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5.5: Sequence Rules for Specifying Configuration

  • Page ID
    67097
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    Objectives

    After completing this section, you should be able to

    1. assign Cahn-Ingold-Prelog priorities to a given set of substituents.
    2. determine whether a given wedge-and-broken-line structure corresponds to an R or an S configuration, with or without the aid of molecular models.
    3. draw the wedge-and-broken-line structure for a compound, given its IUPAC name, complete with R or S designation.
    4. construct a stereochemically accurate model of a given enantiomer from either a wedge-and-broken-line structure or the IUPAC name of the compound, complete with R or S designation.
    Key Terms

    Make certain that you can define, and use in context, the key terms below.

    • absolute configuration
    • R configuration
    • S configuration
    Study Notes

    When designating a structure as R or S, you must ensure that the atom or group with the lowest priority is pointing away from you, the observer. The easiest way to show this is to use the wedge-and-broken-line representation. You can then immediately determine whether you are observing an R configuration or an S configuration.

    To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature.

    Introduction

    The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:

    1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
    2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.

    However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule,at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.

    Stereocenters are labeled R or S

    The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.

    Consider the first picture: a curved arrow is drawn from the highest priority (1) substituent to the lowest priority (4) substituent. If the arrow points in a counterclockwise direction (left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,(Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. For example, (R)-2-Bromobutane or (S)-2,3- Dihydroxypropanal.

     

    Sequence rules to assign priorities to substituents

    Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:

    Rule 1

    First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number.

    1. When dealing with isotopes, the atom with the higher atomic mass receives higher priority.
    2. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away.
    3. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.

    When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated.
    Remember that

    • Wedges indicate coming towards the viewer.
    • Dashes indicate pointing away from the viewer.

    Rule 2

    If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.

    If the chains are similar, proceed down the chain, until a point of difference.

    For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.

    The hydrogen ranks lower than the carbon bases on the first point of difference and their relative molecular weights.svg
    The "-H" (left) ranks lower than the "C-" (right) based on the relative molecular weights at the first point of difference.

     

    Example \(\PageIndex{1}\)

    example of ranking groups at first point of difference.svg

    Worked Exercise \(\PageIndex{1}\)

    For the following pairs of substituents, determine which would have the higher and lower priority based on the Cahn-Ingold-Prelog rules. Explain your answer.

    worked exercise 1.svg

    Answer

    A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:

    1-methylethyl vs ethyl with priorities labeled.svg
    The "C-" (right) ranks higher than the "H-" (left) based on
    the first point of difference and their relative atomic numbers.

    However:

    1-bromoethyl vs isopropyl with priorities labeled.svg
    In this case, even though the bold carbon on the right structure has two
    connections to a non-hydrogen atom (C), it is the lower priority.
    This is because one of the atoms attached to the bold carbon on the left molecule
    ranks higher than any of the atoms attached to the bold carbon on the right structure,
    since Br has a higher atomic number than C.

    Caution!!
    Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.

    worked exercise part 3 with priorities labeled.svg

    When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.

    worked exercise part 4 with priorities labeled.svg

    Rule 3

    If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to.

    • If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority
    • If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority.
    Example 5.5.2

    A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:

    The "C-" (right) ranks higher than the "H-" (left) based on the first point of difference and their relative molecular weights.

    However:

    In this case, even though the one on the right has two connections to an atom (C), it is the lower priority. This is because one of the atoms on the left molecule (Br) ranks higher (it has a bigger atomic number) than any of the atoms on the right.

    Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above.

     

    Caution!!
    Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.

    When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.

    After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S.

    1. Put the lowest priority substituent in the back (dashed line).
    2. Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3)
    3. Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise.

    i) If it is clockwise it is R.
    ii) if it is counterclockwise it is S.

    USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).

    IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.

    References

    1. Schore and Vollhardt. Organic Chemistry Structure and Function. New York:W.H. Freeman and Company, 2007.
    2. McMurry, John and Simanek, Eric. Fundamentals of Organic Chemistry. 6th Ed. Brooks Cole, 2006.

    Exercises

    Exercise \(\PageIndex{1}\)

    Orient the following so that the least priority (4) atom is paced behind, then assign stereochemistry ((R) or (S)).

    5.5qu.png

    Answer

    A = S  and B = R

    5.5sol.png

    Exercise \(\PageIndex{2}\)

    Draw (R)-2-bromobutan-2-ol

    Answer

    5.5(2).png

    Exercise \(\PageIndex{3}\)

    Assign R/S to the following molecule:

     

    5.5(3).png

    Answer

    The stereo center is (R).

    Exercise \(\PageIndex{4}\)

    Identify which substituent in the following sets has a higher ranking.

    1. -H or -CH3
    2. -CH2CH2CH3 or CH2CH3
    3. -CH2Cl or CH2OH
    Answer
    1. -CH3
    2. -CH2CH2CH3
    3. -CH2Cl
    Exercise \(\PageIndex{5}\)

    Identify which substituent in the following sets has a higher ranking.

    1. -NH2 or -N=NH
    2. -CH2CH2OH or -CH2OH
    3. -CH=CH2 or -CH2CH3
    Answer

    a) -N=NH

    b) -CH2OH

    c) -CH=CH2

    Exercise \(\PageIndex{6}\)

    Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)

    1. -NH2, -F, -Br, -CH3
    2. -SH, -NH2, -F, -H
    Answer

    a) -CH3 < -NH2 < -F, < -Br

    b) -H < -NH2 < -F, < -SH

    Exercise \(\PageIndex{7}\)

    Place the following sets of substituents in each group in order of lowest priority (1st) to highest priority (4th)

    1. -CH2CH3, -CN, -CH2CH2OH, -CH2CH2CH2OH
    2. -CH2NH2, -CH2SH, -C(CH3)3, -CN
    Answer

    a) -CH2CH3 < -CH2CH2CH2OH < -CH2CH2OH, < -CN

    b) -C(CH3)3 < -CH2NH2 < -CN < -CH2SH

    Exercise \(\PageIndex{8}\)

    Assign the following chiral centers as (R) or (S).

    5.6.5.svg

    Answer

    a) (S): I > Br > F > H. The lowest priority substituent is going backwards so following the highest priority, it goes left (counterclockwise).

    b) (R): Br > Cl > CH3 > H. Using a model kit, you need to rotate the H to the back position where the Br is. This causes the priority to go to the left (clockwise) when looking at it with the H in the back position. Alternatively, if you do not have a model kit, you can imagine the structure 3-dimensionally and since the lowest priority (H) is facing up (as drawn), if you look at it from below, starting with Br (1st priority) and moving towards Cl (2nd priority), you are moving right (clockwise) which represents (R) stereochemistry.

    c) Neither (R) or (S): Since there are two identical substituents (H’s) the molecule is achiral and cannot be assigned (R) or (S).

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