Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: jena on February 15, 2005, 02:56:27 PM
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Could someone please help me with this question:
A compound with a molecular forumula of C13H19BrO has at most how may rings?
My answer is four, using this forumula:
# of Rings= 1+#C-(#H+#X)/2 + (#B+#N)/2
#C-Carbons
#H-Hydrogens
#X-Halogens
#B-Boron
#N-Nitrogen
Please help and Thank You :)
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this is how i do it
alkanes are normally CnH2n+2
so that makes (13 x 2) + 2 = 28 supposed hydrogens for an alkane
your molecule only has 19 so ---> 28 - 19 = 9 hydrogens less
we know a double bond reduces an alkane by 2 electrons -- alkenes are CnH2n
and we know that a ring constitutes (is like) a double bond in this manner
so you possibly have 4 double bonds, or four rings
(not including triple bonds and others) just making the explanation as simple as possible
in short all I do is get the max possible Hs by (13 x 2) + 2 = 28
deduct that by what the molecule has 28 - 19 = 9
divide it by two -- 9/2 = 4.5
that gives you 4 possible double bonds
if there are other methods, i'm all ears too :)
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So how do you figure in the extra's that are nitrogen or boron based off the formula that I used previously.
Thank you :)
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it just tells me what the max possible is, then i go to H-NMR, C13 NMR, FTIR
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Hi,
Sorry but I don't understand what you mean by H-NMR, C13 NMR, FTIR ???
Again, how do you figure in the extra's that are nitrogen or boron?
Please Help :-\
Thank you :)
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each molecular formula can give you a lot of different molecular configurations, so you can't really figure out the structure by just looking at it. That's when FTIR and NMR helps. I think this site has a link to explanations of FTIR and NMR. We need to ask Mitch