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A_S20 · Jul 16, 2010, 04:06 PM

So on this practice test it says to write the structural formula and give the name for all 11 isomers of this "iodobutenes". I have noooo idea were to start with this question. I get what an isomer is. I just need help starting this. Oh and I am having problems with the naming part when it comes to the cis-trans too.

Unknown008 · Jul 18, 2010, 08:00 AM

It's easy, you'll see :)

Ok, I'll have difficulty in introducing the pictures, so bear with my descriptions.

1. Put all the carbon atoms in a straight line.

The double bond between the first and the second carbon.

Put the iodine on the first carbon.

If the iodine is on the same side as is the rest of the hydrocarbon chain, this will be a 1 iodo cis butene.

Code:

H        H
  \     /
   C = C 
  /     \
I        C2H5

Putting iodine the other way, you get 1 iodo trans butene;

Code:

I        H
  \     /
   C = C 
  /     \
H        C2H5

We now put the I onto the next carbon atom,

Code:

H        I
  \     /
   C = C 
  /     \
H        C2H5

This is 2 iodo butene.

Putting I onto the other carbon atoms give you 3 iodo butene and 4 iodo butene.

Code:

H        I
  \     /
   C = C 
  /     \
H        CHI - CH3

Code:

H        I
  \     /
   C = C 
  /     \
H        CH2 - CH2I

Note that 3 iodo butene also has stereoisomers (or optical isomers). This gives rise to one more isomer. [Note, if you can draw the 3 dimensional drawing, it'll be better. It's really difficult for me to do that on computer. But if you ask me, I'll take some time to show you the isomers.]

Code:

H        I
  \     /
   C = C 
  /     \
H        CIH - CH3

Now, you can have... a methyl group! Like this:

Code:

I        CH3
  \     /
   C = C 
  /     \
H        CH3

This doesn't have any cis trans isomers since there are similar groups (-CH3) on both sides on the carbon atom. This is 1 iodo 2 methyl propene.

Doing as before (that is putting I onto the other carbon atoms), we get one more; 3 iodo 2 methyl butene:

Code:

H        CH2I
  \     /
   C = C 
  /     \
H        CH3

Then, we come to the but-2-enes.

Code:

IH2C        H
     \     /
      C = C 
     /     \
   H        CH3

This is 1 iodo trans but-2-ene.

Code:

IH2C        CH3
     \     /
      C = C 
     /     \
   H        H

This is 1 iodo cis but-2-ene.

How about it? 1 more to go! This one is... a cyclic alkane!

Code:

    H   H
    |   |
H - C - C - H
    |   |
H - C - C - H
    |   |
    H   I

Which is cyclo iodo butane.

I hope it helped! :)