what is this compound, it has a mf of c6h12o i. imgur. com/gCgw0Hi. png
The IR shows a bird wing at about 3500, so we know we have an OH group, and the degree of unsaturation is 2, so we have 2 double bonds or a triple bond.
I am no where near positive, but I believe that the answer is
H
I
CH3-C=C
I I
CH3-C=C CH3
I I
H OH
This accounts for the splitting patterns that I see in H NMR, but some of the placements of the peaks kind of bug me, so I am not sure. I haven't done this in a while, so hopefully someone else comes along and gives their input. Hope this helps.
The IR shows a bird wing at about 3500, so we know we have an OH group, and the degree of unsaturation is 2, so we have 2 double bonds or a triple bond.
I am no where near positive, but I believe that the answer is
CH3
CH3--CH=C--C=C--CH3
OH H
This accounts for the splitting patterns that I see in H NMR, but some of the placements of the peaks kind of bug me, so I am not sure. I haven't done this in a while, so hopefully someone else comes along and gives their input. Hope this helps.
there is only 1 degree of unsaturation
The CH3, H, and OH group are connected to the structure I tried to type in the middle of this mess. The top CH3 goes with the fourth carbon (a covalent bond) from the left, the OH goes with the third carbon from the left, and the hydrogen goes with the fifth carbon from the left.
D of Unsaturation =[(2 +(2 x number of carbons(6))-12]=2
you divide that by 2 so it's 1
your correct.
Like I said I'm not sure, but it looks like this may be
2,3-dimethyl-cyclopropanol. This would account for the splitting problem, but since I am not sure I wouldn't bet money on it.