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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: sparkya10 on November 24, 2009, 12:45:49 AM

Title: C13 NMR Problem
Post by: sparkya10 on November 24, 2009, 12:45:49 AM

Hi, I am really having trouble with this problem:

 Compound A, C7H8, shows the following 13C NMR proton deocoupled
spectrum: δ 143 ppm, CH; 74, CH2; 45, CH. When A undergoes catalytic hydro-
genation, it gives compound B, C7H12.  This saturates alkenes but does not
open rings.  What is compound A?

I know that the 143 ppm indicates a C=C or ring, and 45 ppm indicates C-C, but I am unsure what 74 ppm indicates. I think it indicates C-c triple bond but I cant seem to make this information work with the problem. Any help with this problem would be really helpful, thanks

Title: Re: C13 NMR Problem
Post by: orgopete on November 24, 2009, 12:58:16 AM

Since there are only three signals, there has to be a lot of symmetry. If no double bonds are in the product, there has to be two rings. It isn't cyclopentadiene or acetylene, but may be made from them.

Title: C13 NMR Problem
Post by: Yan_sora on November 24, 2009, 07:13:37 AM

It's (1s,4s)-bicyclo[2.1.1]hex-2-ene  ;)

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Title: Re: C13 NMR Problem
Post by: tmartin on November 24, 2009, 07:50:35 AM

Unless my counting is incorrect, (1s,4s)-bicyclo[2.1.1]hex-2-ene, has only 6 carbons...that cannot be compound A, as the given formula is C7H8.  Also, hydrogenation of (1s,4s)-bicyclo[2.1.1]hex-2-ene would give the chemical forumla C6H8...would it not?

To sparky:  When A undergoes catalytic hydrogenation, it gives compound B, C7H12... this seems good evidence to me that a double or triple bond is present.  It may be helpful to compare the IHD (or degrees of unsaturation) of the starting material and the product to determine how much may be changing.

Title: Re: C13 NMR Problem
Post by: jacy.cn on November 24, 2009, 08:56:06 AM

THIS question is very difficultly to answer
hehe

Title: Re: C13 NMR Problem
Post by: Arctic-Nation on November 24, 2009, 01:34:23 PM

I've got a possible solution, but not one with only 3 equivalent carbon atoms. Degree of saturation and the result of hydrogenation suggests a compound with two rings and two double bonds.

Title: Re: C13 NMR Problem
Post by: DrCMS on November 24, 2009, 02:11:24 PM

I think Yan_sora nearly got it right. 
If you add a carbon to that structure to give 2 double bonds and a symmetrical structure I think you'll have the right answer.

Title: C13 NMR Problem
Post by: Yan_sora on November 24, 2009, 02:30:45 PM

Oh, I'm sorry :(

It's (1s,4s)-bicyclo[2.2.1]hepta-2,5-diene

Reaction:

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ChemNMR 13C Estimation

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Title: Re: C13 NMR Problem
Post by: sparkya10 on November 25, 2009, 12:54:25 PM

Hey yeah I think that is the right answer thanks for all your help it is really appreciated, I worked on this problem so long to no result. I knew there had to be 2 rings but I couldnt think of a structure to make it work. Thanks