LearnChem · Part V — Why that product and not another?
Thirty modules have told you the shapes of molecules. This one asks who looked, and finds that nobody did.
Every picture of a molecule you have met was drawn by somebody. Nobody drew it from life. So how does anyone know it is right?
This page answers with four measurements. None of them looks at a molecule. Section 1 starts with the one that comes closest.
Module 30 told you that one chain of polyethylene holds hundreds to a few thousand units. It took that from a reference sample held by NIST. That sample has two average masses, 52,000 and 18,310 grams per mole. It stated them and stopped. Nobody counted those units. Somebody weighed something, and worked backwards. This page is about the working backwards, and it is the last module of the course.
A molecule is small, and that is the honest reason. The waves of visible light are thousands of times longer than a bond. You cannot measure a thing with a ruler that coarse. So for a century chemists worked out shapes without ever looking at one.
Then, in 2009, a picture arrived. Guess what the instrument was recording.
In 2009 a team at IBM published an image of one pentacene molecule. What did the instrument record?
Predict first. The rest of the section opens when you answer.
The instrument was an atomic force microscope. A probe was held above the molecule, and the tiny force on that probe was measured, point by point. The picture is a map of that force. No light reached the molecule at all.
The conditions were severe. The molecule lay flat on a cold surface, at 5 kelvin, in vacuum. One carbon monoxide molecule was fixed to the tip of the probe to sharpen it. A single image took about twenty hours.
The authors called their paper the chemical structure of a molecule resolved by atomic force microscopy. Resolved, not seen, and the word is careful. Three things the picture does not give you. The probe bends as it passes, which sharpens the lines further. So lengths in the image are not to scale. Only the face turned towards the probe appears, and bonds inside a three-dimensional molecule stay out of reach. And the molecule is stuck to a cold surface, which is not where chemistry happens.
So the closest thing we have to a photograph of a molecule is a map of a force. People who knew what to look for read it. Everything else on this page is further from looking than that. None of it is worse for it.
Picture a parcel you are not allowed to open. You can still weigh it. You can shake it and listen. You can hold it up to a bright light. None of that is opening the parcel, and between them they can tell you a great deal.
The image breaks in two places. A parcel could be opened and the guess checked. A molecule cannot, so each method here is checked against the others. And you handle one parcel, while every measurement on this page is made on trillions of molecules at once. What comes back is an average over all of them.
— The closest thing to a photograph of a molecule is a map of a force.
Start with the crudest question you can ask a molecule. How heavy is it?
Here is how it is asked. The sample is knocked with fast electrons. Each molecule loses one electron of its own and becomes an ion. An electron weighs almost nothing, so the ion weighs what the molecule weighed. That ion is then steered by electric and magnetic fields, and where it lands depends on its mass. Count the ions landing at each mass and you have a spectrum.
The knock is hard enough to do something else as well. It breaks many of the molecules. So the detector sees the whole molecule and its pieces, all at once, sorted by mass. The tallest peak is called the base peak. The peak at the mass of the whole molecule is called the molecular ion.
The useful trick is subtraction. Take the mass of the whole molecule, take the mass of a piece, and the difference is what left. A difference of 15 is a CH3 gone. A difference of 18 is water. Bench 1 has six real spectra, each from a public reference record.
Ethanol is the one to read first. Its molecular mass is 46, and there is a peak at 46. The whole molecule survives the knock often enough to be seen. The tallest peak is at 31. The difference is 15, so a methyl group was knocked off and the rest was caught.
Not every molecule is so obliging. Change the bench to propan-2-ol and the peak at its own mass of 60 almost disappears. The molecule breaks nearly every time. Given only that spectrum, the mass of the molecule would be the hardest thing on it to read.
Propanone has a molecular mass of 58 and its tallest peak is at 43. What left?
Use the subtraction, then check against the bench.
58 minus 43 is 15, and 15 is a methyl group. Propanone has two of them, one on each side of its C=O, so the loss is no surprise.
One caution the bench prints and the page repeats. A peak at 43 tells you a mass of 43 and nothing more. C3H7 weighs 43 and so does CH3CO, and at this resolution the record cannot tell you which one landed. Mass spectrometry weighs pieces. It does not say where in the molecule they were attached.
— Subtract a piece from the whole and the difference names what left.
Module 23 gave you isomers. Two substances, the same atoms in the same numbers, arranged differently. Propan-1-ol and propan-2-ol are a pair: both are C3H8O, both weigh 60. The oxygen sits on an end carbon in one and on the middle carbon in the other.
A balance cannot tell them apart. They weigh the same. So before you look, decide what the spectrum will do.
Propan-1-ol and propan-2-ol have the same formula and the same mass. What will their mass spectra look like?
Predict first. The rest of the section opens when you answer.
They break in different places, and that is the whole of it. Set bench 1 to each in turn. Both spectra came from the same laboratory, on the same instrument, at the same energy. That is what makes the comparison worth anything.
Propan-1-ol's tallest peak is at 31. Propan-2-ol's tallest peak is at 45. In the first the oxygen carries one carbon away with it. In the second it carries two, because that is where it was sitting. The arrangement decided where the molecule gave way, and the spectrum shows it.
Now the part that matters more. Look at the peaks at 27 and at 29. They are nearly the same height in both spectra. They are real peaks, measured on real molecules, and they tell you nothing whatever about which alcohol you have. A method does not become useless because parts of its output are uninformative. It becomes dangerous only if you forget which parts those are.
Some elements come as a mixture of masses. Chlorine is about three parts of mass 35 to one of mass 37. Bromine is almost half and half, 79 and 81. The abundances are measured, and NIST publishes them.
That mixture leaves a mark. A molecule holding one bromine arrives at the detector as two peaks, two units apart. Their heights are near enough equal, because about half the molecules carry the heavier isotope. Set the bench to bromoethane and read the two tall peaks at 108 and 110. Then set it to the compound with two bromine atoms and count the peaks in the group at the top.
1,2-dibromoethane holds two bromine atoms. How many peaks does its molecular ion arrive as, and in what rough proportion?
Each bromine is independently heavy or light. Work out the combinations.
Three peaks. Both bromines light, one of each, or both heavy. There are two ways to get one of each, and only one way to get each of the others. So the middle peak is about twice as tall as its neighbours. The record shows 186, 188 and 190 in roughly that proportion.
This is the happiest thing in the whole method. You are counting bromine atoms in a molecule you have never seen, from the shape of a group of peaks. The arithmetic is arithmetic a reader can do.
— Where a molecule breaks depends on how it is put together.
Weighing the pieces leaves a gap. You know the masses. You do not know which bonds are in there. Two methods answer that, and neither looks either.
The first uses infrared. A bond is two atoms held together, and it can stretch and bend, like a spring. Every kind of bond stretches at its own rate. The masses of the atoms set that rate, and so does the stiffness between them. Shine a whole spread of infrared through the sample, and the frequencies that match a bond's own rate are absorbed. What comes out the other side has gaps in it, and the gaps name the bonds.
The scale used is the wavenumber, in units of one over a centimetre. A bigger number means a faster vibration. Bench 2 carries two published tables of ranges, and one real spectrum.
Ethanol shows the method working. A broad absorption near 3391, which is its O-H. One near 2981, its C-H. One near 1055, its C-O. Three bonds named from three gaps in a beam of light, on a molecule nobody saw.
The bench also shows the method's edge, and it shows it honestly, because the two tables do not agree. Put the bench on the C=O of a ketone and then on the C=O of an aldehyde. The two published ranges overlap over a narrow strip only. A peak sitting inside that strip is consistent with both, and no amount of staring at it will decide. Infrared tells you which kinds of bond are present. It does not tell you how they are joined to each other.
The second method puts the sample in a strong magnet. Hydrogen nuclei behave like tiny magnets themselves. In a field they absorb radio waves, at a frequency that depends on what is around them. A hydrogen next to an oxygen sits in one place. A hydrogen at the far end of the chain sits in another. The scale is set against one reference substance, tetramethylsilane, which is put at zero.
Two things a beginner can read off it without any theory. How many separate signals there are, which is how many different kinds of hydrogen the molecule has. And how large each signal is, which is how many hydrogens of that kind there are.
Ethanol is CH3CH2OH. How many kinds of hydrogen does it have, and in what proportion?
Group the hydrogens by what each one is attached to.
Three kinds, 3 to 2 to 1. Three hydrogens on the CH3, two on the CH2, one on the oxygen. The three on the CH3 sit in identical positions, so they give one signal between them. The proportions add to six, which is the number of hydrogens in C2H6O. The count checks itself.
One warning that comes with the method. The signal from the O-H moves about, because it depends on the solvent and on how concentrated the sample is. Published ranges for it are wide and they disagree. Count the signals and read their sizes; do not read that one's position to the decimal place.
— Infrared names the bonds present; the magnet counts the kinds of hydrogen.
You now have the mass, the pieces, the bonds and the kinds of hydrogen. You still do not have a shape. Module 7 told you molecules have shapes, with angles between the bonds. Nothing so far measures an angle.
For that the sample has to be grown into a crystal, and then hit with X-rays. X-rays are used because their waves are about as long as the gaps between atoms. That is the ruler this job needs. Before the bench, decide what comes back.
A crystal sits in a narrow X-ray beam, with a detector behind it. What does the detector record?
Predict first. The rest of the section opens when you answer.
Spots, in particular directions, with darkness between them. The reason is worth having in full, because it is why the method works at all.
A crystal is a stack of identical layers of atoms. X-rays scatter off every layer. Waves coming back from two neighbouring layers have travelled different distances. So they arrive out of step, and mostly they cancel. At certain angles, and only there, the extra distance is a whole number of wavelengths. There the waves arrive in step, and they add. That condition is Bragg's law: n times the wavelength equals twice the layer spacing times the sine of the angle.
So the output is a list of directions, with a brightness for each. Bench 3 computes them for common salt, whose layers are well known. It sets each computed angle beside the angle a real pattern of salt was measured at.
Every computed angle in the bench lands within a tenth of a degree of the measured one. That is the moment worth pausing on. The spacing came from one number, the edge of the repeating cube of salt. The wavelength came from a table. A piece of school trigonometry then predicted where a real machine found real spots.
Now the part that a picture of a diffraction pattern hides. The detector records where each spot is and how bright it is. It does not record the timing of each arriving wave. The timing is exactly what you need to add the waves back into an image. That missing timing has a name, the phase problem, and getting round it is a separate piece of work. What comes out at the end is not a photograph but a map of where the electrons are dense. Atoms are then placed into that map and adjusted until they fit.
So a diffraction pattern is not a shadow and not a picture. It is a set of measurements from which a picture is built, by people, with an argument.
In 1913 the Braggs, father and son, worked out the structure of common salt from patterns like the bench's. What they concluded was not a shape of a molecule but the absence of one. Each chlorine has six sodiums round it, and each sodium six chlorines, all the way through. So there is no NaCl molecule anywhere in the crystal. A chemical idea was overturned by a pattern of spots.
In 1952 Raymond Gosling, working with Rosalind Franklin at King's College London, recorded a pattern from fibres of DNA. Franklin and Gosling read three things from it. A repeat of 34 angstroms along the fibre. A strong reflection at 3.4 angstroms, lying on the tenth layer line, which points to ten units per turn. And a helix about 20 angstroms across, with the phosphate groups on the outside. Those numbers came off the pattern. The pairing of the bases did not come off the pattern. It came from Watson and Crick building models. It is worth keeping the two apart, because the photograph is often credited with both.
— A crystal returns spots, and a structure is argued out of them.
Four measurements, and not one of them answers the question on its own.
Each method is narrow, and each is narrow in a different direction. That is why they are used together. A chemist with an unknown substance weighs it and breaks it. Then looks for the bonds, and counts the kinds of hydrogen. Then, if a crystal can be grown, measures where the atoms sit. Then one structure is proposed. The test of it is that it accounts for every one of those measurements at once. If it explains three and not the fourth, it is wrong.
This is the move the whole course has been making. Module 4 said atoms have parts, and nobody had seen a part. What had been done was to fire small fast things at a thin foil and watch where they went. A few came almost straight back, which a solid lump of jelly could not explain, so the picture changed. The measurement was of deflections. The conclusion was about structure. Everything in this module is that same move, with better instruments.
You have a white solid, and you want to know the angles between its bonds. Which measurement can answer that?
Look at what each one gives, above.
Only the X-rays, and only with a crystal. That condition is not a footnote. Growing a good crystal is often the hardest part of the whole job. For many substances it has never been managed. Where there is no crystal, there is no map. The shape then has to be argued from the other three, and from chemistry.
Which is where this course ends. Across thirty modules you were given shapes, energies, rates and mechanisms. Every one of them was somebody's conclusion from a measurement of something else. That is not a weakness in the chemistry. It is what knowing anything about a molecule consists of.
— One structure has to account for every measurement at once, or it is wrong.
A colourless liquid is handed to you. Its mass spectrum has a small peak at 60 and a tall peak at {BASE}. Separately, a crystal of a salt sits in an X-ray beam of wavelength {LAM} angstroms. You are asked about its {HKL} layers, whose spacing is {D} angstroms. Modules 23, 7 and 4 are all in this question.
1. Module 23. The liquid is one of the two propanols. Which one, and how do you know from one peak?
2. Put the spacing and the wavelength into Bragg's law, first order. Give the angle from the layers, in degrees, to two decimal places.
3. A machine reports that reflection as twice your angle. Give that figure, to two decimal places.
4. Module 4 fired fast particles at a foil and measured where they went. Module 7 says molecules have shapes. Which sentence describes both that experiment and this whole module?
A classmate has the two propanol spectra side by side. He says the peak at 29 tells them apart. It is there in both, and it is nice and tall.
1. Read both spectra off bench 1 at m/z 29. Then say in one sentence what is wrong with his method, and what he should use instead.
A new substance has been made. It will not crystallise, and every attempt so far has given an oil. You have a mass spectrometer, an infrared machine and a magnet, and you are not getting an X-ray structure.
1. Say what you can still establish about the substance, and what you cannot. Then say how confident a chemist should be in a structure proposed on that evidence. There is no single right answer; there is a right way to argue it.
Nobody looks at a molecule. Every method measures something else, and reasons back.