# Why are the given compounds in a homologous series?

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?

• You got it all wrong: compounds in a homologous series have different chain length. – Ivan Neretin Mar 29 '16 at 17:41
• @Ivan Neretin, I am sorry for that. – 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.
• Could you explain a bit more? @ttdijkstra – Iaamuser user Mar 29 '16 at 17:48
• For the series in the question, another possibility is n=0 – jerepierre Mar 29 '16 at 18:02
• Thanks for the note on formatting @IͶΔ I just noticed myself. – ttdijkstra Mar 29 '16 at 18:02