Corrected exercises: alkyl halides (202-SNU-RE)

1-Bromobutane $CH_{3}CH_{2}CH_{2}CH_{2}Br$: the carbon bearing the halogen is bonded to only ONE other carbon, so it is PRIMARY. 2-Bromobutane $CH_{3}CHBrCH_{2}CH_{3}$: the carbon bearing the halogen is bonded to TWO carbons, so it is SECONDARY. 2-Bromo-2-methylpropane $\left(CH_{3}\right)_{3}CBr$: the carbon bearing the halogen is bonded to THREE carbons, so it is TERTIARY.

First method: FREE-RADICAL HALOGENATION of an alkane with $X_{2}$ under ultraviolet light, poorly selective with chlorine but highly selective with bromine. Second: ADDITION of $HX$ to an alkene, regioselective according to Markovnikov's rule. Third: SUBSTITUTION of the $OH$ group of an alcohol, by concentrated $HX$ or by reagents such as $SOCl_{2}$ or $PBr_{3}$, which are cleaner. This last method is the most widely used in the laboratory because alcohols are cheap and readily available.

$C-F$: $3.98-2.55=1.43$. $C-Cl$: $0.61$. $C-Br$: $0.41$. $C-I$: $0.11$. All these bonds are polarized with the carbon at $\delta^{+}$, which makes it electrophilic, but the polarity DECREASES sharply from fluorine to iodine. The $C-I$ bond is almost nonpolar, since iodine has an electronegativity very close to that of carbon.

Because the reactivity of these compounds is governed not by the polarity of the bond but by its STRENGTH and by the leaving-group ability of the halide ion that is released. The long, weak $C-I$ bond costs only 213 kJ/mol to break, against 485 for $C-F$: that is more than double. In addition, the $I^{-}$ ion, large and highly polarizable, spreads its charge well: it is a very weak base, hence an excellent leaving group, whereas $F^{-}$ is a relatively strong base that refuses to leave. The two factors work in the same direction and far outweigh the polarity effect, which would work the other way. TO REMEMBER: it is the stability of the LEAVING group that decides, not the polarity of the bond.

Precisely because of the strength of the $C-F$ bond, the strongest in all of organic chemistry, and the very poor leaving-group ability of fluoride. No ordinary nucleophile manages to displace a fluorine, and no base causes an elimination: polytetrafluoroethylene is therefore chemically unaffected by acids, bases and solvents. On top of this, the small, highly electronegative fluorine atoms form a tight sheath around the carbon skeleton, which they shield sterically, and they give the material very weak intermolecular forces, hence its non-stick character.

By DECREASING leaving-group ability: $I^{-}>Br^{-}>Cl^{-}\gg F^{-}$. Iodide leaves most readily, fluoride practically never.

A good leaving group is a species that is STABLE once released, hence a WEAK BASE. The strength of a base is read from the $pK_{a}$ of its conjugate acid: the lower this $pK_{a}$, the stronger the acid and the weaker its conjugate base. With $pK_{a}\left(HI\right)=-10$, the $I^{-}$ ion is the weakest base of the four, hence the best leaving group. At the other end, $HF$ has a $pK_{a}$ of $+3.2$: it is a weak acid, so $F^{-}$ is a relatively strong base and a very poor leaving group. The ranking by $pK_{a}$ reproduces exactly the one in part a.

Yes, perfectly consistent. The $C-I$ bond is the weakest, 213 kJ/mol, hence the easiest to break; the $C-F$ bond is the strongest, 485 kJ/mol, hence the hardest. The order of bond energies is exactly the REVERSE of the order of leaving-group ability. The two criteria, bond strength and basicity, say the same thing because both depend on the size of the halogen: a large atom overlaps poorly with carbon, hence a weak bond, and spreads its charge well once it is an anion, hence a low basicity.

NO, $OH^{-}$ is a VERY POOR leaving group: its conjugate acid is water, with a $pK_{a}$ of 15.7, which makes it a strong base. Consequence: an alcohol does not undergo direct nucleophilic substitution, and treating an alcohol with an iodide ion gives nothing. The remedy is to PROTONATE the alcohol first in acid: the leaving group then becomes $H_{2}O$, a neutral molecule and a very weak base, with a conjugate $pK_{a}$ of $-1.7$, which leaves without difficulty. This is why every substitution on an alcohol requires an acidic medium. The modern alternative is to convert the $OH$ into a tosylate or into an alkyl halide with $SOCl_{2}$.

1-IODOBUTANE reacts markedly faster, since iodide is the better of the two leaving groups. The typical gap observed for an $S_{N}2$ under the same conditions is a factor of about 30 to 100 between iodide and chloride, that is one to two orders of magnitude. The gap with fluoride would be far greater still, several thousand, which makes the fluoroalkane unusable as a substitution substrate. This is why, when designing a synthesis where the choice is free, one picks the iodide or the bromide, more expensive but much more reactive, rather than the chloride.

The $C-Cl$ bond shortens from 178 pm on an $sp^{3}$ carbon to 172 or 173 pm on an $sp^{2}$ carbon, and its energy rises from about 339 to about 400 kJ/mol: it is clearly shorter and stronger. Two effects add up. First, hybridization: an $sp^{2}$ carbon has 33 % $s$ character against 25 % for an $sp^{3}$, so its orbital is more contracted, the overlap is closer and the bond is stronger. Second, delocalization: chlorine carries lone pairs in a $p$ orbital parallel to the $\pi$ system of the ring or of the double bond, and this lone pair conjugates with it. The bond thus acquires a partial double-bond character, which shows up in resonance forms where chlorine carries a positive charge and a ring carbon a negative one. And a bond that is partly double is not easily broken.

Second-order nucleophilic substitution requires the nucleophile to attack the carbon from the EXACTLY OPPOSITE side to the leaving group, in line with the $C-Cl$ bond, so as to reach the back lobe of the orbital. This is what produces the inversion of configuration. On chlorobenzene, however, that opposite side is occupied by the aromatic ring itself: the nucleophile would have to pass through the benzene ring to reach the carbon. That is physically impossible. The same argument holds for chloroethene, where the attack would have to come through the plane of the double bond. It is therefore not a matter of rate but of geometry: the mechanism simply has no path available.

First-order nucleophilic substitution requires a carbocation to form first. Here one would have to produce an aryl or vinyl cation, whose empty orbital would be an $sp^{2}$ hybrid orbital lying IN the plane of the ring, hence perpendicular to the $\pi$ system: the delocalization that so effectively stabilizes other cations can contribute absolutely nothing, since the overlap is zero. Moreover, this empty orbital, rich in $s$ character, is close to the nucleus and is costly to leave empty. The aryl cation is therefore among the least stable known, far worse than a primary cation, whereas a tertiary carbocation is, on the contrary, stabilized by three electron-donating groups and by hyperconjugation. The slow step of first-order substitution is thus out of reach under ordinary conditions.

The difference comes from the presence of two nitro groups in the ortho and para positions relative to the halogen. The mechanism is neither first-order nor second-order: it is an ADDITION followed by an ELIMINATION. The nucleophile first adds to the carbon bearing the chlorine, which temporarily breaks the aromaticity and gives a cyclic intermediate carrying a delocalized negative charge, called a Meisenheimer complex; then chloride leaves and aromaticity is restored. The slow step is the addition, so anything that stabilizes the negative charge of the intermediate speeds up the reaction. Nitro groups are powerful electron-withdrawing groups, and in the ortho or para position they receive this charge directly by resonance: the intermediate becomes stable enough to form at room temperature. In the meta position they could not receive it and the effect would be much smaller, which is the experimental check of the mechanism. Without an electron-withdrawing group, as in bare chlorobenzene, the intermediate is so high in energy that the extreme conditions of the industrial Dow process are needed. Modern chemistry bypasses all this with palladium-catalyzed coupling reactions, which make aryl halides react under mild conditions by an entirely different mechanism.

SUBSTRATE: primary favours $S_{N}2$ and $E2$; tertiary favours $S_{N}1$ and $E1$, as well as $E2$ with a strong base; secondary is the ambiguous case where everything is possible. REAGENT: a good nucleophile that is only weakly basic, such as $CN^{-}$, $I^{-}$ or $RS^{-}$, favours SUBSTITUTION; a strong base, especially if it is BULKY, favours ELIMINATION. SOLVENT: polar aprotic, such as dimethyl sulfoxide or acetone, favours second-order mechanisms by freeing the nucleophile; polar protic, such as water or an alcohol, favours first-order mechanisms by solvating the carbocation. TEMPERATURE: when high, it always favours ELIMINATION, because elimination increases the number of molecules and therefore the entropy, which the $-T\Delta S$ term rewards.

PRIMARY substrate, STRONG nucleophile that is only weakly basic, polar APROTIC solvent: the four parameters converge on an $S_{N}2$. The major product is pentanenitrile $CH_{3}CH_{2}CH_{2}CH_{2}CN$. A clean, high-yielding reaction, widely used because it lengthens the chain by one carbon.

TERTIARY substrate, a reagent that is both weakly nucleophilic and weakly basic, PROTIC solvent, high temperature. Ethanol plays the double role of solvent and nucleophile here, a situation called SOLVOLYSIS. The mechanism is first-order, through the tertiary carbocation, and a mixture is obtained of the $S_{N}1$ product, the ether $\left(CH_{3}\right)_{3}C-OCH_{2}CH_{3}$, and the $E1$ product, 2-methylpropene. Since heat favours elimination, the alkene is generally the major product.

SECONDARY substrate and a STRONG, BULKY base: this is the signature of $E2$. The bulk of tert-butoxide prevents it from reaching the electrophilic carbon, but in no way from pulling off a peripheral hydrogen. The product is therefore an alkene. Notably, a bulky base preferentially removes the most ACCESSIBLE hydrogen, the one on the terminal methyl, which gives the LEAST substituted alkene, but-1-ene: this is the Hofmann product, in contrast to Zaitsev's rule, which prevails with a small base such as ethoxide.

A TERTIARY substrate does not undergo $S_{N}2$ for a STERIC reason: the mechanism requires a backside attack, exactly opposite the leaving group, and the three alkyl groups block this access completely. The transition state, with five groups around the carbon, would be far too crowded. A PRIMARY substrate does not undergo $S_{N}1$ for an ELECTRONIC reason: this mechanism requires the prior formation of a carbocation, and a primary carbocation is far too unstable to form, having only one alkyl group to stabilize it. Each class is thus excluded from one mechanism, but for reasons of an entirely different nature, and it is the SECONDARY carbon, excluded from none, that makes predictions tricky.

The alkyl halide is reacted with magnesium metal turnings in anhydrous ether: $R-X+Mg\rightarrow R-MgX$. Magnesium inserts into the bond. Since $EN(C)=2.55$ and $EN(Mg)=1.31$, the difference is $1.24$ IN FAVOUR of carbon: it is therefore the CARBON that carries the partial negative charge and magnesium the positive one. The polarity is exactly the reverse of that of the starting $C-X$ bond, where the carbon was $\delta^{+}$. The carbon then behaves like a carbanion, hence as a nucleophile and a very strong base.

Because the Grignard reagent, the equivalent of a carbanion, is an extremely strong BASE: its conjugate acid is the corresponding alkane, with a $pK_{a}$ close to 50. It therefore instantly pulls a proton off any even weakly acidic species, and water, with a $pK_{a}$ of 15.7, is very much among them. The slightest trace of moisture destroys it by turning it into an alkane, a complete and irreversible reaction. Dried glassware, a solvent distilled over sodium and an inert atmosphere are therefore needed. The same sensitivity applies to alcohols, acids and even amines.

In every case, the nucleophilic carbon attacks the carbon of the carbonyl group, and a final acid hydrolysis gives an ALCOHOL. With METHANAL $HCHO$, which carries no alkyl group, a PRIMARY alcohol is obtained. With an ALDEHYDE $RCHO$, a SECONDARY alcohol. With a KETONE $R_{2}CO$, a TERTIARY alcohol. The rule is easy to remember: the class of the alcohol obtained equals the number of alkyl groups initially on the carbonyl, plus the one brought by the Grignard reagent.

The nucleophilic carbon attacks the carbon of $CO_{2}$, itself strongly electrophilic since it is surrounded by two oxygens. A carboxylate is obtained, which acid hydrolysis turns into a CARBOXYLIC ACID $R-COOH$. This is a method of choice for lengthening a chain by exactly one carbon while installing an acid group, and it is carried out simply by adding the Grignard reagent to dry ice.

2-Methylbutan-2-ol is $\left(CH_{3}\right)\left(C_{2}H_{5}\right)C(OH)CH_{3}$: a TERTIARY alcohol whose carbon bearing the $OH$ is bonded to a methyl, an ethyl and another methyl. We must therefore start from a KETONE and a Grignard reagent supplying the missing group. Butanone $CH_{3}-CO-CH_{2}CH_{3}$ already provides a methyl and an ethyl; a methyl has to be added. Butanone is therefore reacted with METHYLMAGNESIUM BROMIDE $CH_{3}MgBr$ in anhydrous ether, then hydrolyzed in acid. The product is indeed 2-methylbutan-2-ol. A note on method: for this kind of question, identify the carbon bearing the $OH$, mentally cut one of its three alkyl bonds, and assign that fragment to the Grignard reagent, the rest becoming the carbonyl compound.

Ethoxide is a STRONG base but not very bulky, the substrate is SECONDARY, and the temperature is high: these three elements converge on an $E2$ ELIMINATION. The major product is PROPENE. Since the substrate can give only one alkene, Zaitsev's rule does not come into play here. A minor share of substitution, giving ethyl isopropyl ether, accompanies the main product.

Iodide is an EXCELLENT nucleophile but a very weak base, and acetone is a polar aprotic solvent: $S_{N}2$ substitution clearly wins. The product is 2-iodopropane. This reaction serves as a TEST because sodium and potassium iodides are soluble in acetone, whereas the sodium bromide and chloride formed are NOT: the appearance of a white precipitate therefore signals visually that an $S_{N}2$ has taken place. The speed at which the precipitate appears ranks the substrate: immediate for a primary, slow for a secondary, none for a tertiary.

Butan-1-ol has FOUR carbons whereas 1-bromopropane has only three: the chain must therefore be lengthened by one carbon while installing a primary $OH$. Step 1, preparing the Grignard reagent: $CH_{3}CH_{2}CH_{2}Br+Mg$ in anhydrous ether gives $CH_{3}CH_{2}CH_{2}MgBr$. Step 2, addition to METHANAL $HCHO$, which supplies the extra carbon. Step 3, acid hydrolysis. Butan-1-ol, a primary alcohol, is obtained. The choice of methanal is forced: it is the only carbonyl compound that gives a primary alcohol.

Butanoic acid also has four carbons, one of which carries the acid group. Steps 1 and 2 are identical: Grignard reagent, then carbonation with $CO_{2}$, for instance on dry ice. Step 3, acid hydrolysis. $CH_{3}CH_{2}CH_{2}COOH$, butanoic acid, is obtained. A possible variant without a Grignard reagent: substitution with $NaCN$, which gives butanenitrile, followed by acid hydrolysis of the nitrile to the acid, which also lengthens the chain by one carbon.

The two syntheses share EXACTLY the first two operations, the formation of the Grignard reagent, and differ only in the ELECTROPHILE it is made to react with: methanal in one case, carbon dioxide in the other. The general rule is that a Grignard reagent is a UNIVERSAL chain-lengthening reagent, whose final product is entirely determined by the electrophilic partner chosen. In synthesis, one therefore decides first which carbon-carbon bond to create, then chooses the electrophile that brings the desired functional group: methanal for a primary alcohol, an aldehyde for a secondary, a ketone for a tertiary, $CO_{2}$ for an acid.

With $HO^{-}$ one obtains $R-OH$, an alcohol. With an alkoxide $R'O^{-}$, one obtains $R-O-R'$, an ether: this is precisely the Williamson synthesis. With $CN^{-}$, one obtains $R-C\equiv N$, a nitrile, and note that this reaction LENGTHENS the chain by one carbon, which is rare and valuable. With a carboxylate $R'COO^{-}$, one obtains $R'COOR$, an ester. With $HS^{-}$, one obtains $R-SH$, a thiol. With an acetylide, one obtains $R'C\equiv C-R$, an alkyne, and this time the chain lengthens by at least two carbons. These six reactions all follow the same second-order nucleophilic substitution mechanism, which is why they all require a primary substrate.

First disconnection: tert-butoxide ion $(CH_{3})_{3}CO^{-}$ plus iodomethane $CH_{3}I$. The substrate is methyl, hence ideal for a second-order substitution, and the reaction succeeds perfectly despite the bulk of the base. Second disconnection: methoxide ion $CH_{3}O^{-}$ plus 2-bromo-2-methylpropane. The substrate is tertiary, second-order substitution is impossible there because of crowding, and methoxide is a base: the only reaction observed is an elimination, which gives 2-methylprop-1-ene and regenerates methanol. Not a trace of ether is obtained. The rule is absolute: in a Williamson synthesis, the ALKOXIDE can be as bulky as you like, but the HALOALKANE must be primary or methyl. This is a perfect example of two formally equivalent retrosynthetic disconnections of which only one works.

The problem is that the primary amine formed, $R-NH_{2}$, is a nucleophile at least as good as the starting ammonia, and often better since the alkyl group is electron-donating. It therefore reacts in turn with the haloalkane to give the secondary amine, which does the same to give the tertiary amine, which ends up as a quaternary ammonium salt. A mixture of the four products is obtained, with boiling points close together: separating them is a nightmare. Two ways around this. The first, the cruder one, is to use a very large excess of ammonia, so that a haloalkane is statistically far more likely to meet an ammonia molecule than the small amount of amine already formed. The second, the clean one, is to go through an intermediate that CANNOT be alkylated a second time: the azide ion $N_{3}^{-}$ gives $R-N_{3}$, which has no hydrogen and is not nucleophilic, and is then reduced to the primary amine. The Gabriel synthesis, which uses potassium phthalimide, rests on the same idea.

Butan-1-amine has four carbons whereas 1-bromopropane has only three: a step that lengthens the chain is therefore needed, and only one of the reactions in a) does so while installing a nitrogen on the new carbon. The sequence is as follows. First step: $CH_{3}CH_{2}CH_{2}Br+NaCN$ in dimethyl sulfoxide, a polar aprotic solvent that speeds up second-order substitution, gives butanenitrile $CH_{3}CH_{2}CH_{2}C\equiv N$. The substrate is primary, the cyanide ion is an excellent nucleophile and a weak base, so elimination remains minor. The carbon of the nitrile group is the fourth carbon: the count is right. Second step: reduction of the nitrile with lithium aluminum hydride, followed by hydrolysis, which turns the carbon-nitrogen triple bond into $CH_{2}-NH_{2}$ and gives butan-1-amine. The nitrile could also have been hydrogenated catalytically. What to remember is the approach: start from the target, count the carbons, and it is the mismatch in the count that dictates the central step.

The molecule offers two sites to the $HO^{-}$ ion: the acidic proton of the carboxyl group, and the carbon bearing the chlorine.

Between an acid-base reaction and a nucleophilic substitution, it is the acid-base reaction that takes place first: a proton transfer is almost instantaneous, whereas the substitution has to break a carbon-chlorine bond.

It is also very favourable: $pK_{a}$ of chloroethanoic acid $2.9$, $pK_{a}$ of water $15.7$, hence a constant of the order of $10^{12.8}$.

With a single equivalent of $NaOH$, all the base is used up by this neutralization: sodium chloroethanoate is obtained, and the chlorine has not moved.

With a second equivalent, second-order substitution takes place at the primary carbon, which has no beta hydrogen, so no elimination competes with it:

At least two equivalents of $NaOH$ are therefore needed. Sodium 2-hydroxyethanoate is obtained, and a final acidification gives glycolic acid, or 2-hydroxyethanoic acid. This is its industrial route of manufacture.

Pitfall: counting a single equivalent because you only look at the halogen. Any acidic hydrogen on the substrate, from a carboxylic acid or a phenol, uses up its share of base before the substitution.

Protic: water, ethanol and ethanoic acid, which all carry a hydrogen on an oxygen. Polar aprotic: propanone, dimethyl sulfoxide, dimethylformamide and ethanenitrile. The latter all have a high dipole moment, often higher than that of water in the case of dimethyl sulfoxide, but their negative end is exposed while their positive end is buried in the heart of the molecule, surrounded by methyl groups. They therefore solvate CATIONS very well and ANIONS very poorly, and above all they are incapable of donating a hydrogen bond.

In a protic medium, each anion is locked in a cage of solvent molecules that give it hydrogen bonds. The smaller the anion and the more concentrated its charge, the sturdier this cage, and the more energy it takes to pull the anion out before it can attack. The tiny fluoride ion is thus almost paralyzed, whereas the large, diffuse iodide ion is poorly solvated and therefore available: hence the order $I^{-}>Br^{-}>Cl^{-}>F^{-}$, which reflects SOLVATION and not intrinsic reactivity. In a polar aprotic medium, these cages do not exist: the anions are practically bare and their nucleophilicity returns to its natural order, that of charge density, $F^{-}>Cl^{-}>Br^{-}>I^{-}$. It is this second ranking that follows basicity, fluoride being the strongest base of the series since hydrogen fluoride is the weakest of the hydrohalic acids. In other words, nucleophilicity and basicity go hand in hand, except when the solvent gets involved.

The speed-up is $\dfrac{1.3\times 10^{6}}{1}$, that is a factor of $1.3\times 10^{6}$: a reaction that took a week in methanol is over in half a second in dimethyl sulfoxide. It comes entirely from the destabilization of the reactant: in methanol, the azide ion sits at the bottom of a solvation well it must climb out of, which raises the activation energy; in dimethyl sulfoxide, it is already free and has only to attack. The solvent has not made the nucleophile better, it has stopped holding it back. The practical consequence is immediate: any second-order nucleophilic substitution involving an anion should be run in a polar aprotic solvent, dimethyl sulfoxide, dimethylformamide or propanone. This is why exam questions systematically specify the solvent, and it should be read as a hint about the mechanism.

First-order substitution has a slow step that is not an attack but an IONIZATION: the $C-Br$ bond breaks heterolytically and produces, where there was no charge at all, a carbocation and a bromide ion. Any solvent able to stabilize these two nascent charges lowers the energy of the transition state and speeds up the reaction. Only a protic solvent does so completely: its negative end solvates the cation, and its hydrogens donate hydrogen bonds to the departing anion. Water, both highly polar and highly protic, is therefore the best medium, hence the factor of $10^{5}$ relative to the less polar ethanol. What paralyzed the nucleophile in the previous case is exactly what is needed here, because the anion is no longer a reagent to be freed but a product to be stabilized. The rule to remember fits in one sentence: polar APROTIC solvent to favour second-order substitution, polar PROTIC solvent to favour first-order substitution. In practice, a question that says "in hot ethanol" announces a first-order mechanism or an elimination, and one that says "in dimethyl sulfoxide" announces a second-order one.

1-Bromobutane and 1-bromo-2-methylpropane are primary, 2-bromobutane is secondary, 2-bromo-2-methylpropane is tertiary. Silver nitrate test, which follows carbocation stability: the tertiary precipitates immediately in the cold, the secondary slowly and mostly after heating, and the two primaries give nothing in the cold and very little even when hot. Sodium iodide test in propanone, which follows the accessibility of the carbon, hence crowding: the two primaries precipitate rapidly, 1-bromobutane a little faster than 1-bromo-2-methylpropane, whose branching on the neighbouring carbon already hinders the approach; the secondary reacts slowly; the tertiary does not react at all. The two tests thus rank the substrates in EXACTLY OPPOSITE ORDERS, and this is what makes them such an effective pair.

Flask A contains 2-bromo-2-methylpropane. The immediate precipitate with silver nitrate signals a very stable carbocation, hence a tertiary one: the silver ion coordinates to the bromine and helps it leave, the tertiary cation forms without difficulty, and silver bromide drops out at once. The total absence of reaction with sodium iodide confirms the diagnosis from the other end: a second-order substitution requires a backside attack on the carbon, and the three methyl groups block access completely. A tertiary substrate is the best for one of the two mechanisms and the worst for the other, which removes all ambiguity.

The two tests point to a PRIMARY substrate: no first-order reaction for lack of a stable cation, rapid second-order reaction for lack of crowding. Two candidates remain, 1-bromobutane and 1-bromo-2-methylpropane. The degradation settles it. Alcoholic KOH while hot causes an elimination; each of the two candidates has only one neighbouring carbon that carries hydrogens, so each can give only a single alkene, and this information is not enough. Ozonolysis, however, cuts the double bond of the alkene and identifies the two fragments. Obtaining methanal and PROPANONE means that one of the carbons of the double bond carried two methyl groups: the alkene was therefore 2-methylprop-1-ene $CH_{2}=C(CH_{3})_{2}$, and flask B contained 1-bromo-2-methylpropane.

1-Bromobutane would have given but-1-ene $CH_{3}CH_{2}CH=CH_{2}$, whose ozonolysis yields methanal and PROPANAL. Both hypotheses thus lead to the same methanal accompanied by a compound of formula $C_{3}H_{6}O$, and everything depends on the nature of the latter: an aldehyde in one case, a ketone in the other. Tollens' reagent separates them immediately, since propanal reduces the silver ion and deposits a silver mirror whereas propanone does not react; Fehling's solution, with its brick-red precipitate, does the same job. The general logic of this type of identification deserves to be stated. One begins with REACTIVITY tests, which say nothing about the structure but classify the substrate as primary, secondary or tertiary, and which are all the more reliable when two that run in opposite directions are crossed. One finishes with a DEGRADATION followed by identification of the fragments, which reconstructs the carbon skeleton. Neither step would suffice alone, and it is their combination that makes the proof.

Because they offer no reactive site accessible to tropospheric reagents. As seen in exercise 1, $C-F$ bonds are the strongest in organic chemistry and fluoride is a dreadful leaving group; the $C-Cl$ bonds are also strong; there is no $\pi$ bond to add to, no hydrogen to remove, no basic site. CFCs therefore react neither with oxygen, nor with water, nor with atmospheric oxidants, and are not washed out by rain either since they are poorly soluble.

Because a compound that reacts with nothing is not DESTROYED: it accumulates. The molecules released by aerosols and refrigerators therefore have an atmospheric lifetime of several decades, more than enough for atmospheric mixing to carry them up to the STRATOSPHERE, above 20 km altitude. There, conditions are no longer the same: short-wavelength ultraviolet radiation, filtered out lower down by ozone itself, is intense. The inertness merely allowed the compound to reach the one place where it could react.

The $C-Cl$ bond, the WEAKER of the two at 327 kJ/mol against 485 for $C-F$. A sufficiently energetic ultraviolet photon breaks it HOMOLYTICALLY and releases a chlorine radical $Cl\cdot$, while the $C-F$ bonds hold. It is therefore chlorine, and not fluorine, that is responsible for ozone destruction, a crucial observation for what follows.

The cycle has two steps that repeat. First: $Cl\cdot+O_{3}\rightarrow ClO\cdot+O_{2}$, the chlorine radical pulls an oxygen atom off ozone. Second: $ClO\cdot+O\rightarrow Cl\cdot+O_{2}$, chlorine monoxide reacts with a free oxygen atom, abundant at this altitude, and REGENERATES the chlorine radical. The net result is $O_{3}+O\rightarrow 2O_{2}$, and chlorine does not appear in it: it is a CATALYST. Since it is regenerated on every turn, a single atom can start over indefinitely, and it is estimated to destroy of the order of a hundred thousand ozone molecules before finally being trapped in a stable form such as $HCl$. This is exactly the structure of a chain reaction, as in the free-radical halogenation of exercise 1 b), with the same consequence: an effect out of all proportion to the amount of initiator.

They were first replaced by HYDROCHLOROFLUOROCARBONS, or HCFCs, then by HYDROFLUOROCARBONS, or HFCs, which contain no chlorine. The guiding principle is twofold. On the one hand, REMOVING chlorine eliminates the chain carrier: a compound containing only fluorine cannot release $Cl\cdot$, and the $C-F$ bond resists radiation anyway. On the other hand, INTRODUCING hydrogen atoms deliberately creates a weak point: the hydroxyl radical $OH\cdot$, present in the troposphere, can pull off this hydrogen and start the breakdown of the compound BEFORE it reaches the stratosphere. Perfect inertness was thus given up on purpose, by designing a molecule stable enough for its use but fragile enough to disappear. The Montreal Protocol is regarded as the most successful environmental treaty ever signed, and the ozone hole has been closing ever since. HFCs, however, pose another problem, their high global warming potential, which the Kigali Amendment addressed.

Secondary substrate, excellent nucleophile, aprotic solvent: SN2. Backside attack inverts the carbon; since the new group is still an iodine, the priorities do not change, and a molecule that has exchanged is (S).

90 % R and 10 % S. The 10 % of S cancel 10 % of R, leaving 80 % of R uncancelled: the optical rotation is 80 % of its initial value.

The loss of optical rotation is 20 % when the exchange is 10 %: each exchange removes the contribution of two molecules, the one that inverted and another that it cancels. Ratio: 2.

Each exchanged molecule would give R or S at random: of 10 % exchanged, 5 % would be S and 5 % R. There would be 95 % R and 5 % S, an optical rotation of 90 %, a loss of 10 % for 10 % exchange: ratio 1.

A ratio of 2 proves that EVERY substitution is accompanied by an inversion: this is the experimental demonstration of the backside attack of SN2. After a very long time, exchanges continue in both directions, R to S and S to R, until they are equal: the solution becomes racemic and its optical rotation tends to 0.

The bromine must be axial: it is the only position where it is aligned, at 180°, with axial hydrogens on the neighbouring carbons. An equatorial bromine is anti-periplanar only to ring C-C bonds.

In a 1,2-cis cyclohexane, one substituent is axial and the other equatorial: bromine axial, methyl equatorial. C2 then carries an axial hydrogen, anti to the bromine, and so does C6. Two alkenes: 1-methylcyclohexene, trisubstituted, by loss of the H on C2, and 3-methylcyclohexene by loss of an H on C6. The first, the more substituted, is the major product: Zaitsev's rule applies.

In a 1,2-trans cyclohexane, the two substituents are either both equatorial or both axial: bromine axial forces methyl axial. C2 no longer has an axial hydrogen; its hydrogen is equatorial and cannot leave. Only C6 provides an axial hydrogen: only 3-methylcyclohexene is obtained. The geometry of the E2 takes precedence over the stability of the product.

Cis isomer: the two chairs are (Br axial, Me equatorial), 2.0 kJ/mol, and (Br equatorial, Me axial), 7.6 kJ/mol. Gap 5.6 kJ/mol: $e^{5.6/2.48}=e^{2.26}\approx 9.6$, that is $\frac{9.6}{10.6}\approx 90$ % with axial bromine. Trans isomer: diequatorial chair at 0, diaxial chair at $7.6+2.0=9.6$ kJ/mol: $e^{9.6/2.48}\approx 48$, that is $\frac{1}{49}\approx 2.0$ % with axial bromine.

The cis isomer reacts much faster: 90 % of its molecules are ready to eliminate, against 2 % for the trans, which has to go through its less stable chair. With equal reactivity of the reactive chair, the rate gap is of the order of 45.

The Grignard reagent pulls off a deuteron: $CH_{3}CH_{2}CH_{2}MgBr+D_{2}O\rightarrow CH_{3}CH_{2}CH_{2}D+Mg(OD)Br$. 1-Deuteriopropane is obtained, with the deuterium on the carbon that carried the magnesium.

The deuterium has to end up on the central carbon: start from 2-bromopropane. Magnesium in anhydrous ether, then $D_{2}O$: $CH_{3}CHBrCH_{3}\rightarrow CH_{3}CH(MgBr)CH_{3}\rightarrow CH_{3}CHDCH_{3}$. This is a clean method of isotopic labelling at a chosen position.

Diol: $n=\frac{0.450}{90.12}=4.99\times 10^{-3}$ mol. Methane: $n=\frac{0.245}{24.5}=1.00\times 10^{-2}$ mol. Ratio: $\frac{1.00\times 10^{-2}}{4.99\times 10^{-3}}=2.00$. Each molecule carries two mobile hydrogens, those of its two $OH$ groups, which is consistent with a diol; the molar mass matches butane-1,4-diol $C_{4}H_{10}O_{2}$.

Ethanol destroys the Grignard reagent: its $OH$, with a $pK_{a}$ close to 16, gives up its proton and releases the alkane. Hexane and ethoxyethane have no acidic hydrogen. The ether also has lone pairs on the oxygen, which bind to magnesium and stabilize the Grignard reagent in solution: this is why it is used, and not hexane.

No: as soon as one molecule of Grignard reagent forms, it pulls the proton off the $OH$ of another molecule and destroys itself. The alcohol must first be PROTECTED, for example as a silyl ether, the Grignard reagent prepared and reacted, and then the protecting group removed.

The carbon bearing the bromine is bonded to only one carbon: neopentyl bromide is PRIMARY. It is nonetheless the slowest of the primaries, a hundred thousand times slower than bromoethane.

$\frac{30}{1}=30$: a single methyl added on the carbon attacked divides the rate by 30. $\frac{1}{0.025}=40$: a second methyl divides it by a further 40.

$\frac{k_{1}}{k_{2}}=e^{\Delta E_{a}/RT}$, hence $\Delta E_{a}=RT\ln\frac{30}{1\times 10^{-5}}=2.48\times\ln(3\times 10^{6})=2.48\times 14.9\approx 37$ kJ/mol.

For kinetics of the same order and equal concentrations, the half-life is inversely proportional to the rate constant: $10\times\frac{1}{1\times 10^{-5}}=1\times 10^{6}$ min, that is $\frac{10^{6}}{60\times 24}\approx 694$ days, nearly two years. In practice, neopentyl bromide does not undergo SN2.

It is not the carbon attacked that is crowded, but its neighbour: the beta carbon carries three methyls, which stand in the path of the nucleophile's backside approach. 1-Bromo-2-methylpropane, with two methyls on the beta carbon, is already slowed by a factor of about 30; the third almost completely shuts the passage. The class of the carbon is therefore not enough: one has to look at the whole volume around the line of attack.

Bromobenzene: $5.0\times 1.50=7.50$ g, that is $\frac{7.50}{157.0}=4.78\times 10^{-2}$ mol. Magnesium: $\frac{1.20}{24.31}=4.94\times 10^{-2}$ mol, a slight excess. Benzophenone: $\frac{7.30}{182.2}=4.01\times 10^{-2}$ mol. The Grignard reagent, at most $4.78\times 10^{-2}$ mol, reacts mole for mole with the ketone: benzophenone is limiting.

$4.01\times 10^{-2}\times 260.3=10.43$ g at best. Yield: $\frac{7.82}{10.43}=75.0$ %.

Benzene comes from destruction of the Grignard reagent by a trace of water: $C_{6}H_{5}MgBr+H_{2}O\rightarrow C_{6}H_{6}$. Biphenyl comes from coupling between Grignard reagent already formed and a molecule of bromobenzene that has not yet reacted, favoured by a high local concentration of bromobenzene and by heat: this is why it is added dropwise.

Biphenyl is a nonpolar hydrocarbon, very soluble in hexane. Triphenylmethanol carries an $OH$ group: its molecules are held together by hydrogen bonds, and it dissolves poorly in a cold nonpolar solvent. The wash carries off the biphenyl and leaves the product.

2 equivalents are needed. The first adds to the ester, and the intermediate expels methoxide to give benzophenone. But a ketone is more reactive than an ester toward the Grignard reagent: it immediately consumes a second equivalent, before the remaining ester molecules do. The ketone therefore cannot be isolated, and the tertiary alcohol carrying two identical groups is obtained directly.