Here is one small alcohol, CH₃CH₂CH₂CH₂OH, drawn three ways.
The Lewis structure on the left shows every atom and every bond. The condensed formula in the middle is shorter, but it shows no bonds at all. The drawing on the right is a skeletal structure, also called a line-angle structure. It shows the carbon skeleton, the chain of carbons that holds the molecule together, along with the OH group on it. It leaves out every carbon symbol and every hydrogen on carbon, and four rules let you put them back.
That saving matters most for big molecules. Cholesterol, which you met in Foundations, has twenty-seven carbons. Its Lewis structure is a page of Cs and Hs that no chemist could scan, while its skeletal structure shows the rings and the OH at a glance.
Four rules for reading the drawing
1. A line is a bond, and every corner or end of a line is a carbon. Look at the skeletal drawing above. Four lines run from the left end to the OH. The left end and the three corners are the four carbons, and no C is written at any of them. So a zigzag of five lines is a six-carbon chain, because both ends count. Forgetting the ends is the easiest miscount to make.
2. Hydrogens on carbon are not drawn. You work them out instead. Carbon makes four bonds, so subtract the bonds you can see at a carbon from four, and the rest are hydrogens. In the drawing above, the left end has one line, so it is a CH₃. Each corner has two lines, so each is a CH₂.
Rule 2 counts bonds, not lines on the page, and the difference shows up at a multiple bond. A double bond is two of carbon’s four bonds, and a triple bond is three. In CH₃CH=CH₂, the middle carbon has a double bond and a single bond. That is three bonds, so it has one hydrogen. The end carbon has only the double bond, so it has two hydrogens: CH₂.
A triple bond works the same way. In CH₃C≡CCH₃, each carbon of the triple bond also has one single bond. That makes four bonds, so neither carries a hydrogen. In CH₃CH₂C≡CH, the carbon at the very end has only the triple bond, so it keeps one hydrogen. That ≡C–H hydrogen comes back in Alkenes & Alkynes, where it turns out to be unusually easy to remove.
Notice how the triple bonds are drawn in the figure below. The chain runs straight through them instead of turning a corner. The two carbons of a triple bond are sp carbons. Each one holds its two neighbors 180° apart, as the hybridization section of Foundations showed. A corner there would draw a shape the molecule cannot have.
3. Every atom that is not carbon IS drawn, and so are the hydrogens attached to it. Look again at the OH at the top of the page. The O is written, and so is its H. The same goes for N, S, Cl and every other atom that is not carbon, along with any hydrogen on them. An O–H is written OH, and an N–H is written NH or NH₂.
Hydrogen on carbon is hidden, but hydrogen on any other atom is shown. This asymmetry trips people up, and the reason for it is chemical. A hydrogen on O or N is always drawn because it matters to how the molecule behaves: it forms hydrogen bonds, and it often takes part in the reaction itself. A C–H hydrogen on a chain rarely does either. The notation hides what rarely reacts and keeps what does.
4. Lone pairs are usually left off, and you are expected to supply them. The oxygen of the OH above has two bonds, so it has two lone pairs, just as in its Lewis structure. The skeletal drawing simply does not show them. You will need them in the next section, where curved arrows start from lone pairs.
Charged atoms: what a + or a − does to the hidden half
Take methoxide, CH₃O⁻, from Foundations. Its skeletal structure is a single line ending in O⁻. The unlabeled end of that line is the carbon of the CH₃ group. The oxygen has one bond, a minus charge and no written H. An oxygen with one bond and a −1 charge has three lone pairs, so those three pairs are there even though nobody drew them.
A drawn charge is not decoration. It changes how many bonds and lone pairs its atom has, and those are exactly the things a skeletal drawing hides. The arithmetic is the formal-charge formula from Foundations, read backwards. Valence electrons minus unshared electrons minus bonds gives the charge. So once you know the charge and the bond count, the number of lone pairs follows. Four cases cover almost everything:
- Carbon with a + and three lines is a carbocation. It has three bonds and nothing else: no hydrogen and no lone pair. It has an empty p orbital. Drawn with two lines, a carbocation has one hydrogen and still no lone pair.
- Carbon with a − and three lines is a carbanion. It has three bonds plus one lone pair, so it has no hydrogen. The lone pair is what makes a carbanion reactive, and it is usually left off the drawing.
- Oxygen with a − and one line, as in methoxide or acetate, has one bond, no hydrogen and three lone pairs. A neutral oxygen with one bond would carry an H and two lone pairs, so the minus sign has traded a hydrogen for a lone pair.
- Nitrogen with a + and four lines is an ammonium nitrogen. It has four bonds, no hidden hydrogen and no lone pair, because nitrogen used its lone pair to make the fourth bond.
Acetate, CH₃COO⁻, is drawn as a two-carbon skeleton. Its right-hand carbon carries a double-bonded O above it and a single-bonded O⁻ to the right. Go atom by atom.
Left carbon: one line, no charge. 4 − 1 = 3, so it has three hydrogens: CH₃.
Right carbon: one line to the CH₃, two to the double-bonded oxygen and one to the charged oxygen. That is four bonds already, so it has no hydrogen and no lone pair.
Double-bonded oxygen: two bonds and no charge, so no hydrogen and two lone pairs.
Single-bonded oxygen with the minus: one bond and a −1 charge, so no hydrogen and three lone pairs.
That makes six lone pairs in all, and none of them is drawn. When acetate comes back in Resonance, later in this chapter, its curved arrows start from the lone pairs on the charged oxygen. The drawing does not show those pairs, so you have to supply them.
Reading one back
You need to read skeletal drawings as well as draw them, and reading feels harder at first. Work in the same order every time and it stops being hard. On this page, molecules are identified by their condensed formulas. Their names come later, in IUPAC Nomenclature.
Take a zigzag of five carbons with an OH drawn on the second carbon from the left.
Count the corners and ends: five carbons. Read the other atoms: one oxygen, drawn, with its hydrogen drawn. Fill in the hydrogens by counting bonds: the left end has one line, so it is CH₃. The second carbon has three lines, two along the chain and one to the OH, so it has one hydrogen: CH. The third and fourth carbons have two lines each: CH₂. The right end has one line: CH₃.
Put it together: CH₃–CH(OH)–CH₂–CH₂–CH₃. The read takes one pass, and every hydrogen comes from a subtraction, not a guess.
A branched chain. A zigzag of four carbons with a short line up from the second one gives five carbons and no other atoms. Carbon 1 is a chain end: CH₃. Carbon 2 is where the branch leaves the chain. Three lines meet there, so it has one hydrogen: CH. Carbon 3 has two lines: CH₂. Carbon 4 is an end: CH₃. The branch tip is also an end: CH₃. Written as a condensed formula, the molecule is (CH₃)₂CHCH₂CH₃, C₅H₁₂.
A ring with a branch. A hexagon with a line drawn off one corner has seven carbons. The corner with the line has three lines, so it has one hydrogen. The other five corners have two lines each, so each is a CH₂. The far end of the line is a CH₃. The molecule is C₆H₁₁–CH₃, C₇H₁₄. A ring corner is read exactly like a chain corner. The ring needs no special rule; it is simply a zigzag that closes on itself.
A double bond. Take the branched chain again and make the bond between carbons 2 and 3 a double bond. Carbon 1 is an end: CH₃. Carbon 2 has the double bond (two bonds), the bond to carbon 1 and the bond to the branch. That is four already, so it has no hydrogen. Carbon 3 has the double bond and one bond to carbon 4, so it has one hydrogen. Carbon 4 and the branch tip are ends: CH₃. Altogether it is (CH₃)₂C=CHCH₃, C₅H₁₀. It has two hydrogens fewer than the branched chain, because the double bond uses up one extra bond on each of carbons 2 and 3.
One molecule, several drawings
The figure below shows four drawings. The first three are the same compound, CH₃CH(OH)CH₂CH₂CH₃. One is mirrored and one is turned upright, but in each the OH sits on the second carbon from an end of a five-carbon chain. The fourth has its OH on the middle carbon, so it is a different compound.
A skeletal structure records which atoms are joined to which, not where they sit on the page. You can rotate the drawing, bend the zigzag around a corner, start the chain from the other end, or point a branch up instead of down. As long as the same atoms are joined to the same atoms, it is the same compound. This matters more than it sounds. A lot of exam anxiety comes from seeing a structure that looks different from the one in the notes and assuming it is a different molecule.
The zigzag is drawn with 120° angles, but the molecule does not have them. The bonds of an sp³ carbon point about 109.5° apart, as molecular geometry in Foundations showed, so a bent chain is a much better picture than a straight line. Chemists draw the angle at 120° only because that spreads the chain evenly on the page. And because a chain can rotate about every single bond, no one drawn shape is the molecule’s shape. Alkanes & Conformations studies that rotation. Stereochemistry gives its own notation to the parts of a drawing that are not free to move.
Reading the condensed form
Condensed formulas are the other notation you will be handed, often in an exam question where a drawing would be hard to typeset. Foundations covered how to read one in Lewis structures. Here the job is to turn one into a skeletal drawing.
Start with CH₃CH(OH)CH₃. Written left to right, each carbon is followed by its own hydrogens, and a bracket hangs its group on the carbon just written. So this is a three-carbon chain whose middle carbon carries an OH. Drawn skeletally, it is two lines meeting at a corner, with a third line up from that corner to the OH.
A bracket at the front of a formula hangs its group on the carbon that follows it. So (CH₃)₂CH– is a carbon bearing two methyl (CH₃) groups and one hydrogen, with its fourth bond going to whatever comes next. In (CH₃)₂CHCH₂OH, that next atom is a CH₂ carrying the OH. (CH₃)₃C– is the same idea with three methyls and no hydrogen, as in (CH₃)₃COH.
Turning a condensed formula into a skeletal drawing is the skeletal read in reverse. Put a corner or end for every carbon, attach whatever is bracketed after it, and join the rest in the order written.
Rings, and why they are the clearest case for the notation
A hexagon with nothing inside it is six carbons and twelve hydrogens, since every corner has two lines and therefore two hydrogens. The hexagon says that in one stroke, where a Lewis structure needs eighteen atom symbols. The molecule is cyclohexane.
Benzene, C₆H₆, from Foundations, is the same hexagon with three double bonds that alternate with single bonds. Each corner then has three bonds, so it has one hydrogen. Many books draw a circle inside the hexagon instead. Both drawings are the same molecule, and both appear in exams.
The circle reflects a measurement: all six C–C bonds in benzene are identical. So the alternating single and double bonds of the first drawing are not really there, and the circle does not pretend they are. Why they are identical is a question for Resonance, later in this chapter. The alternating drawing stays in use because it lets you count hydrogens the ordinary way and, from the next section on, draw curved arrows on it. The circle does neither.
Rings are also where the notation starts to hint at shape. A hexagon drawn flat is a convenience. The real cyclohexane ring is not flat, and Alkanes & Conformations draws the same six carbons in the shape they actually take.
What carries forward
Every drawing after this section uses this notation, and two later parts of the course depend on reading it fluently. Stereochemistry puts the wedges and dashes from Foundations onto skeletal structures to show which bonds come toward you and which go away, and that is meaningless while the skeleton itself still needs decoding. Mechanisms ask you to draw curved arrows between atoms that are not written down. That only works if you can see the hidden carbons and the undrawn lone pairs as clearly as the drawn ones.