Asked by k
what is this compound, it has a mf of c6h12o i. imgur. com/gCgw0Hi. png
Answers
Answered by
Devron
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.
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.
Answered by
Devron
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.
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.
Answered by
k
there is only 1 degree of unsaturation
Answered by
Devron
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.
Answered by
Devron
D of Unsaturation =[(2 +(2 x number of carbons(6))-12]=2
Answered by
k
you divide that by 2 so it's 1
Answered by
Devron
your correct.
Answered by
Devron
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.
2,3-dimethyl-cyclopropanol. This would account for the splitting problem, but since I am not sure I wouldn't bet money on it.
Answered by
Devron
This gave me a headache. I'm sorry, 1,2,- trimethyl-cyclopropanol.
Answered by
Devron
1,23,- trimethyl-cyclopropanol.
Answered by
Devron
This is bugging me, because I'm not sure. the peaks for d and e should show splitting, but they don't. The only structure that I can think of is 2,3,3-trimethyl-cyclopropanol. Hopefully someone else comes along and is of some assistance. Post the right answer if you can; I am very curious about this one.
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