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Organic : draw structural formulas for isomer help me plz
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 Message 1 of 2 in Discussion 
From: MSN Nicknameiiambow  (Original Message)Sent: 11/4/2007 11:31 AM
i need yr help. i tried to understand it but i didn't understand it.
so can you guys help me please
1. draw structural formulas for 3 isomers of C3H8O
2. draw structural formulas for all possible isomers having the following molecular formulas C3H6, C3H8, C4H10O, C3H7Cl, C4H9F, C2H2I2, C2H4F2, C3H6Cl2
                                                   thank you guys


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 11/4/2007 6:28 PM
The idea is to connect the atoms together in as many ways possible in order to get all the molecules with those formulas.  You just have to be careful that you obeyed the normal bonding patterns for each atom and that no structures are repeated.
 
For the first one, C3H8O, you can connect the carbons and oxygens as follows:
 
 
C–C–C–O   =   CH3CH2CH2OH  (1-propanol)
 
C–C–O–C   =   CH3CH2OCH3   (methyl ethyl ether)
 
C–O–C–C   =   CH3OCH2CH3   (methyl ethyl ether again, so this one doesn't count)
 
  O              OH
  |              |
C–C–C     =   CH3CHCH3     (2-propanol)
 
 
Those are condensed structural formulas on the right.  Structural formulas show all of the bonds.  For example,
 
  H  H  H
  |  |  |
H–C––C––C––O–H
  |  |  |
  H  H  H
 
 
 
Common bonding patterns:
 
Halogens:  one bond   –Cl
 
Oxygen:  two bonds    –O�?nbsp; and  =O
 
Nitrogen:  three bonds  –N�?nbsp; and  =N�?nbsp; and  ≡N
                         |
 
                       |          /
Carbon:  four bonds   –C�?nbsp; and  =C  and  ≡C�?nbsp; and  =C=
                       |          \
 
 
When the lone electron pairs are added to the halogens, oxygen, and nitrogen, the octet rule is satisfied and the formal charge on the atom is zero.  
 
 
Steve