Molecular formula for 4-methyl-1-pentanol?

According to Wikipedia 4-methyl-1-pentanol is $\ce{C\_6H\_{14}O}$.


But here is what I got, which is slightly different.


Methyl = Think of methyl group, which is $\ce{CH\_3}$.


Pentanol = *Penta* means we have 5 carbons, *-ane* ending means we have $2n+2$ hydrogens, i.e. 12 hydrogens and *-ol* ending means we have a hydroxyl group $\ce{-OH}$.


So the molecular formula that I arrived at is: $\ce{CH\_3C\_5H\_{12}O}$ = $\ce{C\_6H\_{15}O}$. 


How did I end up with an extra hydrogen compared to Wikipedia answer?




Actually, the formula you arrived at is $\ce{CH3C5H12OH}$ = $\ce{C6H16O}$. In any case, in your "2n+2" step, you need to reduce that number of hydrogens by 2. One is eliminated where the methyl group is attached, and a second is eliminated where the $\ce{-OH}$ group is attached. So removing 2 hydrogens from $\ce{C6H16O}$, gives you the desired $\ce{C6H14O}$.


 5

You’re applying the $2n +2$ method wrongly. You need to stupidly count *all* the carbons in a fully saturated molecule and take *that* number as $n$.


In your case this is:


* 5 because *pent*anol
* another 1 because methyl.


That gives you six carbons in total. $n = 6 \Rightarrow 2n + 2 = 14$


Alcoholic oxygens do not introduce unsaturation, so you can safely ignore the oxygen here.


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