Finding Isomers-Alcohols+Ethers

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agm06002

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Hi,

I'm working on a specific question for my Organic class and I was wondering if somebody could help me out with it.:confused:

Here's the question: "Find as many isomers as possible of the molecule, C_4H_10O [Alcohols + Ethers]."

I would appreciate any help I can get on this problem!

Many Thanks,

Andrew

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Well there are no degrees of saturation so all of the bonds are singe.

Degrees of saturation = [(Cx2)+2-H]/2 = [(4x2)+2-10]/2 = (8+2-10)/2 = (10-10)/2 = (0/2) = 0

I can only think of two alcohols and two ethers (four isomers total).

H3C-CH2-CH2-O-CH3

H3C-CH2-O-CH2-CH3

H3C-CH2-CH2-CH2-OH

H3C-CH2-CH(OH)-CH3

There are no stereo isomers since there are no chiral carbons.

Are you sure about that last statement? That last isomer looks like a possible R/S around the second carbon from right to me....hmmm C bonded to H, OH, methyl and ethyl???
 
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OK. Maybe if I post my answers you can tell me what I'm doing wrong because I'm not sure what I should be changing.:eek: (i.e. I don't want to change a correct answer to incorrect)

The files can be found at: http://www.flickr.com/photos/11295624@N06/

(Copy and paste the link)

At the site, please look at the photos and let me know what I am doing wrong with my diagrams.

Many Thanks for all of your help,

Andrew
 
Okay, after looking at your list, there are 3 isomers that I missed and 2 that you missed. There should be 8 isomers total:

Ethers (3)

H3C-CH2-CH2-O-CH3

H3C-CH(CH3)-O-CH3

H3C-CH2-O-CH2-CH3

Alcohols (5)

H3C-CH2-CH2-CH2-OH

H3C-CH(CH3)-CH2-OH

H3C-(H3C)C(OH)-H3C

H3C-CH2-CH(OH)-CH3 - 2 stereoisomers (1 R, 1 S)

Hopefully that's all of them...
 
Okay, after looking at your list, there are 3 isomers that I missed and 2 that you missed. There should be 8 isomers total:

Ethers (3)

H3C-CH2-CH2-O-CH3

H3C-CH(CH3)-O-CH3

H3C-CH2-O-CH2-CH3

Alcohols (5)

H3C-CH2-CH2-CH2-OH

H3C-CH(CH3)-CH2-OH

H3C-(H3C)C(OH)-H3C

H3C-CH2-CH(OH)-CH3 - 2 stereoisomers (1 R, 1 S)

Hopefully that's all of them...


Looks good to me.
 
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