LearnChem · Part V — Why that product and not another?
This page follows one nitrogen lone pair through three uses.
A swimming pool smells of chlorine. What you are smelling is not chlorine. So what is it, and where has it come from?
The CDC's Healthy Swimming pages state that what you smell is not chlorine. This page asks what the nitrogen groups do. It follows one lone pair from a pool to a peptide bond.
Module 28 left you one sentence. The neighbour will take the electrons and nothing is in the way. An amine attacking a carbonyl carbon is that sentence in action. The attacker is m25's nucleophile, the electron-rich partner that moves first. Module 28 taught that side of the attack. This module owes the other side. Once the attack is done, the nitrogen's lone pair is spent. It has gone into the new bond, and it cannot go anywhere else.
Walk into an indoor pool and the air has a smell. Most people call it chlorine. The CDC says something else. Its sentence is that healthy pools do not have a strong chemical smell. Swimmers bring their own nitrogen into the water. It arrives as urea, as creatinine and as amino acids, and every one of those carries a nitrogen atom. The water also holds free chlorine, which is the chlorine still available to react. Its active form there is hypochlorous acid. Make a guess before you read on.
A pool gets busier and less well chlorinated. Does the smell get stronger or weaker?
Answer from your gut. The reveal follows.
The smell gets stronger. Free chlorine does not sit still in a pool. It reacts with the nitrogen that swimmers bring in. The unspent lone pair on that nitrogen is what attacks. That is m25's nucleophile again, retrieved and not taught afresh. The products are chloramines, and a chloramine is a nitrogen carrying one or more chlorine atoms. Health Canada attributes the geranium-like odour to trichloramine in particular. So the smell is evidence that a lone pair was spent on something. The CDC's own sentence is that healthy pools do not have a strong chemical smell. This page does not stretch that sentence any further. The three equations below show the steps.
HOCl + NH3 → NH2Cl + H2O
HOCl + NH2Cl → NHCl2 + H2O
HOCl + NHCl2 → NCl3 + H2O
Read the three equations above in order. Each step swaps one hydrogen on the nitrogen for one chlorine. The first step gives NH2Cl. The second gives NHCl2 and the third gives NCl3. Every step is the same attack by the same kind of lone pair. Each equation is balanced by releasing one molecule of water.
Here is the picture to carry through the page. Draw the lone pair as a lobe on the nitrogen. Sometimes the lobe sticks out where a proton can reach it. Sometimes it is already swallowed by the ring the nitrogen sits in. Sometimes it is swallowed by the C=O group next door, a carbon double bonded to an oxygen. The pair is the same one thing in every case. Only its place changes. One case has no lobe left to draw at all. That case is the quaternary ammonium ion, and the picture simply ends there.
Nitrogen's lone pair can only be spent once, and what it was spent on decides the rest.
The rest of the page asks a narrower question. How strong a base is a nitrogen, measured rather than guessed? Section 2 puts real numbers on a bench and reads them.
— The smell at a pool is a clue about nitrogen, not only about chlorine.
Module 19 gave you one definition, and this page uses it rather than repeating it. A larger pKa of the conjugate acid means a stronger base. The table below heads that column pKaH, which is simply the pKa of the conjugate acid. Every claim about basicity on this page is a comparison between named rows on stated numbers. No compound here gets a one-word label instead. Bench 1 holds measured pKaH values for four simple amines in water. They are ammonia, methylamine, dimethylamine and trimethylamine. Ammonia carries no alkyl group on its nitrogen. An alkyl group is a small chain of carbon and hydrogen. Methylamine carries one, dimethylamine two and trimethylamine three. Predict before the table opens.
Of ammonia, methylamine, dimethylamine and trimethylamine, which is the strongest base in water?
Predict, then read the bench.
The table is open now, all from one compilation, all in water. The column headed pKaH is the pKa of the conjugate acid. A larger number in that column means a stronger base. Dimethylamine sits highest of the four amines, so it is the strongest base of them here. Ammonia sits lowest of the four. Trimethylamine sits below dimethylamine, which is not what a simple count of alkyl groups would suggest. The two boxes above the table add two further sets of numbers for the same four amines.
| compound | pKaH in water | temperature | source |
|---|
One more thing about this table, and it matters. The source states no temperature anywhere. The temperature column says so on the table's own face. So this page does not assume a standard laboratory temperature for these values. Where a source is silent, the page stays silent with it.
Tick the first box above the table. A second compilation gives its own numbers for the same four amines. Its digits are not the digits of the first compilation. Both sets are now printed side by side on the bench. Neither set has been corrected to fit the other. Neither has been averaged into the other. Compare them row by row before the check below.
Now tick the second box. It adds gas-phase proton affinities for the same four amines. A proton affinity is the energy given out when the amine takes a proton with no water present. A larger proton affinity means the amine holds that proton more firmly in the gas phase. Read the gas-phase column downwards and the value rises at every step, from ammonia to methylamine to dimethylamine to trimethylamine. Now read the water column and the rise stops before the last row. Trimethylamine falls below dimethylamine, and the readout works the gap out from those two rows. The keep-sentence explains the large moves on this bench, the ones worth a million-fold or more. It does not explain this last small reversal among the simple alkylamines. One sourced statement names the degree of solvation of the protonated amine as a factor in aqueous basicity. That statement is a general one. No source found here says why trimethylamine in particular falls where it does. So the page names the factor and stops.
The two compilations now sit side by side on the bench. One question about them is worth asking before the page moves on. It takes a step past reading either table on its own.
Do the two compilations rank the four amines in the same order?
The order survives the disagreement. In the first compilation dimethylamine is highest of the four and ammonia is lowest. In the second compilation dimethylamine is highest and ammonia is lowest again. Trimethylamine sits below dimethylamine in both. The digits differ from row to row, by small amounts, and the page prints both sets as they stand. The ranking does not move with the digits. The top two rows sit close together, within a few hundredths of a unit. The clear move is trimethylamine's drop. So the fall of trimethylamine in water is not an artefact of one compilation's rounding. Neither set has been averaged into the other.
Section 3 adds two more rows to the same bench. Both are nitrogens in six-membered rings. Both keep a lone pair that has not been spent on anything. The bench will put them next to each other.
Carbon monoxide is a gas you cannot see or smell. It binds haemoglobin far more tightly than oxygen does. Haemoglobin is the protein in blood that carries oxygen. The line below prints the range the sources give. This page never prints a single value in place of that range. Carbon monoxide appears here as a hazard note. This page does not teach how it binds.
Sources, verified at build time. ATSDR Toxicological Profile for Carbon Monoxide, chapters 2 and 3. NIH/NCBI StatPearls, Carbon Monoxide Poisoning.
— Every basicity claim on this page is a comparison between two named rows, on stated numbers.
Bench 1 now gains two rows from the same source. Both are nitrogen atoms sitting inside a six-membered ring. Piperidine is a saturated ring, which means every bond in the ring is a single bond. Pyridine is an aromatic ring, which is module 24's idea, named here and not taught again. Neither nitrogen has given its lone pair to anything, and that is what unspent means here. The pair is still on the nitrogen and still available to a proton. One thing about the scale before you predict. One pKa unit is a factor of ten. Two units are a factor of a hundred. So a small-looking difference in this column is not a small difference at all.
Both nitrogens keep an unspent lone pair. Are piperidine and pyridine equally strong bases, or is one much weaker?
Predict, then look at the new rows on bench 1, above.
Pyridine is the weaker of the two, and by a long way. Both nitrogens keep an unspent lone pair, so that is not what separates them. Piperidine's pKaH is 11.22 and pyridine's is 5.14. Subtract 5.14 from 11.22 and the gap is 6.08 pKa units. Ka is the acid constant that module 19 built the pKa scale on. Each pKa unit is a factor of ten in Ka, so 6.08 units is a factor of about 1.2 million. The readout below does that subtraction live on the bench's own two rows. The reason for the gap is not written anywhere on the table.
Here is an image for the lone pair, and here is where it fails. Think of the pair as a coin you can spend exactly once. You can spend it into a ring, or into a C=O group, or into a fourth bond. Once it is spent it is committed. The image breaks in two places and both matter. A spent coin is gone, but a spent pair is only spread out or committed. It can be bought back at a price, and a pKa is what that price measures. A coin is also worth the same in any pocket. An unspent pair is not, because the orbital it sits in changes what it is worth. The image does not carry that second point at all.
Now the reason, and it is not the keep-sentence's reason. Pyridine's lone pair is unspent, exactly as piperidine's lone pair is unspent. Pyridine is still about a million times the weaker base of the two. So the difference cannot be where the pair went, because it went nowhere in either case. The difference is which orbital the pair sits in. An orbital is the region around an atom where a pair of electrons is found. Piperidine's pair sits in an sp3 orbital and pyridine's sits in an sp2 orbital. An sp2 orbital carries more s-character, which means a larger share of the round s orbital in its make-up. More s-character holds the pair closer to the nucleus. A pair held closer is harder for a proton to take. So this page names a limit. The keep-sentence reaches what a pair was spent on. It does not reach which orbital an unspent pair sits in. That one thing is named here rather than papered over. The sentence still accounts for the large moves on this bench, between a spent pair and an unspent one.
The same honesty is owed one section back. Among the simple alkylamines the order in water is not the order in the gas phase. Trimethylamine is the weaker base than dimethylamine in water, on both compilations. Nothing about those nitrogens was spent differently between the two columns. The molecules are the same molecules, so this is the second named limit. What a pair was spent on is what the keep-sentence reaches, and the solvent is not that. What changed is the solvent. One sourced statement names the degree of solvation of the protonated amine as a factor in aqueous basicity. That is a general statement about amines, not an account of this one comparison. No source found here explains why trimethylamine in particular sits where it does, and this page does not invent one.
The orbital account is now on the page, and an account is worth testing. One more nitrogen sits further along the same axis. A nitrile has a carbon triple bonded to the nitrogen. See whether the account still predicts what the numbers do.
In a nitrile, the nitrogen's pair sits in an sp orbital. Is it a stronger or weaker base than pyridine?
The nitrile is covered by the limit already written on this page, one step along it. Its pair is unspent, and it sits in an sp orbital rather than an sp2 one. This page does not set out to teach nitriles. It does not claim that the keep-sentence predicts them either. The figure printed above comes from one source only. The page prints it as that one source's value and does not rank it against any other.
Section 4 changes the ring. Pyrrole is a five-membered ring with one nitrogen in it. The page will set it beside pyridine on the same bench. Both rings are module 24's material.
— Two numbers from one bench, compared as a ratio and not as a verdict.
Pyrrole is a ring of five atoms, four carbons and one nitrogen. That nitrogen carries one hydrogen, written N-H, and it has one lone pair. Pyridine has one ring nitrogen and one lone pair too. Both rings are aromatic, which is module 24's idea, named here and not taught again. On paper the two molecules look like close relatives. They share an element in the ring, one lone pair each and one aromatic ring each. The bench is about to put both pKaH values on the table. Predict first, because the prediction is the whole point of this section.
Pyrrole and pyridine each have one ring nitrogen and one lone pair. Which is stronger, and by how much?
Predict, then look at the new rows on bench 1, above.
Pyridine is the stronger base, and by a very long way. Pyrrole's lone pair is two of the six ring electrons that module 24 called the aromatic sextet. It is spent, and it was spent on the ring itself. Pyridine's lone pair sits in the plane of the ring, in an sp2 orbital, outside the ring's pi system. The pi system is the shared cloud of electrons module 24 drew. It is unspent. So here are two nitrogens with the same element and the same single lone pair. Each of them sits in an aromatic ring. They behave in opposite ways. No sentence about how much nitrogen a molecule holds can tell them apart. Where the pair went tells them apart in one move. The readout below works the gaps out from the rows themselves.
Pyrrole is not one number on this table, and that is deliberate. Two figures are printed for it and both of them stay. One is minus 3.8, and the source that gives it says plainly that the proton goes on C2. C2 is the ring carbon next to the nitrogen. The other figure is 0.4, and it comes from two independent sources. Neither of those two names a protonation site at all. So the sources say different things about where the proton lands. This page does not average the two figures. It does not rank one above the other, and it does not quote either one alone. Both sit on the table together.
Now a case that belongs to module 24, retrieved by name. Aniline is a benzene ring carrying an NH2 group. Its lone pair is still there, and aniline still takes its proton on the nitrogen. So the pair has not gone missing. It is spread into the ring, which is exactly the delocalisation module 24 built. That module's answer is the answer here, and this page does not derive it again. Cyclohexylamine carries the same NH2 group on a saturated ring, with nothing to spread into. Cyclohexylamine's pKaH is 10.64 and aniline's is 4.62. Subtract 4.62 from 10.64 and the gap is 6.02 pKa units. At a factor of ten per unit, that is a factor of about a million. The readout below does the same subtraction live on the bench's own rows. So aniline is about a million times the weaker base of the two. That is what spending looks like as a matter of degree. It is a number rather than a label, and it is large but finite.
Two figures for pyrrole now sit on the table together. Before section 5, it is worth being exact about what they let you say. The check below asks precisely that.
Pyrrole has two printed pKaH figures. One names the site C2, and one names none. What may you conclude?
So the honest statement is this one. The two figures are not confirmed to be the same equilibrium. One of them names its site and the other names no site at all. The site question therefore stays open for the 0.4 figure. The page quotes neither figure as the pKaH of pyrrole, and it picks no single value. A number that does not exist is worth saying out loud. Now a prediction, using the rule you already have. Imidazole is a five-membered ring with two nitrogens in it. One is bonded to two ring carbons and keeps its pair unspent. The other carries an N-H, and its pair is inside the ring's electron count. Which nitrogen takes a proton? Answer from where each pair went, with no new number. Decide before you read the next paragraph.
The nitrogen bonded to two ring carbons takes the proton. Its pair is unspent and lies in the plane of the ring, outside the ring's electron count. So a proton can reach it. The other nitrogen's pair is already inside that count, spent on the ring, exactly as pyrrole's pair is. This page has no pKaH figure for imidazole and does not need one. The rule settled the question without a number.
Section 5 turns to the nitrogen inside a peptide bond. That nitrogen is called an amide nitrogen, and it is bonded to a carbonyl carbon. It is the case this whole module is built on.
— Two figures for one compound are printed together, and neither is averaged away.
An amide nitrogen is a nitrogen bonded straight to a carbonyl carbon. A carbonyl carbon is the carbon of a C=O group, module 28's own subject. When an amine attacks such a carbon that also carries a leaving group, an amide is what remains. Inside a protein, that amide is called a peptide bond. The two pieces it joins are called residues, one amino acid unit each. Bench 2 builds two bonds, joining three residues, and three residues make a tripeptide. Two bonds is where this bench stops, and the bench says so on its own face. An amide nitrogen still has its lone pair. Predict what that means for it as a base.
An amide nitrogen keeps its lone pair. Compared with an amine, how strong a base is an amide?
Predict, then look at the new row on bench 1, above.
Bench 1 now carries an amide row, and that row has a number on it. The source gives the amide's pKaH as a range, from about minus 0.5 down to about minus 1. Methylamine's row on the same table sits a little above 10.6. Subtract the amide's pKaH from methylamine's, remembering that the amide's value is negative. Subtracting a negative adds it, so the gap runs from about 11.1 to about 11.6 pKa units. One pKa unit is a factor of ten in Ka. So the gap is a hundred billion to a few hundred billion. The readout below does that subtraction live on the bench's own rows. The amide nitrogen's pair is already fed into the carbonyl beside it. It cannot be spent again on a proton. So the proton goes onto the oxygen rather than onto the nitrogen. That is the keep-sentence's own prediction and not an exception to it. Module 28's attack made this bond, and it left the pair spent.
Bench 2 prints two measured facts about the bond it builds. The first is a length. The peptide C-N bond is 1.33 angstroms, and an ordinary carbon to nitrogen single bond is 1.47 angstroms. Subtract one from the other and the peptide bond is 0.14 angstroms the shorter. The canvas draws both lengths as bars starting from zero. A bond shortened like that carries double-bond character, and a double bond does not turn freely. This page reads that together with the pKaH. The pair fed into the carbonyl is shared with the C-N bond. A short C-N is what that sharing looks like on the bench. The second fact is a population ratio. Trans and cis say which side of the bond the two neighbouring residues sit on. Opposite sides is trans and the same side is cis. In peptide bonds the ratio runs at about 1000 to 1 in favour of trans. Before proline it runs at about 30 to 1 instead. Proline is an amino acid whose side chain loops back onto its own nitrogen. A thousand to one shows which of the two flat forms is preferred. The bond length is what carries the hindered rotation. This bench builds two bonds at most, a tripeptide, and stops there. No rotation barrier figure is quoted on this page.
Two facts sit side by side here, and the page keeps them side by side. The first fact is sourced. The N-H of an amide is a good hydrogen-bond donor, which means it gives its hydrogen to a hydrogen bond. The second fact is sourced as well. The amide nitrogen's lone pair is spent into the carbonyl. What is not sourced is the link between them. No source found here says that the spending is why the N-H donates well. So the page states both facts and joins them with nothing. The N-H is marked on the canvas, so you can see which hydrogen is meant.
One more place a lone pair can go, and it is the end of the road. Spend it on a fourth bond to a fourth group and nothing is left over. The nitrogen then has four bonds and a positive charge, and that is a quaternary ammonium ion. Choline and acetylcholine each carry one. No proton is taken at that nitrogen, because no pair is there to take it. No hydrogen bond is accepted at that nitrogen either. The charge stays whatever the pH is changed to. There is also no lobe left to draw there. That is where the anchor image from section 1 ends, and the page says so plainly.
The amide row now sits on the bench with a number beside it. One more question is worth asking about how to describe that row. It is a question about wording as much as about chemistry.
Why is a one-word label the wrong description of how an amide compares with an amine?
So here is the wording this page keeps. The amide has a measurable pKaH, somewhere between about minus 0.5 and about minus 1. That number sits far below methylamine's number, and the gap is on the readout above. The amide also takes its proton on the oxygen rather than on the nitrogen. A number and an atom name can carry both of those facts. The phrase not basic carries neither of them. That phrase is the loose claim a reader walks away with, and the keep-sentence does not make it. Basicity on this page is always a comparison between named rows on stated numbers. No compound here is sorted into a yes box or a no box.
Section 6 takes the same two kinds of nitrogen somewhere else. They sit along the edges of four rings, adenine, thymine, guanine and cytosine. The bench there shows every atom on those edges.
— A number on the scale is a better answer than a label.
Four rings sit at the heart of DNA. They are adenine, thymine, guanine and cytosine. Each ring carries nitrogen atoms, and some of those nitrogens carry a hydrogen as an N-H. In a hydrogen bond, one atom donates and another accepts. The atom that donates gives up its hydrogen to the bond. The atom that accepts takes that hydrogen towards its own lone pair. Labels such as N6 and O4 name a position in the base, using the numbering the rings already carry. Predict before the bench opens.
Adenine with thymine makes 2 hydrogen bonds, guanine with cytosine 3. Is that size matching, or donors and acceptors?
Predict, then read the atoms on bench 3.
Two kinds of nitrogen from the earlier sections turn up along these edges. A pyridine-type nitrogen is a ring nitrogen whose pair lies in the plane of the ring. A pyrrole-type nitrogen is one whose pair has been spent. The counts below are not recalled, they are read off the atoms. Pick adenine with thymine on bench 3. Adenine's N6-H donates to thymine's O4, and that is one bond. Thymine's N3-H donates to adenine's N1, and that is two. There is no third bond on that edge. Now pick guanine with cytosine and count the arrows there. Three appear, each one running from an N-H that donates to an atom that accepts. Read the atom names for those three from the table rather than from here. The distances shown come from the paper's own X-ray tables. Adenine with thymine appears there as two independent copies, and their distances differ by a few hundredths of an angstrom.
| donates from | accepted by | distance, angstrom | nitrogen reading |
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Now the join, and this page owns it. The general distinction is sourced. A ring nitrogen whose pair is unspent and lies in the ring plane can accept a hydrogen bond. A nitrogen whose pair is spent into the ring carries an N-H instead, and that N-H is observed to donate. Section 5 refused to say that the spending is why an N-H donates well, and that refusal holds here too. The pattern is observed and the cause is not sourced. The atom table on bench 3 is sourced as well, from a published set of X-ray distances. What is not sourced anywhere found is the join between the rule and these four rings. No paper was found that calls adenine's N1 or guanine's N1-H pyridine-type or pyrrole-type in those words. That join is this module's own argument, built here from the general rule and the atom table. It is the page's reading rather than a quotation, and it should be held as such.
So the code on this page is a pattern. It is made of spent and unspent nitrogens read along an edge. An unspent ring nitrogen accepts. A nitrogen whose pair went into the ring carries an N-H, and that N-H is seen to donate. On this page the code means that pattern and nothing else. Read one edge and you get a sequence of donors and acceptors. Read the partner's edge and you get the sequence that has to match it. Adenine with thymine matches at two places, and guanine with cytosine matches at three. That is the third thing the keep-sentence decides, and none of it is about basicity.
Bench 3 now has every atom on it, with an arrow for every bond. One question takes a step past counting those arrows. It asks where a single named arrow lands, atom by atom.
In guanine with cytosine, which atom accepts from cytosine's N4-H, guanine's O6 or cytosine's O2?
That is the argument of the module. Three exercises follow, and they use only what is already on this page. Below them sits the one sentence worth keeping, printed in the box at the foot.
— Read the atoms one at a time before you read the total.
Module 19's definition is the one you need here. A larger pKa of the conjugate acid means a stronger base. Bench 1 gives cyclohexylamine a pKaH of {CHX} and aniline a pKaH of {ANI}. Work out how many pKa units stronger a base cyclohexylamine is than aniline. Then turn that gap into a factor in Ka, remembering that one unit is a factor of ten.
1.
2. Now the harder half of it. Using module 24's idea and nothing else, why is aniline the weaker of the two? Say it in your own words before you open the worked answer.
Subtract aniline's pKaH from cyclohexylamine's and the gap is about six pKa units. One pKa unit is a factor of ten, so six units are a factor of about a million. Cyclohexylamine is therefore about a million times the stronger base of the two. Now module 24's answer, given back. Aniline's lone pair is spread into the benzene ring. Taking a proton onto that nitrogen costs that spreading, and the pKa measures the cost. The pair never left the nitrogen, and aniline still protonates there. So spending is a matter of degree, and a million-fold is what this degree is worth.
3. Here is a third nitrogen, sitting in a ring, and it has no name on this page. Its lone pair sits in an sp3 orbital, and that orbital is not part of any ring pi system. Compare it with what you now know about pyrrole's nitrogen and pyridine's nitrogen. Use only where the pair went. No new number is given, and none is needed. Should it behave like piperidine's nitrogen, or like pyrrole's?
This nitrogen's lone pair is unspent. It is not part of the ring's electron count, and it is not fed into a neighbouring C=O group. It sits in an sp3 orbital, which is where piperidine's pair sits as well. So it should behave as piperidine's nitrogen does rather than as pyrrole's. Pyrrole's pair is inside the ring's own six electrons and is therefore spent. The rule transfers on where the pair went, and no figure was needed to apply it.
4. Here is a made-up ring called X. Along its pairing edge it has one N-H that donates, and two ring nitrogens that accept. Its partner Y has two donors and one acceptor on its own edge. Only these counts are given, and no structure is drawn. At most how many hydrogen bonds can X and Y make between them? Then open the second question, which asks why X cannot pair fully with another X.
X and Y can make at most three. X has one donor and Y has one acceptor, and that pairs off to give one bond. Y has two donors and X has two acceptors, and that pairs off to give two more. One plus two is three, with nothing left unmatched on either edge. Now take X against a second X. Each X brings one donor and two acceptors. The first X's donor meets the second X's acceptor, and the second X's donor meets the first X's acceptor. That is two bonds, and one acceptor on each edge faces nothing at all. So a pair of X rings manages two bonds and not three. The edges do not match count for count.
Nitrogen's lone pair can only be spent once, and what it was spent on decides the rest.