Answers to Assignment #5

Transcription

1 Answers to Assignment #5 A. 9 8 l 2 5 DBE (benzene + 1 DBE) ( 9 2(9) = 10 ˆ 5 DBE) nmr pattern of two doublets of equal integration at δ7.4 and 7.9 ppm means the group (the δ7.9 shift) IR band at 825 cm -1 confirms the para disubstituted benzene ring, EW That leaves 3 4 l 2 and 1 DBE: IR shows = stretch at <1700 cm -1, this must be a conjugated ketone as there is no 2750 cm -1 band to indicate an aldehyde, nmr pattern of two triplets at δ3.5 and 3.9 ppm indicate a fragment where both of the 2 are directly bonded to either a halogen or other electron withdrawing group l 2 2 l B DBE (1 benzene + 1 DBE) 1 nmr pattern of multiplet at δ6.9 and triplet at δ7.2 ppm with an integration ratio of 3:2, means the benzene is mono substituted with the substituents being an electron donating group (the δ6.9 shift) IR band confirmation of two bands just below 700 cm -1 is not obvious due to overlapping =- bends but we don t need it, ED- 6 5 That leaves and 1 DBE: IR shows broad - band at cm -1 and = stretch at 1710 cm -1, this must be a non-conjugated acid, - nmr pattern of two triplets at δ2.7 and 4.2 ppm indicate a fragment where one of the 2 is directly bonded to an oxygen (δ4.2) and the other 2 is bonded to a carbonyl (δ2.7) 2 2

2 DBE (benzene = 1 DBE) 1 nmr pattern of two doublets of equal integration at δ7.4 and 7.9 ppm means the group (the δ7.9 shift) IR band confirmation is not obvious but we don t need it, EW That leaves and 1 DBE: IR shows broad - band at cm -1 and = stretch at 1690 cm -1, this must be a conjugated acid, -Ar- nmr pattern of one singlet at δ1.3 with an integration ratio of 9 indicates a t-butyl group, ( 3 ) D l 5 DBE (benzene + 1 DBE) 1 nmr pattern of two doublets of equal integration at δ7.3 and 8.0 ppm means the group (the δ8.0 shift), EW That leaves 4 7 l and 1 DBE: IR shows a strong = band at 1770 cm -1, this must be a conjugated acid chloride, -Ar-l 1 nmr pattern of triplet-sextet-triplet at 1.0, 1.7, 2.7 ppm indicates a fragment bonded to a benzene ring, Ar- l 2 2 3

3 E DBE (benzene + 1 DBE) 1 nmr pattern of two doublets of equal integration at δ7.3 and 8.0 ppm means the group (the δ8.0 shift), EW That leaves and 1 DBE: IR shows a very broad band from cm -1 and a = band <1700 cm -1, this must be a conjugated acid, -Ar- 1 nmr pattern of doublet and septet at 1.3 and 3.0 ppm with ratios of 6:1 protons indicates an isopropyl fragment bonded to a benzene ring, ( 3 ) 2 -Ar- 3 3 F N 4 DBE (benzene) 1 nmr pattern of two doublets of equal integration at δ6.9 and 6.6 ppm means the benzene is para disubstituted with both of the substituents being electron donating groups (the shifts are < 7.2 ppm), ED ED That leaves 2 7 N and 0 DBE: IR shows two broad bands at ~3400 cm -1, this must be an amine, -N 2 1 nmr shows a broad singlet at δ3.5 ppm with an integration of 2 protons, this must be the amine, -N 2 1 nmr pattern of triplet and doublet at 1.2 and 2.5 ppm with ratios of 3:2 protons indicates an ethyl fragment bonded to a benzene ring, 3 2 -Ar- 2 N 2 3

4 G DBE (benzene + 1 DBE) 1 nmr pattern of two doublets of equal integration at δ7.0 and 7.8 ppm means the group (the δ7.8 shift) and one of the substituents being an electron donating group (the δ7.0 shift), EW ED That leaves and 1 DBE: IR shows a = stretch <1700 cm -1 and a =- stretch at 2750 cm -1, this must be an aldehyde, - nmr shows a singlet at δ9.9 ppm with an integration of 1 proton, this must be the aldehyde and it has no neighbors, - 1 nmr pattern of triplet and doublet at 1.4 and 4.1 ppm with ratios of 3:2 protons indicates an ethyl fragment bonded to an oxygen, DBE (benzene ring) 1 nmr pattern of two doublets of equal integration at δ6.8 and 7.1 ppm means the benzene is para disubstituted with one of the substituents being an electron donating group (the δ6.8 shift), ED That leaves 3 8 and 0 DBE: IR shows a broad - stretch at 3300 cm -1, this must be an alcohol, - 1 nmr shows a singlet at δ5.3 ppm with an integration of 1 proton, this must be the alcohol, - 1 nmr pattern of doublet and septet at 1.2 and 2.8 ppm with ratios of 6:1 protons indicates an isopropyl fragment bonded to a benzene ring, ( 3 ) 2 -Ar- 3 3

5 I. 4 7 Br 1 DBE We know from DBE that we have a double bond or a ring: 1 nmr shows two peaks > 5 ppm, each with an integration of 1 proton, so a 2 2 with a double bond is likely. Since there is no observed multiplicity this would have to be a gem-alkene 2 =< (geminal coupling of 2 z can t be observed without expanding the spectrum, cis coupling of ~8 z or trans coupling of ~16 z would be readily apparent). IR shows a strong band at 900 cm -1 confirming a 1,1-disubstituted alkene R 2 = 2 That leaves 2 5 Br and 0 DBE: nmr shows a singlet at δ3.9 ppm with an integration of 2 protons, this must be a 2 next to an electronegative substituent and no proton neighbors, - 2 Br 1 nmr shows a singlet at δ1.9 ppm with an integration of 3 protons, this must be a 3 with no proton neighbors and could be next to a double bond to explain the slightly increased chemical shift (compared to 3 at 1.0 ppm), -= Br J. 5 7 N 3 DBE We know from DBE that we have three double bonds, one triple bond + one double bond, two double bonds + a ring, or one triple bond + a ring: 1 nmr shows two peaks > 5 ppm, each with an integration of 1 proton, so a 2 2 with a double bond is likely. The observed multiplicity is complex, so we can rule out a gem-alkene but can t see easily if the alkene protons are cis or trans. IR shows a strong band at 970 cm -1 confirming a trans-disubstituted alkene R=R IR shows a strong band at 2250 cm -1 indicating a nitrile, -/N which accounts for 2 DBE for all 3 DBE are now accounted for. That leaves 2 5 and 0 DBE: nmr shows a doublet at δ3.1 ppm with an integration of 2 protons, this must be a 2 next to and the downfield shift must be due to being adjacent to both multiple bonds, N/- 2 -= nmr shows a doublet at δ1.8 ppm with an integration of 3 protons, this must be a 3 with one proton neighbors and could be next to a double bond to explain the slightly increased chemical shift (compared to 3 at 1.0 ppm), -= N

6 K DBE We know from DBE that we have two double bonds, one triple bond, or one double bonds + a ring: 1 nmr shows no peaks > 5 ppm, so a double bond is unlikely unless it was fully substituted. IR shows a weak band at 2100 cm -1 indicating an alkyne, -/- which accounts for both DBE are now accounted for. IR shows a strong band at 3300 cm -1 indicating the alkyne is terminal, -/ That leaves 4 9 and 0 DBE: IR shows a broad - stretch at 3400 cm -1, this must be an alcohol, - nmr shows a singlet at δ2.4 ppm which exchanges with D 2 and has an integration of 1 proton, this must be the alcohol, - 1 nmr pattern of triplet and doublet at 1.0 and 1.7 ppm with ratios of 3:2 protons indicates an ethyl fragment, nmr shows a singlet at δ1.5 ppm with an integration of 3 protons, this must be a 3 with no proton neighbors and could be two bonds away from the electronegative oxygen atom to explain the slightly increased chemical shift, nmr shows a singlet at δ2.3 ppm which has an integration of 1 proton, this must be the terminal alkyne, -/ L N 2 1 DBE We know from DBE that we have one double bond or a ring: 1 nmr shows no peaks > 5 ppm, so a carbon double bond is unlikely unless it was fully substituted. IR shows a strong = band at 1650 cm -1, this must be a conjugated amide, but since there is not another double bond to conjugate with that means the conjugation must be to a lone pair on another nitrogen, >N:--NR 2 IR shows two broad bands at ~3400 cm -1, this could be from one 1 o amine, -N 2, or two 2 o amines, >N nmr shows a broad singlet at δ4.9 ppm which exchanges with D 2 and has an integration of 2 protons, this could be an -N 2 or two equivalent >N groups. That leaves 4 10 and 0 DBE: nmr pattern of triplet and doublet at 1.2 and 3.2 ppm with ratios of 6:4 protons indicates two equivalent ethyl fragments next to an electronegative atom, 3 2 -N:- 2 N N or 3 2 N N 2 3

7 M. 4 7 Br 1 DBE We know from DBE that we have one double bond or a ring: 1 nmr shows two doublets > 5 ppm with integration of one proton each and coupling of ~5-7 z, so a cis disubstituted alkene is likely, -=- IR shows a strong band at 700 cm -1 confirming the cis alkene That leaves 2 5 Br and 0 DBE: nmr pattern of triplet and doublet at 1.3 and 4.0 ppm with ratios of 3:2 protons indicates an ethyl fragment next to an electronegative atom conjugated to a double bond for the chemical shift of the quartet to be as high as 4 ppm, = Br 2 3 N DBE (benzene + 1 DBE) We know from DBE that we have a benzene plus another double bond or ring: 1 nmr integration of 4 in the aromatic region shows the benzene is disubstituted 1 nmr pattern of a doublets and a triplet, each with an integration of one, means that there must be at least three adjacent protons on the benzene ring. The other two protons have overlapping signals so the multiplicity can t be used to distinguish ortho from meta. 1 nmr shows a singlet at δ10.5 ppm with an integration of 1 proton, this must be the aldehyde and it has no neighbors, - 1 nmr doublet at δ7.8 means the proton is ortho to an electron withdrawing group, which must be the aldehyde. That leaves 2 5 and 0 DBE: 1 nmr pattern of a quartet at δ4.1 ppm with an integration of 2 and a triplet at δ1.5 ppm with an integration of 3 must be due to an ethoxy group on the benzene ring, Ar the options are now ortho- or meta-substituted -Ar- 2 3 so we need to calculate the shifts of both options to determine that the ortho isomer is the best match to the observed δ7.8d (1) 7.5t (1) and overlapping signals at 7.0 (2) in the aromatic region. 7.43t 7.01d d s 7.06t 7.75d correct answer 7.37t 7.44d chemical shifts do not match

8 DBE (benzene + 1 DBE) We know from DBE that we have a benzene plus another double bond or ring: 1 nmr integration of 4 in the aromatic region (δ ppm) shows the benzene is disubstituted nmr pattern at δ shows overlapping multiplets and a singlet, this means that there must be at least one proton with no ortho neighbors on the ring so the disubstitution pattern must be meta. nmr pattern of δ6.7dd (1), 5.7d (1), 5.2d (1) is diagnostic of a mono-substituted alkene, 2 =-, and the shifts confirm that the alkene is bonded to the benzene ring. That leaves 3 and 0 DBE: nmr pattern of a singlet at δ2.3 ppm with an integration of 3 must be due to a methyl group on a double bond, -Ar-- 3 or =- 3 -Ar-- 3 is the only real option because if the terminal methyl was bonded to the terminal mono-sub alkene, 3 -= 2, we would have nowhere left to connect the Ar. d (J=16 z) d (J=8 z) d t d s dd (J=16, 8 z) 3