Corrected exercises: organic molecules, functional groups and isomers (202-SN1-RE)

Ethanol: ALCOHOL, $-OH$ group carried by a carbon. Ethanoic acid: CARBOXYLIC ACID, $-COOH$ group. Propanone: KETONE, $C=O$ group flanked by two carbons. Ethanal: ALDEHYDE, $-CHO$ group at the end of the chain. Ethyl ethanoate: ESTER, $-COO-$ group between two chains. Ethylamine: AMINE, $-NH_{2}$ group. Diethyl ether: ETHER, an oxygen INSERTED between two carbons.

Ethanal and propanone. Both do carry a carbonyl $C=O$, but its POSITION differs: in the aldehyde, the carbonyl carbon is at the end of the chain and still carries a hydrogen, which is written $-CHO$; in the ketone, it is in the middle, flanked by two carbons, and carries no hydrogen. This difference of a single hydrogen has a major chemical consequence: the aldehyde can still be oxidized to a carboxylic acid, whereas the ketone cannot, having no hydrogen to give up. This is exactly what Fehling's solution and Tollens' reagent test for.

Ethanol: $2$ carbons, $6$ hydrogens, $1$ oxygen, that is $C_{2}H_{6}O$. Dimethyl ether: $2$ carbons, $6$ hydrogens, $1$ oxygen, that is $C_{2}H_{6}O$ as well. They have the SAME molecular formula and yet they are two different substances: ethanol is a liquid that boils at $78$ °C, dimethyl ether a gas that boils at $-24$ °C. They are called functional isomers, and this is the first reason why a molecular formula never suffices to designate an organic molecule.

Ethanoic acid $CH_{3}COOH$, that is $C_{2}H_{4}O_{2}$: $M=2(12.01)+4(1.01)+2(16.00)=24.02+4.04+32.00=60.06$ g/mol. Counting the hydrogens takes care: three on the methyl group, one on the $-OH$ group of the acid, and none on the carbonyl carbon.

Ethyl ethanoate $CH_{3}-COO-CH_{2}-CH_{3}$: $4$ carbons, $8$ hydrogens, $2$ oxygens, that is $C_{4}H_{8}O_{2}$. $M=4(12.01)+8(1.01)+2(16.00)=48.04+8.08+32.00=88.12$ g/mol. This ester is the solvent in nail polish remover, and its fruity smell is the signature of the family: esters account for most of the aromas of banana, pear and pineapple.

Count: $CH_{3}$ gives $1$ carbon and $3$ hydrogens, the three $CH_{2}$ give $3$ carbons and $6$ hydrogens, the $OH$ gives $1$ oxygen and $1$ hydrogen. Total: $C_{4}H_{10}O$. The usual check is the valence: $4$ carbons with four bonds each make $16$ bonds leaving the carbons, of which $6$ are used by the three carbon-carbon bonds counted twice, so $10$ remain for the hydrogens and the oxygen.

Carbon-carbon bonds: $3$, since four carbons in a chain are joined by three bonds. Carbon-hydrogen bonds: $9$, that is the three of the terminal methyl plus two for each $CH_{2}$. Carbon-oxygen bond: $1$. Oxygen-hydrogen bond: $1$. Total: $3+9+1+1=14$ bonds. Note that the tenth hydrogen of the molecular formula is the one on the $-OH$ group, and that it is NOT bonded to a carbon: it is the one that does all the chemistry of the alcohol.

Four: three vertices and one end, the end being the terminal methyl group. Each vertex and each end of the broken line stands for a carbon, and the hydrogens are implied in just the number needed to complete a valence of $4$. The $-OH$ group, on the other hand, is written out: the rule is that every atom other than carbon, together with the hydrogens it carries, stays visible. The carbon bonded to the oxygen counts as a vertex, since the line bends there.

2-Methylbutan-1-ol has $5$ carbons. An extra carbon at the end of a branch carries three hydrogens, but it removes one from the carbon that carries it: the net change is $+CH_{2}$. So we go from $C_{4}H_{10}O$ to $C_{5}H_{12}O$. Check with the alkane rule: a saturated acyclic alcohol follows $C_{n}H_{2n+2}O$, and $2(5)+2=12$.

Because several distinct molecules share the same molecular formula: they are isomers. $C_{4}H_{10}O$ has SEVEN, four alcohols and three ethers, which have neither the same boiling point, nor the same solubility, nor the same reactivity. The molecular formula only gives the count of the atoms; it says nothing about how they are connected, and it is this connectivity that determines the properties. That is the reason for the other two ways of writing, and the subject of exercises 4 and 5.

The valence of $4$ is shared between carbon neighbours and hydrogens. Bonded to ONE carbon, it has $3$ bonds left, so $3$ hydrogens: a primary carbon, the terminal $CH_{3}$. Bonded to TWO carbons, $2$ hydrogens: a secondary carbon, $CH_{2}$. Bonded to THREE carbons, $1$ hydrogen: a tertiary carbon, $CH$. Bonded to FOUR carbons, $0$ hydrogens: a quaternary carbon. This scale is then used to classify alcohols, whose reactivity depends on the carbon that carries the $-OH$ group.

A saturated acyclic alkane follows $C_{n}H_{2n+2}$. The reasoning: each carbon brings $4$ bonds, that is $4n$ in all; the $n-1$ carbon-carbon bonds use up $2(n-1)$ of them; this leaves $4n-2n+2=2n+2$ bonds available for the hydrogens. Butane, $n=4$: $C_{4}H_{10}$, and $2(4)+2=10$. Octane, $n=8$: $C_{8}H_{18}$, and $2(8)+2=18$.

A cycloalkane follows $C_{n}H_{2n}$. Cyclohexane is therefore $C_{6}H_{12}$. Closing the ring costs exactly TWO hydrogens: one more carbon-carbon bond is needed to close the chain, and this bond ties up one valence at each of the two ends. The same reasoning holds for a double bond, which also costs two hydrogens. This is why a cyclohexane and a hexene have the same molecular formula $C_{6}H_{12}$ while having nothing else in common.

Three: pentane, a straight chain of five carbons; 2-methylbutane, a chain of four with a methyl group on the second carbon; 2,2-dimethylpropane, a chain of three with two methyl groups on the central carbon, which is then quaternary. There is no fourth possibility: a methyl group on the third carbon of a four-carbon chain gives 2-methylbutane again, seen from the other end, which is the most frequent counting error. Their boiling points, $36$, $28$ and $10$ °C, decrease with branching, the molecule becoming more compact and offering less contact surface.

Because oxygen opens a possibility that carbon alone does not have: it can sit AT THE END of a chain, giving an alcohol, or be INSERTED between two carbons, giving an ether. To the skeletal isomerism, the only kind available to $C_{5}H_{12}$, we therefore add positional isomerism of the $-OH$ group and FUNCTIONAL isomerism between alcohol and ether. Three sources instead of one, hence seven isomers instead of three. General rule to remember: a heteroatom multiplies the isomers much faster than an extra carbon.

ETHANOL $CH_{3}-CH_{2}-OH$, an alcohol, and DIMETHYL ETHER $CH_{3}-O-CH_{3}$, an ether. In both cases $2$ carbons, $6$ hydrogens, $1$ oxygen. The only difference is the place of the oxygen: at the end of the chain and carrying a hydrogen in the alcohol, inserted between the two carbons in the ether.

Ethanol boils at $78$ °C and dimethyl ether at $-24$ °C, a gap of $102$ degrees for the same molecular formula. The reason is HYDROGEN bonding: ethanol carries a hydrogen directly bonded to the oxygen, very polarized, which latches onto the oxygen of the neighbouring molecule. This network has to be broken to vaporize, and that is costly. Dimethyl ether has no hydrogen on its oxygen: its molecules are held together only by dipole-dipole forces, much weaker, and it is a gas at room temperature. A gap of a hundred degrees for a single hydrogen moved is the clearest demonstration that connectivity determines properties.

PROPANAL $CH_{3}-CH_{2}-CHO$, an aldehyde; PROPANONE $CH_{3}-CO-CH_{3}$, a ketone; prop-2-en-1-ol $CH_{2}=CH-CH_{2}-OH$, an unsaturated alcohol. All three have $3$ carbons, $6$ hydrogens and $1$ oxygen. The third deserves a word: it needs a carbon-carbon double bond to account for the two hydrogens that the carbonyl of the other two saves, which illustrates the rule of the previous exercise, namely that an unsaturation costs two hydrogens, whether it is $C=C$ or $C=O$.

$M=3(12.01)+6(1.01)+1(16.00)=36.03+6.06+16.00=58.09$ g/mol. It is necessarily the same for all three, since the molar mass depends only on the count of the atoms. This is precisely why an elemental analysis, which gives only the molecular formula, can never settle the question alone: a chemical test or a spectroscopy is needed.

Fehling's test, or Tollens' test. The aldehyde, whose carbonyl carbon still carries a hydrogen, is oxidized to a carboxylic acid: blue Fehling's solution gives a brick-red precipitate of copper(I) oxide, and Tollens' reagent deposits a silver mirror on the wall of the tube. Propanone, whose carbonyl carbon carries no hydrogen, reacts with neither. A single hydrogen thus separates two entirely different chemical behaviours, and it is the same hydrogen that already decided exercise 1.

A saturated acyclic alcohol or ether follows $C_{n}H_{2n+2}O$, because the oxygen is inserted into a bond without costing any hydrogens. For $n=4$: $2(4)+2=10$, exactly the count in the statement. The molecule is therefore saturated and acyclic. If there were a ring or a double bond, two hydrogens would be missing and the formula would be $C_{4}H_{8}O$. This check takes ten seconds and avoids looking for isomers that do not exist.

Four. On a straight chain of four carbons, the $-OH$ group can go on carbon $1$, giving BUTAN-1-OL, or on carbon $2$, giving BUTAN-2-OL; positions $3$ and $4$ give the first two again, seen from the other end. On the branched skeleton, a three-carbon chain carrying a methyl group, the $-OH$ can go at the end, giving 2-METHYLPROPAN-1-OL, or on the central carbon, giving 2-METHYLPROPAN-2-OL. Two skeletons, two positions each: four alcohols.

Three. An ether splits the four carbons into two pieces on either side of the oxygen. The possible splits are $2+2$, giving ETHOXYETHANE $CH_{3}CH_{2}-O-CH_{2}CH_{3}$, and $1+3$, but the three-carbon piece exists in two forms, straight or branched: hence 1-METHOXYPROPANE $CH_{3}-O-CH_{2}CH_{2}CH_{3}$ and 2-METHOXYPROPANE $CH_{3}-O-CH(CH_{3})_{2}$. The split $0+4$ makes no sense, since the oxygen must lie between two carbons. Total: $4+3=7$ isomers.

The class of an alcohol is read from the number of carbons attached to the carbon that carries the $-OH$. Butan-1-ol: this carbon is bonded to only one other carbon, so PRIMARY. Butan-2-ol: bonded to two carbons, SECONDARY. 2-Methylpropan-1-ol: bonded to only one carbon, PRIMARY as well, despite its branching. 2-Methylpropan-2-ol: bonded to three carbons, TERTIARY. Summary: two primary, one secondary, one tertiary. The classic error is to believe that branching determines the class; it is the position of the group that does.

An ETHER: ETHOXYETHANE, at $35$ °C, is the one given, and all three ethers behave alike. No ether has a hydrogen bonded to the oxygen, so none can form a hydrogen bond WITH ITS NEIGHBOUR: its molecules are held together only by dipole-dipole and London forces. The four alcohols, on the other hand, all form them, and boil from $82$ to $118$ °C, at least $47$ degrees higher. Seven molecules, the same atoms, and a boiling gap of more than $80$ degrees between the extremes: this is the whole point of the chapter, and it is what Part B builds on.

The two carbons of the double bond each carry one methyl group and one hydrogen. Since the double bond prevents any rotation, two distinct arrangements are possible: the two methyl groups on the SAME side of the plane of the double bond, which is the CIS isomer, or on opposite sides, which is the TRANS isomer. These are two different substances, with their own melting and boiling points, and not two drawings of the same molecule: no rotation can turn one into the other without breaking the $\pi$ bond.

Because one of the two carbons of the double bond carries TWO hydrogens, hence two identical substituents. Swapping two identical substituents produces nothing new: the two drawings we think are distinct are the same molecule flipped over. This is the rule: EACH of the two carbons of the double bond must carry two different substituents, and it takes only one that fails for the isomerism to disappear.

Butane has no double bond. Its carbon-carbon bonds are single, and a single bond rotates FREELY about its axis at room temperature. The arrangements we think are frozen when drawing interconvert billions of times per second: there is only one molecule of butane. The two causes are therefore quite distinct, and they must be named separately. But-1-ene has the rigidity but not the asymmetry; butane has the potential asymmetry but not the rigidity.

Two conditions, both necessary. First a RIGIDITY, that is a carbon-carbon double bond or membership of a ring, which prevents rotation. Second an ASYMMETRY: each of the two atoms involved must carry two substituents that differ from each other. If either one is missing, there is only one molecule.

The TRANS. Its two methyl groups are opposite, which gives an elongated and nearly symmetrical molecule that packs regularly in the crystal: the molecules touch along their whole length, the London forces add up, and the temperature has to go higher to break up the structure, up to $-106$ °C. The cis has its two methyl groups on the same side, which gives it a bent, bulky shape that packs poorly: the crystal is looser and melts as low as $-139$ °C, that is $33$ degrees lower. The lesson goes beyond this case: for the same composition, a more symmetrical molecule melts higher, and this is also why trans fats are solid at room temperature while cis oils stay liquid.

Butane: $4(12.01)+10(1.01)=48.04+10.10=58.14$ g/mol. Propanone: $3(12.01)+6(1.01)+16.00=36.03+6.06+16.00=58.09$ g/mol. Propan-1-ol: $3(12.01)+8(1.01)+16.00=36.03+8.08+16.00=60.11$ g/mol. All three fall within an interval of $2$ g/mol, less than $4\%$: molar mass therefore cannot explain any of the differences that follow, and that is the whole point of this trio.

Butane $-0.5$ °C: a nonpolar molecule, held only by LONDON forces. Propanone $56$ °C: the carbonyl $C=O$ is very polar and the molecule has a permanent dipole moment, hence DIPOLE-DIPOLE forces in addition to London forces. Propan-1-ol $97$ °C: the $-OH$ group carries a hydrogen bonded to a very electronegative atom, hence HYDROGEN BONDING between neighbouring molecules, the strongest of the three. The order of the temperatures follows exactly the order of the forces, with no exception.

Butane to propanone: $56-(-0.5)=56.5$ °C. Propanone to propan-1-ol: $97-56=41$ °C. Both jumps are considerable for molecules that differ only in their functional group. The first measures what a permanent dipole contributes, the second what hydrogen bonding adds on top. Remember the order of magnitude: changing functional group at constant mass shifts a boiling point by several tens of degrees, whereas adding a $CH_{2}$ to a chain shifts it by only about twenty.

Propan-1-ol and propanone are both MISCIBLE with water in all proportions, whereas butane is practically INSOLUBLE in it. The rule is that like dissolves like: water is polar and a hydrogen-bond donor, so it readily takes in molecules that can do the same. Propan-1-ol donates and accepts hydrogen bonds; propanone cannot donate any, having no hydrogen on its oxygen, but it can ACCEPT them from water, which is enough here. Butane can do neither, and dissolving it would require breaking hydrogen bonds of water while offering nothing in exchange.

As the chain gets longer, the molecule becomes more and more NONPOLAR over most of its length, whereas the $-OH$ group that provides the affinity for water stays the same. Methanol has only one carbon, the $-OH$ dominates, and it is miscible in all proportions. Butan-1-ol has four, the balance already tips the other way and the solubility falls to $73$ g/L. Octan-1-ol has eight: its chain behaves like a hydrocarbon and the solubility collapses to $0.5$ g/L. The ratio of the last two is $\dfrac{73}{0.5}=146$, for a difference of only four carbons. The same reasoning explains why a soap has a polar head and a nonpolar tail, and why vitamins fall into two families, water-soluble and fat-soluble.

FALSE. Ethanol and dimethyl ether share the formula $C_{2}H_{6}O$ and boil $102$ degrees apart, one being a liquid and the other a gas at room temperature. The molecular formula only gives the count of the atoms; the properties depend on the CONNECTIVITY, that is, on the way these atoms are joined. The correct sentence: two isomers have the same molecular formula and, in general, different properties.

TRUE. Both carry a carbon doubly bonded to an oxygen. The difference is its position: at the end of the chain for the aldehyde, which then keeps a hydrogen on this carbon, and between two carbons for the ketone, which keeps none. This single hydrogen decides everything else, since it allows the aldehyde to be oxidized to a carboxylic acid whereas the ketone cannot be.

FALSE. But-1-ene $CH_{2}=CH-CH_{2}-CH_{3}$ does have a double bond, hence the necessary rigidity, but its first carbon carries TWO hydrogens, hence two identical substituents. The two drawings we think are distinct are the same molecule flipped over. It is but-2-ene, whose double-bond carbons each carry one methyl group and one hydrogen, that exists as cis and trans. The correct sentence: but-1-ene has no geometric isomerism, because one of its doubly bonded carbons carries two identical substituents.

FALSE, and the order is reversed. The alcohol boils much higher, because its hydrogen bonded to oxygen allows hydrogen bonds between neighbouring molecules, which the ether cannot form. Ethanol $78$ °C against dimethyl ether $-24$ °C; butan-1-ol $118$ °C against ethoxyethane $35$ °C. The correct sentence: an alcohol boils markedly higher than the ether that is its isomer, because of hydrogen bonding.

FALSE. This is true for the first members of the series, methanol, ethanol and propan-1-ol, which are miscible in all proportions, but the solubility collapses as soon as the chain gets longer: $73$ g/L for butan-1-ol, $0.5$ g/L for octan-1-ol. The molecule behaves more and more like a hydrocarbon, the $-OH$ group no longer being enough to offset a long nonpolar chain. The correct sentence: short-chain alcohols are miscible with water, and the solubility decreases sharply as the number of carbons increases.

$M=3(12.01)+6(1.01)+2(16.00)=36.03+6.06+32.00=74.09$ g/mol.

CARBOXYLIC ACID, and the three observations back each other up. Indicator paper turning red signals an acid, which rules out the ester, the alcohol and the ether right away. The gas that turns limewater cloudy is carbon dioxide, and only an acid strong enough to displace carbonic acid from its salt makes hydrogen carbonate fizz: this is the test that distinguishes a carboxylic acid from a simple alcohol, which does nothing. Finally the fruity smell obtained with an alcohol in an acidic medium is the signature of an esterification, a reaction that requires precisely a carboxylic acid and an alcohol.

ONLY ONE: propanoic acid $CH_{3}-CH_{2}-COOH$. Once the $-COOH$ group is placed, using one carbon and both oxygens, only two carbons remain to be put in a chain, and there is only one way to do it. The other isomers of $C_{3}H_{6}O_{2}$ in the families of this chapter are ESTERS, two of them: ethyl methanoate $HCOO-CH_{2}CH_{3}$ and methyl ethanoate $CH_{3}COO-CH_{3}$. Three isomers in all among these families, only one of them an acid, and that is exactly why the three tests are enough to conclude.

$CH_{3}CH_{2}COOH+NaHCO_{3}\rightarrow CH_{3}CH_{2}COONa+H_{2}O+CO_{2}$. The acid gives up its proton to the hydrogen carbonate ion, which turns into carbonic acid, which immediately decomposes into water and carbon dioxide: it is this decomposition that produces the fizzing. The ratio is $1$ to $1$, and only one molecule of gas is released per molecule of acid.

$n=\dfrac{3.70}{74.09}=4.994\times 10^{-2}$ mol of acid. Since the ratio is $1$ to $1$ and the hydrogen carbonate is in excess, $4.994\times 10^{-2}$ mol of carbon dioxide forms. Volume: $V=4.994\times 10^{-2}\times 24.5=1.22$ L. Sanity check: $50$ millimoles of gas at $24.5$ L/mol give a little over a litre, the volume of a small bottle, which fits the spectacular fizzing observed for a few grams of acid.

Apply the usual rule, hydrogen at $+1$ and oxygen at $-2$, and solve for carbon. Ethanol $C_{2}H_{6}O$: $2x+6(+1)+(-2)=0$ gives $2x=-4$ and $x=-2$. Ethanal $C_{2}H_{4}O$: $2x+4(+1)+(-2)=0$ gives $2x=-2$ and $x=-1$. Ethanoic acid $C_{2}H_{4}O_{2}$: $2x+4(+1)+2(-2)=0$ gives $2x=0$ and $x=0$. The progression $-2$, $-1$, $0$ is perfectly regular, and it is what defines an oxidation chain.

Each step raises the average oxidation number by ONE unit, and the molecule has TWO carbons: each step therefore costs $2\times 1=2$ electrons, and the two steps together $4$ electrons per molecule. This can be checked directly between the extremes: from $-2$ to $0$ is $2$ units per carbon, that is $4$ electrons for two carbons. The constant number of carbons is what makes the calculation so simple here, and also what allows the three compounds to be compared directly.

Volume of pure ethanol: $750\times 0.12=90.0$ mL. Mass: $90.0\times 0.789=71.0$ g. Amount: $n=\dfrac{71.0}{46.07}=1.541$ mol. The trap in this step is the alcohol strength, which is a percentage by VOLUME and not by mass: you must go through the density of pure ethanol, never through that of the wine.

Oxidation keeps the two carbons: one molecule of ethanol gives one molecule of ethanoic acid, a ratio of $1$ to $1$. So $n=1.541$ mol of acid, and $m=1.541\times 60.06=92.6$ g. The mass increases although nothing was added to the skeleton: it is the oxygen taken from the air that makes the molecule heavier, from $46.07$ to $60.06$ g/mol.

$92.6$ g in $750$ mL, that is $\dfrac{92.6}{750}\times 100=12.3$ g per $100$ mL: the theoretical vinegar would be $12.3$ degrees. This is nearly double commercial vinegar, which is $6$ to $8$ degrees. There are two reasons for the gap, both instructive. The first is biological: acetic acid bacteria do not survive beyond about $10$ to $12$ degrees of acidity, and the fermentation stops by itself well before everything has been converted. The second is commercial: the vinegar sold is diluted to a standard strength. The calculation therefore gives the theoretical ceiling, and the gap with the real value is information here, not an error.