FINDIING THE NUMBER OF ALCHOL AND ETHER ISOMERS

ISOMERISM TRICK For Alchols and Ethers

REMEMBER THIS TABLE FIRST

Carbon Number	No. of Isomers
1	1
2	1
3	2
4	4
5	8

 ALCOHOLS AND ETHERS

If carbon and hydrogen are in this relation CnH_2n+2O then be sure the given molecular formula gives alcohol and ether isomers.

We know general formula for alcohol is R-OH and that of ether is R-O-R
[Since alchols and ethers are functional isomers  so from the provided molecular formula both the isomers of alchols and ethers can be obtained.]
First you should place the provided molecular formula in the above format.

Eg.

a)  CH₄O
CH₃-OH
 Here there is only one carbon so from above table it is clear that this formula contains only one alcohol isomer.

 To be ether two carbon must be present but there is only one carbon so no ether isomer exists.

b)  C₂H₆O
C₂H₅-OH 
Carbon no. 2  so, 1 alcohol isomer

Now, for ether
CH₃-O-CH₃
Rem: See the carbon no. on right hand side only, keep the carbon no. on left hand side constant [you will understand why I am doing it other examples]
 You should keep the least carbon no. always 1 on Left hand side at first.

Carbon no. 1 so, 1 ether isomer.

Total no. of isomers = 1 alcohol + 1 ether = 2 isomers

So from the above given molecular formula 2 isomers can be formed, one is alcohol and the other is ether isomer.

 c)  C₃H₈O
C₃H₇-OH 
Carbon no. 3  so, 2 alcohol isomers
To find 1⁰alcohol isomer: Divide the total no. of alcohol isomer by 2.
In this case : No. of 1⁰alcohol isomer is :  2/2 = 1


Now, for ether
CH₃-O-C₂H₅
See carbon no. on R.H.S.

Carbon no. 2 so, 1 ether isomer.

Total no. of isomers = 2 alcohols + 1 ether = 3 isomers

 So from the above given molecular formula 3 isomers can be formed, two is alcohol and one is ether isomer.

d)  C₄H₁₀O
C₄H₉-OH 
Carbon no. 4  so, 4 alcohol isomer
 No. of 1⁰alcohol isomer is :  4/2 = 2

Now, for ether
CH₃-O-C₃H₇
See carbon no. on R.H.S.

Carbon no. 3 so, 2 ether isomers.

In case of these types of ethers having unequal no. of carbon atoms on R.H.S and L.H.S increase the no. of carbon atom on L.H.S by 1 at a time. 
[ You may have ques why I didn't increase the no. in case of CH₃-O-C₂H_5. Yes! consider increasing then you will obtain this ether C₂H5-O-CH₃ this ether and the ether before they are same so we didn't increase the carbon no. Got it.]

Increasing the carbon no. on L.H.S we get this ether
C₂H₅-O-C₂H₅
See carbon no. on R.H.S.
Carbon no. 2 so, 1 ether isomer.

Total no. of ether isomers = 2 + 1 = 3

Total no. of isomers = 4 alcohols + 3 ethers = 7 isomers

 So from the above given molecular formula 7 isomers can be formed, 4 are alcohol and 3 are ether isomers.

e)  C₅H₁₂O
C₅H₁₁-OH 
Carbon no. 5 so, 8 alcohol isomers
 No. of 1⁰alcohol isomer is :  8/2 = 4

Now, for ether
CH₃-O-C₄H₉
See carbon no. on R.H.S.

Carbon no. 4 so, 4 ether isomers.

Increase the no. of carbon atom on L.H.S by 1 at a time. 

Increasing the carbon no. on L.H.S we get this ether
C₂H₅-O-C₃H₇
See carbon no. on R.H.S.
Carbon no. 3 so, 2 ether isomer.

[You may think of increasing again the carbon no. on L.H.S. but it will give the same ether as:
C₃H₇-O-C₂H₅

Total no. of ether isomers = 4 + 3 =7

Total no. of isomers = 8 alcohols + 7 ethers = 14 isomers

 So from the above given molecular formula 14 isomers can be formed, 8 are alcohol and 7 are ether isomers.