splitting pattern of 1,4 Dimethoxybenzene

[suhshi]

1,4-Dimethoxybenzene



Why does this one have 2 singlets?

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[saDDS]

the methyl hydrogens are equivalent and the 4 hydrogens on the benzene are all equivalent with no neighboring hydrogens.

 

[saDDS]

see the methyl h's are the same the 4 benzene h's are the same. all have no neighboring h's so they are both singlets.

 

[suhshi]

thanks for your help
so for acetone there would only be one singlet correct

 

[saDDS]

yeah it would the methyl h's again are equivalent with no neighboring h's. be wary of symmetry I had one of my DAT that was asking about the splitting patter of a symmetrical ether.

 

[suhshi]

thanks alot

 

[jhanago]

How are all 4 H's not neighboring each other when they are 1 sigma bond away?

 

[yappy]

equivalent H's dont split each other.

How are all 4 H's not neighboring each other when they are 1 sigma bond away?

 

[jhanago]

equivalent H's dont split each other.

Fail. It's too late to think efficiently, thanks!!

 

[invictusx]

The hydrogen on the -OCH3 is identical to the other -OCH3 on the other side. The carbon adjacent to the carbon with the substituent has only one hydrogen and that hydrogen is identical to every carbon on the benzene ring so the NMR machine can't differentiate them from one another, that's why there's only two singlets. It's a perfectly symmetrical molecule.

Whenever you do these, just see if the hydrogen or carbon (depends on what machine you're using) can see any other DIFFERENT hydrogens or carbons around them.

 

[rav4182]

its a symmetrical compound, there are no c-h bonds next to each other. therefore only 1 singlet