Corrected exercises: SN1, SN2, E1 and E2, substitution and elimination (202-SNU-RE)

Pair 1: $I^{-}$ is an excellent leaving group; $CN^{-}$ is a good nucleophile and a weak base; primary carbon; polar aprotic solvent.

$S_{N}2$: hexanenitrile $CH_{3}(CH_{2})_{4}CN$, the chain gains one carbon.

Pair 2: tertiary bromide, neutral and weak methanol, protic solvent, 25 °C.

Neither $S_{N}2$ (the carbon is too crowded) nor $E2$ (no strong base): the tertiary carbocation forms on its own. $S_{N}1$ is major: 3-ethyl-3-methoxypentane, with a little 3-ethylpent-2-ene by $E1$.

Pair 3: same substrate, but a strong, concentrated base, at reflux: $E2$. Product: 3-ethylpent-2-ene, the only possible alkene, since the three neighbouring carbons are equivalent.

Pair 4: tertiary alcohol, concentrated acid, heat: protonation, loss of water, tertiary carbocation, loss of a proton. $E1$: 1-methylcyclohexene is major, with a trisubstituted double bond.

Pitfall: pairs 2 and 3, same substrate, two mechanisms. It is the REAGENT that decides between $E1$ and $E2$, not the class of the carbon.

Pair 5: no reaction. $F^{-}$ leaves far less readily than $Cl^{-}$ ($pK_{a}$ of $HF$: 3.2, against $-7$ for $HCl$), and above all the $C-F$ bond, 485 kJ/mol, is the strongest of the carbon-halogen bonds.

Pair 6: no reaction. An alkoxide ion, $CH_{3}O^{-}$ or $C_{6}H_{11}O^{-}$, would have to be expelled, and it is a strong base ($pK_{a}$ of the corresponding alcohol close to 16). The ethanoate, a weak nucleophile, changes nothing.

Pair 7: acid-base only. An amine is a BASE and $HBr$ a strong acid: the instantaneous proton transfer gives pentylammonium bromide $CH_{3}(CH_{2})_{4}NH_{3}^{+}\,Br^{-}$.

Pair 8: no reaction. Without acid, the leaving group would be $HO^{-}$, a strong base ($pK_{a}$ of water: 15.7). Bromide is a good nucleophile all the same: it is the DEPARTURE that blocks, not the attack.

Pitfall: writing “no good leaving group” for pair 7. Question 4 comes first: an amine and an acid always react, by proton transfer.

Pair 5: replace the fluorine with a bromine or an iodine. 1-Bromohexane undergoes $S_{N}2$ with sodium hydroxide and gives hexan-1-ol.

Pair 6: protonate the oxygen of the ether. With concentrated, hot $HBr$ or $HI$, the leaving group becomes a neutral alcohol, and the ether is cleaved.

Pair 8: protonate the oxygen. With $NaBr$ and concentrated $H_{2}SO_{4}$, or with $HBr$ directly, water becomes the leaving group and 1-bromohexane is obtained by $S_{N}2$. One can also go through a tosylate, or use $PBr_{3}$.

Pair 7: nothing is missing, the acid-base reaction takes place. For the amine to act as a nucleophile, it must be paired with an alkyl halide, not an acid.

Methanol: an $E2$ requires a base strong enough to pull a proton off a neutral molecule WHILE the bromide leaves, in a single step. Methanol is a very weak base: it can react only once the carbocation has formed, that is, by $S_{N}1$ or by $E1$.

Potassium hydroxide: concentrated $HO^{-}$ pulls off a neighbouring hydrogen at the first encounter. The rate of the $E2$, $v=k[RBr][HO^{-}]$, grows with the concentration of the base, that of the ionization does not: at high concentration the $E2$ wins the race, and the $S_{N}2$ stays blocked on a tertiary carbon.

A leaving group is better the WEAKER a base it is, so the STRONGER the conjugate acid, with a low $pK_{a}$: it carries off the electron pair and stays stable once gone.

Best: $I^{-}$; worst: $NH_{2}^{-}$. Between $H_{2}O$ and $HO^{-}$ there are 17.4 $pK_{a}$ units: that is the whole gain from protonating an alcohol.

The tosylate $TsO^{-}$ ($pK_{a}$ of $TsOH$: $-2.8$), whose charge is spread over three oxygens, leaves as well as a heavy halide: it is the second way to make an OH leave (exercise 4).

$F^{-}$ and $CN^{-}$: low basicity is not enough, the bond to carbon must also break. The $C-F$ bond, 485 kJ/mol, is the strongest of the carbon-halogen bonds; $C-CN$ is a carbon-carbon bond, strong as well.

Pitfall: “weak base, therefore good leaving group” holds for $I^{-}$, $Br^{-}$, $Cl^{-}$, water and the sulfonates, not for $F^{-}$ or $CN^{-}$.

The bromine is carried by an $sp^{2}$ carbon, of the ring or of the double bond. Substitution and elimination take place only at an $sp^{3}$ carbon.

$S_{N}1$: an $sp^{2}$ carbocation would be far too unstable; the more s character an orbital has, the more tightly it holds its electrons near the nucleus, and the less the carbon can bear a positive charge.

$S_{N}2$: backside attack is blocked by the ring or by the double bond, and the shorter $C-Br$ bond is stronger.

Rule: look for a good leaving group ON an $sp^{3}$ carbon. An exception worth knowing: sodium amide $NaNH_{2}$, a very strong base, eliminates a vinyl halide and gives an alkyne.

Arrow 1: on the student's work, it starts at the carbon and points to the sulfur. That is backwards: a $\delta^{+}$ carbon has no electron pair to give.

Correct: the arrow starts at a lone pair on SULFUR and points to the $CH_{2}$ carbon.

Arrow 2: it starts at the bromine and points to the $C-Br$ bond. Again backwards: the bond breaks and its two electrons leave with the bromine.

Correct: the arrow starts at the MIDDLE of the $C-Br$ bond and points to the bromine atom.

Pitfall: an arrow does not say who “goes toward” whom. It says where the electrons go, so it always starts at an electron pair, never at an atom that has none to give.

Primary carbon, strong nucleophile that is only weakly basic, polar aprotic solvent: $S_{N}2$.

Two arrows, a single step: sulfur attacks the carbon from the side opposite the bromine, and the $C-Br$ bond breaks in the same motion.

Product: [(ethylsulfanyl)methyl]cyclohexane $C_{6}H_{11}CH_{2}SCH_{2}CH_{3}$, a thioether.

First error: the thiolate is CONSUMED. It has gone into the product, so it cannot reappear on the right.

Second error: $NaBr^{+}$ does not exist. The $Na^{+}$ ion is a spectator from start to finish, and the bromide leaves as $Br^{-}$: together they form a neutral salt.

Charges: on the left $0-1+1=0$; on the right $0+1-1=0$. An equation whose charges are not conserved is wrong, whatever the mechanism.

Arrow 1: from a lone pair on the oxygen of $CH_{3}CH_{2}O^{-}$ to a hydrogen carried by a carbon NEXT TO the one bearing the bromine.

Arrow 2: from the middle of the $C-H$ bond to the middle of the $C-C$ bond, where the double bond forms.

Arrow 3: from the middle of the $C-Br$ bond to the bromine.

Products: cyclopentene, ethanol and $Br^{-}$, with $Na^{+}$ a spectator. Three arrows, one step; the hydrogen removed and the bromine must lie on opposite faces of the ring (anti).

$S_{N}1$: step 1, one arrow, from the $C-X$ bond to $X$; step 2, one arrow, from a lone pair on $I^{-}$ to the carbocation. Two arrows in all.

$E1$: step 1, the same arrow; step 2, two arrows, from the base to H and then from the $C-H$ bond to the $C-C$ bond. Three arrows in all.

The rule of thumb holds: a substitution counts two arrows, an elimination three, in one step or in two. With a neutral nucleophile such as water, the substitution adds a final deprotonation.

Ethanol is a PROTIC solvent: its O-H hydrogen forms hydrogen bonds with the $CH_{3}CH_{2}S^{-}$ anion, which becomes surrounded by a cage of solvent that must be broken before it can attack.

DMF, polar but APROTIC, dissolves the salt by surrounding $Na^{+}$ and leaves the anion nearly bare, hence more reactive.

Rule: $S_{N}2$ by an anion, polar aprotic solvent (DMF, DMSO, propanone, acetonitrile); $S_{N}1$ and $E1$, protic solvent, which stabilizes the carbocation and the leaving anion.

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.

Tertiary carbon, weak and neutral water, protic solvent: $S_{N}1$. Water attacks the planar carbocation on both faces, then the oxonium ion loses an $H^{+}$.

Product: 3-methylhexan-3-ol, a racemic mixture of (R) and (S). The polarimeter reads zero optical rotation, or nearly: a slight excess of the inverted product appears when the bromide ion, still close by, shields one face. A few alkenes also form by $E1$.

Pitfall: writing “(S)” amounts to applying the $S_{N}2$ rule to an $S_{N}1$. The mark is lost twice: wrong mechanism and wrong stereochemistry.

Before: $Br>CH_{2}OH>$ propyl $>H$. The $CH_{2}OH$ carbon carries $(O,H,H)$, that of the propyl $(C,H,H)$: O beats C.

After, $CH_{2}OH$ and $C\equiv N$ must be separated. Both begin with a carbon, so we compare their substituents: $(O,H,H)$ for $CH_{2}OH$, $(N,N,N)$ for $CN$, the triple bond counting three times.

We compare the heaviest atom of each list first: O beats N, so $CH_{2}OH>CN$.

After: $CH_{2}OH>CN>$ propyl $>H$.

The $S_{N}2$ inverted the geometry, but the entering group took only rank 2, and $CH_{2}OH$ rose to rank 1: the first two ranks swapped places. The two reversals cancel. Product: (S)-2-(hydroxymethyl)pentanenitrile.

Pitfall: we compare atom by atom, at the first point of difference. A single oxygen beats three nitrogens.

The sentence is false. An $S_{N}2$ always inverts the GEOMETRY, not necessarily the letter.

Correct sentence: the letter flips when the entering group takes, in the priorities, the rank of the leaving group (case A); it may stay the same otherwise (case C). Redraw the product, then rank its four groups again.

A, $S_{N}2$: a single step, in which the two reactants meet. $v=k[RI][N_{3}^{-}]$: doubling $[N_{3}^{-}]$ doubles the rate, and so does doubling $[RI]$.

B, $S_{N}1$: the slow step, the breaking of the $C-Br$ bond, involves only the substrate. $v=k[RBr]$: doubling the substrate doubles the rate; adding a nucleophile does not change it.

The added azide changes only the PRODUCT: part of the carbocation gives the azide instead of the alcohol.

The same logic holds for eliminations: $E2$, $v=k[RX][B]$; $E1$, $v=k[RX]$. That is the origin of the numbers 1 and 2 in the names.

1) Protonation: a lone pair on the oxygen picks up the $H^{+}$ from $HBr$. This gives the oxonium ion $R-OH_{2}^{+}$ and $Br^{-}$.

2) Loss of water, the slow step: the $C-O$ bond breaks, water carries off the electron pair, and a planar tertiary carbocation remains.

3) The bromide attacks the carbocation: 1-bromo-1-ethylcyclohexane.

Three steps: one acid-base step, then the two steps of the $S_{N}1$.

$S_{N}1$ because the carbon is tertiary: backside attack is blocked by the three carbon groups, and those same groups stabilize the carbocation.

1) Protonation, as for A.

2) The carbon is primary: a primary carbocation would be far too unstable to form on its own. Water therefore leaves only at the moment the bromide attacks from behind: $S_{N}2$, in a single step.

Two steps in all. Product: 1-bromopentane. That is why heat is needed: the $S_{N}2$ by a bromide, a moderate nucleophile, is slow.

Only the OH group reacts: $HCl$ protonates the oxygen, water leaves, and the tertiary carbocation is captured by $Cl^{-}$. $S_{N}1$, as in a).

The fluorine does not leave: the $C-F$ bond, 485 kJ/mol, is the strongest of all the carbon-halogen bonds.

Product: $(CH_{3})_{2}CCl-CH_{2}CH_{2}CH_{2}F$.

Name: with the OH gone, nothing forces the numbering. All the substituents are prefixes, so we look for the lowest locants: $\{1, 4, 4\}$ starting from the fluorine, against $\{2, 2, 5\}$ from the other end. The product is 4-chloro-1-fluoro-4-methylpentane.

Pitfall: taking the halogen for the leaving group because it is a halogen. The leaving group is the one that leaves as a weak base: here, water.

The 1-phenylethyl cation is secondary, but BENZYLIC: the positive carbon is bonded to the ring, and its empty orbital combines with the $\pi$ system. This is the resonance effect.

The charge is spread over the carbon outside the ring and over three carbons of the ring, the two ortho carbons and the para carbon: four resonance forms, of which three place the charge on the ring.

A charge spread over four atoms costs far less than a charge concentrated on one: a benzylic alcohol undergoes $S_{N}1$ like a tertiary one.

Configuration: the carbocation is planar, and $Cl^{-}$ attacks it on both faces. The starting (R) is lost: a racemic mixture of (R)- and (S)-(1-chloroethyl)benzene.

Step 1: tosyl chloride $TsCl$ in pyridine. The oxygen of the alcohol attacks the sulfur: the $C-O$ bond is untouched, so the geometry is retained. The tosylate is (R): $OTs>$ butyl $>CH_{3}>H$, as for the alcohol.

Step 2: sodium cyanide in DMSO. $TsO^{-}$, a very weak base, leaves like a bromide; $CN^{-}$ undergoes $S_{N}2$, with inverted geometry.

$CN$, a carbon $(N,N,N)$, outranks butyl, a carbon $(C,H,H)$: it takes rank 1, so the letter changes. Product: (S)-2-methylhexanenitrile.

Why not $HBr$ then $NaCN$: on a secondary alcohol, protonation partly opens the $S_{N}1$ route, and the configuration is partly lost. The tosylate gives a single, clean inversion.

Protonation, then loss of water: a SECONDARY carbocation on carbon 2.

Its neighbour, carbon 3, is tertiary and carries an H. That hydrogen slides with its electron pair toward carbon 2: this is a 1,2-hydride shift. The carbocation becomes TERTIARY, on carbon 3.

The bromide attacks this new carbocation: 2-bromo-2-methylbutane.

Rule: as soon as a carbocation forms, in $S_{N}1$ as in $E1$, look at its neighbours. An H or a methyl that can migrate toward a more stable carbocation does so. An $S_{N}2$, with no carbocation, never rearranges.

1) Protonation: the acid $H_{3}PO_{4}$ gives up an $H^{+}$; a lone pair on the oxygen of the alcohol, the base, picks it up. This gives the oxonium ion and $H_{2}PO_{4}^{-}$.

2) Loss of water, the slow step: the $C_{3}-O$ bond breaks, leaving a tertiary carbocation. Neither acid nor base in this step.

3) A base, water or $H_{2}PO_{4}^{-}$, removes a hydrogen carried by a carbon next to the carbocation; the electrons of the $C-H$ bond form the double bond. The carbocation plays the role of the acid here.

Carbon 3 has three neighbours that carry hydrogens: $C_{2}$, $C_{4}$ and the methyl.

$C_{2}$ and $C_{4}$ are equivalent, the molecule being symmetric: they give the same alkene, 3-methylpent-2-ene, which exists as (E) and (Z).

The methyl gives 2-ethylbut-1-ene.

Three alkenes in all.

Zaitsev's rule: elimination gives mainly the most substituted alkene, which is the most stable.

3-Methylpent-2-ene carries three carbon groups on its double bond (trisubstituted); 2-ethylbut-1-ene carries only two (disubstituted).

Major: 3-methylpent-2-ene. Each additional alkyl group stabilizes the double bond.

On $C_{2}$: $CH_{3}>H$. On $C_{3}$: ethyl, a carbon $(C,H,H)$, $>$ methyl, a carbon $(H,H,H)$.

Methyl of $C_{2}$ and ethyl of $C_{3}$ on the same side: (Z); on opposite sides: (E).

Both are trisubstituted, and each keeps two groups facing each other: methyl against methyl in the (E), methyl against ethyl, slightly bulkier, in the (Z). The difference is small: both are obtained, and the steric argument predicts a slight excess of (E).

Pitfall: “trans” is not “E”. When a carbon carries two different carbon groups, only the priorities decide.

The student's equation is not balanced in charge: $H_{2}PO_{4}^{-}$ is left alone on the right, without the $H^{+}$ it lost. That $H^{+}$ went to water in step 3, then returns to $H_{2}PO_{4}^{-}$.

The acid is consumed in step 1 and RETURNED in step 3: it does not appear in the balanced equation.

$C_{6}H_{14}O\rightarrow C_{6}H_{12}+H_{2}O$, with $H_{3}PO_{4}$ written above the arrow.

Secondary substrate, strong and concentrated base, high temperature: $E2$ elimination.

The base removes a hydrogen from carbon 3, placed anti to the chlorine. In one step, the $C_{2}=C_{3}$ double bond forms and $Cl^{-}$ leaves.

Three arrows: from a lone pair on $HO^{-}$ to H; from the $C_{3}-H$ bond to the $C_{2}-C_{3}$ bond; from the $C_{2}-Cl$ bond to Cl.

Why carbon 3: Zaitsev's rule. Its hydrogen gives pent-2-ene, disubstituted; a hydrogen on carbon 1 would give pent-1-ene, monosubstituted and less stable. With a small base such as $HO^{-}$, the more stable alkene wins.

Of the two pent-2-enes, the conformation in which carbon 1 and carbon 4 are anti is the least crowded: it gives (E)-pent-2-ene, the major product.

(Z)-pent-2-ene: $E2$ from the conformation in which carbons 1 and 4 are on the same side.

Pent-1-ene: $E2$ through a hydrogen on carbon 1, the Hofmann product.

Pentan-2-ol: $S_{N}2$ of $HO^{-}$ at carbon 2.

Neither $S_{N}1$ nor $E1$: with a strong, concentrated base on a secondary carbon, the carbocation has no time to form.

Four organic products in all. $H_{2}O$, $Na^{+}$ and $Cl^{-}$ are not counted among them.

Pitfall: forgetting the substitution product. The question says “substitution and elimination”: $HO^{-}$ is also a nucleophile.

Pent-2-ene: on $C_{2}$, $CH_{3}>H$; on $C_{3}$, ethyl $>H$. Priority groups on opposite sides: (E); on the same side: (Z).

Pent-1-ene: $C_{1}$ carries two H, so there is no stereoisomerism.

Pentan-2-ol: the $S_{N}2$ inverts the geometry. Before, $Cl>$ propyl $>CH_{3}>H$; after, $OH>$ propyl $>CH_{3}>H$. The entering group takes the rank of the leaving group: the letter changes, (R) becomes (S). Product: (S)-pentan-2-ol.

$CN^{-}$ is a very good nucleophile but a weak base ($pK_{a}$ of $HCN$: 9.2); aprotic DMSO makes it more nucleophilic still; 25 °C does not favour elimination.

$S_{N}2$ almost alone: 2-methylpentanenitrile.

Priorities: $CN$, a carbon $(N,N,N)$, $>$ propyl, a carbon $(C,H,H)$, $>CH_{3}>H$. The entering group again takes rank 1: geometry and letter both invert, (R) becomes (S).

Product: (S)-2-methylpentanenitrile.

Bond formed: $C-O$, on a primary carbon. An $S_{N}2$ is needed, hence a strong nucleophile: sodium ethoxide $CH_{3}CH_{2}O^{-}Na^{+}$. This is the Williamson synthesis.

Ethanol alone: too weak a nucleophile for a primary carbon, a very slow reaction.

Sodium hydroxide: it would give 2-cyclohexylethan-1-ol, not the ether.

The tert-butoxide: a strong, bulky base, it would mainly do the $E2$ and give vinylcyclohexane.

Bond formed: $C-O$, on a secondary and BENZYLIC carbon.

Methoxide, a strong base, opens the $E2$ route, all the more easily because the alkene formed, (E)-1-phenylprop-1-ene, is conjugated with the ring: a good part of the substrate is lost as alkene.

Methanol, weak and neutral, lets the bromide leave: the benzylic cation, stabilized by resonance as in exercise 4, is captured by the solvent. $S_{N}1$, then loss of an $H^{+}$.

Reagent: methanol, which also serves as the solvent. Racemic product, with $HBr$ released.

The desired alkene is the LESS substituted one: this is the Hofmann product. A strong, bulky base is needed, potassium tert-butoxide $(CH_{3})_{3}CO^{-}K^{+}$. Too large to reach the hydrogen on carbon 3, it removes one from the methyl, the most accessible.

Ethoxide, a small base, would follow Zaitsev: 4-methylpent-2-ene, mostly (E).

Another way round: start from 1-bromo-4-methylpentane, which can give only one alkene, and treat it with tert-butoxide.

Pitfall: “strong base” is not enough. Small, it gives Zaitsev; bulky, it gives Hofmann.

At 25 °C in water, $HO^{-}$ attacks the primary carbon: the $S_{N}2$ gives mainly 2-phenylethan-1-ol.

First change: replace the sodium hydroxide with a strong, bulky base, potassium tert-butoxide in tert-butanol. It can hardly undergo $S_{N}2$ any more.

Second change: heat. Elimination makes three species from two; its reaction entropy is more favourable, and the $-T\Delta S$ term favours it all the more as the temperature rises.

Styrene forms all the better because its double bond is conjugated with the ring.

Methoxide: strong base, small, hence also a good nucleophile. $S_{N}2$ on a primary carbon, $E2$ on a secondary or tertiary one.

tert-Butoxide: an equally strong base, but its three methyls keep it from approaching a carbon. Bulky base: Hofmann $E2$.

Iodide and cyanide: weak bases ($pK_{a}$ of $HI$: $-10$; of $HCN$: 9.2), good nucleophiles: $S_{N}2$.

Water: weak in every respect, it waits for the carbocation. $S_{N}1$ or $E1$, and only on a tertiary, benzylic or allylic carbon.

Pitfall: nucleophilicity and basicity are not the same thing. Basicity is an equilibrium with $H^{+}$; nucleophilicity is a rate of attack on a carbon.

Step 1, acid-base: $NaH$, a very strong base, removes the H of the OH group. An alkoxide forms, and dihydrogen is evolved.

Step 2, INTRAMOLECULAR $S_{N}2$: the charged oxygen attacks carbon 5, primary, which carries the bromine, at the other end of its own chain. A six-membered ring, five carbons and one oxygen, closes.

Product: oxane, or tetrahydropyran, $C_{5}H_{10}O$, with $NaBr$.

A five- or six-membered ring closes quickly, because the two ends of the chain often meet: this is the Williamson synthesis inside a single molecule.

FALSE. The polarity of the bond does not decide: it is the DEPARTURE that counts.

The $C-F$ bond, 485 kJ/mol, is the strongest of the carbon-halogen bonds: fluorine does not leave.

Correct sentence: 1-fluorohexane hardly reacts at all with sodium hydroxide; 1-chlorohexane undergoes $S_{N}2$ and gives hexan-1-ol.

FALSE. The class of the carbon does not choose alone between $E1$ and $E2$: it is the BASE.

Methoxide is a strong base: it removes a hydrogen before the carbocation, secondary and not very stable, has had time to form.

Correct sentence: a secondary halide and a strong base when hot react by $E2$, with a little $S_{N}2$; the $E1$ requires a weak base and a stable carbocation.

FALSE, twice. Methoxide is consumed: it is in the ether. And sodium chloride is neutral: $Na^{+}$, a spectator, and $Cl^{-}$, the leaving group.

Correct sentence: $RCl+CH_{3}O^{-}+Na^{+}\rightarrow ROCH_{3}+Na^{+}+Cl^{-}$, total charge zero on each side.

FALSE. Before looking for a leaving group, look for an acid-base reaction: an amine is a base, $HCl$ a strong acid.

Correct sentence: the amine is protonated instantly; an alkylammonium chloride $RNH_{3}^{+}\,Cl^{-}$ is obtained.

FALSE. The thiol attacks the carbocation and does form a sulfonium ion, but a third step follows: a base in the medium removes its $H^{+}$.

Correct sentence: with a neutral nucleophile, the $S_{N}1$ ends with a deprotonation. Products: 2-(ethylsulfanyl)-2-methylbutane and $HCl$, both neutral.

Nucleophile: the lone pair on the nitrogen of the $NH_{2}$ group.

Carbon attacked: that of the methyl bonded to sulfur, made electrophilic by the neighbouring positive charge.

Leaving group: S-adenosylhomocysteine, a NEUTRAL thioether $R^{1}-S-R^{2}$.

As it leaves, the sulfur takes the electron pair of the $S-CH_{3}$ bond and loses its charge: it leaves uncharged, as a very weak base. This is exactly what is asked of a leaving group; the halogen is only a special case.

The carbon attacked is a METHYL. The $CH_{3}^{+}$ cation would be the least stable of all: no alkyl group stabilizes it, and it never forms in solution.

On the other hand, a methyl is the least crowded carbon there is: backside attack is ideal. $S_{N}2$, and nothing else.

Step 1, $S_{N}2$, two arrows: from the lone pair on the nitrogen to the methyl carbon; from the $C-S$ bond to the sulfur.