how would I draw this compound?
2-bromo-6-chloro-4,4-dimethylheptane
I've tried it again and again but I can't seem to get the Cl on the 6th carbon while still having a 6 carbon chain of heptane
I drew it
so that
it was
CH3 CBr CH2 CCH3CH3 CH2 Cl
I just can't figure it out..can someone help me?
Thanks
heptane is a seven chain...
I know but the problem is when I have to fit the Cl on the 6th carbon...but I can't seem to do that without putting in too much carbons.
Take your structure in the first post and simply add CH3 at the end (substitute CH3 for one of the Hs on the CH2Cl. I THINK your problem is you are counting one of the dimethyl groups as one of the carbons in the chain for heptane.
CH3
|
CHBr
|
CH2
|
C(CH3)2
|
CH2
|
CHCl
|
CH3
Check out my try and see if it has any flaws.
To draw the compound 2-bromo-6-chloro-4,4-dimethylheptane, you will need to consider the location of the bromine (Br) and chlorine (Cl) atoms, as well as the arrangement of the methyl (CH3) groups and the heptane chain.
Here's how you can properly draw it:
1. Start with a straight line representing the heptane chain, which consists of 7 carbon atoms (C).
2. Number the carbons in the chain from left to right, starting with 1 at the end closest to the bromine (Br) atom.
3. Place the bromine (Br) atom on the second carbon (C2) of the heptane chain. Draw a bond connecting the bromine (Br) atom to the second carbon (C2).
4. For the chlorine (Cl) atom, it should be placed on the sixth carbon (C6) of the heptane chain. Draw a bond connecting the chlorine (Cl) atom to the sixth carbon (C6).
5. Add the methyl (CH3) groups. There are two methyl groups, and they are both attached to the fourth carbon (C4) of the heptane chain. Draw two bonds connecting the fourth carbon (C4) to the methyl (CH3) groups.
Now, the final structure of 2-bromo-6-chloro-4,4-dimethylheptane should look like this:
CH3 CH3
| |
CH3 - C - C - C - C - C - Br
| |
Cl CH3
Make sure to double-check your drawing to ensure that the bonds and atoms are properly placed.