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?

  • 10
    $\begingroup$ 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

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.
  • $\begingroup$ Could you explain a bit more? @ttdijkstra $\endgroup$ – Iaamuser user Mar 29 '16 at 17:48
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ I cleaned up the formatting and the wording a bit, but @ttdijkstra nailed the correct idea. $\endgroup$ – MaxW Mar 29 '16 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.