,~~n\geq 0 \tag{n-alkanes} $$

Alternatively, this series is defined by the absence of any particular functional group other than methyl and methylene.

Ethane, $\ce{C2H6}$ is the member of the series for $n=0$. Technically, one could exclude methane, $\ce{CH4}$, from the series, since it doesn't have two capping methyl groups. However, to my knowledge it's usually considered part of the series as a special case.

For the 1-alkenes and 1-alkynes, the double and triple bonded carbons are the functional groups defining the series:

$$ \ce{H3C-(CH2)_n-\underline{\ce{CH=CH2}}} ~,~~n\geq 0 \tag{1-alkenes} $$

$$ \ce{H3C-(CH2)_n-\underline{\ce{C#CH}}} ~,~~n\geq 0 \tag{1-alkynes} $$

Parallel to the alkanes, the shortest "official" members of the two series are technically propene $\left(\ce{H3C-CH=CH2}\right)$ and propyne $\left(\ce{H3C-C#CH}\right)$, respectively. As with methane above, though, ethylene/ethene $\left(\ce{H2C=CH2}\right)$ and acetylene/ethyne $\left(\ce{HC#CH}\right)$ are typically grandfathered in.

The case of non-terminal linear alkenes/alkynes is less straightforward. Depending on one's philosophy, they can either be lumped in with the terminal alkene/alkyne series, or they can be considered as distinct homologous series:

$$ \ce{H3C-(CH2)_n-\underline{\ce{CH=CH-CH3}}}~,~~n\geq 0 \tag{2-alkenes} $$

$$ \ce{H3C-(CH2)_n-\underline{\ce{C#C-CH2-CH3}}}~,~~n\geq 1 \tag{3-alkynes} $$

$$ etc. $$

Note that $n\geq 1$ is strictly required for the 3-alkynes, since $n=0$ would give 2-pentyne $\left(\ce{H3C-C#C-CH2CH3}\right)$, a 2-alkyne. This requirement is general: e.g., for the x-alkenes, $n \geq x-2$ must hold for $x\geq 3$.

Branched alkanes/alkenes/alkynes complicate things even further. There are at least two common alkane series that I can think of off the top of my head:

$$ \ce{H3C-(CH2)_n-\underline{\ce{CH(CH3)2}}} ~,~~n\geq 0 \tag{iso-alkanes} $$

$$ \ce{H3C-(CH2)_n-\underline{\ce{C(CH3)3}}} ~,~~n\geq 0 \tag{neo-alkanes} $$

You can mix and match branching, double bonds and triple bonds in countless ways to make new homologous series. I would personally argue you could define a homologous series to contain multiple variable-length segments:

$$ \ce{\underline{\ce{(H3C)2C=C}}-(CH2)_n-\underline{\ce{C#C}}-(CH2)_m-\underline{\ce{CH(CH3)}}-(CH2)_r-\underline{\ce{CH=CH2}}} \tag{?} $$

Of course, the practical utility of any correspondence of properties among the various species in such a convoluted homologous series may be ... limited.

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  • $\begingroup$ the homologous series could also be defined as the compounds have the same general formula, in that case for eg. Alkanes would have a genral formula of CnH2n+2 , and menthane would fit in as when n=1 the compound produced is CH4. Are the two definitions of the homologous series contradicting each other or perhaps another reason why they are defined in two different ways? $\endgroup$ – Kaberi Ghosh Nov 25 '16 at 10:59
  • $\begingroup$ @KaberiGhosh They're just two different approaches to defining the same concept. To the best of my knowledge, "homologous series" doesn't have a rigorous definition, so neither one is strictly "correct." To note, a significant downside to, e.g., the $\ce{C_nH_{n-2}}$ definition for alkynes is that the formula also matches the dienes: butadiene's formula is $\ce{C4H6}$, the same as butyne's. You also have to specify structural features along with the $\ce{C_xH_y}$ formula in order to accurately define a given homologous series. $\endgroup$ – hBy2Py Nov 25 '16 at 12:32
  • $\begingroup$ then that gets us back to the same point, that neither of the definitions seem to completely encompass the concept. So which definition would be the most accurate? $\endgroup$ – Kaberi Ghosh Nov 26 '16 at 12:35
  • $\begingroup$ @KaberiGhosh Neither? It depends on the circumstances, and on who you're asking about the definition. Unfortunately, I don't think you're going to be able to pin this down as neatly as you'd like. $\endgroup$ – hBy2Py Nov 26 '16 at 16:02

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