The+Naming+Scheme

=__ Rules for naming the alcohols __= 1) Find the longest chain containing the hydroxide. Carbons branched off of this chain are known as substituents and use the ending -yl. They are numbered according to their position on the parent chain.   EX. CH3-CH2-CH2 -OH

====The Highlighted Carbons in this case form the longest chain of 3 known as Propane .There are no substituents. ====

2) Place the OH (Hydroxide) on the lowest possible number of the chain. The hydroxy groups get __first priority for naming.__
 EX. CH3-CH2-CH2- OH

3) Is it a cyclic structure? (Extra details below)
 EX. CH3-CH2-CH2-OH

This example is not a cyclic structure therefore we shall continue on to step 4.

===4) Remove e from the parent alkane chain and add ol. When multiple alcohols are being used use di, tri, etc before the ol but after the parent name. ex. 2,3-hexandiol. ===

 EX. CH3-CH2-CH2-OH

Seeing as there is only one hydroxide being used, to finish naming we must simply replace the e from propan__e__ (Our longest chain) with the ending ol. So, propanol.

Your Turn:

More than one hydroxide:
<span style="font-family: Arial,Helvetica,sans-serif; font-size: 140%;">OH-CH2-CH2-OH

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 140%;">In this case the longest chain is 2 (ethane)

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 140%;">Since there are two hydroxides we use the prefix di instead of replacing the e and then add the ol afterwards. (Note the locations of the hydroxides still need to be stated.)

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 140%;">So, in this example the answer would be 1,2-ethanediol

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 140%;">There are...hydroxides located at positions...and.....

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 140%;"> Therefore the name is...

Naming with cyclic structures:

Naming a cyclic structure is quite simple. All we have to do is establish the number of sides of the carbon ring, which is six in this case. Therefore we have have a hexanol, but to account for the cyclic structure we add a cyclo on the front. So, Cyclohexanol.

Your turn: Photo from:[]

This ring consists of ... carbons

And since the OH receives first priority for numbering the methyl attachment is located at ...

Lastly the Bromide attachment is at...

(Note: When naming alcohols the extensions for a name go in alphabetical order. )

Accounting for double and triple bonds:

Naming to show double and triple bonds is the most difficult occurrence we will encounter. When naming, it is mandatory to state the location and type of the bond using either en or yn within the name.

--OH--Br ---l-l

<span style="font-family: Arial,Helvetica,sans-serif;">CH3-CH2-CH=CH-CH2-CH2-CH3
---l --Br

We know that the OH is located at the 4 position no matter which end you begin with, so
Br we can narrow it down to 4-heptanol.

At the 6 there are 2 bromide attachments, so we can add the 6,6-dibromo on to what we have so far. So at the moment we have 6,6 dibromo-4-heptanol.

Now we must show the presence of the double bond following the third carbon. (Double/triple bonds are the exception for having the lowest number on a chain instead of the OH). We do so by changing heptanol into heptenol. But we still need to state the position. We do so by placing the 3 after the hept and before that enol.

So in the end we are left with 6,6-dibromo-4-hept-3-enol