2 quick questions

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2 quick questions.

One is on naming compounds (hydrocarbons)

a) CH3 - C = CH - CH2 - CH3
|
CH3

(the CH3 should be under the C)

what's the key? is number one counting up how many C and H's there are? which element is first?

my second question is on structural and geometric isomers. I've read up on both but just when I think I understand I read something else and it confuses me. In one or two lines, what's the difference?
I thought it was simply a geometric isomer would have to be GEOMETRICALLY shaped differently, but apparently that is not true.

Thanks
 

Answers and Replies

  • #2
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Well you should have a table or whatever that says what the affixs are. Meth-Eth-prop-but-pent-hex-hept

The key i supose is find the longest chain of C's

So it's the horizontal one. So you have 5 which is PENT

So you have Pent so far. Next the double bond between 2nd=3rd carbon which means it's an ene instead of ane. We also have to show where the double bond is located. since it's between the 2nd and 3rd carbon on that chain we signify it.

SO, we've got 2-Pentene sofar.

Next that other little addition on the C, CH3 that's alone is called an Methyl and we also again have to signify where it's attached.

So the final name I think is 2-Methyl 2-Pentene
If it's some sort of test or something I suggest not using my answer; very unlikely correct.

For your second question I don't get what ur asking lol.

if your talking about the way they look.
Where one thing is like a small triangle looking bond and that means coming toward you. Dotted means away from you. then double line would mean it's flat with the page, and it's a double bond.

On the otherhand i don't know really what ur asking.
 
  • #3
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munky99999 said:
Well you should have a table or whatever that says what the affixs are. Meth-Eth-prop-but-pent-hex-hept

The key i supose is find the longest chain of C's

So it's the horizontal one. So you have 5 which is PENT

So you have Pent so far. Next the double bond between 2nd=3rd carbon which means it's an ene instead of ane. We also have to show where the double bond is located. since it's between the 2nd and 3rd carbon on that chain we signify it.

SO, we've got 2-Pentene sofar.

Next that other little addition on the C, CH3 that's alone is called an Methyl and we also again have to signify where it's attached.

So the final name I think is 2-Methyl 2-Pentene
If it's some sort of test or something I suggest not using my answer; very unlikely correct.

For your second question I don't get what ur asking lol.

if your talking about the way they look.
Where one thing is like a small triangle looking bond and that means coming toward you. Dotted means away from you. then double line would mean it's flat with the page, and it's a double bond.

On the otherhand i don't know really what ur asking.

note: the branched of are always under or below C.

ok, I have a few more here, which I'll attempt:

CH3
|
CH3 - CH2 - C - CH - CH3
| |
CH3 CH3

so I identify the longest chain of C..which is 5, so once againt its pent, but its pentane because there is only a single bond? correct?
Then we look at whats not on the straight line..we have CH3, three times...

so is it 3-methyl pentane?

another one:

Cl
|
CH3-C = C - CH2 - CH3
|
Cl

We have a chain of 5 once again, its pent, but its a double bond betweent C and C, so its pentene.
Finally, we have 2Cl's on the side.

So, its 2-Chloride Pentene?

Finally, I have

CH3-CH2-CtriplebondC-CH2-CH3

not sure about the last one..can someone help there.

if someone can look over this, I'd appreciate it, alot.
 
  • #4
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Watch yourself on the first example. Just because it looks like the carbons are branching off doesn't mean that can't be the central carbon backbone. The longest chain is a 7-C chain making it hept-. And according to the way you wrote it down, it has a double bond, making it an alkene, making your example a heptene.

Next I will label all of the carbons in the molecule

C
|
3 - 4 - 5 - 6 - 7
||
2 - 1

As you can see, the double bond is in the second bond space (always want to have the fewest numbers possible in terms of simplification). But you also have the [tex]CH_3[/tex] molecule on the third carbon in the chain. Therefore this should be:

3-methyl,2-pentene
 
  • #5
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bross7 said:
Watch yourself on the first example. Just because it looks like the carbons are branching off doesn't mean that can't be the central carbon backbone. The longest chain is a 7-C chain making it hept-. And according to the way you wrote it down, it has a double bond, making it an alkene, making your example a heptene.

Next I will label all of the carbons in the molecule

C
|
3 - 4 - 5 - 6 - 7
||
2 - 1

As you can see, the double bond is in the second bond space (always want to have the fewest numbers possible in terms of simplification). But you also have the [tex]CH_3[/tex] molecule on the third carbon in the chain. Therefore this should be:

3-methyl,2-pentene


did u see my note at the top..I said that all the branched off ones are under C..so none are on the 1st (the reason its on the first is because somehow when I post it they always move to the 1st position)
 
  • #6
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Starting off, isomers are molecules with the same molecular formula but different shape.

A structural isomer differs in the way the atoms are arranged.

For instance:

Take [tex]C_6H_1_4[/tex]

It could be written as:

| | | | | |
- C - C - C - C - C - C -
| | | | | |

OR
|
- C -
| | | | |
- C - C - C - C - C -
| | | | |

Both have 6 carbons and 14 hydrogens (I left the H's out for simplicity) but they are arranged differently. This is an example of a structural isomer.

For a geometric isomer you usually have something like a double bond which prevents rotation. The molecule will have the same bonding pattern, but because of the three dimensional shape of molecules they end of being different.

Take for example:
H Cl
\ /
C = C
/ \
Br Br
And

Br Cl
\ /
C = C
/ \
H Br

Both are [tex]C_2HBr_2Cl[/tex] and both have a carbon attached to a hydrogen and bromine; and a carbon attached to a bromine and a chlorine. But how can you make them the same...you can't unless you were able to rotate the double bond (you can't do this without breaking and reforming). This would be an example of geometrical isomerism.
 

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