Unsaturation are implied by the molecular formula C8H10ClNO?

[useful_idiot]

How many elements of unsaturation are implied by the molecular formula C8H10ClNO? Draw Three structure, one containing ether and amine group, one containing amide group, one containing amine and ketone group.

I have no clue where to begin and read the chapter. I will be very thankful...

Regards
-James

 

[TriTertButoxy]

Hi, useful_idiot . This seems to be a schoolwork related question, and so it should have been posted in the appropriate section. Anyway, I haven't heard of the term "elements" of unsaturation, but degree of unsaturation is something of which I have heard. It is defined as the number of rings plus number of double bonds plus twice the number of triple bonds in the system.

Recall from late General Chemistry or Organic Chemistry lectures that for an organic molecule containing carbon, oxygen, nitrogen and halogens, the degree of unsaturation is given by

\text{D.O.U}=\frac{2n_\text{C}+2-n_\text{H}+n_\text{N}-n_\text{X}}{2},​

where n_\text{C}, n_\text{H}, and n_\text{N} are the number of carbon, hydrogen and nitrogen atoms in the molecule of interest, respectively, and where n_\text{X} is the total number of halogens. Notice that the expression for the degree of unsaturation does not involve the number of oxygen atoms.

(as a rather good exercise, try to rationalize the above expression based on the defenition of the degree of unsaturation)

So, for your molecule, \text{C}_8\text{H}_{10}\text{ClNO}, we have

n_\text{C}=8,
n_\text{H}=10,
n_\text{N}=1,\text{ and}
n_\text{X}=1,

So, a direct substitution of these quantities into the expression for the degree of unsaturation yields

\text{D.O.U}=\frac{2(8)+2-10+1-1}{2}=4.​

Therefore, we shuold expect a combination of 4 rings, double bonds, triple bonds. (Be sure to weight the triple bonds accordingly since each counts as two degrees of unsaturation).

Based on this, we shuold expect to see a myriad molecules with 4 degrees of unsaturation. See if you can come up with the molecules on your own. If you're stuck; I'd be glad to assist you. Take care.

 

[useful_idiot]

Thanks a lot mate! I had that question posted last night when I was doing the assignement. I kind of procrastinated. Finally I did find the formula posted above and found an answer. Thanks once again...