GLOBAL KNOWLEDGE ARK

Chemistry · Lesson

Conformation, rings and molecular flexibility

Follow bond rotations and ring motions while retaining atom identity. Compare individual conformational basins with their combined populations under one explicitly invented equilibrium model.

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By the end, you can

  • State the viewing direction and dihedral used for a Newman comparison.
  • Keep axial/equatorial positions distinct from up/down configuration.
  • Normalize a three-basin population model with its multiplicity counted once.
  • Avoid inferring a rate from equilibrium populations alone.
01

The same connectivity can have many spatial conformations

Conformation describes spatial arrangements accessible by motions such as rotation about selected single bonds or ring deformation. The connectivity can stay fixed while distances between nonbonded groups change. A continuous torsion angle has infinitely many positions; a list of named low-energy basins is not a list of every possible geometry.

For a Newman view, explicitly say which bond you look along and which atom is nearer. For butane labeled a1–a2–a3–a4, look from a2 toward a3 and define the a1–a2–a3–a4 dihedral. In ideal schematic projections, staggered methyl relationships occur at 60°, 180° and 300°: the 180° relationship is anti and the two others gauche. Eclipsed reference positions occur at 0°, 120° and 240°. A complete 360° turn returns the same geometry.

02

A population difference and an interconversion barrier differ

Steric contacts, torsional effects, electronic interactions and the environment influence conformational energetics. A minimum is a local feature of a potential or free-energy landscape under its stated definition. Comparing equilibrium basins needs their free energies or compatible statistical weights, including the internal configurations represented by each basin.

A barrier controls possible dynamical pathways; a free-energy difference constrains an equilibrium ratio. The ratio does not supply the forward and reverse time scales. A gas-phase electronic-energy difference, a solution free-energy difference and a measured equilibrium ratio also need not be numerically identical. The illustrative population model below therefore declares its weights and does not borrow an unmeasured reaction barrier.

03

Normalize individual basins before adding a family

Consider one invented equilibrium model with a lower basin L and two equivalent higher basins H₁ and H₂. Each higher basin has stipulated enthalpy difference ΔH=2.00 kJ mol⁻¹, independent of temperature over 200–400 K. Internal statistical weights and effective basin volumes are equal, with no other entropy difference. Here R is the molar gas constant, 8.314462618 J mol⁻¹ K⁻¹, and T is absolute temperature.

The relative weight of each higher basin is w=exp[−2000/(RT)], so the normalizer is Q=1+2w. Fractions are p_L=1/Q, p_H₁=p_H₂=w/Q and p_H,total=2w/Q. The factor two counts the two basins once. If an input free-energy difference already represents their aggregate, applying the same multiplicity again would double count. These parameters are not measured butane data, even though the example illustrates a common conformational counting idea.

Two higher basins can collectively outweigh one lower basin

Equilibrium population fraction

Original three-basin equilibrium toy with ΔH=2.00 kJ mol⁻¹ for each H basin, equal internal weights/effective volumes, no other entropy difference and constant ΔH over 200–400 K. Markers distinguish the lower basin, one higher basin and the combined pair. The dashed curve shows only p_L. No measured butane parameters, transition barrier or interconversion time is represented.20025030035040000.250.50.751

Absolute temperature / K

  • Lower basin L
  • Each higher basin H₁ or H₂
  • Combined higher pair H₁+H₂
  • — Lower-basin fraction in the stipulated model
Underlying data
Absolute temperature / KEquilibrium population fraction
2000.6247071604267103 · Lower basin L
2500.5668551634264464 · Lower basin L
3000.5271404270899531 · Lower basin L
3500.49853085914901496 · Lower basin L
4000.4770697445141916 · Lower basin L
2000.18764641978664487 · Each higher basin H₁ or H₂
2500.21657241828677684 · Each higher basin H₁ or H₂
3000.23642978645502347 · Each higher basin H₁ or H₂
3500.2507345704254925 · Each higher basin H₁ or H₂
4000.26146512774290426 · Each higher basin H₁ or H₂
2000.37529283957328974 · Combined higher pair H₁+H₂
2500.4331448365735537 · Combined higher pair H₁+H₂
3000.47285957291004693 · Combined higher pair H₁+H₂
3500.501469140850985 · Combined higher pair H₁+H₂
4000.5229302554858085 · Combined higher pair H₁+H₂

Original three-basin equilibrium toy with ΔH=2.00 kJ mol⁻¹ for each H basin, equal internal weights/effective volumes, no other entropy difference and constant ΔH over 200–400 K. Markers distinguish the lower basin, one higher basin and the combined pair. The dashed curve shows only p_L. No measured butane parameters, transition barrier or interconversion time is represented.

04

A ring flip changes axial/equatorial while preserving up/down

A cyclohexane chair has axial bonds alternating up and down around its six carbons, and equatorial directions lying more around the ring’s perimeter. At any one carbon, its axial and equatorial substituent positions have opposite up/down directions. A chair inversion swaps which position is axial or equatorial, while retaining whether the attached group is up or down relative to the ring.

For the usual chair reference, a diequatorial pair is trans at 1,2, cis at 1,3 and trans at 1,4. A ring flip preserves that cis/trans relationship. For a substituted ring, conformational preferences depend on the substituents and conditions; larger groups often favour equatorial placement in familiar simple examples, but a qualitative chair label alone is not a quantitative free-energy calculation. Rings can also have other shapes, so a two-chair cartoon is not the full continuous motion.

Worked examples

Count the two equal higher basins once

At 300 K, find the three individual fractions and the combined higher fraction for the stated model.

  1. Each higher/low relative weight is exp[−2000 / (8.314462618 × 300)] ≈ 0.448514.
  2. Q=1+2w≈1.897028.
  3. p_L≈0.527140 and each p_H≈0.236430.
  4. The combined higher population is 2p_H≈0.472860; all three individual fractions sum to one.

About 52.7% lower, 23.6% in each higher basin and 47.3% in the higher pair. A pair total must not be added as a fourth independent basin.

Compare the aggregate at a second temperature

At 400 K, does the combined higher pair exceed the lower basin in this same toy?

  1. Evaluate p_L = 1 / [1 + 2 exp(−2000 / (R × 400))] ≈ 0.477070.
  2. Each higher fraction is about 0.261465, so their combined fraction is 0.522930.
  3. The lower basin still exceeds either higher basin individually, while the pair collectively exceeds it.

Yes: approximately 52.3% in the combined higher pair versus 47.7% lower. No kinetic conclusion follows.

Preserve configuration through a chair flip

At carbon 1 of a chosen chair, an up methyl group is axial. Where does it go on chair inversion?

  1. At the same carbon in the inverted chair, the axial direction becomes down.
  2. The attached methyl group remains up.
  3. It must therefore occupy the up equatorial position.

Axial up becomes equatorial up at that same carbon; it does not become a down substituent.

Try it yourself

  1. For the stated butane dihedral reference, classify the 180° staggered methyl relationship.

    Show the answer and reasoning

    Anti.

    The selected methyl directions are opposite in the declared Newman reference.

  2. At 300 K the individual higher/low ratio is 0.448514. What is the combined higher/low ratio?

    Show the answer and reasoning

    Approximately 0.897028.

    Two equivalent higher basins each have that weight, so the aggregate ratio is twice it.

  3. Does a chair flip change a cis-1,3 substitution into trans-1,3?

    Show the answer and reasoning

    No.

    Up/down at each substituted carbon is preserved while axial/equatorial changes.

  4. Do p_L=0.527 and a combined higher fraction 0.473 determine an interconversion time?

    Show the answer and reasoning

    No.

    The equilibrium weights contain no specified kinetic coefficients, barriers or dynamics model.

What to carry forward

  • Declare the bond, viewing direction and torsion reference.
  • Continuous conformations and named basins have different counting roles.
  • Count multiplicity once in a stated equilibrium model.
  • Chair inversion retains up/down and cis/trans configuration.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.

  1. Blue Book P-9: Configuration and conformationHenri A. Favre; Warren H. Powell (prepared for publication); G. P. Moss (HTML preparation)Sourceiupac.qmul.ac.ukRead source
  2. Michigan State University2 sources
    1. Alkene stereoisomers and ethane/butane conformationsWilliam ReuschSourcewww2.chemistry.msu.eduRead source
    2. Ring conformations, chirality and optical activityWilliam ReuschSourcewww2.chemistry.msu.eduRead source
  3. Yale University2 sources
    1. Lecture33: Conformational energy and molecular mechanicsJ. Michael McBrideSourceoyc.yale.eduRead source
    2. Lecture36: Bond energies, Boltzmann factor and entropyJ. Michael McBrideSourceoyc.yale.eduRead source
  4. 2022 CODATA fundamental constants complete listingCODATA Task Group on Fundamental Constants; NIST reference compilationSourcephysics.nist.govRead source