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5.7: 2D NMR Solutions

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    195117

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    Exercise 5.1.1:

    ethyl butanoate

    Exercise 5.1.2:

    clipboard_e5bce9169fe35edfe2e742607479c4ac2.png

    Exercise 5.1.3:

    clipboard_e0a55e4f8927b4a344906679baf3724ce.png

    Exercise 5.1.4:

    clipboard_e5c8eb5872217c6a37dc54a46a1cece06.png

    Exercise 5.1.5:

    clipboard_e131a2c6553392448a37c52cefad9ca39.png

    Exercise 5.1.6:

    clipboard_e91cb0902ded3fe6d6016a6e90904cc73.png

    Exercise 5.1.7:

    clipboard_e4f185c633376c33f11be9fb4b624d8e8.png

    Exercise 5.1.8:

    clipboard_e149839961c6b6f8ece1e8aa92a9dff6d.png

    Exercise 5.1.9:

    clipboard_ed732c46ac92279d066de4afabb32c67f.png

    Exercise 5.1.10:

    clipboard_e159a77b7ed8e4aed8fae4e198e327ff4.png

    Exercise 5.1.11:

    clipboard_eb49fb4ba8680407a9c48b7bf17832431.png

    Exercise 5.1.12:

    clipboard_eb8ed32490193c2bd15e66c52f88928f7.png

    Exercise 5.2.1:

    clipboard_eb683178c9efdf33e94ccc8276a2a1b34.png

    Exericse 5.2.2:

    clipboard_ec87334451597a5f3089c2a4f5086c9dc.png

    Exercise 5.2.3:

    clipboard_e7170d3bfbddde006e0ac8c617189d6cb.png

    Exercise 5.3.1:

    clipboard_e79fa11dcffa2c55970907873efe1490c.png

    clipboard_e9a3d68a590edfadf68a488773ab625f0.png

    Exercise 5.3.2:

    clipboard_ed2167ce2fd4aea3b41e8562dc586dfd5.png

    Exercise 5.3.3:

    clipboard_e4ab16a94d63a6a7a66683336ede0bcbc.png

    Exercise 5.3.4:

    clipboard_ebfe4df4eb6bcb9b6e8b97c407575a0c5.png

    Exercise 5.3.5:

    clipboard_ec7429761ebe51660bfd28c935a19e0c5.png

    Exercise 5.3.6:

    clipboard_e983813900870425b62351af85842affd.png

    Exercise 5.3.7:

    clipboard_e6a3a6762ea975520602039c70a28feeb.png

    Exercise 5.3.8:

    clipboard_ec90f22dbec0a4036733e2afa70748526.png

    Exercise 5.4.1:

    clipboard_e115f65a1954962d47b7b4724beb1b062.png

    Exercise 5.4.2:

    clipboard_eb09b142af6ada34bf442f48c13d3641a.png

    Exercise 5.4.3:

    a) Due to resonance, there is substantial pi character to the amide bond which restricts free rotation around that bond.

    clipboard_e8f9f0c13c066f1b0e37eca9333383e38.png

    b)

    clipboard_e24040690bca999328393858d7f95883f.png

    c)

    clipboard_eabd8ea4824122a3821133017f70cfc9b.png

    Exercise 5.4.4:

    clipboard_e5bb2d7eeed288196625b44fe1c4740f1.png

    Exercise 5.4.5:

    clipboard_e677749b87acd5f607caec90faf92609e.png

    Exercise 5.4.6:

    a)

    clipboard_e1132907af51226d308a2550626a3c263.png

    b)

    clipboard_ee861b4de6f1bf97eb06b1810c40a8eab.png

    c)

    clipboard_e9a5222cf3a2ca0b4a13f3b09b209f783.png

    Exercise 5.4.7:

    clipboard_ea80cf78e7ce9bdf8121dc2382f8ff6e8.png

    Exercise 5.5.1:

    clipboard_ed08238390a0cc40d72875de6ae309ecb.png

    Exercise 5.5.2:

    clipboard_eea8326bb672e0338f44462196d908d14.png

    Exercise 5.5.3:

    clipboard_eb1c060cb0ad01773cfcd5a25a53c6986.png

    Exercise 5.5.4:

    clipboard_e1df5210bbcd87ccfae21bf93aab4f492.png

    Exercise 5.5.5:

    clipboard_e8ff99c14c1b8e042e9bf494e8d12d983.png

    Exercise 5.6.1:

    1H NMR

    Chemical shift (ppm) Integration Multiplicity Partial Structure
    5.54 2H multiplet CH=C (x 2)
    4.17 2H doublet O-CH2-CH
    2.08 2H quintet CH-CH2-CH3
    1.47 1H boad singlet OH
    0.95 3H triplet CH2-CH3

    * total # H: 10

    13C NMR

    Chemical shift (ppm)

    		 Type of carbon
    134 sp2
    128

    sp2
    58 sp3-O
    21 sp3
    14 sp3

    *total # C: 5

    COSY

    Assignment 1H COSY
    A 0.95 2.08
    B 2.08 0.95, 5.54
    C 4.17 5.54
    D 5.54 2.08
    E 5.54 4.17

    *HMQC indicates two hydrogens at 5.54 are in two different environments

    HMQC

    Assignment 13C 1H
    A 14 0.95
    B 21 2.08
    C 58 4.17
    D 128 5.54
    E 134 5.54

    Formula:

    C5H10O (1 O indicated from shift in 13C, 1H NMR)

    FW = \(5 \times 12) + (10 \times 1) + (1 \times 16) = 86\)

    Compare C5H10 ratio to C5H12 in hydrocarbon

    Degrees of unsaturation = \(\frac{(2 \times 5) + 2 - 10}{2} = 1 unit (1 double bond)

    The data tables should be consitent with this structure:

    pent-2-en-1-ol (could be cis or trans based on this analysis)

    Exercise 5.6.2:

    1H NMR:

    Chemical shift (ppm) Integration Multiplicity Partial structure
    4.7 5H singlet solvent
    3.93 1H triplet CH2-CH-N
    2.40 2H multiplet CH2-CH2?
    2.09 2H multiplet CH2-CH2?
    1.4 9H singlet C(CH3)3

    *Total number of H: 19 H

    13C NMR:

    Chemical shift (ppm) Type of carbon
    170 sp2 (C=O)
    80 sp3 (C-O)
    52 sp3 (C-N)
    32 sp3
    28 sp3
    26 sp3

    *Total number of C: 6 apparent, but two more suggested by symmetry (3 methyl groups in 1H NMR) for 8 C; a third extra suggested by MW fit for 9 C

    COSY:

    Assignment 1H COSY
    Solvent 4.7 --
    B 3.93 2.40
    D 2.40 2.09
    C 2.09 2.40, 2.09
    A 1.4 --

    Formula:

    C9H18O3N2 (extra O indicated from shift in 13C, 1H NMR; second O suggested by C=O in 13C NMR; additional CO needed to fit MW)

    FW = \((9 \times 12) + (18 \times 1) + (3 \times 16) + (2 \times 14) = 202\)

    FW = ((9 \times 12) + (18 \times 1) + (3 \times 16) + (2 \times 14) = 202\)

    Compare C9H18 to C9H22 for the corresponding hydrocarbon corrected for two nitrogens (therefore two extra hydrogens)

    Degrees of unsaturation = \(\frac{(2 \times 9) + 2 + 2 - 10}{2} = 2\) units (2 double bonds)

    The data tables should be consistent with this structure:

    clipboard_ead408645f2f062bc2fa72c1030ec8294.png

    Exercise 5.6.3:

    The data should be consistent with this structure:

    clipboard_ece9c5ea821dd696fe8402a6b611a9260.png

    Exercise 5.6.4:

    1H NMR:

    Chemical shift (ppm)

    		 Integration Multiplicity Partial Structure
    4.71 -- singlet solvent
    4.17 1H doublet? CO-CH-N
    3.75 3H singlet O-CH3
    2.25 1H multiplet CH-CH-(CH3)2
    0.92 6H triplet? 2 x CH3

    13C NMR:

    Chemical shift (ppm) Type of carbon
    170 sp2 C=O
    60 sp3 C-N
    52 sp3 C-O
    30 sp3 C
    19 sp3 C

    COSY:

    Assignment 1H COSY
    A 4.17 2.25
    B 3.75 --
    C 2.25 4.17
    D 0.92 2.25

    Formula:

    C6H13O2N (1 O indicated from shift in 13C, 1H NMR)

    FW = \((6 \times 120 + (13 \times 1) + (2 \times 16) + (1 \times 14) = 131\)

    Degrees of unsaturation = \(\frac{(2 \times 6) + 2 + 1 - 13}{2} = 1\) unit (1 double bond)

    The data should be consistent with this structure:

    clipboard_ed8329eeae8d4515423ef3e4828369129.png

    Exercise 5.6.5:

    The data should be consistent with this structure:

    clipboard_e94eb1b86f7320e2ed8c8b25b77b559fe.png

    Exercise 5.6.6:

    The data should be consistent with this structure:

    clipboard_e61a6b7aa2d4738f631bde31840f44066.png

    Exercise 5.6.7:

    The data should be consistent with this structure:

    clipboard_e5d98b582ca345f3d0767e7b6de1c3297.png

    Exercise 5.6.8:

    The data should be consistent with this structure:

    clipboard_e2f638d4b318bc1bb740dd38cfe4e09bc.png

    Exercise 5.6.9:

    The data should be consistent with this structure:

    clipboard_e9bfbabdd794d8e1f2095ccbca8ad834c.png

    Exercise 5.6.10:

    The data should be consistent with this structure:

    clipboard_ee6134fbc46addb5e4d88525de6b9109f.png

    Exercise 5.6.11:

    The data should be consistent with this structure:

    clipboard_e97eab53cd7a5d4700a2f147ad5418308.png

    Exercise 5.6.12:

    The data should be consistent with this structure:

    clipboard_ee218fdeeb8b8fb51af16489d140aad2f.png

    Exercise 5.6.13:

    1H NMR:

    Chemical shift (ppm) Integration Multiplicity Partial structure
    5.7 1H singlet C=CH-CO
    2.77 1H multiplet CH2-CH-CO
    2.62 1H multiplet C-CH-C
    2.39 1H multiplet C-CH-C
    2.05 1H doublet? C-CH-CH?
    1.96 3H singlet C-CH3
    1.48 3H singlet C-CH3
    0.98 3H singlet C-CH3

    13C NMR:

    Chemical shft (ppm) Type of carbon
    204 sp2 C=O
    170 sp2
    121 sp2
    59 sp3
    55 sp3
    50 sp3
    41 sp3
    28 sp3
    24 sp3
    22 sp3

    COSY:

    Assignment 1H COSY
    1 2.39 2.77, 2.62?
    3 5.7 2.05?
    5 2.62 2.62?
    7a 2.77 2.05
    7b 2.05 2.77
    8 1.48 --
    9 0.98 --
    10 1.96 --

    Formula:

    C10H14O (1 O indicated from shift in 13C, 1H NMR)

    FW = \((10 \times 12) + (14 \times 1) + (1 \times 16) = 150 \)

    Degrees of unsaturation = \(\frac{(2 \times 10) + 2 - 14}{2} = 4 units (e.g. 2 rings, 2 double bonds)

    The data should be consistent with this structure:

    clipboard_ee27a8905599737f54cd81e4f0bed4801.png

    Exercise 5.6.14:

    The data should be consistent with this structure:

    clipboard_e458251c20f61c1f7000b2fdfc6c83d93.png

    Exercise 5.6.15:

    The data should be consistent with this structure:

    clipboard_ef584d1112d799ff07a6f73c068271dde.png

    This page titled 5.7: 2D NMR Solutions is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Chris Schaller via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.