A shortcut of Homologous Series

Hello there in Organic Chemistry there is a term "Homologous Series" that means those compounds that have the same physical and chemical properties due to their formula . In homologous series all the compounds have a standard formula such as alkoholes are specified by C_n H_2n+1 O H. Here a list iof all homologous standard formula.

Homologous	Standard Formula	Example
Alkane	Cn H2n+2	Methane (CH4)
Alkene	Cn H2n	Ethene (C2H4)
Alkyne	Cn H2n-2	Propyne (C3H4)
Alkohol	Cn H2n+1OH	Butanol (C4H9OH
Aldehyde	Cn H2n	Propanal(C3H7)CHO
Carboxylic Acids	Cn H2n+1COOH	Hexanoic acid (CH3(CH2)4COOH

In some times it seems very hard to remember this tabble. For this we have found an shortcut to memorise it. Here it's
Here if we take all the standard formula of all homologous we cam see a common tern in all homologous here it's.

So here C_n H_2n is common in all homolohous and after that there were added an suffix. Now for our conveniency of understanding we are rearranging the list. Here it's:
So noew firstly after C_n H_2n there were added -2 in Alkynes, secondly there were added 0 n Alkenes, thirdly there added +2 in Alkanes forthly there were added +1 in Alkohol, Aldehyde, Carboxylic Acids with their formila.
So now think about a line of numbers where there are $ points. First of them is -2 then 0 then +1 and then +2.
Now if we put C_n H_2n at the point -2 we will get the homologous standrad formula of Alkynes, if we put C_n H_2n at the point 0 we will get the homologous standrad formula of Alkenes, if we put C_n H_2n at the point +2 we will get the homologous standrad formula of Alkanes and if we put C_n H_2n at the point +1 with the formula of Alkohol, Aldehyde and Carboxylic Acids we will get the homologous standrad formula of them. So finally we get:

Now a short line to remember it is that "Alkyne is at -2, Alkene is at 0, Alkane is at +2 and all the others are at the point +1 with their formula.
Hope that you have understand it and it seems easy to you. Of you have any question aboutit comment us or you vacn use ask us on our Ask A Question page

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