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Why are the given compounds in one homologous series?

$$\ce{CH3-OH}$$ $$\ce{CH3-CH2-OH}$$ $$\ce{CH3-CH2-CH2-OH}$$ $$\ce{CH3-CH2-CH2-CH2-OH}$$

I know that homologous series can be represented by a common general formula and have the same chain length. But I could not answer this question. Can anyone help me?

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    $\begingroup$ You got it all wrong: compounds in a homologous series have different chain length. $\endgroup$ – Ivan Neretin Mar 29 '16 at 17:41
  • $\begingroup$ @Ivan Neretin, I am sorry for that. $\endgroup$ – Iaamuser user Mar 29 '16 at 17:49
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Because they can all be described by the general formula $\ce{CH3-(CH2)_n-OH}$. The only varying parameter in this case is the number of middle $-\ce{CH2}- $ groups.

So while the the number of middle $-\ce{CH2}- $ groups may vary $\left(n = 0,1,2,\ldots\right)$ they all fit in this general formula.

The formula $\ce{CH3-(CH2)_n-OH}$ tells you that:

  • The chain starts with a $\ce{CH3}-$ group
  • The chain ends with an $-\ce{OH}$ group
  • Contains a variable number, $n$, of $-\ce{CH2}-$ groups in between.
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  • $\begingroup$ Could you explain a bit more? @ttdijkstra $\endgroup$ – Iaamuser user Mar 29 '16 at 17:48
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    $\begingroup$ For the series in the question, another possibility is n=0 $\endgroup$ – jerepierre Mar 29 '16 at 18:02
  • $\begingroup$ Thanks for the note on formatting @IͶΔ I just noticed myself. $\endgroup$ – ttdijkstra Mar 29 '16 at 18:02
  • $\begingroup$ Please visit this page, this page and this ‎one on how to format your future posts better with MathJax and Markdown. Welcome to chem.SE! $\endgroup$ – M.A.R. Mar 29 '16 at 18:05
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    $\begingroup$ I cleaned up the formatting and the wording a bit, but @ttdijkstra nailed the correct idea. $\endgroup$ – MaxW Mar 29 '16 at 18:15

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