Why do homologous series differ by CH2, and why does this not affect the compound’s properties?

Question

OK, so this might be a dumb question, might be not. Im a 10th grader so bare with me please.

So I just learned what homologous series is. According to the definition that the teacher told me:

A homologous series is a series of compounds in which each member differs from the next/previous by $\ce{CH2}$ or 14 mass units. Physical properties change in a homologous series but chemical properties remain almost the same because the functional group does not change.

Now, I got a couple of question on this:

1. Why $\ce{CH2}$? Why not $\ce{CH4}$? What makes $\ce{CH2}$ so special that we have an entire term dedicated to a series in which every compound differs by $\ce{CH2}$?

2. It says that there is no change in chemical properties as the functional group does not change. Well … the hydrocarbon chain does change. So does that mean that a hydrocarbon has no unique chemical properties and all its chemical properties comes from the functional groups in it?

3. This is related the the 2nd question. It says that the functional group remains the same. I drew a chemical diagram thing:

   

   Now I don't know if this is 100% correct but what I imagined is that the $\ce{CH2}$ replaced the $\ce{OH}$ and the carbon bonded with the carbon chain, then it took the $\ce{H}$ of the $\ce{OH}$ and we are left with $\ce{O}$ alone. If I am correct and if this can really happen then doesn't this mean that we are changing the chemical properties of a homologous series (which can't practically happen)?

   If this example of mine is incorrect then there may be some other example like this where the $\ce{CH2}$ replaces the functional group itself thus changing the chemical properties?

This 3rd question may be illogical and stuff but I am curious.

Answer 1

I’m going to answer these questions backwards, because there is a serious misconception in the third that must be addressed as soon as possible.

Concerning the third question, $\ce{CH2}$ is not added to an existing molecule. Rather, the idea of homologous compounds is a purely empirical and comparative one. You cannot go from butan-1-ol to pentan-1-ol in one step by adding a mysterious $\ce{[CH2]^\dagger}$ compound. There is no such reaction and therefore there is no chance to introduce ‘side reactions’ like the one you imagined. There are homologisation reactions, e.g. the Arndt-Eistert homologisation for carboxylic acids but they are significantly more complicated than inserting a $\ce{CH2}$ group (and not taught until university).

On the second question, saying that chemical properties will not change is a mild simplification — depending on where you draw the line between physical and chemical properties. You may know already that the heavier a molecule is, the more London interactions hold it together and thus the higher melting and boiling points become. Additionally, assuming only one single functional group, the longer a molecule is the less effect a functional group has on the entire thing. So while methanol (chain length 1) is infinitely miscible with water, pentan-1-ol (chain length 5) is not.

The hydrocarbon chain in itself actually does not display a lot of interesting chemistry except for combustion. However, research efforts are being strongly directed towards what is known as $\ce{C-H}$ activation, meaning forcing a (seemingly arbitrary) carbon–hydrogen bond to react under defined conditions. If that is pushed far enough, the hydrocarbon chain can be understood to have interesting chemistry but at present we are not there yet.

Finally, your first question. I’ll just quickly reiterate that the idea of a homologous series is purely empirical, comparative and descriptive. Taking alcohols as an example, we know methanol $\ce{CH3OH}$, ethanol $\ce{C2H5OH}$, propan-1-ol $\ce{C3H7OH}$, butan-1-ol $\ce{C4H9OH}$, etc. Upon carefully analysing the reactions of each of these compounds, it was soon established that they share a similarity in spite of their different physical properties and their notably different effects on the human body. When elemental analysis was established, it could be shown that they indeed all have different carbon contents and the increase is always $\ce{CH2}$.

And why $\ce{CH2}$ and not something else? Well, remember that carbon is typically tetravalent, i.e. is most stable if it has formed four bonds. In a formal (and paper chemical) exercise, take a random alkane for example ethane. Its structure is given below.

  H H
  | |
H-C-C-H
  | |
  H H

Now assume you want to draw a new structure with one more carbon atom. For example, assume this carbon atom to be added at the very right. I’m first going to formally pull apart the $\ce{C-H}$ bond, then formally put in the carbon, connecting it to the left and right. Notice what will be missing:

  H H          H H            H H            H H H
  | |          | |            | | .          | | |
H-C-C-H  =>  H-C-C· ·H  =>  H-C-C-C-H  =>  H-C-C-C-H
  | |          | |            | | ˙          | | |
  H H          H H            H H            H H H

Note how it requires two hydrogens to satisfy carbon’s valence. This is why the group formally added is $\ce{CH2}$: Assuming a carbon inserted into something it needs two additional hydrogens to keep satisfied.

Answer 2

1. $\ce{CH2}$ because the chain is 1 $\frac{longer}{shorter}$. Remember that carbon tends to form 4 bonds. The other 2 bonds that would be attached to Hs in methane, for example, are in this case attached to other carbons in the chain. So $\ce{CH2}$ (plus a bond, or 2 half bonds) is really all that you are adding.

2. Its chemical properties do not change significantly, but its physical properties might. Boiling point, density, etc are physical properties. Formic acid, acetic acid, and propanoic acid are quite similar, but have differing chain lengths.

3. The $\ce{CH2}$ is just added to the chain. ie $\ce{HO-CH2-CH2-CH3}$ vs $\ce{HO-CH2-CH3}$. $\ce{O2}$ should not be released.

Your initial answer as linked does some additional work beyond what is necessary. No need to change the $\ce{OH}$ at all, just add the $\ce{CH2}$.

Answer 3

Formulas of alkane alkene and alkyne itself results in the difference of CH2 unit. As if I talk about methane. It has formula of CH4 as it is alkane and following the formula of CnH2n+2 .Similarly Ethane it is having formula C2H6 .And if you see the difference between the two you will automatically get CH2 .The same thing implies to all other members of series alkane . Even if you see alkenes and alkynes you will find the same difference i.e. CH2 between all the members .