Self-check: 202-SNU-RE, Organic Chemistry

c) $sp^{3}$ hybridization is the mathematical combination of the $2s$ orbital and the THREE $2p$ orbitals. It produces FOUR identical hybrid orbitals, of equal energy, intermediate between those of $s$ and $p$, and pointing toward the four corners of a tetrahedron. Each holds one unpaired electron and can form a $\sigma$ bond. The name $sp^{3}$ gives the recipe: one part $s$, three parts $p$.

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

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b) Four resonance structures: the charge on oxygen (the major structure, with the charge on the most electronegative atom), then three structures where it enters the ring, on the two ORTHO carbons and on the PARA carbon. Never at meta: the conjugation gives no access to it. This ortho-ortho-para trio is a signature that will come back throughout aromatic chemistry.

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c) 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.

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b) 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.

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a) Secondary carbon, $N_{3}^{-}$ an excellent nucleophile and a weak base, aprotic DMSO: $S_{N}2$. Before: $I>$ pentyl chain $>CH_{3}>H$. After: $N_{3}>$ pentyl chain $>CH_{3}>H$. The entering group takes rank 1, like the leaving group: inverted geometry gives an inverted letter. Product: (R)-2-azidoheptane.

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a) ACTIVATING: $-CH_{3}$, weakly, by an electron-donating inductive effect; $-OH$ and $-NH_{2}$, strongly, by the electron-donating resonance effect of the lone pair. DEACTIVATING: $-NO_{2}$ and $-COOH$, strongly, by an electron-withdrawing effect; $-Cl$, moderately.

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b) First disconnection: phenoxide ion $C_{6}H_{5}O^{-}$ and bromoethane. The carbon attacked is primary, and the $S_{N}2$ proceeds without difficulty. Second disconnection: ethoxide ion and bromobenzene. It is ruled out, because the carbon to be attacked belongs to the aromatic ring, where neither the $S_{N}2$ nor the $S_{N}1$ is possible, for the reasons detailed in d). It remains to form the phenoxide. The equilibrium $C_{6}H_{5}OH+HO^{-}\rightleftharpoons C_{6}H_{5}O^{-}+H_{2}O$ has the constant $K=10^{pK_{a}(H_{2}O)-pK_{a}(\text{phenol})}=10^{15.7-10.0}=10^{5.7}\approx 5\times 10^{5}$: the deprotonation is complete, and sodium hydroxide is enough. For ethanol, the same constant is $10^{15.7-16.0}=10^{-0.3}\approx 0.50$: the equilibrium stays split, a good part of the alcohol is not deprotonated, and the remaining $HO^{-}$ ions also attack the haloalkane, giving an alcohol instead of the ether. An alkoxide must therefore be prepared by an IRREVERSIBLE reaction: sodium metal, $2\,CH_{3}CH_{2}OH+2\,Na\rightarrow 2\,CH_{3}CH_{2}O^{-}Na^{+}+H_{2}$, or sodium hydride $NaH$, whose dihydrogen escapes and cannot take the proton back. The rule to remember: a phenol is deprotonated by sodium hydroxide, an alcohol requires sodium or $NaH$; and the oxygen of the ether always comes from the partner whose carbon cannot be attacked.

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c) The percentage hydrated is $\frac{K}{1+K}$. For methanal, $\frac{2200}{2201}=99.95$ percent. For ethanal, $\frac{1.4}{2.4}=58.3$ percent. For propanone, $\frac{0.0020}{1.0020}=0.20$ percent. The gap between the extremes is a factor of $1.1\times 10^{6}$ on the constants, and it is entirely explained by the two reasons of exercise $1$: the two methyl groups of propanone crowd the tetrahedral carbon to be formed and at the same time lower its positive charge, whereas methanal has neither handicap. One can also read the equilibrium from the other end: the more stable the carbonyl, the less it hydrates. Finally, the hydrate is almost never isolable because the reaction is an EQUILIBRIUM in aqueous solution: as soon as the water is evaporated to recover it, Le Chatelier's principle shifts the equilibrium toward the carbonyl and the hydrate decomposes between your fingers. The only exceptions are those where the equilibrium is crushed on the hydrate side, like the chloral of exercise $1$.

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b) With $4.00$ mol of alcohol, the relation becomes $K=\frac{x^{2}}{(1-x)(4-x)}=4.0$, that is $x^{2}=4(4-5x+x^{2})$, then $3x^{2}-20x+16=0$. The discriminant is $400-192=208$, so $x=\frac{20-\sqrt{208}}{6}=\frac{20-14.42}{6}=0.930$ mol, the root with the plus sign being rejected since it would exceed the initial amount of acid. The yield relative to the acid therefore rises from $66.7$ to $93.0$ percent. The expected comment is twofold. On the one hand, Le Chatelier's principle works exactly as announced: adding a reactant shifts the equilibrium. On the other hand, the yield is calculated here relative to the LIMITING reactant, the acid, and it would be absurd to calculate it relative to the alcohol, of which more than three quarters remains unused. In practice the cheaper reactant is used in excess, and the surplus is recovered by distillation.

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a) The four isomers are propan-1-amine $CH_{3}CH_{2}CH_{2}NH_{2}$ and propan-2-amine $\left(CH_{3}\right)_{2}CHNH_{2}$, both PRIMARY; $N$-methylethanamine $CH_{3}CH_{2}NHCH_{3}$, SECONDARY; and $N,N$-dimethylmethanamine $\left(CH_{3}\right)_{3}N$, TERTIARY. The counting is radically different from that of alcohols, and this is the most common pitfall of the chapter. For an alcohol, you look at the CARBON bearing the $-OH$ group and count how many other carbons are bonded to it. For an amine, you look at the NITROGEN and count how many carbons are bonded to it. The consequence is immediate: propan-2-amine is a PRIMARY amine, whereas propan-2-ol is a SECONDARY alcohol, even though the skeleton is the same. You must never carry the reasoning over from one chapter to the other.

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b) For $1$-bromopropane $CH_{3}CH_{2}CH_{2}Br$: the $CH_{2}$ carried by the bromine comes out near $3.4$ ppm, it has two neighbours so it is a TRIPLET integrating for $2$; the central $CH_{2}$ comes out near $1.9$ ppm, it has $2+3=5$ neighbours so it is a SEXTET integrating for $2$; the methyl comes out near $1.0$ ppm, it has two neighbours so it is a TRIPLET integrating for $3$. That makes three signals, $2:2:3$. For $2$-bromopropane $\left(CH_{3}\right)_{2}CHBr$: the six hydrogens of the two methyls are EQUIVALENT by symmetry and have only one neighbour, so a DOUBLET integrating for $6$ near $1.7$ ppm; the central hydrogen has six neighbours, so a SEPTET integrating for $1$ near $4.3$ ppm. That makes only two signals, $6:1$. It is $2$-bromopropane that NMR identifies faster: two signals instead of three, an integration of $6:1$ that cannot be mistaken, and a septet whose sight alone is enough to conclude. This is the ISOPROPYL pattern, the second major pattern to recognize at once after the ethyl pattern.

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a) First convention: the carbon chain is VERTICAL and the most oxidized carbon, here the carbonyl, is placed at the TOP. Second convention: each asymmetric carbon sits at the intersection of a cross, its two side substituents being carried by the horizontal lines. The spatial rule, which is the heart of the convention and is often forgotten, is that the HORIZONTAL bonds point TOWARD the viewer, in front of the plane of the page, while the VERTICAL bonds point away from the viewer, toward the back. A Fischer projection is therefore not a flat drawing but the flattened representation of a molecule in which each carbon is seen in an imposed orientation. A strict handling rule follows: a projection can be rotated by $180$ degrees in the plane of the page without changing anything, but a rotation by $90$ degrees or simply flipping the page over INVERTS the configuration of every center, which is the classic error.

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a) $-OCH_{3}$: oxygen donates a lone pair to the ring by resonance, an effect that outweighs its inductive withdrawal, hence strongly ACTIVATING, ORTHO and PARA directing. $-CH_{3}$: inductive donor only, weakly activating, ortho and para. $-Cl$: strong inductive acceptor but resonance donor, hence WEAKLY deactivating and yet ortho and para directing. This is the exception to know, and it is understood by separating the two questions: the inductive effect decides the RATE, resonance decides the POSITION, because only resonance stabilizes the carbocation intermediate of ortho or para attack. $-COCH_{3}$, $-NO_{2}$ and $-SO_{3}H$: acceptors, DEACTIVATING, META directing. The grid in the solution ranks the six groups.

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b) The density is $\frac{79.6}{60.0}=1.327$ g/mL. The table gives dichloromethane at $1.326$: it is that one. Methanol was not a possible candidate, and for a reason that has nothing to do with its density: it is MISCIBLE with water in all proportions, so it never forms a second layer. This is the trap of the table, which lists extraction solvents side by side with a solvent that extracts nothing. An extraction solvent must satisfy three conditions at once: be immiscible with water, dissolve the product better than water does, and boil low enough to be evaporated afterwards without heating the product. Methanol, ethanol and acetone fail the first; ether, dichloromethane, ethyl acetate and toluene satisfy it. The field test, when in doubt about which layer is which, is to add a drop of water at the top of the funnel and watch where it goes: if it passes through the upper layer, that layer is organic.

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