Aromatic Side Chain Assignment Abroad

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Opening Essay

Our modern society is based to a large degree on the chemicals we discuss in this chapter. Most are made from petroleum. In Chapter 7, we noted that alkanes—saturated hydrocarbons—have relatively few important chemical properties other than that they undergo combustion and react with halogens. Unsaturated hydrocarbons—hydrocarbons with double or triple bonds—on the other hand, are quite reactive.

In fact, they serve as building blocks for many familiar plastics—polyethylene, vinyl plastics, acrylics—and other important synthetic materials (e.g., alcohols, antifreeze, and detergents).

Figure 8.1 Common polymers made using alkene building blocks. Upper left, a stainless steel and ultra high molecular weight polyethylene hip replacement. The polyethylene repeating unit is shown in the lower left. Upper middle, shatterproof acrylic plexiglas used to build a large indoor aquarium. The methylacrylate repeating unit is shown in the lower middle. Upper right, common PCV piping used as material being used for sewage and drains. The polyvinylchloride repeating unit is shown in the lower left.

Hip replacement photo provided by:The Science Museum London / Science and Society Picture Library. Plexiglas aquarium photo provided by:Leonard G.PVC pipe installation photo provided by:Steve Tan.


Aromatic hydrocarbons are defined by having 6-membered ring structures with alternating double bonds (Fig 8.2).

Figure 8.2: Aromatic Hydrocarbons.Aromatic hydrocarbons contain the 6-membered benzene ring structure (A) that is characterized by alternating double bonds. Ultradur, PBT is a plastic polymer that contains an aromatic functional group. The repeating monomer of Ultradur is shown in (B). Ultradur can be found in showerheads, toothbrush bristles, plastic housing for fiber-optics cables, and in automobile exterior and interior components. Biologically important molecules, such as deoxyribonucleic acid, DNA (C) also contain an aromatic ring structures.


Thus, they have formulas that can be drawn as cyclic alkenes, making them  unsaturated.  However, due to the cyclic structure, the properties of aromatic rings are generally quite different, and they do not behave as typical alkenes. Aromatic compounds serve as the basis for many drugs, antiseptics, explosives, solvents, and plastics (e.g., polyesters and polystyrene).

The two simplest unsaturated compounds—ethylene (ethene) and acetylene (ethyne)—were once used as anesthetics and were introduced to the medical field in 1924. However, it was discovered that acetylene forms explosive mixtures with air, so its medical use was abandoned in 1925. Ethylene was thought to be safer, but it too was implicated in numerous lethal fires and explosions during anesthesia. Even so, it remained an important anesthetic into the 1960s, when it was replaced by nonflammable anesthetics such as halothane (CHBrClCF3).

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8.1 Alkene and Alkyne Overview

By definition, alkenes are hydrocarbons with one or more carbon–carbon double bonds (R2C=CR2), while alkynes are hydrocarbons with one or more carbon-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturatedhydrocarbons, which are defined as hydrocarbons having one or more multiple (double or triple) bonds between carbon atoms. As a result of the double or triple bond nature, alkenes and alkynes have fewer hydrogen atoms than comparable alkanes with the same number of carbon atoms. Mathematically, this can be indicated by the following general formulas:

In an alkene, the double bond is shared by the two carbon atoms and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious, ie the condensed formula for ethene is CH2CH2. The double or triple bond nature of a molecule is even more difficult to discern from the molecular formulas. Note that the molecular formula for ethene is C2H4, whereas that for ethyne is C2H2. Thus, until you become more familiar the language of organic chemistry, it is often most useful to draw out line or partially-condensed structures, as shown below:

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8.2 Properties of Alkenes

The physical properties of alkenes are similar to those of the alkanes. Table 8.1 shows that the boiling points of straight-chain alkenes increase with increasing molar mass, just as with alkanes. For molecules with the same number of carbon atoms and the same general shape, the boiling points usually differ only slightly, just as we would expect for substances whose molar mass differs by only 2 u (equivalent to two hydrogen atoms). Like other hydrocarbons, the alkenes are insoluble in water but soluble in organic solvents.

Some representative alkenes—their names, structures, and physical properties—are given in Table 8.1.

Table 8.1 Physical Properties of Some Selected Alkenes

The first two alkenes in Table 8.1 —ethene and propene, are most often called by their common names—ethylene and propylene, respectively. Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.

Figure 8.3. Ethene and Propene.The ball-and-spring models of ethene/ethylene (a) and propene/propylene (b) show their respective shapes, especially bond angles.

Looking Closer: Environmental Note

Alkenes occur widely in nature. Ripening fruits and vegetables give off ethylene, which triggers further ripening. Fruit processors artificially introduce ethylene to hasten the ripening process; exposure to as little as 0.1 mg of ethylene for 24 h can ripen 1 kg of tomatoes. Unfortunately, this process does not exactly duplicate the ripening process, and tomatoes picked green and treated this way don’t taste much like vine-ripened tomatoes fresh from the garden.

Other alkenes that occur in nature include 1-octene, a constituent of lemon oil, and octadecene (C18H36) found in fish liver. Dienes (two double bonds) and polyenes (three or more double bonds) are also common. Butadiene (CH2=CHCH=CH2) is found in coffee. Lycopene and the carotenes are isomeric polyenes (C40H56) that give the attractive red, orange, and yellow colors to watermelons, tomatoes, carrots, and other fruits and vegetables. Vitamin A, essential to good vision, is derived from a carotene. The world would be a much less colorful place without alkenes.

Figure 8.4 The bright red color of tomatoes is due to lycopene. 

Photo from : © Thinkstock; Lycopene structure from:Jeff Dahl


Concept Review Exercises
  1. Briefly describe the physical properties of alkenes. How do these properties compare to those of the alkanes?

  2. Without consulting tables, arrange the following alkenes in order of increasing boiling point:

Answers
  1. Alkenes have physical properties (low boiling points, insoluble in water) quite similar to those of their corresponding alkanes.

  2. ethene < propene < 1-butene < 1-hexene

Key Takeaway
  • The physical properties of alkenes are much like those of the alkanes: their boiling points increase with increasing molar mass, and they are insoluble in water.
Exercises
  1. Without referring to a table or other reference, predict which member of each pair has the higher boiling point.

    1. 1-pentene or 1-butene                                                                                                                                                  
    2. 3-heptene or 3-nonene                                                                                                           
  2. Which is a good solvent for cyclohexene? pentane or water?                                                          

Answer
    Concept Review Exercises
    1. Briefly identify the important distinctions between a saturated hydrocarbon and an unsaturated hydrocarbon.

    2. Briefly identify the important distinctions between an alkene and an alkane.

    3. Classify each compound as saturated or unsaturated. Identify each as an alkane, an alkene, or an alkyne.

      1. CH3CH2C≡CCH3
    Answers
    1. Unsaturated hydrocarbons have double or triple bonds and are quite reactive; saturated hydrocarbons have only single bonds and are rather unreactive.

    2. An alkene has a double bond; an alkane has single bonds only.

      1. saturated; alkane
      2. unsaturated; alkyne
      3. unsaturated; alkene
    Key Takeaway
    • Alkenes are hydrocarbons with a carbon-to-carbon double bond.

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    8.3 Alkynes

    The simplest alkyne—a hydrocarbon with carbon-to-carbon triple bond—has the molecular formula C2H2 and is known by its common name—acetylene (Fig 8.5). Its structure is H–C≡C–H.

    Figure 8.5 Ball-and-Spring Model of Acetylene. Acetylene (ethyne) is the simplest member of the alkyne family.

    Note

    Acetylene is used in oxyacetylene torches for cutting and welding metals. The flame from such a torch can be very hot. Most acetylene, however, is converted to chemical intermediates that are used to make vinyl and acrylic plastics, fibers, resins, and a variety of other products.

    Alkynes are similar to alkenes in both physical and chemical properties. For example, alkynes undergo many of the typical addition reactions of alkenes. The International Union of Pure and Applied Chemistry (IUPAC) names for alkynes parallel those of alkenes, except that the family ending is –yne rather than –ene. The IUPAC name for acetylene is ethyne. The names of other alkynes are illustrated in the following exercises.

    Concept Review Exercises
    1. Briefly identify the important differences between an alkene and an alkyne. How are they similar?

    2. The alkene (CH3)2CHCH2CH=CH2 is named 4-methyl-1-pentene. What is the name of (CH3)2CHCH2C≡CH?

    3. Do alkynes show cis-trans isomerism? Explain.

    Answers

    1. Alkenes have double bonds; alkynes have triple bonds. Both undergo addition reactions.

    2. No; a triply bonded carbon atom can form only one other bond. It would have to have two groups attached to show cis-trans isomerism.

    Key Takeaway
    • Alkynes are hydrocarbons with carbon-to-carbon triple bonds and properties much like those of alkenes.
    Exercises
    1. Draw the structure for each compound.

      1. acetylene
      2. 3-methyl-1-hexyne
    2. Draw the structure for each compound.

      1. 4-methyl-2-hexyne
      2. 3-octyne
    3. Name each alkyne.

      1. CH3CH2CH2C≡CH
      2. CH3CH2CH2C≡CCH3

    Answers

      1. H–C≡C–H

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    8.4 Aromatic Compounds: Benzene

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    Aromatic Side-chain Resonance Assignment

    Overall, aromatic side-chain assignment is very similar to aliphatic side-chain assignment, only one uses aromatic (H)CCH-COSY and aromatic 13C-resolved NOESY.

    Tyrosine residues are the easiest to assign, since they only have HD/CD and HE/CE spin pairs, and the chemical shifts of CD and CE are quite different. Histidine residues are relatively easy, while tryptophan residues have a very complex side chain. Phenylalanines can be especially bad, since all HD/CD, HE/CE and HZ/CZ spin pairs have very similar chemical shifts, causing so-called strong coupling effects in the 13C dimensions, and overlap of NOE peaks.

    Therefore, the recommended order would be Tyr -> His -> Trp -> Phe.

    For aromatic side chain assignment you would need 7 windows open at the same time:

    1. Open aliphatic 2D [13C, 1H]-HSQC with Open PolyScope and select aliphatic 13C-resolved NOESY for the strip display.
    2. Open aromatic 2D [13C, 1H]-HSQC with Open PolyScope and select aromatic 13C-resolved NOESY for the strip display.
    3. Open aromatic 3D (H)CCH-COSY with Open SystemScope (rotated).... In the rotation matrix pop-up window choose the following orientation: Hinept dimension must be Z*, *Ccosy must be Y*, and *Cinept must be *X*. Thus, the left panel of SystemScope will contain a C-C plane (with Ccosy dimension vertical), and the right panel will contain the H-C plane. This issue is addressed in detail in this FAQ question: http://www.cara.ethz.ch/Wiki/FAQ#III10
    4. Open 2D [15N, 1H]-HSQC with Open PolyScope and select 15N-resolved NOESY for the strip display.
    5. Open long-range 2D [15N, 1H]-HSQC with Open PolyScope (optional - for His assignment).
    6. Open 15N-resolved NOESY with Open SystemScope
    7. Open 13C-resolved NOESY with Open SystemScope

    Tyr

    For Tyr you have to assign HD/CD and HE/CD spin pairs.

    1. Display HB2/CB or HB3/CB strip in window 1.
    2. Pick an HD spin candidate - this should be a low-to-medium strength peak in both strips. Make sure that you don't mistake it for a long-range amide NOE peak. Note down the chemical shift of this HD spin.
    3. In the HSQC plane of window 2 find a peak at the chemical shift of HD. Since there are usually not that many aromatic residues in a given protein, it will likely be a unique choice. Place the cursor on the peak and pick the CD1 candidate: right-click and select Extend Vertically, then select the HD spin of the residue you are trying to assign.
    4. If the HD/CD1 pair is correctly assigned you should see HB2-HD and HB3-HD NOE peaks in the strip panels of window 2. Adjust the HD1/CD1 position to center the peaks in the strip panels.
    5. In window 3 display the HD/CD1 strip and use Show Depth. Pick the CE1 spin candidate.
    6. Select the CE1 spin and use right-click -> Show Orthogonal to display the orthogonal plane. In the orthogonal plane pick the HE spin.
    7. in window 2 verify and adjust HE/CE1 position. You should see a strong HD-HE NOE peak.

    Under normal conditions Tyr rings are almost always flipping fast enough to make HD1/HD2, HE1/HE2, CD1/CD2 and CE1/CE2 chemical shifts degenerate. Therefore, pseudoatom labels HD and HE should be used. The CD and CE labels are not compatible with CYANA, therefore, CD1 and CE1 should be used instead.

    Very rarely a stalled Tyr ring is encountered, which flips so slowly that the degeneracy breaks down. In this case, separate HD1/CD1, HD2/CD2, HE1/CE1 and HE2/CE2 spin pairs are observed.

    Ring flipping at an intermediate rate leads to line broadening - in the worst case no peaks are observed.

    Note that HD and HE can have very similar chemical shifts, even though CD and CE shifts are different.

    His

    For His you have to assign HD2/CD2 spins pairs. If you have a long-range 2D [15N, 1H]-HSQC spectrum, you can also determine protonation states and assign HE1/CE1 spin pairs.

    1. Display HB2/CB or HB3/CB strip in window 1.
    2. Pick an HD2 spin candidate - this should be a low-to-medium strength peak in both strips. Make sure that you don't mistake it for a long-range amide NOE peak. Note down the chemical shift of this HD2 spin.
    3. In the HSQC plane of window 2 find a peak at the chemical shift of HD2. Since there are usually not that many aromatic residues in a given protein, it will likely be a unique choice. Place the cursor on the peak and pick the CD2 candidate: right-click and select Extend Vertically, then select the HD2 spin of the residue you are trying to assign.
    4. If the HD2/CD2 pair is correctly assigned you should see HB2-HD2 and HB3-HD2 NOE peaks in the strip panels of window 2. Adjust the HD2/CD2 position to center the peaks in the strip panels.

    If you have recorded a long-range 2D [15N, 1H]-HSQC you can assign more spins in a histidine side-chain. (Picture needed).

    1. Note the chemical shift of the HD2 spin.
    2. In long-range 2D [15N, 1H]-HSQC (window 5) find the spin system that with the matching HD2 shift. Depending on the protonation state you will encounter one of the following:
    a. For a charged His form you should see two cross-peaks from HD2 at around 170-190 ppm. Use Extend Vertically to pick them as ND1 and NE2. a. For an ε-protonated neutral form you should see a cross-peak to NE2 at about 160-180 ppm. Use Extend Vertically to pick it. a. For a δ-protonated neutral form you should see a cross-peak to ND1 at about 200-260 ppm. Use Extend Vertically to pick it.
    1. Use Extend Horizontally to pick the HE1 spin
    2. Pick the missing ND1 or NE2 if it is a neutral form.
    3. Note the chemical shift of HE1.
    4. In aromatic 2D [13C, 1H]-HSQC plane (window 2) pick the CE1 spin using Extend Vertically.
    5. Adjust the position of the HE1/CE1 pair according to the aromatic 13C-resolved NOESY spectrum.

    Note that the His-tag gives rise to the strongest HD2/CD2 and HE1/CE1 in aromatic 2D [13C, 1H]-HSQC, which overlap closely.

    The HD1/ND1 and HE2/NE2 peaks are rarely observed in 2D [15N, 1H]-HSQC because the protons exchange with the solvent. If they are visible, then usually in the neutral forms. The proton chemical shifts are unique at about 11-12 ppm.

    Trp

    1. Display HB2/CB or HB3/CB strip in window 1.
    2. Pick an HD1 spin candidate - this should be a low-to-medium strength peak in both strips. Make sure that you don't mistake it for a long-range amide NOE peak. Note down the chemical shift of this HD2 spin.
    3. In the aromatic 2D [13C, 1H]-HSQC plane of window 2 find a peak at the chemical shift of HD1. Since there are usually not that many aromatic residues in a given protein, it will likely be a unique choice. In constant-time HSQC this will be on of the peaks in the middle with a different sign than most. Place the cursor on the peak and pick the CD1 candidate: right-click and select Extend Vertically, then select the HD1 spin of the Trp residue you are trying to assign.
    4. If the HD1/CD1 pair is correctly assigned you should see HB2-HD1 and HB3-HD1 NOE peaks in the strip panels of window 2. Adjust the HD1/CD1 position to center the peaks in the strip panels.
    5. In the same strip pick an HE1 candidate. This spin has a distinctive chemical shift of about 10 ppm.
    6. In window 6 select the H-N strip of the Trp residue. Select the HE1 spin, right-click and use Show Orthogonal to display the orthogonal plane. Pick the NE1 candidate in the orthogonal plane. Look for the HE1-HB2 and HE1-HB3 peaks at the NE1 position. (You can also pick NE1 in the 2D [15N, 1H]-HSQC plane of window 4).
    7. In window 4 adjust the position of the HE1/NE1 pair according to the 15N-resolved NOESY.
    8. In the strip of window panel pick an HZ2 candidate.
    9. Display the HD1/CD1 strip in window 7. Select the HZ2 spin, right-click and select Show Orthogonal from the pop-up menu. In the orthogonal plane pick the CZ2 spin.
    10. Adjust the HZ2/CZ2 position to center the peaks in the strip panels of window 2. The correct HZ2/CZ2 pair must exhibit the HZ2-HE1 peak.
    11. Display the HZ2/CZ2 strip in window 3 and show the depth plane. Pick a CH2 spin candidate.
    12. Select the CH2 spin, right-click and select Show Orthogonal. In the orthogonal plane pick the HH2 candidate.
    13. Adjust the HH2/CH2 position to center the peaks in the strip panels of window 2.
    14. Display the HH2/CH2 strip in window 3 and show the depth plane. Pick a CZ3 spin candidate.
    15. Select the CZ3 spin, right-click and select Show Orthogonal. In the orthogonal plane pick the HZ3 candidate.
    16. Adjust the HZ3/CZ3 position to center the peaks in the strip panels of window 2.
    17. Display the HZ3/CZ3 strip in window 3 and show the depth plane. Pick a CE3 spin candidate.
    18. Select the CE3 spin, right-click and select Show Orthogonal. In the orthogonal plane pick the HE3 candidate.
    19. Adjust the HE3/CE3 position to center the peaks in the strip panels of window 2. Verify that you see weak HE3-HB2 and HE3-HB3 peaks in the strip panels.

    Phe

    1. Display HB2/CB or HB3/CB strip in window 1.
    2. Pick an HD spin candidate - this should be a low-to-medium strength peak in both strips. Make sure that you don't mistake it for a long-range amide NOE peak. Note down the chemical shift of this HD2 spin.
    3. In the aromatic 2D [13C, 1H]-HSQC plane of window 2 find a peak at the chemical shift of HD1. Since there are usually not that many aromatic residues in a given protein, it will likely be a unique choice. Place the cursor on the peak and pick the CD1 candidate: right-click and select Extend Vertically, then select the HD1 spin of the Phe residue you are trying to assign.
    4. If the HD/CD1 pair is correctly assigned you should see HB2-HD1 and HB3-HD1 NOE peaks in the strip panels of window 2. Adjust the HD/CD1 position to center the peaks in the strip panels.
    5. Display the HD/CD1 strip in window 3 and show the depth plane. Pick a CE1 spin candidate.
    6. Select the CE1 spin, right-click and select Show Orthogonal. In the orthogonal plane pick the HE candidate.
    7. Adjust the HE/CE position to center the peaks in the strip panels of window 2.
    8. Display the HE/CE1 strip in window 3 and show the depth plane. Pick a CZ spin candidate.
    9. Select the CZ spin, right-click and select Show Orthogonal. In the orthogonal plane pick the HZ candidate.
    10. Adjust the HZ/CZ position to center the peaks in the strip panels of window 2.

    The above procedure works only for ideal case. Very often it would be difficult to pick 13C spins, since they will all have similar chemical shifts. In this case it may be easier picking 1H spins based on NOE peaks.


    The starting point is to assgin HD spins of Phe and Tyr and HD2 of His in aliphatic 13C-resolved NOESY from the cross-peaks to HB2/3. For Trp you have to follow weaker correlations HB2/3 -> HD2 -> HE1 -> HZ2 or HB2/3 -> HE3.

    Then directly attached 13C spins are assigned in aromatic [13C, 1H]-HSQC and aromatic 13C-resolved NOESY using PolyScope.

    The procedure to assign remaining spins is similar to that of aliphatic side-chain assignment. The LUA script to calculate chemical shifts is GFT_aroHCCH_calc.

    The assignment of other His spins is normally not pursued at this stage. The exchngeable HD1 and HE2 protons are not normally seen in NMR spectra. The HE1 spins can often be assigned during structure refinement based on preliminary structure. It is also possible to use HCN experiments (developed for RNA/DNA) or HMBC and long-lange HSQC experiments to assign HE1.

    Important! Normally CD/HD and CE/HE spin pairs of Tyr and Phe are completely degenerate due to fast ring flippping. They should ultimately be labeled as CD1/HD and CE1/HE, respectively. Even though in this case CD and CE are valid labels for Phe and Tyr in CARA (as 13C atom groups aka. pseudoatoms), they are not compatible with CYANA 2.1. Also see the related issue with CD and CG spins of Leu and Val.


    -- Main.AlexEletski - 20 Jul 2007

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