hydrogen environments of benzophenone

The hydrogen in $\ce{C4}$, $\ce{C6}$, $\ce{C10}$ and $\ce{C8}$ belong to same hydrogen environment.  

The hydrogen in $\ce{C3}$, $\ce{C1}$, $\ce{C11}$ and $\ce{C13}$ belong to the another hydrogen environment.  

Hydrogen in $\ce{C12}$ and $\ce{C2}$ belong to the once again, another hydrogen environment. 


$\ce{C3}$, $\ce{C1}$ , $\ce{C11}$ $\ce{C13}$, $\ce{C12}$ and $\ce{C2}$ all have two hydrogen on the neighboring carbon atoms. 
So my question is why aren't they classified as one hydrogen environment? Instead, why are the categorized under two different environments? 


![benzenephenone](https://i.stack.imgur.com/bGOaSb.png)




Whether environments are the same or not is based on **symmetry** considerations and **not** the number of hydrogens attached to the adjacent carbon atom.


Molecules and atoms within a molecule can be of three types:


* **identical** - they can be interconverted by simple rotations or translations.
* **enantiotopic** - they can be interconverted by reflection in a mirror or about a center of symmetry
* **diastereotopic** - they cannot be interconverted.


Molecules or atoms within a molecule that are found to be identical are said to be in the same environment. Molecules or atoms that are identical, or in the same environment, **must** be identical with respect to **all** chemical and physical properties. These properties would include chemical reactivity, spectroscopic environment, *etc.*


In benzophenone, I can rotate the benzene ring about the $\ce{C5-C7}$ bond. This rotation interconverts $\ce{C4}$ and $\ce{C6}$. If I told you to look away and I performed this rotation (and we removed the numeric subscripts), there is no way that you could tell if the original carbon or the other carbon was now in the $\ce{C4}$ location. These two carbons (and their attached hydrogens) are therefore said to be equivalent by the above definition. The same applies to $\ce{C1}$ and $\ce{C3}$, they are also **equivalent to each other, or in the same environment.**


I can also rotate the molecule around the $\ce{C7=O}$ carbonyl axis. If I do this I can interconvert $\ce{C4, C6, C8, C\_{10}}$. This same rotation also (separately) interconverts $\ce{C1, C3, C\_{11}, C\_{13}}$, and also (separately) $\ce{C2}$ and $\ce{C\_{12}}$.


Therefore, these 3 sets of carbons


* $\ce{C4, C6, C8, C\_{10}}$
* $\ce{C1, C3, C\_{11}, C\_{13}}$
* $\ce{C2, C\_{12}}$


are different (diastereotopic) from one another (I can't interconvert $\ce{C4}$ and $\ce{C\_{12}}$, for example). So we have 3 different chemical environments, but **within each environment** there are several identical atoms. The same methodology and reasoning applies to the hydrogens attached to these carbons. They also exist in 3 different environments, and within each of these 3 environments there are several equivalent hydrogens.


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