Corrected exercises: nomenclature and isomerism in organic chemistry (202-SNU-RE)

Method: 1) find the LONGEST chain; 2) number from the end that gives the lowest locants; 3) list the substituents in ALPHABETICAL order; 4) assemble. Here the longest chain has 6 carbons (hexane), with two methyl groups and one ethyl group. Numbering from the left: methyls at 2 and 3, ethyl at 4, that is the locants {2, 3, 4}; from the right we would get {3, 4, 5}, which is worse. Alphabetical order: ethyl before methyl (the prefixes di- and tri- do not count). Name: 4-ethyl-2,3-dimethylhexane. Molecular formula: 10 carbons, acyclic alkane, so $C_{10}H_{22}$ ($C_{n}H_{2n+2}$).

“2-Ethyl-3-methylpentane”: a chain of 5 carbons with an ethyl group on C2 and a methyl group on C3. But going THROUGH the ethyl group gives a chain of 6 carbons: the longest-chain rule is broken, and pentane was not the true parent chain. Renumbering on the hexane, the two methyl groups fall on 3 and 4: the correct name is 3,4-dimethylhexane (same formula $C_{8}H_{18}$). This is THE classic trap: a substituent drawn as a “branch” can be part of the parent chain.

There are FIVE: hexane (straight chain), 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane. Beware of false duplicates: “4-methylpentane” does not exist (it is 2-methylpentane numbered backwards) and neither does “3,3-dimethylbutane” (it is 2,2- read from the other end). Five skeletons, five different boiling points: the more branched the isomer, the lower it boils.

8 carbons, an alkane: $C_{8}H_{18}$, the heavily branched isomer of octane. The two possible numberings give the locants {2, 2, 4} or {2, 4, 4}: we compare at the FIRST point of difference: 2 = 2, then 2 < 4, so the set {2, 2, 4} wins. The rule is not “the lowest sum” but “the lowest locant at the first point of difference”, and it is the one that decides here.

The first is 1-ethyl-3-methylcyclohexane, the second 1,1-dimethylcyclopentane, the third 1-bromo-2-chlorocyclobutane. The numbering rule is this: start at a carbon that bears a substituent, give it locant 1, and go around in the direction that gives the substituents the lowest set of locants. In case of a tie, alphabetical order decides, and the substituent that comes first alphabetically gets the lowest locant: that is why the first compound is named 1-ethyl-3-methyl and not 1-methyl-3-ethyl, ethyl coming before methyl in the alphabet. The third illustrates the same rule, bromo coming before chloro.

The first is propylcyclohexane and the second cyclopropylheptane. The rule is that of the larger number of carbons: the parent structure is the one that has more, the other becoming a substituent. Cyclohexane has six carbons against three for the chain, so it is the parent; heptane has seven against three for the ring, so heptane becomes the parent and the ring is only a cyclopropyl substituent. In a strict tie, current nomenclature gives priority to the ring. Remember that a ring is nothing special: it competes with the chain exactly as a longer side chain would.

The compound $C_{6}H_{5}CH_{2}Cl$ is called benzyl chloride if the ring and its $CH_{2}$ are treated as a block, or (chloromethyl)benzene if benzene is taken as the parent structure. Both are correct, and the first is common usage. Next: $C_{6}H_{5}OH$ is phenol, $C_{6}H_{5}CH_{3}$ toluene and $C_{6}H_{5}NH_{2}$ aniline. These three are retained names that official nomenclature keeps, and you must know them because they serve as parent names themselves: we speak of cresol, nitrotoluene or chloroaniline without ever going back to benzene.

Ortho corresponds to positions 1,2; meta to positions 1,3; para to positions 1,4. So 1,4-dimethylbenzene is also called para-xylene, and 1,2-dichlorobenzene is called ortho-dichlorobenzene. For a benzene disubstituted by two IDENTICAL groups there are exactly three positional isomers, ortho, meta and para, because positions 1,5 and 1,6 reduce by rotation to positions 1,3 and 1,2. For a benzene trisubstituted by three identical groups there are also three: 1,2,3, 1,2,4 and 1,3,5. Counting is a classic exercise where the mistake is to count as distinct arrangements that superimpose by a simple rotation of the ring.

A hydroxyl group $-OH$ (alcohol) and a carboxyl group $-COOH$ (carboxylic acid). The acid has PRIORITY: it imposes the suffix (-oic acid) and its carbon is necessarily C1; the alcohol is demoted to the rank of prefix “hydroxy”. Chain of 4 carbons, OH on C3: 3-hydroxybutanoic acid (the “ketone body” β-hydroxybutyrate of biologists).

$CH_{3}CH(OH)CH_{2}CH_{3}$: butan-2-ol, a SECONDARY alcohol (the carbon bearing the OH is bonded to two carbons). $CH_{3}COCH_{2}CH_{3}$: butan-2-one, a ketone. $CH_{3}CH_{2}CH_{2}CHO$: butanal, an aldehyde (a group that is necessarily at the end of the chain, so no locant is needed). Oxidation: butan-2-ol (a secondary alcohol) is oxidized to butan-2-one; butanal comes from the mild oxidation of butan-1-ol, and is itself oxidized to butanoic acid. A tertiary alcohol, with no H on its carbon, is not oxidized.

The two esters: ethyl methanoate $HCOOCH_{2}CH_{3}$ (smell of rum) and methyl ethanoate $CH_{3}COOCH_{3}$ (smell of glue). Degree of unsaturation: $DU=\frac{2(3)+2-6}{2}=1$: it is the $C=O$ double bond, present in all three isomers. Three molecules, three different functional groups or connectivities, a single $DU$: the degree of unsaturation counts the unsaturations but does not say where they are.

Chain of 4 carbons. Groups present: carboxylic acid (priority: suffix -oic acid, C1), alcohol (prefix hydroxy-, on C4), double bond (infix -en-, between C2 and C3). Configuration: on C2, $COOH$ has priority over H; on C3, $CH_{2}OH$ has priority over H; the two priority groups are on opposite sides: E. Assembly: (E)-4-hydroxybut-2-enoic acid. The name reads like a sentence: every piece has its place imposed by the priorities.

$CH_{3}CH_{2}NH_{2}$ is ethanamine, a primary amine, the nitrogen carrying only one carbon group. $(CH_{3})_{2}NH$ is N-methylmethanamine, or dimethylamine, a secondary amine. $(CH_{3}CH_{2})_{3}N$ is N,N-diethylethanamine, or triethylamine, a tertiary amine. $(CH_{3})_{3}C-NH_{2}$ is 2-methylpropan-2-amine: it is indeed a PRIMARY amine, because its nitrogen carries only one carbon group, even though that carbon is itself tertiary. This is the classic trap of the chapter: for alcohols and alkyl halides, the class is read on the FUNCTIONAL CARBON, but for amines it is read on the NITROGEN. Confusing the two conventions leads to errors about reactivity, since a primary amine and a tertiary alcohol have nothing in common.

$CH_{3}CH_{2}CONH_{2}$ is propanamide, an amide. $CH_{3}CH_{2}C\equiv N$ is propanenitrile, a nitrile; the carbon of the nitrile group counts in the parent chain and has locant 1, which explains why $CH_{3}CH_{2}C\equiv N$, with three carbons in all, the nitrile one included, is named on the root “propane” and not “ethane”. This is the most common mistake on this functional group. $CH_{3}COOCH_{2}CH_{3}$ is ethyl ethanoate, an ester, whose name is built in two words, the acid giving the first part and the alcohol the second. $CH_{3}CH_{2}OCH_{3}$ is methoxyethane, an ether; since an ether has no suffix of its own, it always appears as an alkoxy prefix.

The carboxylic acid has the highest priority of the three: it gives the suffix, so the molecule is named as an “...oic acid” and its carbon has locant 1. The ketone, of lower priority, appears as the prefix “oxo”, and the alcohol as the prefix “hydroxy”. For $HOCH_{2}-CO-CH_{2}-COOH$, the chain has four carbons; numbering from the acid carbon, the ketone is on 3 and the alcohol on 4. The name is therefore 4-hydroxy-3-oxobutanoic acid, the prefixes being listed in alphabetical order. The method is always the same: identify the priority group, make it the suffix, number from it, and relegate everything else to prefixes.

Systematic nomenclature gives 2-aminoethanoic acid, the common name being glycine. This compound carries both an acid group and a basic group, and the carboxylic acid is quite strong enough to protonate the amine: the molecule reacts with itself and exists in solution as a ZWITTERION, $H_{3}N^{+}-CH_{2}-COO^{-}$, electrically neutral overall but carrying two opposite charges. The consequences are considerable. A species carrying two charges is strongly solvated by water: glycine is very soluble in it, much more than an acid of similar size such as propanoic acid. Above all, in the solid state, the zwitterions attract each other through genuine electrostatic forces and not through mere hydrogen bonds: the crystal behaves almost like an ionic solid. Glycine thus melts around 233 °C with decomposition, whereas propanoic acid, of comparable molar mass, melts at $-21$ °C. A gap of more than 250 degrees between two molecules of similar size is entirely explained by an internal acid-base reaction.

Closing a ring means creating a C-C bond between two carbons that each carried an H: two hydrogens disappear. Creating a $\pi$ bond (going from C-C to C=C) also removes one H from each carbon. Each “unsaturation” therefore costs exactly 2 H relative to the maximum $2c+2$, hence the division by 2. Oxygen, being divalent, is INSERTED into an existing bond (C-O-H or C-O-C) without changing the hydrogen count: it is invisible to the formula. Nitrogen, being trivalent, brings one more place instead, hence the $+n$.

$DU=\frac{2(4)+2-6}{2}=2$. Three very different readings of the same number: buta-1,3-diene (two $\pi$ bonds), but-1-yne (one triple bond, worth two $\pi$ bonds), and cyclobutene (one ring + one $\pi$ bond). Same molecular formula, unrelated chemistries: the $DU$ constrains the structure, it does not give it.

$DU=\frac{2(7)+2-8}{2}=4$: exactly the signature of a benzene ring (1 ring + 3 $\pi$ bonds), which uses up all four unsaturations: the rest of the molecule is saturated and acyclic. Possible functional isomers: benzyl alcohol $C_{6}H_{5}CH_{2}OH$, the cresols $CH_{3}C_{6}H_{4}OH$ (phenols), or anisole $C_{6}H_{5}OCH_{3}$ (an ether): an alcohol, a phenol and an ether on the same molecular formula.

Limonene: $DU=\frac{2(10)+2-16}{2}=3$; two unsaturations are the double bonds given, leaving $3-2=1$ RING (the six-membered ring of limonene). Caffeine: $DU=\frac{2(8)+2+4-10}{2}=\frac{12}{2}=6$, and the inventory indeed gives $2$ rings $+2$ ($C=O$) $+2$ ($C=N$) $=6$. The formula, apart from exotic atoms ($+n$ for nitrogen), holds for any molecule of biochemistry.

Butan-1-ol / butan-2-ol: POSITION isomers (same alcohol group, same skeleton, different carbon). Butan-1-ol / ethoxyethane: FUNCTIONAL isomers (same $C_{4}H_{10}O$, alcohol against ether). Butane / 2-methylpropane: CHAIN isomers (straight skeleton against branched). Propanal / propanone: FUNCTIONAL isomers (aldehyde against ketone, same $C_{3}H_{6}O$). The first three types make up constitutional isomerism: the connectivity of the atoms differs.

$DU=1$: one double bond OR one ring. Alkenes: but-1-ene, (Z)-but-2-ene, (E)-but-2-ene, 2-methylprop-1-ene. Rings: cyclobutane and methylcyclopropane. Six molecules in all: five different CONSTITUTIONS (but-1-ene, but-2-ene, 2-methylpropene, cyclobutane, methylcyclopropane), and within but-2-ene, the (Z)/(E) pair: same connectivity, different spatial arrangement, these being the only two stereoisomers in the list.

On each carbon of the double bond of but-2-ene, we must compare the two groups carried: here $CH_{3}$ against H, and $CH_{3}$ has priority (C > H). Molecule 1: the two methyl groups on the SAME side of the plane of the double bond: (Z)-but-2-ene. Molecule 2: methyl groups on opposite sides: (E)-but-2-ene. But-1-ene has two identical H on its terminal carbon, and 2-methylpropene has two identical $CH_{3}$ on the same carbon: as soon as a carbon of the double bond carries two identical groups, swapping sides changes nothing, so there is no Z/E.

Going from Z to E requires rotating about the $C=C$ bond, and therefore BREAKING the $\pi$ bond: about 260 kJ/mol, out of reach at room temperature. The two molecules can therefore be separated and stored in different bottles: they are DIASTEREOMERS, with different properties: (Z), whose two methyl groups on the same side create a small dipole moment, boils at about 4 °C, while (E), more symmetrical, boils at about 1 °C, and their melting points differ even more. About a SINGLE bond, rotation breaks nothing: a barrier of a few kJ/mol, crossed billions of times per second: the conformers interconvert constantly and will never be two distinct products.

Carbon 2 of butan-2-ol carries four groups, ALL different: $OH$, $CH_{3}$, $C_{2}H_{5}$ and H: it is an asymmetric carbon (stereocentre). Priorities: $OH$ first (O, atomic number 8), then ethyl and methyl, two tied carbons, must be ranked: we look at their neighbours: ethyl carries (C, H, H), methyl (H, H, H); C > H at the first comparison, so ethyl wins. Ranking: $OH>C_{2}H_{5}>CH_{3}>H$.

A single carbon, four atoms directly comparable by atomic number: $Br(35)>Cl(17)>F(9)>H(1)$. Procedure: orient the molecule with group 4 (here H) pointing to the BACK, then follow 1→2→3 (Br→Cl→F): clockwise = R, counterclockwise = S. If the drawing puts group 4 toward the front, there is no need to twist the molecule: read the apparent direction and REVERSE the answer. This molecule, the smallest example of an asymmetric carbon, does exist as two enantiomers.

The rule predicts $2^{2}=4$ combinations: (R,R), (S,S), (R,S) and (S,R). But the molecule is symmetrical (the two $-CHOH-COOH$ halves are identical): the (R,S) form is superimposable on the (S,R) form after flipping, it is ONE SINGLE molecule, achiral despite its two stereocentres: the MESO compound, which has an internal mirror plane. So only 3 stereoisomers exist: the pair of enantiomers (R,R)/(S,S), and the meso compound, a DIASTEREOMER of the first two (melting point 140 °C against 170 °C, no optical rotation).

Two enantiomers share ALL scalar properties: same melting point, same solubility, same spectra: only one physical measurement distinguishes them, the direction in which they rotate the plane of polarized light (+α and −α). But in an environment that is ITSELF chiral they become different: our olfactory receptors are chiral proteins, and (R)-limonene smells of orange while (S) leans toward lemon and turpentine. The same logic, with much higher stakes, applies to drugs: biological targets being chiral, one enantiomer may heal while the other is inactive or harmful: modern pharmacopoeias require testing them, and often producing them, separately.

The Fischer convention is not symmetrical: the horizontal comes out of the sheet and the vertical goes into it. A rotation by 90° therefore swaps the two directions, and turns the bonds that pointed toward the viewer into bonds pointing away: the configuration represented is REVERSED, and without noticing you obtain the drawing of the enantiomer. A rotation by 180°, on the other hand, brings the horizontal back onto the horizontal and the vertical onto the vertical: both directions keep their meaning and the configuration is intact. This is trap number one of the Fischer projection, and it does not show on the drawing, which remains perfectly plausible.

First rank the four groups by the Cahn, Ingold and Prelog rules, from the highest atomic number to the lowest. Then read the direction of rotation from 1 to 2 to 3, as it appears on the cross. If the priority 4 group is VERTICAL, it points to the back, which is the standard position: the direction read is the right one, clockwise gives $R$ and counterclockwise gives $S$. If group 4 is HORIZONTAL, it points toward the viewer, that is, opposite to the standard position: we read the direction in the same way, then REVERSE the conclusion. The justification is simple: looking at a molecule from the side of the lowest-priority group, instead of from the opposite side, amounts to looking at it in a mirror as far as the direction of rotation is concerned, so the apparent direction is reversed.

Swapping two groups on an asymmetric carbon produces its enantiomer, and therefore reverses the descriptor; two successive swaps, whichever they are, bring us back to the starting configuration. We can therefore, without changing the molecule, make two well-chosen swaps to bring the priority 4 group to the bottom position. Once this is done, we are always in the standard case, where the direction read translates directly into $R$ or $S$ with no reversal to remember. The benefit is that it removes the conditional rule of b), which is the most frequent source of errors on exams: it is better to make two swaps on scratch paper than to try to apply a mental reversal under time pressure.

The priorities were established in exercise $7$: $OH>C_{2}H_{5}>CH_{3}>H$. Wedge-and-dash drawing: the hydrogen is carried by the hashed bond, so it points to the back, the standard position. We read the path $OH$ at the top, ethyl at the bottom right, methyl at the bottom left: it is CLOCKWISE, with no reversal to make, so $R$. Fischer projection: the hydrogen is on the horizontal, on the right, so directed TOWARD the viewer. We read $OH$ on the left, ethyl at the bottom, methyl at the top: from the left we go down to the bottom, then back up on the right toward the top, which is COUNTERCLOCKWISE. Since group 4 points toward us, we reverse: $R$. Check by two swaps. We swap the hydrogen and the ethyl: the hydrogen goes to the bottom, the ethyl to the right, and the configuration is reversed. We then swap the $OH$ and the methyl: $OH$ at the top, methyl on the left, and the configuration is restored. Group 4 is now vertical, toward the back; the path $OH$ at the top, ethyl on the right, methyl on the left is clockwise, so $R$ with no reversal to remember: the two methods agree. Conclusion: the two drawings, which look nothing alike, show the SAME molecule, $(R)$-butan-2-ol. That is the lesson of the question: we never compare two representations by eye, we compare their descriptors.

The concentration is $c=\dfrac{10.0}{100}=0.100$ g/mL. Biot's law gives $\alpha=66.5\times 1.00\times 0.100=+6.65^{\circ}$. In a 2.00 dm cell, the light passes through twice as many molecules and the angle doubles: $+13.3^{\circ}$. With a solution twice as dilute in the original cell, the angle is halved: $+3.33^{\circ}$. The measured angle therefore depends entirely on the setup, whereas the specific rotation $[\alpha]$ is a constant of the substance, which is precisely why it is defined that way. The solvent and the temperature at which it was measured must always be stated.

The enantiomeric excess is the ratio of the observed rotation to that of the pure enantiomer: $ee=\dfrac{50.0}{62.5}\times 100=80.0\,\%$. Let $x$ be the percentage of the $R$ enantiomer, the major one since the sign is negative, and $y$ that of $S$; we have $x+y=100$ and $x-y=80.0$, so $x=90.0\,\%$ and $y=10.0\,\%$. The sample contains 90 % of (R)-carvone and 10 % of (S)-carvone. The reasoning is worth understanding rather than memorizing: the 10 % of $S$ cancel 10 % of $R$, leaving 80 % of $R$ rotating alone, hence an optical rotation equal to 80 % of that of the pure enantiomer.

If all the reagents, the solvent and the catalyst are achiral, the reaction medium has no way of telling left from right. Creating an asymmetric carbon means choosing between two faces of a plane or between two sides of an intermediate, and these two possibilities are strictly equivalent in energy in an achiral medium: the two transition states leading to them are mirror images of each other, hence of the same energy, hence taken at the same rate. The two enantiomers are thus formed in strictly equal amounts, and the optical rotation of the mixture is zero by exact cancellation. This zero is misleading: every individual molecule is indeed chiral and optically active, but for each one that deflects the light one way, another deflects it the other way. Obtaining a single enantiomer requires introducing chiral information into the system, in the form of a chiral catalyst, an enzyme or an auxiliary reagent.

In an achiral environment, two enantiomers have exactly the same boiling points, the same solubilities and the same adsorption constants: no distillation, no crystallization, no ordinary chromatography can separate them. Reacting the mixture with a pure CHIRAL base changes everything: we obtain two salts that are no longer enantiomers but DIASTEREOMERS, since they differ in the configuration of only one centre, the other being identical. Now diastereomers have different physical properties: one crystallizes, the other stays in solution, and a simple filtration separates them. All that remains is to treat each salt with a strong acid to regenerate the starting acid, this time enantiomerically pure, and to recover the chiral base. The whole strategy rests on this idea: temporarily turning an undetectable difference into an exploitable one. The stakes go far beyond the laboratory, because living matter is itself chiral. Receptors and enzymes recognize only one enantiomer, which explains why the two forms of carvone smell one of caraway and the other of spearmint, and above all why two enantiomers of a drug can have different effects, the thalidomide tragedy having given in 1960 the most dramatic demonstration. Regulatory agencies now require the activity of each enantiomer to be documented separately.

A conformation is a spatial arrangement obtained by simple ROTATION about single bonds, without breaking anything. Anti: methyl groups opposite, angle 180°, the absolute minimum of the curve (0 kJ/mol): the two big groups are as far apart as possible. Gauche: the two local minima at 60° and 300°, at 3.8 kJ/mol: staggered but with neighbouring methyl groups. Eclipsed: the three maxima (0°, 120°, 240°), where the bonds overlap. The molecule lives essentially at the bottom of the three wells and jumps from one to another constantly.

At 120° and 240°, each methyl group eclipses a hydrogen: this is the “ordinary” eclipsing repulsion (16 kJ/mol). At 0°, the TWO METHYL GROUPS eclipse each other: on top of the torsional repulsions comes the direct steric crowding of the two biggest groups, hence the peak at 19 kJ/mol. The hierarchy of the maxima reads directly the size of the groups that pass each other.

Boltzmann factor per gauche conformation: $e^{-3800/(8.314\times 298)}=e^{-1.53}\approx 0.216$. With two gauche conformations for one anti: $f_{anti}=\frac{1}{1+2(0.216)}\approx 0.70$. At room temperature, butane is anti about 70 % of the time and gauche 30 %: the energy curve translates directly into populations, and what we measure is indeed the average of all these forms.

The chair of cyclohexane places all six C-C bonds in a perfectly staggered arrangement (no eclipsing, angles close to 109.5°): it is strain-free, unlike the boat form, which is eclipsed and crowded. A methyl group there has the choice between the AXIAL position (vertical, hindered by the two axial H on the same side of the ring) and the EQUATORIAL position (pointing outward, unhindered): the equatorial wins. Boltzmann ratio: $e^{+7600/(8.314\times 298)}\approx 21.5$, that is $\frac{21.5}{22.5}\approx 96$ % of the molecules with the methyl group equatorial. Golden rule of all the chemistry of sugars and steroids: big groups settle in the equatorial position.

The longest chain containing the double bond has five carbons. Numbering from the right, the double bond starts at C2 and the methyl group is on C3; from the left, the double bond would start at C3. We keep the first numbering: 3-methylpent-2-ene.

The chain has five carbons. From the end with the double bond, the double bond is on 1 and the triple bond on 4; from the other end, the triple bond on 1 and the double bond on 4. The two sets of locants are equal, {1, 4}: the tie rule then gives the lowest locant to the double bond. The name is pent-1-en-4-yne. The trap is to write “pent-4-en-1-yne” in the belief that the triple bond, being more unsaturated, comes first.

The parent chain contains the double bond and has six carbons: $C_{1}H_{3}-C_{2}H=C_{3}H-C_{4}H(CH_{3})-C_{5}H_{2}-C_{6}H_{3}$. Name: 4-methylhex-2-ene. C4 carries four different groups, a hydrogen, a methyl group, an ethyl group and the propenyl group: it is an asymmetric carbon. The molecule therefore has two independent stereogenic elements, the double bond ($Z$ or $E$) and the asymmetric carbon ($R$ or $S$): $2\times 2=4$ stereoisomers, (2Z,4R), (2Z,4S), (2E,4R) and (2E,4S).

3,3-Dimethylbut-1-yne: $HC\equiv C-C(CH_{3})_{3}$, that is $C_{6}H_{10}$ and $DU=\frac{2(6)+2-10}{2}=2$, the two $\pi$ bonds of the triple bond. Cyclohexa-1,3-diene: a six-carbon ring bearing two conjugated double bonds, $C_{6}H_{8}$, and $DU=\frac{2(6)+2-8}{2}=3$, one ring plus two $\pi$ bonds.

“Pent-3-ene” is wrong: numbered from the other end, it is pent-2-ene. “2-Methylbut-3-yne” is wrong for the same reason: the triple bond must receive the lowest locant, before the methyl group, and the correct name is 3-methylbut-1-yne. “3-Methylbut-1-ene” is correct: the double bond is already on 1.

The first four atoms are carbons: we compare the attached sets. $-COOH$: (O, O, O), the oxygen of the $C=O$ being duplicated. $-CHO$: (O, O, H). $-CH_{2}OH$: (O, H, H). $-CH(CH_{3})_{2}$: (C, C, H). Ranking: $-COOH>-CHO>-CH_{2}OH>-CH(CH_{3})_{2}$. A big carbon group loses against a small oxygen group: it is the atomic number that counts, never the size.

At the first rank, ethynyl and tert-butyl carry (C, C, C), vinyl and isopropyl (C, C, H). To separate ethynyl from tert-butyl, we go down one rank: the real terminal carbon of ethynyl carries (C, C, H) thanks to its duplicates, whereas each methyl group of tert-butyl carries only (H, H, H). Same reasoning for vinyl, whose $CH_{2}$ carries (C, H, H), against (H, H, H) for the methyl groups of isopropyl. Ranking: $-C\equiv CH>-C(CH_{3})_{3}>-CH=CH_{2}>-CH(CH_{3})_{2}$.

C3 carries $OH$, a vinyl group, an isopropyl group and a hydrogen: four different groups, so it is the asymmetric carbon. Ranking: $OH$ (oxygen), then vinyl, which beats isopropyl at the second rank as in b), then isopropyl, then H.

With group 4 toward the back, the path 1, 2, 3 is clockwise: the enantiomer is $R$. The student who swaps vinyl and isopropyl reads the path $OH\rightarrow$ isopropyl $\rightarrow$ vinyl, counterclockwise, and concludes $S$: a single priority error reverses the descriptor.

The vinyl group becomes an ethyl group: $CH_{3}CH_{2}-CH(OH)-CH(CH_{3})_{2}$, 2-methylpentan-3-ol. C3 remains asymmetric, but the priorities have changed: ethyl (C, H, H) now ranks AFTER isopropyl (C, C, H). Groups 2 and 3 swap ranks without anything moving in space: the path becomes counterclockwise and the product is $S$. No bond of the asymmetric carbon was broken, there is no inversion: only the name has changed.

On C2, we compare $CH_{3}$ and H: the methyl group wins. On C3, we compare $CH_{3}$ and $CH_{2}CH_{3}$: the ethyl group wins, (C, H, H) against (H, H, H). The two methyl groups are on the same side, so the methyl group of C2 and the ethyl group of C3 are on opposite sides: it is $(E)$-3-methylpent-2-ene. The trap is to look at the two identical groups and conclude “cis, so $Z$”.

On C1, Cl beats H; on C2, Br beats $CH_{3}$. Cl and Br on the same side: $(Z)$-2-bromo-1-chloroprop-1-ene. The hydrogen and the methyl group are then necessarily on the same side as well, on the other side of the double bond.

First rank: $-CH_{2}OH$ carries (O, H, H) and $-CH(CH_{3})_{2}$ carries (C, C, H). We compare the first atom of each set: O beats C, and the comparison stops there. $-CH_{2}OH$ has priority, even though it is smaller.

Hexa-2,4-diene has two stereogenic double bonds, so at most $2^{2}=4$ combinations: (2Z,4Z), (2E,4E), (2Z,4E) and (2E,4Z). But the molecule is symmetrical: (2Z,4E) read from the other end is (2E,4Z), the same molecule. There are therefore 3 stereoisomers. Hepta-2,4-diene lacks this symmetry, one end finishing with a methyl group and the other with an ethyl group: the 4 combinations are distinct.

In a six-membered ring, joining the two carbons of the double bond to two opposite positions would impose an enormous twist of the $\pi$ bond: trans-cyclohexene cannot exist as a stable molecule. A cyclic chain long enough to go around is needed; the smallest trans cycloalkene isolable at room temperature is trans-cyclooctene, with eight carbons.

$DU=\frac{2(5)+2-12}{2}=0$: neither ring nor multiple bond. The isomers are all acyclic and saturated, the oxygen being either in an $OH$ or in a $C-O-C$ bond.

We shorten the chain one carbon at a time. Chain of five: pentan-1-ol, pentan-2-ol, pentan-3-ol. Chain of four with one methyl group: 2-methylbutan-1-ol, 2-methylbutan-2-ol, 3-methylbutan-2-ol, 3-methylbutan-1-ol. Chain of three with two methyl groups: 2,2-dimethylpropan-1-ol. That makes $3+4+1=8$ alcohols. The false duplicates to avoid: “pentan-4-ol” is pentan-2-ol, “2-methylbutan-3-ol” is 3-methylbutan-2-ol.

Primary, the carbon of the $OH$ bonded to a single carbon: pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol, that is 4. Secondary: pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol, that is 3. Tertiary: 2-methylbutan-2-ol, alone, that is 1.

Pentan-2-ol: C2 carries OH, H, $CH_{3}$ and propyl, it is asymmetric. 2-Methylbutan-1-ol: C2 carries H, $CH_{3}$, ethyl and $CH_{2}OH$, asymmetric. 3-Methylbutan-2-ol: C2 carries OH, H, $CH_{3}$ and isopropyl, asymmetric. The other five have no carbon with four different groups: C3 of pentan-3-ol carries two ethyl groups, C2 of 2-methylbutan-2-ol two methyl groups. Three chiral alcohols, each as two enantiomers: $5+3\times 2=11$ bottles.

Secondary and chiral: pentan-2-ol and 3-methylbutan-2-ol; the only one with a branched chain is 3-methylbutan-2-ol. The only achiral secondary alcohol is pentan-3-ol, whose functional carbon carries two identical ethyl groups.

Six ring carbons, three from the isopropyl group and one from the methyl group: 10 carbons. A saturated ring bearing an $OH$: $C_{10}H_{20}O$. Check: $DU=\frac{2(10)+2-20}{2}=1$, exactly the ring.

The alcohol is the only functional group: it gives the suffix -ol and its carbon receives locant 1. We then go around in the direction that gives the lowest locants to the substituents: toward the isopropyl group, we find isopropyl on 2 and methyl on 5, that is {2, 5}; in the other direction, {3, 6}. The substituents are listed in alphabetical order: 5-methyl-2-(propan-2-yl)cyclohexan-1-ol, which is also written 2-isopropyl-5-methylcyclohexanol.

C1 carries OH, H and two different ring paths; C2 carries the isopropyl group, H and two different paths; C5 carries the methyl group, H and two different paths, one leading toward the $OH$, the other toward the isopropyl group. Three asymmetric carbons. The molecule has no internal symmetry, so no meso form: $2^{3}=8$ stereoisomers, that is 4 pairs of enantiomers, menthol, neomenthol, isomenthol and neoisomenthol.

The enantiomer reverses all three centres: (+)-menthol is (1S,2R,5S). (−)-Menthol has 8 stereoisomers in all; apart from itself and its enantiomer, the other 6 are its diastereomers. The trap is to invert a single centre and call the result an enantiomer: it is a diastereomer.

The ring flip turns each equatorial position into an axial one: in the other chair, the 3 substituents are axial, which is very unfavourable. (−)-Menthol, and its enantiomer, are the only stereoisomers that can place all three groups equatorial at once; all the others must put at least one axial in their best chair.