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Consider two alcohols:

$$\rm{CH_3 - CH_2 - OH }$$

and

$$\rm{CH_3 - CHOH - H }$$

In the second alcohol, the hydroxyl radical instead of sticking out from the molecule, is bonded below the terminal carbon, like this $$\rm{H}$$

$$|$$ $$\rm{R- C- H}$$ $$|$$ $$\rm{OH}$$

Now, this doesn't affect the nomenclature (as far as I know), both will be called $\rm{1-ethanol}$. Clearly, these are isomers. But IUPAC nomenclature is supposed to be completely unambiguous. Or did I go wrong somewhere and there actually is a difference in their naming?

PS: Is there any way to produce structural formulas more efficiently in latex on chemistry stackexchange?

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  • $\begingroup$ I never tried to draw structures in LaTeX here on Chemistry.SE and to my knowledge, chemfig isn't supported. For me, BKChem with export to png does the job. $\endgroup$ – Klaus-Dieter Warzecha Jan 29 '14 at 8:16
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Actually, both of these molecules are exactly the same. The way you write it on paper is just an approximate representation and does not represent the actual structure of the molecule.

By writing the hydroxyl group below the carbon or jutting out from it with a single bond makes absolutely no difference to the actual structure of ethanol.( Ethanol has no isomeric alcohol.) Therefore, it does not matter how you represent it on paper, it will be the same molecule.

But in certain chiral compounds, by using some predetermined rules of representation, writing different groups on different sides does change the compound. But this is true only in cases where stereoisomeric forms of the compound exist. Then, writing the group in a different way changes the compound to one of its stereoisomer. For example, Z form is written with higher priority groups on one side and the opposite is followed while representing the E form. Fischer projection, Sawhorse projection and Newman projections to represent molecules are a few examples where the representation of molecules on paper is sensitive to the way in which different groups are attached to the compound, But not here.

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They are not isomers, they are the same molecule, ethanol (the "1" is superfluous, there is no 2-ethanol), CH3CH2OH

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