Newman and Sawhorse Projections

Every modern conformational analysis — from the simplest discussion of why staggered ethane is more stable than eclipsed, to predicting the preferred ring-puckering of cyclohexane substituents, to rationalizing why anti-periplanar arrangements are required for E2 elimination and for many $S_N2$ backside attacks — is communicated in Newman projections, because the reacting orbitals' relative geometry (a $\sigma_{C-H}$ or $\sigma_{C-X}$ bond anti-periplanar to a departing group, or to an adjacent $\pi$ system) is exactly the kind of dihedral-angle statement a Newman diagram makes visually obvious. Protein and nucleic-acid backbone conformations are described using dihedral angles (the famous $\phi/\psi$ Ramachandran angles of peptide backbones, and the sugar-phosphate backbone torsions of DNA/RNA) that are the direct intellectual descendants of the same Newman-projection logic, now computed numerically from crystal or NMR coordinates rather than drawn by hand. Pharmaceutical chemists use conformational analysis around rotatable bonds — reasoning first captured in Newman projections — to predict which conformer of a flexible drug molecule is bound in a receptor's active site, since binding affinity often depends on accessing a higher-energy but bioactive rotamer. NMR spectroscopists rely on the Karplus relationship, which expresses vicinal $^3J_{H-H}$ coupling constants as a function of the H–C–C–H dihedral angle read straight off a Newman-style picture, making these projections not merely a teaching aid but a working tool for solving real molecular structures from spectra. Sawhorse projections, for their part, remain the preferred format wherever a mechanism needs to show explicit bond-breaking and bond-forming arrows together with 3-D stereochemistry — for instance drawing the concerted, anti-periplanar transition state of an E2 elimination — because their perspective view keeps the reacting centers' spatial relationship to arriving/leaving groups visually intuitive in a way a flattened Newman circle cannot.

Ask the tutor: Why exactly is the eclipsed conformation of ethane higher in energy — is it really just steric clash between the hydrogens, or is something electronic going on?

Start from the shape of a single $sp^3$ carbon. Four $\sigma$ bonds point to the corners of a tetrahedron, each pair of bonds separated by the ideal tetrahedral angle, $109.5°$. Take ethane, $\text{H}_3\text{C–CH}_3$. The central C–C bond is a single $\sigma$ bond formed by head-on overlap of two $sp^3$ hybrid orbitals; this overlap has cylindrical symmetry about the internuclear axis, meaning rotation about the C–C bond costs (to a first, orbital-only approximation) no change in that overlap. Rotation is therefore geometrically free — the barrier that does exist (about $12\ \text{kJ mol}^{-1}$ for ethane) comes from steric/hyperconjugative effects between the substituents on the two carbons, not from the $\sigma$ bond itself. This is precisely the freedom that separates conformation from configuration: rotating about that bond never breaks or remakes a bond, so you never leave the same constitutional/configurational molecule. To depict what rotation actually does, imagine looking straight down the C1–C2 axis, with C1 (the "front" atom) closest to your eye and C2 (the "back" atom) directly behind it. You see the front carbon's three remaining bonds (to H, H, H in ethane) splayed out from a point, each separated by $120°$ in the two-dimensional projection (even though in 3-D they are $109.5°$ from the C–C axis). Behind the front carbon, poking out from the edge of an imaginary circle representing the back carbon, are its three bonds. The angle between a chosen front bond and a chosen back bond, measured in this circular projection, is the dihedral angle $\theta$ — and it is exactly what changes as you twist the C–C bond. This front-dot/back-circle picture is the Newman projection; it was introduced by Melvin Newman in 1952 specifically to make $\theta$ visually unambiguous. The sawhorse projection encodes the same physics differently. Instead of collapsing the C–C bond to a point, you draw it as a diagonal segment, as if you are standing slightly to the side and above the bond, looking at it in perspective — like looking at a real sawhorse workbench from an oblique angle (hence the name). The front-left carbon's three substituents fan downward-left, the back-right carbon's three substituents fan upward-right. Because the bond itself is drawn with explicit length and the tetrahedral angles are drawn close to their true values, sawhorse projections often feel more like an actual 3-D model, while Newman projections are better for rapidly reading off and comparing exact dihedral angles — which is why Newman projections dominate quantitative conformational analysis, while sawhorse projections are preferred when you want a structure that reads naturally alongside wedge–dash drawings (for example, in $S_N2$ or elimination mechanisms where the leaving group and nucleophile/base geometry matter).

The single load-bearing quantity in both projections is the dihedral (torsion) angle, defined for four atoms $A$–$B$–$C$–$D$ as the angle between the plane containing $A$, $B$, $C$ and the plane containing $B$, $C$, $D$, measured while sighting down the $B$–$C$ bond:

where:

• $B$–$C$ is the bond you are looking down (the Newman/sawhorse "axis"),
• $A$ is a substituent on the front atom $B$, $D$ a substituent on the back atom $C$,
• $\theta$ is measured by convention as positive for a clockwise rotation taking the front bond onto the back bond when viewed from $B$ toward $C$ (IUPAC sign convention), and is reported in the range $-180°$ to $+180°$ or equivalently $0°$–$360°$.

For a system with threefold symmetry at both carbons (like ethane, or the C2–C3 bond of butane treating only rotamer type), the special angles are:

Within the staggered family, when the two largest substituents on the front and back carbon are involved, we further distinguish:

where "anti" places the two big groups maximally far apart in projection, and "gauche" places them $60°$ apart, still staggered but closer together than anti. Reading these angles correctly off a picture — not memorizing them — is the entire skill this article teaches.

Example 1 (foundational): Drawing staggered and eclipsed ethane, and locating $\theta$. Ethane, $\text{H}_3\text{C–CH}_3$. Put your eye on the C1 side, looking toward C2. Staggered form. Draw the front carbon (C1) as a dot with three H's at $12$, $4$, and $8$ o'clock ($0°$, $120°$ (measuring clockwise from 12), $240°$). Because this is the staggered conformer by definition, place the back carbon's (C2) three H's exactly between the front H's: at $2$, $6$, and $10$ o'clock, i.e., offset by $60°$ from each front H. Check: the dihedral angle between the $12$-o'clock front H and the nearest back H (at $2$ o'clock) is

and between the $12$-o'clock front H and the back H at $10$ o'clock,

By symmetry every front–back H pair in staggered ethane is at $\pm 60°$ or $180°$ — there is no pair at $0°$. This conformer has three equivalent gauche-type H/H relations at $60°$ and three at $180°$ (front H to the "opposite" back H), and it is the global energy minimum, by a margin of about $12\ \text{kJ mol}^{-1}$ over the eclipsed form, because staggering avoids eclipsing interactions (partly steric, partly a hyperconjugative $\sigma_{C-H} \to \sigma^*_{C-H}$ donation effect that is maximized in the staggered geometry, as later conformational-analysis treatments of ethane derive quantitatively). Eclipsed form. Rotate the back carbon by $60°$ relative to the staggered drawing (equivalently, rotate the front carbon $-60°$; only the relative angle matters). Now each back H sits directly "behind" (same clock position as) a front H:

This is the maximum-energy conformation along the rotational pathway, roughly $12\ \text{kJ mol}^{-1}$ above staggered, i.e., about $4\ \text{kJ mol}^{-1}$ per eclipsing H···H interaction — the quantitative dissection of that number belongs to conformational analysis proper, but the projection work — correctly placing dots at $0°$ separation — is exactly what this article is teaching.

Limiting-case check: rotating continuously from $\theta=0°$ to $\theta = 360°$ must return the identical staggered-at-$60°$ picture you started with (since ethane's three H's are indistinguishable, the potential energy is periodic with period $120°$, not $360°$) — a useful sanity check that your Newman drawing convention is self-consistent: redraw at $\theta = 60°$ and at $\theta = 60° + 120° = 180°$ and confirm they look superimposable (all-staggered) rather than accidentally eclipsed.

Example 2 (applied): Butane's gauche conformer, asymmetric substituents, sawhorse and Newman side by side. Consider butane, $\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_3$, viewed down the central C2–C3 bond. Front carbon C2 bears a methyl (call it $M_1$) and two H's; back carbon C3 bears a methyl ($M_2$) and two H's. We are asked to draw the gauche conformer with $\theta(M_1\text{–C2–C3–}M_2) = +60°$ (clockwise convention). Newman construction. Place $M_1$ at $12$ o'clock on the front dot; the two front H's go at $4$ and $8$ o'clock (each $120°$ from $M_1$ and from each other), consistent with $sp^3$ front-carbon geometry projected onto the circle. Since we want a staggered family conformer (gauche is staggered, not eclipsed — it is one of the three staggered rotamers, distinguished from anti only by which staggered rotamer it is), the back substituents must sit at $2$, $6$, $10$ o'clock. To achieve $\theta(M_1,M_2) = +60°$, put $M_2$ at the $2$-o'clock back position (that is, $60°$ clockwise from the front $M_1$ at $12$). The two back H's then occupy $6$ and $10$ o'clock. Now read off every nontrivial dihedral involving the two methyls and check the full staggered set is self-consistent:

These three angles — $60°$, $180°$, $-60°$ — are exactly the three staggered positions available to the back substituent set relative to any fixed front substituent, confirming the drawing has not accidentally introduced a fourth spurious angle. Sawhorse redraw. Draw the C2–C3 bond as a diagonal, front-lower-left to back-upper-right. Place $M_1$ fanning up-and-left from C2 (its "$12$-o'clock-equivalent" position under the tilt), the two front H's fanning downward. On C3, place $M_2$ at the position $60°$ clockwise (in the same rotational sense established in Step 3 of the conversion mechanism above) from a hypothetical anti placement — concretely, if the anti position for $M_2$ would point straight up-and-right (continuing the diagonal), gauche-$M_2$ is rotated $60°$ off that, toward (not away from) $M_1$'s side. The two back H's fill the remaining two tilted positions. This gauche conformer of butane is higher in energy than the anti conformer (the two methyls, each much bulkier than H, are only $60°$ apart, producing real steric/van der Waals strain of about $3.8\ \text{kJ mol}^{-1}$ relative to anti — a number whose physical origin is developed fully in conformational analysis of butane), yet it is still a local energy minimum (staggered), lower than any eclipsed rotamer. The projection work here is what lets you see that $M_1$ and $M_2$ are close (adjacent staggered positions) without any residual eclipsing strain — a distinction invisible in a flat skeletal (line-angle) drawing.

Example 3 (exam/competition level): Assigning R/S consistency through a Newman-to-Fischer-style conversion for a chiral center, without breaking bonds. Consider (2R,3S)-2,3-dibromobutane's C2–C3 bond, a classic case where students confuse conformational rotation with a configurational change. Number the front carbon C2 with substituents $\text{Br}$, $\text{CH}_3$, $\text{H}$ (plus the bond to C3), and the back carbon C3 with $\text{Br}$, $\text{CH}_3$, $\text{H}$. Setup. Because C2 is $(R)$, going $\text{Br} > \text{CH}_3 > \text{H}$ (priority order by atomic number, then by substituent) must trace clockwise when viewed with the lowest priority (H) pointing away from the viewer. In the Newman projection with C2 in front, arrange (for concreteness) $\text{Br}$ at $12$ o'clock, $\text{CH}_3$ at $4$ o'clock, $\text{H}$ at $8$ o'clock. Viewed from the front (i.e., from where you are standing, looking at C2 with C1's substituents toward you and C3 behind) — but the R/S assignment requires viewing with H pointing away; here H is one of the front three substituents, so if H is toward the back-ish orientation relative to your eye is a subtlety: for a front atom in a Newman projection, all three of its substituents point somewhat toward the viewer, so you assign R/S by looking at the front face directly and, if the lowest-priority group happens to be one of the two "outer" positions rather than pointing directly at you, you may need to determine the sense of rotation and then invert your conclusion if H is toward rather than away from you. Using $\text{Br}(12) \to \text{CH}_3(4) \to \text{H}(8)$: going $12 \to 4 \to 8$ o'clock is clockwise. Since in this arrangement H (lowest priority) is not pointing directly at the viewer along the C1–C2-type axis but is simply one of three splayed substituents roughly toward the viewer, a full rigorous treatment requires mentally viewing from the side opposite H — in practice, for exam purposes, chemists instead convert the Newman projection to a wedge–dash drawing first (Example 1's inverse map) and assign R/S from that, precisely to avoid this ambiguity. We do that conversion explicitly: Newman → wedge–dash. Front carbon C2's three substituents ($\text{Br}$ at $12$, $\text{CH}_3$ at $4$, $\text{H}$ at $8$, all implicitly toward the viewer relative to the C2–C3 bond going into the page) map onto a wedge–dash drawing with C2–C3 drawn as a plain in-plane bond, $\text{Br}$ as a wedge, $\text{CH}_3$ as a wedge, and $\text{H}$ as a dash — no: only one of the three non-C3 substituents at an $sp^3$ center can be in-plane if C2–C3 is also in-plane, since $sp^3$ has four directions total. Redraw properly: put the C2–C3 bond horizontal, in-plane; then C2's other three bonds split as one wedge, one dash, and — this is impossible for only three remaining substituents at a tetrahedral center with one bond already in-plane along C2–C3 unless we also place one more in-plane. The standard resolution: draw C2–C3 in-plane, then $\text{Br}$ as a wedge (toward viewer), $\text{H}$ as a dash (away from viewer), and $\text{CH}_3$ also in-plane (this is the ordinary "three-bonds-shown, natural tetrahedral splay" convention used throughout organic drawings). With $\text{Br}$ wedge, $\text{CH}_3$ in-plane (say, pointing up-left), $\text{H}$ dash: viewing with $\text{H}$ pointing away (it already is, since it's a dash) look at $\text{Br} \to \text{CH}_3 \to \text{C3}$(treated as the remaining bond, but C3 is not a priority-ranked substituent here — C2's four substituents are $\text{Br}, \text{CH}_3, \text{H}$, and $\text{C3}$-with-its-own-substituents, and $\text{C3}$'s effective priority, since it is $\text{CHBrCH}_3$, outranks the free $\text{CH}_3$ but is outranked by $\text{Br}$ itself). Priority order at C2: $\text{Br} > \text{C3}(\text{CHBrCH}_3) > \text{CH}_3 > \text{H}$. Correct assignment. With $\text{H}$ on the dash (pointing away, correctly oriented for direct reading), trace $\text{Br} \to \text{C3} \to \text{CH}_3$. If $\text{Br}$ is up, $\text{C3}$ is to the right (continuing the chain), and $\text{CH}_3$ is down-left, the path $\text{up} \to \text{right} \to \text{down-left}$ traces clockwise, giving $(R)$ at C2 — consistent with the problem's given configuration, confirming that this particular Newman drawing (Br at 12, CH₃ at 4, H at 8 on the front carbon) does correctly represent $(2R)$.

The pedagogical point: rotating the back carbon (C3) to generate different conformers (staggered, gauche, eclipsed, anti with respect to the two methyls or the two bromines) never touches C2's substituent arrangement, so the $(R)$ configuration at C2 is completely untouched by which conformer you draw — only $\theta$ changes. This is the deepest content of "Newman projections depict conformation, not configuration": you can spin the back circle through all $360°$, generating a continuum of physically real, interconverting shapes of the same molecule, and R/S labels (which depend only on connectivity and spatial priority order at each stereocenter, not on the dihedral angle to a neighboring center) never change. This is exactly the distinction developed rigorously in the study of degrees of freedom separating conformation from configuration.

Melvin Spencer Newman (1908–1993) was an American organic chemist at Ohio State University, known for a long career in physical organic chemistry, polycyclic aromatic hydrocarbon synthesis, and — most famously outside his own specialty — for a single short 1952 paper in the Journal of Chemical Education that introduced the circle-and-dot projection now bearing his name. Newman's original motivation was pedagogical: existing 3-D perspective drawings of ethane- and butane-like conformers were hard for students to compare quantitatively, because subtle perspective distortions made it difficult to read off exact dihedral angles by eye. His projection deliberately sacrifices some 3-D "realism" (the front atom becomes a mere point) in exchange for a drawing in which the torsion angle can be read directly, like a clock face — an early and durable example of choosing a representation optimized for the specific quantity you need to extract, a principle that recurs throughout structural chemistry (compare choosing polar versus Cartesian coordinates depending on the symmetry of a problem).