Corrected exercises: hydrocarbons (202-SNU-RE)

$DoU=\dfrac{2C+2-H}{2}$. Pentane: $\dfrac{12-12}{2}=0$. Pentene: $\dfrac{12-10}{2}=1$. Pentyne: $\dfrac{12-8}{2}=2$. The alkane is SATURATED: it has no ring and no multiple bond. The alkene has one degree, which corresponds to a double bond OR a ring, something the molecular formula alone cannot settle: cyclopentane also has the formula $C_{5}H_{10}$. The alkyne has two degrees: a triple bond, two double bonds, or a ring and a double bond.

The two molecules have the same formula, hence the same number of electrons and the same overall polarizability; only London forces hold them together. What differs is the CONTACT AREA: linear pentane lines up against its neighbours along its whole length, multiplying the points of interaction, while nearly spherical neopentane touches only over a small cap. Less contact means weaker London forces, hence a boiling point about 26 degrees lower. General rule: for the same formula, the more branched the molecule, the lower it boils.

Propane: $C_{3}H_{8}+5O_{2}\rightarrow 3CO_{2}+4H_{2}O$. Octane: since the coefficient $\dfrac{3\times 8+1}{2}=12.5$ is not an integer, double everything: $2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O$. Oxygen check: $2\times 25=50$ atoms on the left, $2\times 16+18=50$ on the right.

With too little oxygen, combustion is INCOMPLETE and produces carbon monoxide $CO$, even solid carbon in the form of soot. The danger of $CO$ comes from three properties together: it is odourless, colourless, and it binds to hemoglobin with an affinity about 240 times that of oxygen, blocking oxygen transport almost irreversibly. The victim notices nothing and falls asleep. This is why a combustion appliance, whether a stove, a heater or a generator, must never run in an enclosed space, and why a $CO$ detector is mandatory.

One litre of octane weighs $0.703\times 1000=703$ g. Its molar mass is $8\times 12.01+18\times 1.008=96.08+18.14=114.2$ g/mol, so $n=\dfrac{703}{114.2}=6.154$ mol. Each mole of octane gives 8 moles of $CO_{2}$, that is $8\times 6.154=49.23$ mol, and a mass of $49.23\times 44.01=2167$ g. One litre of gasoline therefore releases about 2.17 kg of $CO_{2}$, three times its own mass. This result always surprises, and it has a simple explanation: most of the mass of the $CO_{2}$ comes from the oxygen of the air, not from the fuel.

INITIATION: ultraviolet light or heat breaks the $Cl-Cl$ bond HOMOLYTICALLY, each atom taking one electron, which gives two $Cl\cdot$ radicals. PROPAGATION, in two steps: first $Cl\cdot+CH_{4}\rightarrow HCl+CH_{3}\cdot$, the chlorine radical abstracting a hydrogen; then $CH_{3}\cdot+Cl_{2}\rightarrow CH_{3}Cl+Cl\cdot$, which forms the product AND regenerates a chlorine radical. TERMINATION: any two radicals meet and pair up, for example $Cl\cdot+Cl\cdot\rightarrow Cl_{2}$ or $CH_{3}\cdot+CH_{3}\cdot\rightarrow C_{2}H_{6}$, which ends the chain.

Because each propagation cycle CONSUMES one radical and REGENERATES one: a single initial radical can therefore trigger thousands of cycles before it is destroyed by termination. A minimal initiation is enough to convert a large amount of reactants, which explains the very high quantum yield of these reactions. To start it, you need a source of energy able to break $Cl_{2}$ homolytically: ultraviolet light, strong heat, or a chemical initiator such as a peroxide. In the dark and in the cold, nothing at all happens.

By increasing stability: methyl < primary < secondary < TERTIARY. The reason is the same as for carbocations: the radical is an electron-deficient centre, and the neighbouring alkyl groups stabilize it by their inductive electron-donating effect and by hyperconjugation. The more alkyl substituents the radical carbon carries, the more stable the radical, so the more easily it forms. Direct consequence: abstraction of a TERTIARY hydrogen is the fastest.

Propane $CH_{3}-CH_{2}-CH_{3}$ has SIX primary hydrogens, on the two methyl groups, and TWO secondary hydrogens, on the central carbon. Weight each type by its reactivity: primary $6\times 1=6$, secondary $2\times 3.8=7.6$. The total is 13.6. The proportions are therefore $\dfrac{6}{13.6}=44$% of 1-chloropropane and $\dfrac{7.6}{13.6}=56$% of 2-chloropropane. The secondary product wins even though there are three times fewer hydrogens available: reactivity largely outweighs statistics, but without crushing them, which gives a mixture that is hard to separate.

That bromine is MUCH more selective than chlorine: its tertiary-to-primary reactivity ratio is of the order of 1600, against 5 for chlorine. The reason lies in the Hammond postulate: hydrogen abstraction by bromine is ENDOTHERMIC, so its transition state resembles the radical being formed, and the differences in stability between radicals are fully expressed in it. With chlorine, the same step is exothermic, the transition state resembles the reactants, and the differences fade. Bromine is therefore preferred in synthesis because it gives ONE nearly exclusive major product, usable without separation, whereas chlorination gives a mixture.

Because the $\pi$ bond is the WEAKEST in the system: about 267 kJ/mol, against 347 for the carbon-carbon $\sigma$ bond and 414 for a $C-H$ bond. An addition breaks only this $\pi$ bond and forms TWO new $\sigma$ bonds in exchange, which is very favourable energetically. In an alkane there is no weak bond: every reaction requires breaking a strong $\sigma$ bond, which is only possible by a high-energy radical pathway, and the result is then a substitution since the skeleton stays saturated.

With $H_{2}$, in the presence of a platinum, palladium or nickel catalyst: $CH_{3}-CH=CH_{2}+H_{2}\rightarrow CH_{3}-CH_{2}-CH_{3}$, propane. With $Br_{2}$: addition across both carbons, giving 1,2-dibromopropane, and the addition is ANTI, the two bromines ending up on opposite faces of the plane. With $HBr$: the major product is 2-bromopropane, in accordance with Markovnikov's rule.

MARKOVNIKOV'S RULE: when an $HX$ compound adds to an unsymmetrical alkene, the hydrogen attaches to the carbon that ALREADY carries the most hydrogens. The reason is that this mode of addition generates the more SUBSTITUTED carbocation, hence the more stable one. For 2-methylpropene $\left(CH_{3}\right)_{2}C=CH_{2}$ hydrated in acid medium: the proton attaches to the $CH_{2}$, generating a TERTIARY carbocation, which water then attacks. The product is 2-methylpropan-2-ol, a tertiary alcohol.

Two methods. First, HYDROBORATION-OXIDATION: borane $BH_{3}$ adds in a concerted fashion, boron attaching to the LESS hindered carbon for steric reasons, and the subsequent oxidation by $H_{2}O_{2}$ in basic medium replaces the boron with an $OH$. This gives the primary alcohol, here 2-methylpropan-1-ol, by syn addition. Second, for $HBr$ only, RADICAL addition in the presence of peroxides: the mechanism then goes through a radical rather than a carbocation, and it is the more stable radical that controls the outcome, which reverses the regioselectivity. This effect works only with $HBr$, not with $HCl$ or $HI$, for energetic reasons.

Add a few drops of orange bromine water to the sample and shake. If the colour DISAPPEARS immediately, the bromine has been consumed by addition: the sample contains an unsaturation, an alkene or an alkyne. If the colour persists, the compound is saturated, or aromatic, which does not undergo addition. The test therefore distinguishes UNSATURATED hydrocarbons from saturated and aromatic ones, which makes it the fastest qualitative test of the organic chemistry laboratory.

The total strain is obtained by multiplying the excess per $CH_{2}$ by the number of $CH_{2}$ groups in the ring. Cyclopropane: $3\times(697.1-658.7)=3\times 38.4=115$ kJ/mol. Cyclobutane: $4\times 27.5=110$ kJ/mol. Cyclopentane: $5\times 5.3=26.5$ kJ/mol. Cyclohexane: $6\times 0=0$ kJ/mol. Cycloheptane: $7\times 3.7=25.9$ kJ/mol. Cyclopropane is by far the most strained, and cyclohexane is the only ring completely free of strain: this is why it is everywhere in natural molecules, from steroids to sugars.

An $sp^{3}$ carbon wants angles of 109.5°. Cyclopropane, whose three carbons necessarily form an equilateral triangle, imposes 60°: the gap is almost 50°, the orbital overlap is off-axis, along the side, hence weak bonds and an unusual reactivity for an alkane, since cyclopropane opens under hydrogenation or on treatment with bromine. Cyclobutane, at 88°, suffers somewhat less angle strain. The decisive difference lies elsewhere: three points always define a plane, so cyclopropane CANNOT deform out of the plane, and its six $C-H$ bonds remain necessarily eclipsed in pairs. Cyclobutane has four vertices and can pucker slightly, which increases the angle strain a little but relieves the eclipsing strain even more: the balance is favourable, which is why real cyclobutane is not planar.

The residual strain of cyclopentane is not angular, it is TORSIONAL: a planar pentagon would indeed have good angles, but all its neighbouring $C-H$ bonds would be eclipsed, which costs a lot. Cyclopentane therefore adopts a slightly twisted envelope shape, which restores some staggering at the price of a little angle strain; 26.5 kJ/mol remain. Cyclohexane escapes this dilemma altogether because it has a conformation, the chair, in which all six angles are about 111° and in which all neighbouring bonds are perfectly staggered when viewed along a $C-C$ axis. No angle strain, no torsional strain, no crowding between substituents: the 120° of a planar hexagon is only a drawing artifact, the real molecule is never planar.

The ratio of populations follows $K=e^{\Delta E/RT}=e^{7300/(8.314\times 298)}=e^{2.95}=19.0$. The equatorial fraction is therefore $\dfrac{19.0}{1+19.0}=0.950$, that is 95% of the molecules. A difference of 7.3 kJ/mol, negligible on the scale of a chemical bond, is enough to tip the equilibrium in a ratio of 19 to 1: this is the exponential sensitivity of the Boltzmann factor, the same one that governs reaction rates in kinetics. The physical cause of this preference is the steric crowding between the axial methyl group and the two axial hydrogens on the same side of the ring, called the 1,3-diaxial interaction. For 1,2-dimethylcyclohexane, we must examine what each diastereomer allows. The trans isomer allows a chair in which both methyl groups are equatorial at the same time, hence without any 1,3-diaxial interaction. The cis isomer never allows this: whichever chair is chosen, one of the two methyl groups is necessarily axial, and the ring flip merely swaps their roles. The trans isomer is therefore the more stable, by about 7 kJ/mol, which is precisely the cost of one axial methyl group.

Alkane: $sp^{3}$ carbon, one part $s$ to three parts $p$, that is $\dfrac{1}{4}=25$% $s$ character. Alkene: $sp^{2}$, that is $\dfrac{1}{3}=33$%. Alkyne: $sp$, that is $\dfrac{1}{2}=50$%.

An $s$ orbital is spherical and centred on the nucleus, whereas a $p$ orbital has lobes that extend away from it. The more $s$ character a hybrid orbital contains, the CLOSER to the nucleus the electrons it holds are on average, hence the more strongly held and stabilized. Now the acidity of an $H$ is measured by the stability of the CONJUGATE BASE, that is of the carbanion left after the proton leaves. The lone pair of the carbanion occupies precisely the hybrid orbital of the carbon: in an $sp$ anion, with 50% $s$ character, it is much more stabilized than in an $sp^{3}$ anion with 25%. The terminal alkyne therefore gives up its proton much more readily.

The $pK_{a}$ gap is $50-25=25$ units, that is TWENTY-FIVE orders of magnitude, a factor of $10^{25}$. This is a colossal gap, comparable to the one separating hydrochloric acid from water, and it does not come from any functional group: the hybridization of the carbon alone is enough to produce it. That said, a $pK_{a}$ of 25 is still that of a very weak acid in absolute terms, much weaker than water: the terminal alkyne is acidic only by comparison with other hydrocarbons.

The amide YES, the hydroxide NO. An acid-base reaction is favourable only if it produces the WEAKER acid, hence the one with the larger $pK_{a}$. With $NaNH_{2}$, we would form $NH_{3}$, with $pK_{a}$ 38, from an alkyne of $pK_{a}$ 25: we go from 25 to 38, the reaction is favourable and quantitative. With $NaOH$, we would form water, with $pK_{a}$ 15.7, from the same alkyne: we would go from 25 toward 15.7, in the wrong direction, and the equilibrium would stay very largely on the left. This is why sodium amide in liquid ammonia is used and never sodium hydroxide.

The acetylide anion $R-C\equiv C^{-}$ is an excellent NUCLEOPHILE and a strong base. Its main use is ALKYLATION: it attacks a primary alkyl halide by an $S_{N}2$ mechanism, forming a new carbon-carbon bond and lengthening the chain. It is one of the few simple methods for BUILDING a carbon skeleton, which makes it a central tool of synthesis. Two limits: the alkyl halide must be primary, otherwise the very basic anion causes an elimination instead of the substitution; and the alkyne must be TERMINAL, since an internal alkyne carries no acidic hydrogen.

The proton attaches to a terminal carbon, which is the only choice that leaves a conjugated carbocation: $CH_{2}=CH-CH=CH_{2}+H^{+}\rightarrow CH_{3}-CH^{+}-CH=CH_{2}$. This carbocation is allylic, that is directly next to a double bond, and the double bond can shift: the second resonance form is $CH_{3}-CH=CH-CH_{2}^{+}$. Reality is neither of the two: the positive charge is delocalized and spread over the two ends of the system, carbons 2 and 4 of the original chain. This delocalization is what makes the cation much more stable than an ordinary secondary carbocation, and what explains why protonation takes place there and nowhere else.

The bromide ion can attack either of the two carbons carrying the charge. Attack on the carbon next to the one that received the proton gives $CH_{3}-CHBr-CH=CH_{2}$, 3-bromobut-1-ene, called the 1,2-addition product because $H$ and $Br$ have attached to two adjacent carbons of the diene. Attack on the other end gives $CH_{3}-CH=CH-CH_{2}Br$, 1-bromobut-2-ene, the 1,4-addition product, in which $H$ and $Br$ occupy the two ends of the system and the double bond has MOVED to the middle. This shift of the double bond is the signature of 1,4-addition, and it always comes as a surprise at first.

At low temperature, the 1,2-product largely dominates; at higher temperature, the 1,4-product does. The 1,2-product is therefore the KINETIC product: it forms faster, because the bromide ion is statistically closer to the carbon that has just received the proton and because the activation barrier leading to it is lower. The 1,4-product is the THERMODYNAMIC product: it is more stable, since its internal double bond is disubstituted whereas that of the 1,2-product is terminal, hence monosubstituted. Temperature decides because it determines whether or not the reaction is reversible. At $-80$ °C, the available energy is not enough to send the product back once formed: the composition reflects the RATES of formation, hence the barriers. At $+40$ °C, both products can revert to the carbocation and interconvert: the system reaches equilibrium and the composition reflects the STABILITIES. This is the canonical example of kinetic control versus thermodynamic control, and the general lesson is that the major product is not necessarily the most stable one.

The overall equation is $C_{4}H_{6}+C_{2}H_{4}\rightarrow C_{6}H_{10}$. Degrees of unsaturation: buta-1,3-diene has $DoU=\dfrac{2\times 4+2-6}{2}=2$, ethylene $DoU=1$, that is 3 in all; cyclohexene has $DoU=\dfrac{2\times 6+2-10}{2}=2$. One degree is indeed lost, as in any addition: the reactants carry three $\pi$ bonds, the product one $\pi$ bond and one ring, so two $\pi$ bonds have been traded for two new carbon-carbon $\sigma$ bonds. What makes this reaction exceptional is that it forms TWO carbon-carbon bonds simultaneously, in a single concerted step, with no intermediate, no catalyst and often just gentle heating. It builds a six-carbon ring in one go, the most common motif of organic chemistry, and it is stereospecific: the geometry of the reactants is carried intact into the product. Forming a carbon-carbon bond is the hardest and most valuable operation in organic synthesis; forming two in a single move earned Diels and Alder their 1950 Nobel Prize.

Propanal is $C_{3}H_{6}O$ and pentan-3-one is $C_{5}H_{10}O$, together $C_{8}H_{16}O_{2}$. The starting alkene, $C_{8}H_{16}$, has indeed gained exactly the two oxygen atoms brought by the cleavage: the balance is consistent, and it should always be done before drawing anything. We then reconstruct by erasing the two oxygens and rejoining the two carbons that carried them with a double bond. The carbon of propanal carried one hydrogen, that of pentan-3-one carried two ethyl groups. The alkene is therefore $CH_{3}CH_{2}CH=C(CH_{2}CH_{3})CH_{2}CH_{3}$. The longest chain through the double bond starts from the propanal part and runs through one of the two ethyl groups: $CH_{3}-CH_{2}-CH=C-CH_{2}-CH_{3}$, six carbons, the other ethyl group remaining as a substituent. The double bond is at position 3 from either end, and the ethyl group then receives the lowest locant: this is 3-ethylhex-3-ene. The trap is to stop at a five-carbon chain, “3-ethylpent-2-ene”, which would only account for seven carbons. Check of the formula: 8 carbons and 16 hydrogens.

Permanganate cuts at the same place, but it oxidizes any carbon bearing a hydrogen all the way to the carboxylic acid stage. We would therefore obtain propanoic acid $CH_{3}CH_{2}COOH$ and, unchanged, pentan-3-one, which cannot be oxidized further since its carbon is fully substituted. What is lost is the distinction between an aldehyde and an acid: ozonolysis tells you there was a hydrogen on that carbon, permanganate tells you too but less directly, and if the alkene had been terminal, the $CH_{2}=$ fragment would have given methanal, identifiable with ozonolysis, whereas permanganate pushes it on to carbon dioxide, which escapes and tells you nothing more. For a structure determination, reductive ozonolysis is therefore always preferable.

$DoU=\dfrac{2\times 6+2-10}{2}=2$: there are two unsaturations. Yet ozonolysis consumed only one, that of the double bond. The second can therefore only be a RING. Obtaining a single product rather than two is the decisive piece of information: if the double bond had been carried by an open chain, the cleavage would have released two separate fragments. A single product, carrying two carbonyl groups at its two ends, means that the two ends were already joined to each other by the rest of the molecule: cutting the ring at one point does not split it into two pieces, it opens it. The compound is therefore cyclohexene, and hexanedial is the ring unfolded. This reasoning, one product for a ring and two products for a chain, is the reflex to remember.

The simplest test is bromine water: cyclohexene decolourizes it immediately in the cold by addition to the double bond, whereas benzene leaves it orange. Dilute, cold permanganate gives the same verdict, with the purple colour disappearing and a brown precipitate of manganese dioxide appearing for cyclohexene, and nothing for benzene. The reason is that benzene has no true localized double bonds: its six $\pi$ electrons are delocalized over the whole ring and give it a resonance energy of about 150 kJ/mol, which any addition would destroy. Benzene therefore prefers substitution, which restores aromaticity, and it needs a catalyst for that. As for ozonolysis, it does cut a double bond, but it would be a very poor laboratory test: ozone is toxic and unstable, the setup is cumbersome, and the result is not a colour change but products that must then be identified. It is a tool for STRUCTURE DETERMINATION, not a recognition test.

Initiation: an initiator, typically a peroxide that splits on heating, releases a radical $R\cdot$ that adds to an ethylene molecule and gives $R-CH_{2}-CH_{2}\cdot$. Propagation: this new radical adds in turn to an ethylene molecule, which regenerates a radical at the end of the chain, now two carbons longer, and the operation repeats. Termination: two radicals meet and pair up, by combination or by disproportionation, which ends two chains at once. The chain grows for a very long time because propagation is fast and above all because the radicals are extremely dilute in the medium: a radical meets thousands of monomer molecules before it has a chance to meet another radical. This is a general principle of chain reactions, already encountered with the radical halogenation of alkanes.

The degree of polymerization is $\dfrac{1.4\times 10^{5}}{28.05}=4991$, or about 5000 units per chain. Mass percentage of carbon in the monomer: $\dfrac{2\times 12.01}{28.05}\times 100=85.6\%$; in the polymer, the chain is just the repetition of the same unit $-CH_{2}-CH_{2}-$, so the percentage composition is rigorously the same, 85.6%. This is the characteristic of ADDITION polymerizations: nothing is eliminated, the polymer has exactly the molecular formula of the monomer multiplied by $n$, and an elemental analysis can therefore never distinguish ethylene from polyethylene. A molar mass measurement is needed for that. Condensation polymerizations, such as those of nylon or polyester, behave differently since they eliminate a small molecule with each bond formed.

Chloroethene $CH_{2}=CHCl$ gives the repeating unit $-CH_{2}-CHCl-$: this is polyvinyl chloride. Styrene $CH_{2}=CH-C_{6}H_{5}$ gives $-CH_{2}-CH(C_{6}H_{5})-$: this is polystyrene. In both cases, as with propylene, which gives $-CH_{2}-CH(CH_{3})-$, each unit carries a substituent on every second carbon, and this substituent can point to one side or the other of the plane of the chain. If they all point to the same side, the polymer is isotactic; if they alternate regularly, it is syndiotactic; if they are arranged at random, it is atactic. Now a regular chain can pack with its neighbours into ordered crystalline regions, whereas an irregular chain cannot. Isotactic polypropylene is therefore a rigid, tough solid that melts at about 165 °C and is used to make bumpers and housings, whereas atactic polypropylene is a soft wax of no structural use. Same monomer, same molecular formula, materials with nothing in common: geometry decides, and it was the control of this geometry by the Ziegler-Natta catalysts that earned their inventors the Nobel Prize.

Low-density polyethylene is produced under very high pressure by a radical route, a process in which the growing radical sometimes abstracts a hydrogen from its own chain and creates a branch. Branched chains pack poorly, crystallinity stays low, and the material is of low density, soft and translucent: this is the plastic bag and the food wrap. High-density polyethylene is produced at low pressure over a catalyst, which gives linear chains without branching, able to pack tightly into crystalline regions: the material is denser, rigid, opaque and strong, this is the milk jug and the gas can. Same formula, same molar mass, everything depends on branching. As for degradation, it runs into a fundamental difficulty: a polyethylene chain is just an immense alkane molecule, and alkanes are chemically inert, with no polarized bond, no electrophilic or nucleophilic site, no functional group to which an enzyme could bind. On the scale of evolution, no microorganism has had the opportunity to develop an enzyme for a substrate that did not exist before 1930. The chemical stability that makes the material valuable is exactly what makes it a persistent waste, as was already the case for fluorinated and chlorinated compounds.

From ALKANE TO ALKENE: you must first install a leaving group, by radical halogenation, then bring about an ELIMINATION with a strong base, usually $KOH$ in hot alcoholic solution. Thermal cracking of petroleum achieves the same result on an industrial scale. From ALKENE TO ALKANE: simple catalytic HYDROGENATION with $H_{2}$ over platinum, palladium or nickel, a quantitative reaction with no by-product.

Two steps. First, addition of $Br_{2}$ to the alkene, which gives a vicinal DIBROMOALKANE, the two bromines sitting on neighbouring carbons. Second, DOUBLE ELIMINATION with two equivalents of sodium amide, each elimination removing $HBr$ and creating a $\pi$ bond. This gives the alkyne. The base must be very strong, because the second elimination, from a bromoalkene, is difficult.

Step 1, radical halogenation of propane with $Cl_{2}$ under ultraviolet light: a mixture is obtained, from which 2-chloropropane is isolated, the major product at 56% according to exercise 2. Step 2, elimination with hot alcoholic $KOH$: propene is obtained. Step 3, hydration in acid medium, $H_{2}O$ with $H_{2}SO_{4}$: Markovnikov's rule places the $OH$ on the central, more substituted carbon, and propan-2-ol is obtained. One could also, more directly, hydrate the propene obtained by cracking, but the statement requires starting from propane.

Pent-2-yne is $CH_{3}-C\equiv C-CH_{2}-CH_{3}$. It is built from propyne $CH_{3}-C\equiv CH$, which supplies three carbons, by adding an ethyl fragment. Step 1, deprotonation of propyne with $NaNH_{2}$ in liquid ammonia, which gives the acetylide anion $CH_{3}-C\equiv C^{-}$. Step 2, alkylation with BROMOETHANE, a primary alkyl halide, by an $S_{N}2$ mechanism: the anion attacks the carbon bearing the bromine and forms the carbon-carbon bond. Pent-2-yne is indeed obtained.

Because 2-bromopropane is a SECONDARY alkyl halide, and the acetylide anion is not only a good nucleophile: it is also a very strong BASE, its conjugate acid having a $pK_{a}$ of 25. On a hindered secondary substrate, $S_{N}2$ substitution is slowed by steric hindrance while $E2$ elimination, which needs only a hydrogen in the beta position and access to it, becomes competitive and wins. The major product would therefore be propene, along with regenerated propyne, and not the desired alkyne. RULE: acetylide alkylations only work with PRIMARY alkyl halides, preferably unbranched at the beta position.

On the difference in BOILING POINTS, which rise steadily with chain length because of London forces, as established in exercise 1. When the crude oil is heated at the bottom of a column, the lightest fractions, gases and gasolines, rise highest before condensing, while the heavy ones, gas oils and then bitumen, condense at the bottom. The separation is therefore purely PHYSICAL: no bond is broken, we merely exploit a property that varies regularly with molecular size.

Cracking breaks long chains into shorter fragments, by heating to a high temperature, thermal cracking, or over an acid catalyst, catalytic cracking. It is done because demand does not match the composition of crude oil: the market mostly wants gasoline, chains of 5 to 10 carbons, whereas petroleum contains an insufficient proportion of it. Cracking therefore converts the surplus heavy fractions into the light fractions in demand. Since not enough hydrogen is formed to saturate every fragment, it also produces ALKENES, an essential raw material of the petrochemical and polymer industries.

Because autoignition, or knocking, occurs when the mixture ignites spontaneously under compression before the spark, causing a destructive detonation. Autoignition goes through RADICALS, and we saw in exercise 2 that tertiary radicals, favoured by branching, form most easily. A linear alkane such as heptane readily generates secondary radicals in a chain along its molecule, which propagates the ignition: it detonates very easily, hence the rating of 0. Iso-octane, highly branched, offers mostly primary hydrogens, not very reactive, and its tertiary radicals are too stable to propagate the chain efficiently: it resists, hence the rating of 100. BRANCHING is therefore the key to the octane rating.

Reforming increases precisely the proportion of species that resist autoignition: branched compounds, for the reason given in c), and aromatics, whose resonance stability of about 150 kJ/mol makes them very hard to attack by a radical route. The octane rating of the fuel therefore rises, which allows higher compression ratios and, consequently, better engine efficiency. This is the operation that replaced the addition of tetraethyllead, since banned for health reasons.

The annual consumption is $\dfrac{8.0\times 15\,000}{100}=1200$ L. At 2.17 kg of $CO_{2}$ per litre, this gives $1200\times 2.17=2604$ kg, or about 2.6 tonnes of $CO_{2}$ per year for a single vehicle. This order of magnitude is worth remembering: it corresponds roughly to what a mature tree sequesters in a century, and it represents a significant share of an individual's carbon footprint in a region where heating is already largely electric, such as Quebec.

Step 1: $CH_{3}-H$ is broken and $H-Cl$ is formed: $439-432=+7$ kJ/mol. Step 2: $Cl-Cl$ is broken and $CH_{3}-Cl$ is formed: $243-350=-107$ kJ/mol. Overall: $7-107=-100$ kJ/mol, the sum of the two steps being the equation $CH_{4}+Cl_{2}\rightarrow CH_{3}Cl+HCl$.

Step 1: $439-366=+73$ kJ/mol. Step 2: $193-294=-101$ kJ/mol. Overall: $-28$ kJ/mol. The difference lies in step 1: slightly endothermic for chlorine, clearly endothermic for bromine. Its transition state is therefore late, close to the radical formed, which makes bromine slow and very selective.

Step 1: $439-570=-131$ kJ/mol. Step 2: $159-460=-301$ kJ/mol. Overall: $-432$ kJ/mol. Both steps are strongly exothermic and the $F-F$ bond is very weak: the chain runs away, the heat released speeds up the reaction, and the mixture can explode. An alkane is never fluorinated with pure fluorine.

Step 1: $439-298=+141$ kJ/mol. Step 2: $151-239=-88$ kJ/mol. Overall: $+53$ kJ/mol. The reaction is endothermic overall, and its first step is so costly that the chain does not propagate: the radical iodination of methane does not occur.

Initiation supplies only a few radicals, a tiny quantity compared with the thousands of propagation cycles they sustain; it is offset by termination. The overall balance of the reaction is the sum of the two propagation steps, which gives back exactly the overall equation.

Markovnikov's rule places the $OH$ on the more substituted carbon, the central one: the enol formed is prop-1-en-2-ol, $CH_{2}=C(OH)-CH_{3}$. It tautomerizes to propanone, $CH_{3}-CO-CH_{3}$. A terminal alkyne thus always gives a methyl ketone, never an aldehyde.

The enol and the ketone are tautomers: they differ by the position of a hydrogen and a double bond. The keto form, with its very strong $C=O$ bond, is much more stable, and the exchange is catalyzed by the acid in the medium: the equilibrium lies entirely on the side of the ketone.

Boron attaches to the terminal, less hindered carbon; oxidation replaces the boron with an $OH$ and gives the enol $CH_{3}-CH=CH-OH$, which tautomerizes to an aldehyde: propanal, $CH_{3}-CH_{2}-CHO$. Same alkyne, two methods, two different functional groups.

But-2-yne is symmetrical: whichever carbon receives the $OH$, butan-2-one is obtained, a single product. Pent-2-yne is not: the two carbons of the triple bond carry a methyl and an ethyl group, equally substituted, and a mixture of pentan-2-one and pentan-3-one is obtained, so 2 products. Hydration is useful in synthesis only on a terminal or symmetrical alkyne.

The first $HBr$ follows Markovnikov's rule and gives 2-bromopropene. The second adds to this alkene: the proton goes to the terminal $CH_{2}$, and the carbocation forms on the carbon that already carries the bromine, stabilized by a lone pair on it. Both bromines end up on the same carbon: 2,2-dibromopropane, a geminal dihalide.

The diene supplies four carbons, the anhydride the two carbons of its double bond and its anhydride ring: $C_{4}H_{6}+C_{4}H_{2}O_{3}\rightarrow C_{8}H_{8}O_{3}$. $M=8\times 12.01+8\times 1.008+3\times 16.00=152.15$ g/mol. Amount of anhydride: $\frac{10.0}{98.06}=0.1020$ mol, which would give at best $0.1020\times 152.15=15.52$ g. Yield: $\frac{13.2}{15.52}=85.1$%.

Penta-1,4-diene has no conjugated double bonds: they are separated by a $CH_{2}$, so it cannot react. Cyclopenta-1,3-diene is locked in the s-cis conformation by its ring: it does not have to pay the cost of the rotation, and it is the most reactive, so much so that it reacts with itself at room temperature. Buta-1,3-diene must first go from the s-trans conformation, the major one, to the s-cis.

The reaction is concerted and stereospecific: the geometry of the dienophile is retained in the ring. Maleate, cis, gives the cis diester; fumarate, trans, gives the trans diester. Two diastereomeric dienophiles give two diastereomeric products.

The $CHO$ group withdraws electrons from the double bond of propenal and lowers the energy of its empty orbital, which strengthens the interaction with the electron-rich diene. Propenal is a much better dienophile than ethylene, which reacts only at high temperature and under pressure.

The six-membered ring carries a double bond between C3 and C4: carbons C2, C3, C4 and C5 come from the diene, C1 and C6 from the dienophile. The methyl group on C4 comes from 2-methylbuta-1,3-diene, isoprene, and the $CHO$ on C1 from propenal. Formula: $C_{5}H_{8}+C_{3}H_{4}O\rightarrow C_{8}H_{12}O$, and the product does have 8 carbons, 12 hydrogens and one oxygen.

$DoU=\frac{2(5)+2-8}{2}=2$: the two conjugated double bonds. $M=5\times 12.01+8\times 1.008=68.11$ g/mol. 1,4-polymerization consumes one $\pi$ bond per unit to create the links between units, and moves the other to the middle: one double bond remains per unit, between C2 and C3.

The chain is written $\cdots CH_{2}-C(CH_{3})=CH-CH_{2}-CH_{2}-C(CH_{3})=CH-CH_{2}\cdots$. Cutting each double bond releases the fragment between two cuts: $=CH-CH_{2}-CH_{2}-C(CH_{3})=$, which becomes $OHC-CH_{2}-CH_{2}-CO-CH_{3}$, 4-oxopentanal. A single product means that all the units are linked in the same way, head to tail; irregular linkages would also give dialdehydes or diketones.

$\frac{1.0\times 10^{6}}{68.11}\approx 1.47\times 10^{4}$: about fifteen thousand units per chain.

A $Z$ double bond bends the chain: the rubber chains do not line up, stay disordered and coiled, and stretch when pulled before returning. The $E$ double bonds of gutta-percha give nearly straight chains that pack into crystalline regions: the material is hard. Same formula, same linkage, only the configuration changes.

Units: $\frac{100}{68.11}=1.468$ mol. Sulfur: $\frac{3.0}{32.07}=0.0935$ mol. Ratio: $\frac{1.468}{0.0935}\approx 15.7$, that is about one sulfur atom for sixteen units. A few bridges are enough: they keep the chains from sliding past one another, and the rubber returns to its shape instead of flowing.

$8+2+13=23$ carbons. An acyclic alkene: $C_{23}H_{46}$, $DoU=\frac{2(23)+2-46}{2}=1$. $M=23\times 12.01+46\times 1.008=322.6$ g/mol.

The $Z$ alkene comes from the alkyne with the same skeleton, tricos-9-yne $CH_{3}(CH_{2})_{7}C\equiv C(CH_{2})_{12}CH_{3}$. Ethyne supplies the two carbons of the triple bond; an octyl and a tridecyl group must be attached to it, hence 1-bromooctane and 1-bromotridecane.

The hydrogen of a terminal alkyne has a $pK_{a}$ close to 25: a base stronger than hydroxide is needed, amide, to form the nucleophilic acetylide. Each alkylation consumes the acetylide, and the monoalkylated alkyne keeps a terminal hydrogen that must be deprotonated again. Since the acetylide is also a very strong base, a secondary or tertiary bromoalkane would undergo elimination; only a primary one gives substitution.

The Lindlar catalyst: it adds both hydrogens on the same side and gives the $Z$ alkene, the pheromone. Sodium in ammonia would give (E)-tricos-9-ene, a diastereomer that the fly's receptors do not recognize.

Overall yield: $0.85\times 0.80\times 0.90=0.612$, that is 61.2%. $M(C_{8}H_{17}Br)=8\times 12.01+17\times 1.008+79.90=193.1$ g/mol, so $n=\frac{10.0}{193.1}=0.0518$ mol. Pheromone: $0.0518\times 0.612\times 322.6=10.2$ g. Yields multiply: three respectable steps lose nearly 40% of the material.