Compound A (C12H11NO2)
IR : 2225, 1740, 1645, 1600 and 1505 cm-1
2.05δ (3H, dd, J = 7.10, 1.80 Hz)
3.65δ (2H, s)
6.10δ (1H, dq, J = 15.50, 1.80 Hz)
6.50δ(2H, d, J = 7.80 Hz)
7.40δ (2H, d, J = 7.80 Hz)
7.65δ (1H, dq, J = 15.50, 7.10 Hz

also i have to get 2 isomers that will have the same ir spectrum . but ir cant distinguish isomers right?

SO I PREDICTED MY MOLECULE TO HAVE A NITRILE GROUP AND 1 BENZENE RING and its 1,4 di substituted WHEN I DREW IT OUT I AM NOT GETTING A 2 H SINGLET . can you help me predict this molecule please. im guessing the 2 h singlet will be with a nh2 but the molecule has a double bond equivalence of 8. help is greatly appreciated

Do you happen to have images of the spectra or just the peaks?

Okay, I think I figured it out. I would agree that you have a para substituted nitrile system. That takes care of 6 of your DBE's. You clearly have some other carbonyl, which you can hopefully figure out based on the IR carbonyl peak. It would be helpful to have a picture of the spectra though, because there are two carbonyl groups that the peak could correspond to, and the presence and appearance of another peak would let you rule the wrong one out straight away. At any rate, you are then left with 1 DBE.

 

Going through your NMR and looking at the coupling constants you should be able to figure out peaks from from carbons carbons that are connected. Your 2H singlet peak is quite a standout here, because it implies that the carbon those protons are bound to are not next door to any other carbon with protons. This leaves you with only two possible connections given what you already know. The other stand outs are the two 1H peaks at 6.10 and 7.65ppm. Their coupling suggests they are next door, but there is only 1 hydrogen involved on each carbon, leaving you with only one realistic possibility for what functional group they're in.

 

I'll leave that as a starting point for you work from. Let me know how you go.

hi thanks! i believe the carbonyl is an ester and we didnt get spectrum

with an alkene double bond. so a saturated ester.

the dq has to be bonded to a ch3 and h .

so where does this 2h singlet come in to play? is it between the ch3 and carbonyl

Edited November 7, 20169 yr by chemgirl101

I believe you are right about the ester, however I am confused as to why you think the ester is saturated. you are right about the dq being one of the alkene CH's bonded to the CH3 and other alkene CH, but that does beg the question of why you would then think the CH2 singlet is bonded between the ester and CH3? The CH2 is a singlet; it is bonded to carbons containing no hydrogens.

 

Also, I came up with 4 possible structures based on your OP. I don't think I could distinguish between them without looking them up, or looking a bit more into your aromatic NMR peaks.

i think its unsaturated because dbe is 8 the 3 pie bonds =3 nitrile =2 ring =1 and c=o= 1 and c=c is 1 total of 8. i think the ch2 is bonded between the ester and ring. made a mistake typing ch3

i meant unsaturated .not saturated sometimes i mix them up. alkenes are unsaturated !

Edited November 7, 20169 yr by chemgirl101

Yep, that's what I would think also. The question is whether the oxygen of the ester is on the CH2 end or the alkene end. I also considered that the CH2 might be between the ring and the nitrile. I don't know exactly how the shifts would change in that case, but perhaps you could look them up.