Steric hindrance is important for understanding regio-selectivity and stable conformation of molecules. I know the general definition, but can somebody explain to me in detail how does phenomenon of steric hindrance come about?

Like, to give a simple, hypothetical example, $\ce{CH4}$ is said to have no steric hindrance. But why does attacking molecule have no trouble squeezing past $\ce{H}$ atoms and attack center carbon atom. I know hydrogen atom is said to be small, but seriously how small is it?

I am asking this because when examples go beyond $\ce{H}$ atom and alkyl atoms I can't judge whether steric hindrance is present or not and thus what products are formed. Like in $\ce{(CH3)3CHOH^+}$, is there steric hindrance. It is a primary carbon cation but instead of two hydrogen then is $\ce{OH^-}$ group?

  • $\begingroup$ what do you mean by $\ce{CH3CH3CH3CHOH^+}$ ? $\endgroup$
    – ParaH2
    Nov 28, 2016 at 17:52
  • $\begingroup$ Oh... I was giving a example similar like you know how u have tertiary carbon cartion, nucleophile can't attack it simultaneously with leaving group. So now instead of tertiary carbon cation what if u had a carbon cartion attached to propyl, hydrogen and OH- group. If the Oh- group was H carbon cartion will be primary and SN2 can occur but with Oh- doesn't that add steric hindrance like as if carbon cation was tertiary? $\endgroup$
    – TLo
    Nov 28, 2016 at 17:59
  • $\begingroup$ It not answers my question. I just said what do you mean by this $\ce{(CH3)3CHOH^+}$ because written like this it doesn't exist. Also this is not carbon cartion but carbocation. $\endgroup$
    – ParaH2
    Nov 28, 2016 at 18:06
  • $\begingroup$ Okay. this chemical formula is a hypothetical example. It may or may not exist. But the bigger point I wanted to make is which groups provide more steric hindrance OH-, CH3 and so on.... $\endgroup$
    – TLo
    Nov 28, 2016 at 18:09
  • $\begingroup$ $\ce{(CH3)3C^+}$ is a carbocation (which exists) it can be from $\ce{(CH3)3CCl}$ which will do an E1 in a polar solvant for example $\ce{EtOH/H2O}$. If you want to know the size of a hydrogen atom you can look on google, even myself, I don't know exactly the size but it's small. The more you add atoms the more it will be big and create steric hindrance. So $\ce{H^+ < CH3^+ < (CH3)3C^+ }$ and so on ... $\endgroup$
    – ParaH2
    Nov 28, 2016 at 18:18

1 Answer 1


Normally we don't discuss quantum mechanics in this context, but an answer to your question requires it.

Taken from:



What we see here is the LUMO for CH3Cl. This is where an incoming nucleophile will 'stick' its electrons to form the new bond. Since the orbital is antibonding, electrons in the orbital will destabilize a bond, in this case the C-Cl bond. The actual meaning of the size of this orbital, rigorously, is a little unclear. However you can roughly infer that if a nuclephile gets its electrons here, a reaction is 'likely'. The reddish region to the left encompasses the hydrogens. That is also where the nucleophile has to go.

Compare the next from:



This is for ethyl chloride. There is an extra minor node on one of the hydrogens. Because it is small, it is less important than the other regions. Notice the methyl group is partly obscuring the backside of the orbital lobe near the electrophilic carbon. There are less ways a nucleophile can approach this to cause a reaction.

I'm no longer practicing so I do not have software to produce analagous images, so you will have to imagine the extrapolation. 1-chloroopropane would not add much more to the steric hindrance, because the 'extra' methyl group would be dangling on the same of the molecule as the methylene group. However, if we had two methyl groups, such as 2-chloropropane, the hinderance would be much more. Three groups, as in t-butyl chloride? Give it up, to much hinderance for a backside attack reaction.


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