đáp án bài tập phổ hữu cơ IR NMR

Cho đến nay các phương pháp phổ học được lựa chọn là: hồng ngoại, tử ngoại, khối lượng và cộng hưởng hạt nhân. Nhiệm vụ của phổ tử ngoại là phát hiện các hệ thống tiếp cách (conjugated), bởi vì sự kích thích các điện tử từ trạng thái cơ bản lên trạng thái kích thích của những hệ thống như thế mới gây nên sự hấp thu trong vùng tử ngoại. Nhưng sự phát triển hiện nay của các kĩ thuật cộng hưởng từ hạt nhân đã đạt đến mức có thể giảm tối đa vai trò của phổ tử ngoại. Hơn nữa, nhiều dao động của nhiều liên kết hoá học không hoạt động trong phổ hồng ngoại, mà hoạt động trong phổ Raman, do đó phổ học Raman được chọn bổ sung cho việc nghiên cứu dao động của các phân tử hữu cơ

Problem 1

C5H10O

MW 86 The band at 1716 indicates a carbonyl, probably a ketone The bands at 3000-2850 indicate C-H alkane stretches

Problem 1, NMR intrepreted

Problem2

C7H14O

MW 114 The band at 1718 indicates a carbonyl, probably a ketone The bands at 3000-2850 indicate C-H alkane stretches

Prob 2 NMR, interpreted

Problem 3

C4H10O

MW 74 The broad band at 3339 indicates O-H stretch, probably an alcohol The bands at 3000-2850 indicate C-H alkane

stretches The band at 1041 is C-O stretch, consistent with an alcohol

Prob 3 NMR, interpreted

Problem 4

C6H14O

MW 102 The broad band at 3350 indicates O-H stretch, probably an alcohol The bands at 3000-2850 indicate C-H alkane

stretches The bands from 1320-1000 indicate C-O stretch, consistent with an alcohol

Prob 4 NMR, interpreted

Note that the structure has a chiral center and the mixture is racemic Methylene protons adjacent to a chiral center

may not be identical

Problem 5

C4H8O2

MW 88 Prob 5, IR answer The band at 1743 indicates a carbonyl, probably a saturated aliphatic ester The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 could be due to C-O stretch, consistent with an ester

Prob 5 NMR, interpreted

Problem 6

C5H10O2

MW 102 Prob 6, IR answer The band at 1740 indicates a carbonyl, probably a saturated aliphatic ester The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 could be due to C-O stretch, consistent with an ester

Prob 6 NMR, interpreted

Problem 7

C5H10O

MW 86 Prob 7, IR answer The band at 1728 indicates a carbonyl, probably an aldehyde; an aldehyde is also suggested by the band at 2719 which is likely the C-H stretch of the H-C=O group The bands at 3000-2850 indicate C-H alkane stretches

Prob 7 NMR, interpreted

Problem 8

C8H8O

MW 120 Prob 8, IR answer The band at 1703 indicates a carbonyl; that the carbonyl is an aldehyde is suggested by the bands at 2828 and

2733 (C-H stretch of the H-C=O group) The bands at 3000-2850 indicate C-H alkane stretches The band

(unmarked on the graph below) just to the left of 3000 indicates aromatic C-H stretch Aromatics also show bands

in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane)

Problem 9

C9H10O2

MW 150 Prob 9, IR answer The band at 1697 indicates an alpha, beta-unsaturated carbonyl; that the carbonyl is an aldehyde is suggested by the bands at 2828 and 2739 (C-H stretch of the H-C=O group) The bands at 3000-2850 indicate C-H alkane stretches The band (unmarked on the graph below) just to the left of 3000 indicates aromatic C-H stretch Aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-

of-plane)

Prob 9 NMR, interpreted

Problem 10

C3H6O2

MW 74 Prob 10, IR answer The band at 1716 indicates a carbonyl The wide band from 3300-2500 is characteristic of the O-H stretch of carboxylic acids The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000

indicate the C-O stretch of carboxylic acids

Prob 10 NMR, interpreted

Problem 11

C11H14O2

MW 178 Prob 11, IR answer The band at 1684 indicates a carbonyl The wide band from 3300-2500 is characteristic of the O-H stretch of carboxylic acids The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 indicate the C-O stretch of carboxylic acids The band (unmarked on the graph below) just to the left of 3000 indicates aromatic C-H stretch Aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring

stretch), and 900-675 (C-H out-of-plane)

Prob 11 NMR, interpreted

Problem 12

C8H8O2

MW 136 Prob 12, IR answer The band at 1684 indicates a carbonyl The wide band from 3300-2500 is characteristic of the O-H stretch of carboxylic acids The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1320-1000 indicate the C-O stretch of carboxylic acids The band just to the left of 3000 (3065) indicates aromatic C-H stretch Aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675

(C-H out-of-plane)

Prob 12 NMR, interpreted

Problem 13

C7H9N

MW 114 Prob 13, IR answer The two bands at 3433 and 3354 indicate a secondary amine The bands at 3000-2850 indicate C-H alkane stretches The band at 3034 indicates aromatic C-H stretch; aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane) C-N stretch of aromatic amines would

show up at 1335-1250 (there is a band in that region)

Prob 13 NMR, interpreted

Problem 14

C5H13N

MW 87 Prob 14, IR answer The two bands at 3388 and 3292 indicate a secondary amine The bands at 3000-2850 indicate C-H alkane

stretches

Prob 14 NMR, interpreted

Problem 15

C8H14O

MW 126 Prob 15, IR answer The band at 1718 indicates a carbonyl, probably a ketone The bands at 3000-2850 indicate C-H alkane stretches Since the compound is an alkene, one would expect to see C=C stretch at 1680-1640; these weak bands are not

seen in this IR (according to Silverstein, "the C=C stretching mode of unconjugated alkenes usually shows

moderate to weka absorption at 1667-1640") Since the compound is an alkene, C-H stretch should appear above

3000 (not seen: the absorption for this single hydrogen must be too weak)

Prob 15 NMR, interpreted

Problem 16

C4H9Br

MW 137 Prob 16, IR answer The bands at 3000-2850 indicate C-H alkane stretches The bands in the region 1300-1150 could indicate C-H

wag (-CH2Br) of an alkyl halide

Prob 16 NMR, interpreted

Problem 17

C10H14O

MW 150 Prob 17, IR answer The bands at 3000-2850 indicate C-H alkane stretches; the small (unmarked) band just to the left of 3000

indicates aromatic C-H stretch Aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane) The wide band in the region 3500-3200 indicates the O-H stretch of an

alcohol or phenol

Prob 17 NMR, interpreted

Problem 18

C13H10O3

MW 214 Prob 18, IR answer The band at 1682 indicates a carbonyl, probably an ester The band at 3192 indicates C-H aromatic stretch; aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane) The wide band just to the left of the 3192 band indicates O-H stretch (alcohols and phenols)

Prob 18 NMR, interpreted

Note: the correct placement of the -OH on the ring and the arrangement of the ester may be beyond the scope of

your spectroscopy course See this explanation

Problem 18, structure

Problem 19

C8H14

MW 110 Prob 19, IR answer The bands at 3000-2850 indicate C-H alkane stretches There really aren't many other bands in the spectrum to indicate functional groups The compound is an alkyne; we would expect to see a carbon-carbon triple bond

stretch at 2260-2100, however, this is a weak band at best and often does not show up on IR

Prob 19 NMR, interpreted

Note: The integral values for A, B, and C need to be multiplied by 2 to get a total of 14 protons in the molecule

By convention, integral values on an nmr are reported as the lowest common multiple

Problem 20

C9H13N

MW 135 Prob 20, IR answer The bands at 3000-2850 indicate C-H alkane stretches The band at 3028 indicates C-H aromatic stretch;

aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H of-plane) The bands in the region 1250-1020 could be due to C-N stretch The weak, broad banc at about 3500 could be amine N-H stretch or it could be a slight contamination of an impurity (water) in the sample

out-Prob 20 NMR, interpreted

Problem 21

C9H10

MW 118 Prob 21, IR answer The bands at 3000-2850 indicate C-H alkane stretches The band at 3060 indicates C-H aromatic stretch;

aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H

out-of-plane)

Prob 21 NMR, interpreted

Look for absorption bands in decreasing order of importance:

1.the C-H absorption(s) between 3100 and 2850 cm-1 An absorption above 3000 cm-1 indicates C=C, either alkene or aromatic Confirm the aromatic ring by finding peaks at 1600 and 1500 cm-1 and C-H out-of-plane bending to give substitution patterns below 900 cm-1 Confirm alkenes with an absorption generally at 1640-1680 cm-1 C-H absorption between 3000 and 2850 cm-1 is due to aliphatic

hydrogens

2 the carbonyl (C=O) absorption between 1690-1760cm-1; this strong band indicates either an aldehyde, ketone, carboxylic acid, ester, amide, anhydride or acyl halide The an aldehyde may be confirmed with C-H absorption from 2840 to 2720 cm-1.

3.the O-H or N-H absorption between 3200 and 3600 cm-1 This indicates either an alcohol, N-H containing amine or amide, or carboxylic acid For -NH2 a doublet will be observed.

4 the C-O absorption between 1080 and 1300 cm-1 These peaks are normally rounded like the O-H and N-H peak in 3 and are prominent Carboxylic acids, esters, ethers, alcohols and anhydrides all containing this peak.

5 the CC and CN triple bond absorptions at 2100-2260 cm-1 are small but exposed.

7 structure of aromatic compounds may also be confirmed from the pattern of the weak overtone and combination tone bands found from

2000 to 1600 cm-1

b

c d

e

f

1H NMR

IR

4000 3000 2000 1500 1000 500

8.07 ppm (d, J=7.9 Hz, 2H) 7.32 ppm (d, J=7.9 Hz, 2H) 2.35 ppm (s, 3H)

C

N

C N

O

• para substituted

• 2.35 ppm: Methyl ketone or Me-aromatic

1H NMR

IR

4000 3000 2000 1500 1000 500

8.07 ppm (d, J=7.9 Hz, 2H) 7.32 ppm (d, J=7.9 Hz, 2H) 2.35 ppm (s, 3H)

C

N

C N

1H NMR

IR

4000 3000 2000 1500 1000 500

8.07 ppm (d, J=7.9 Hz, 2H) 7.32 ppm (d, J=7.9 Hz, 2H) 2.35 ppm (s, 3H)

1596

1512 1353

only one structure

3109

C N

C

N

C N

O OMe MeO O OMe

1H NMR

IR

4000 3000 2000 1500 1000 500

7.22 ppm (d, J=7.9 Hz, 2H) 7.45 ppm (d, J=7.9 Hz, 2H) 2.41 ppm (s, 3H)

2229

1609 1509

2 Circle the structure of the product that corresponds to the 1H NMR and IR data below Circle only one structure

C N

C

N

C N

O

• para substituted

• 2.41 ppm: Methyl ketone or Me-aromatic

1H NMR

IR

4000 3000 2000 1500 1000 500

7.22 ppm (d, J=7.9 Hz, 2H) 7.45 ppm (d, J=7.9 Hz, 2H) 2.41 ppm (s, 3H)

2229

1609 1509

2 Circle the structure of the product that corresponds to the 1H NMR and IR data below Circle only one structure

C N

C

N

C N

O

• para substituted

• 2.41 ppm: Methyl ketone or Me-aromatic

• IR: arylcyanide, not a ketone

1H NMR

IR

8.17 ppm (t, J=2.1 Hz, 1H) 7.96 (dt, J=8.0, 2.1 Hz, 1H) 7.67 (dt, J=7.7,2.1 Hz, 1H) 7.31 (dd, J=8.0, 7.7 Hz, 1H) 3.91 ppm (s, 3H)

1436 1571

3 Circle the structure of the product that corresponds to the 1H NMR and IR data below Circle only one structure

1729

C N

C

N

C N

O

R

R' H H

H

H

2.1 2.1

2.1

7.7

• meta substituted • 3.91 ppm: Methyl ester

1H NMR

IR

8.17 ppm (t, J=2.1 Hz, 1H) 7.96 (dt, J=8.0, 2.1 Hz, 1H) 7.67 (dt, J=7.7,2.1 Hz, 1H) 7.31 (dd, J=8.0, 7.7 Hz, 1H) 3.91 ppm (s, 3H)

1436 1571

3 Circle the structure of the product that corresponds to the 1H NMR and IR data below Circle only one structure

1729

C N

C

N

C N

O

R

R' H H

H

H

2.1 2.1

1H NMR

IR

8.17 ppm (t, J=2.1 Hz, 1H) 7.96 (dt, J=8.0, 2.1 Hz, 1H) 7.67 (dt, J=7.7,2.1 Hz, 1H) 7.31 (dd, J=8.0, 7.7 Hz, 1H) 3.91 ppm (s, 3H)

1436 1571

3 Circle the structure of the product that corresponds to the 1H NMR and IR data below Circle only one structure

1729

C N

C

N

C N

O

R

R' H H

H

H

2.1 2.1

D D

5 McLafferty rearrangements of the molecules depicted below will give rise to fragments that

can be detected by mass spectrometry Circle the fragments that are observed.

You may need to circle more than one answer for each! (24 points)

Name

OH

D D

OH H D

OH H H

OH H H

OH H H OH H H OH H H

OD H H

OD H H

The McLafferty Rearrangment:

Y

O

R H H

Y

OH

4000 3000 2000 1500 1000 500

1603 1585

Organic Spectroscopy 1 Michaelmas 2011

Lectures 7 and 8 – Worked Problems

Dr Rob Paton

robert.paton@chem.ox.ac.uk

http://paton.chem.ox.ac.uk

2

Recap of Lecture 6

IR Spectroscopy

Useful for identifying functional gropus present in organic molecules

The X-H region ( > 2500cm-1) contains diagnostic absorptions from N-H and O-H stretches On occasion C-H stretches may also be a useful diagnostic tool

O-H stretches are broadened by H-bonding: intermolecular and intramolecular H-bonding are distinguished by taking

spectra at different concentrations

The triple bond region (2000-2500 cm-1) contains diagnostic information from e.g nitriles and alkynes

The double bond region (1600-2000 cm-1) contains absorptions from C=C (albeit weakly, around 1650) and from C=O (strongly, around 1700)

Carbonyls are strengthened by a neighbouring EWG and weakened by a neighbouring EDG

Conjugation weakens carbonyls (approx minus 30cm-1)

Ring strain stiffens carbonyls (approx 40cm-1 for successively smaller rings)

3

Solving Structures

Double-bond equivalents (DBE):

If the molecular formula has been obtained from the mass spectrum, the number of double-bond equivalents (DBE) may be calculated If the molecule contains only C, H, N and O atoms and is neutral, then:

The DBE is the number of double bonds and rings in the molecule (it is useful to remember that benzene has a total of four double-bond equivalents: three C=C double-bonds and one ring)

The above formula works since (2 a + 2) is the number of hydrogens in a saturated hydrocarbon and so subtracting b , the actual number of hydrogens present and dividing by two gives the total number of double bonds and rings The number of divalent atoms (e.g O, S, etc) does not affect the DBE, but the number of mono- and trivalent atoms does All monovalent atoms (e.g F, Cl, Br, etc) count as hydrogens so should be added to b , whilst all trivalent atoms (e.g N, trivalent P, etc) count towards c

NH HN

N Cl

OMe

C

H2

H2C

CH3

H2C

CH3b

CH3b

δ 42 (d - CH2)

δ 16 (d - CH2)

δ 11 (u - CH3)

δ 27 (u - CH3)

δ 52(d - CH2)

δ 22(u - CH)

δ 23(u - CH3)

H2

H2C

CH3b

e

O H a

b

d

H2C C

H2

H2C d

c b

δ 45 (d - CH2)

δ 25 (d - CH2)

δ 23 (d - CH2)

δ 15 (u - CH3)

CH b3

a O

δ 36 (d - CH2)

δ 212 (absent - no CH) O

3.14 3.52

180 160 140 120 100 80 60 40 20ppm 0 200

f H

O

H

H H

b a

H O

H

CH3C

e

c O

H

a

CH3H

f

δ 14(u - CH3)

O

δ 205 (u - CH)

δ 24(d - CH2)

δ 12(u - CH3)

180 160 140 120 100 80 60 40 20ppm 0 200

O

δ 27 (u - CH3)

δ 41 (u - CH)

δ 18 (u - CH3)

δ 212 (absent - no CH)

δ 22(u - CH3)

δ 42(absent - no CH)

19

Solving Structures

1

H NMR spectrum (500 MHz): d 0.86 (6H, d, J= 6.6 Hz), 1.34 (3H, d, J=7.1 Hz), 1.81 (1H, 9lines, J=6.6 Hz), 2.41 (2H, d, J=6.6 Hz), 3.63 (1H, q, J = 7.1 Hz), 7.10 (2H, d, 8.1 Hz), 7.19 (2H, d, J = 8.1 Hz) 12.25 (1H, broad s)

1H NMR spectrum: ( 8 peaks, so 8 proton environments)

O

H h

OH

δ 129.8 (u - CH)

δ 19.4 (u - CH3)

δ 45.1 (d - CH2)

δ 23.0 (u - CH3)

CH3

CO2H

δ 176.3 (absent - no CH)

δ 128.0 (u - CH)

δ 139.35 (no CH) (no CH)δ 140.4

δ 30.5 (u - CH)

δ 45.15 (u - CH)