[jena]

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

[dexangeles]

this is how i do it

alkanes are normally C_nH_2n+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 C_nH_2n
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

[jena]

So how do you figure in the extra's that are nitrogen or boron based off the formula that I used previously.

Thank you  

[dexangeles]

it just tells me what the max possible is, then i go to H-NMR, C13 NMR, FTIR

[jena]

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

[dexangeles]

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