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Chapter 11: Reactions of Alkyl Halides There are two basic types of reactions that alkyl halides undergo: 1. Substitutions: A-B + C-D → A-C + B-D Ex. CH3I + NaOH → CH3OH + NaI Or: OH Br + PBr3 + HO-PBr2 Or: H CH3 1. NaNH2 2. CH3I + NH3 + NaI 2. Eliminations: X → Y + Z Ex. Br H B- + B-H + Br- or: OH + H2O SUBSTITUTIONS: Alkyl halides are AWESOME at doing substitution reactions – they have this attractive carbon group (electron-poor, due to electronegative or polarizable halide attached, thus attractive for nucleophiles) and they have the halide, which easily stabilizes an anion so it can readily FALL OFF from the carbon group, allowing the nucleophile to bond to the carbon: δ+ R electrophile attracts nucleophile δ− X "leaving group" - group that gets displaced General Equation: Displacement of halide by Nucleophile R-X + Nuc⊖ → Nuc-R + X⊖ Ex. + OH Br OH + Br There are two general types of substitution mechanisms, that are dependent upon what type of alkyl halide is undergoing reaction, what type of nucleophile is involved, what reaction conditions are present. 1. SN2 (Substitution, Nucleophilic, Bimolecular) Bimolecular Rate Equation: Rate of reaction is dependent upon TWO species in the ratedetermining step, including [RX] and [Nuc]. Double the concentration of one and the rate doubles. Halve the concentration of one and the rate is halved. Rate = k [RX][Nuc] Concentration of both species affects the number of times collisions can occur. The more times they can collide, the more times they can react. For the reaction to occur, two things have to fall into place: • Need enough energy during collision to overcome e- repulsions • Need correct orientation for collision for bond formation/breakage to occur correctly Let’s look at the mechanism (that detailed description of how and where the e- are moving during a reaction). There are two possible modes of attack by a nucleophile for the SN2 reaction. Let’s look at each… a. Front-side attack: H3CH2C H CH3 (R) Br OH H3CH2C H CH3 (R) OH + Br "retention" of configuration b. Back-side attack: H3CH2C H CH3 (R) Br OH HO CH3 CH2CH3 (S) H + Br "inversion" of configuration Which one is favored? OR more appropriately: which one ISN’T favored? Which one is bad for both steric and electronic reasons? Backside attack (inversion) is favored. Frontside attack (retention of configuration) is never seen. Why? a. Steric interference between incoming Nuc and out-going leaving group H3CH2C CH3 (R) H OH Br b. Electrostatic repulsions between negatively charged (e- rich) Nuc and departing leaving group (X-). H3CH2C H CH3 (R) OH δ− Br The SN2 type reaction is a one-step, concerted process (make new bonds, break old bonds simultaneously). There aren’t any intermediates that form, although we can surmise what the transition state of the reaction must look like: H3CH2C H CH3 (R) !" Br OH H3CH2C CH3 HO Br HO CH3 CH2CH3 (S) H H + Br Note that the transition state involves both the alkyl halide and the nucleophile – THUS – bimolecular reaction! Remember: Transition states are not anything that can be isolated. They are a theoretical summary of what’s probably occurring, based on all information presented before and after the reaction. Factors that affect the SN2 Reaction: Remember – whatever stabilizes the transition state will lower the activation barrier and cause the rate of the reaction to increase. (i.e. smaller mountain, faster to climb over!) 1. Steric Congestion of Alkyl Halide - The more congested and hindered the alkyl halide, the slower the reaction. Remember – ONLY alkyl halides with leaving groups on sp3 hybridized carbons can do this reaction Fastest CH3-X > RCH2-X Slower Never > R2CH-X >>>>>>>>>>>>>>> R3C-X As you can probably surmised, tertiary halides DO NOT do the SN2 type reactions. The Transition State is too crowded: R R' Nuc X R" Now, consider these two alkyl halides. Both are primary but they are not the same. Determine which one will do an SN2 faster: or Br Br The second one is the slower of the two. With the extra branch attached at the β-carbon, the second one has more sterics in the vicinity of the reacting carbon center. 2. Leaving Group – Almost always expelled with a full negative charge in an SN2 reaction. The best leaving groups will be those that can best stabilize the anion (i.e. the weakest bases). Halides: I⊖, Br⊖ > Cl⊖ >>> F⊖ (HO⊖, R2N⊖) (best) (worst) Two other GREAT leaving groups: Mesylates (methane sulfonate groups): (⊖OMs) O O S CH3 O Tosylates (toluene sulfonate groups): (⊖OTs) O O S CH3 O The anions for these sulfonates are resonance-stabilized: O S O O O O R S O O R O S R O and thus EXTREMELY stable anions and VERY weak bases – even better leaving groups than I⊖. So – the Updated List for Leaving Groups: MsO⊖, TsO⊖ > I⊖ , Br⊖ > Cl⊖ >>> F⊖ (HO⊖, R2N⊖) (best) (worst) The formation of these excellent leaving groups from alcohols can be discussed in more detail in Ch. 17 in CHEM 332. 3. Solvent – affects rate by affecting reagent’s energy levels 2 Types of Solvents: a. Protic (ROH, H2O) – Solvation occurs where the solvent molecules’ acidic proton will Hbond to the nucleophile, thus surrounding it (“solvating it”), shielding the nucleophile and preventing it from being able to react. R O δ+ δ+ H δ+ O H R O R H Nuc δ+ H O δ+H O R R b. Polar, Aprotic – Solvation also occurs, but this time the polar solvent molecules align themselves so the negative end of the solvent molecule faces and therefore surrounds (“solvates”) the corresponding metal cation initially associated with the nucleophile (i.e. HO- typically comes as KOH or NaOH). Once the counter ion is solvated, the nucleophile is “exposed” to react, no longer associated with a cation, no longer paired in an ionic salt. Examples of Polar, aprotic solvents: CH3CN (acetonitrile), DMSO (dimethylsulfoxide, (CH3)2SO), DMF (N, N-dimethylformamide, (CH3)2NCHO). H3C H3C C N δ- C N δ H3C C - δ- N δ- N C CH3 Nuc δ- N M δ N C - C CH3 CH3 4. Nucleophile – “Nucleophilicity” is the measure of a compound’s ability to cause a substitution reaction to occur (to be a good nucleophile) Some things to keep in mind: a. Nucleophilicity parallels basicity typically – Ex. HO versus HOH CH3O versus CH3OH b. Nucleophilicity is reduced by the presence of EWG (electron-withdrawing groups) via the inductive effect. If e- density is removed from the attacking atom, its nucleophilicity is reduced. Ex. Cl Cl H C versus H Cl C H Electron density pulled away from central, attacking atom, when electronegative chlorines are attached! While still technically negatively charged (Formal Charge = -1), the electron cloud has migrated away from the carbon atom to the chlorine atoms: Cl C Cl Cl c. TRENDING: Nucleophilicity increases for atoms in the same family (column) on the periodic table as they become larger (as you move DOWN a column on the periodic table). Larger atoms will be more polarizable. Ex. F⊖ < Cl⊖ < Br⊖ < I⊖ d. Resonance stability of a nucleophile results in lower nucleophilicity. Negative charge of nucleophile is delocalized, spread over two or more atoms, reducing its strength. Ex. O OCH2CH3 versus O C CH3 Anion averages over BOTH oxygen atoms, splitting up the strength of the anion: O O C O CH3 O C CH3 1/2 1/2 O O C Examples of SN2 reactions: Draw the products CH3 H H OH Cl H Br I I Br Cl OCH3 Compare the following – Which reaction occurs faster via the SN2 reaction and Why? a. Br b. I vs Br I OTs Cl CH3 CH3 vs c. Br OH Br vs OCH2CH3 [Side Note, for your reading pleasure: The inversion of the stereochemistry in an SN2 reaction is rather like an umbrella flipping inside out in a bad storm. This inversion process is called the Walden Inversion (Paul Walden, 1863-1957), who first noticed the inversion of (+) malic acid to form (-) malic acid, and back again, using PCl5 and Silver (I) oxide in water: O 1. PCl5 OH OH HO O (+) Malic Acid 2 Ag2O, H2O O OH OH HO O (-) Malic Acid He originally thought he’d made the same compound by using a leaving group and then substitution but after checking the optical rotation, he realized he’d made the enantiomer –inverting the chiral center in the process. Emil Fischer (of the Fischer Projection Fischers) said it was “the most remarkable observation made in the field of optical activity since the fundamental observations of Pasteur.” Wondrous things, most assuredly… A Walden cycle is considered to be any series of reactions that converts one enantiomer into its mirror image. A series of studies took off in the 1920s and 1930s, with one of the first being the conversion of (+)-1-phenyl-2-propanol to (-)-1-phenyl-2propanol. Originally it was thought that the same compound had been synthesized by using a leaving group followed by the substitution reaction, but after checking the optical rotation, it was discovered that the enantiomer had been synthesized. The chiral center had been inverted the chiral center in the process: An example of a Walden Cycle is shown below: TsCl, pyridine OTs OH (+)-1-phenyl-2-propanol [α]D = +33.0 NaOAc NaOH OAc OH (-)-1-phenyl-2-propanol [α]D = -33.2 The key to recognizing the inversion was noting that the optical activity was, although equal in size, now OPPOSITE in direction! The first reaction added something to the oxygen to make it leave. The third reaction chopped off something from the oxygen so it would return to normal. So it must be the second reaction where the fun occurred… End Side Note] 2. SN1 (Substitution, Nucleophilic, Unimolecular) Unimolecular rate equation – Rate of this reaction is dependent on the concentration of the alkyl halide, which is the only species involved in the rate-determining step. Mech: the reaction can only go as fast as its slowest step, otherwise known as the “ratedeterming step”, shown below as “r.d.s.”: C X r.d.s. (slow) C + X Nuc (fast) C Nuc The rate of reaction is ONLY dependent on formation of carbocation intermediate (slowest step of the reaction). It is the rate of the “spontaneous dissociation” that slows down this reaction or speeds it up. Note that the Nucleophile is NOT involved in this step. Rate = k [RX] Question: How does the carbocation intermediate affect the stereochemistry of this reaction? Answer: Racemization (formation of a racemic mixture) occurs at the reaction center when the carbocation forms: Ph H3C OH H Br H3CH2C Ph H3 C Ph H3 C OH2 H3CH2C CH2CH3 Br Ph H3C Br CH2CH3 + Ph H3 C CH2CH3 Br Question: How do you think someone figured out that the product was a racemic mixture? Someone measured the optical activity of the product and found out that it didn’t have any optical activity! Factors that affect the rate of an SN1 Reaction: 1. Alkyl Halide: the more stable the carbocation, the faster the reaction will occur (more stable intermediate = lower activation barrier). The rate-determining step is the carbocation formation – the faster this occurs, the faster the overall reaction occurs. Recall that alkyl groups stabilize carbocations via: 1. Inductive Effect 2. Hyperconjugation More stable carbocations have lower activation barriers and thus faster reactions Order of stability – 3º > 2º >>>>>> 1º or methyl Methyl and primary carbocations are so unstable, they cannot form and thus these generally will NOT do SN1 reactions. Another stabilizing factor: Resonance Stabilization – the more the charge can be spread out over multiple atoms, the more stable the charge will be (“delocalization”). Allylic carbocations: Both (or all, if appropriate) resonance forms are reactive. Ex. Benzylic carbocations: Only the resonance form that maintains the aromatic ring reacts (the other three are much higher energy without the stability of the aromatic system) Ex. Of course, systems that have multiple stabilizations of resonance (i.e. a carbocation that might be both allylic AND benzylic) will be even MORE stable. New Order of Stability: 3º or 2º with resonance > 2º or 1º with resonance >>>>>>>>>>> 1º and/or methyl CH3 DID YOU NOTICE…? Those that are the most stable (3º) carbocations do SN1 but do not do SN2 and those that do SN2 best (1º) will not do SN1. BUT WATCH those in the middle that do BOTH!!! 2. Leaving Group: Same order of reactivity seen for SN2 reaction TsO⊖, MsO⊖ > I⊖ > Br⊖ > Cl⊖ ≈ H2O > F⊖, ⊖OH, ⊖NH2 SN1 reactions are often run under acidic conditions, therefore H2O is often seen as a Leaving Group for alcohols (refer to Ch 10, formation of tertiary halides from tertiary alcohols): (CH3)3C-OH H-X (CH3)3C (CH3)3C-OH2 X (CH3)3C-X 3. Nucleophile: Does NOT affect the rate of the reaction because it is not part of the rate-determining step Generally, though, nucleophiles in SN1 reactions are WEAK bases, usually neutral molecules like H2O or ROH or perhaps H2S and RSH (sulfides and thiols). If you choose a stronger nucleophile, it is probably a strong base also… then you will see either an SN2 reaction (from stronger nucleophile) or an elimination reaction (E2) (from strong base) will occur instead! 4. Solvent: affects the rate of SN1 reaction by stabilizing the carbocation intermediate as it forms (lowering the activation barrier, speeding the reaction), thus increasing the rate of reaction. Any POLAR solvent, protic or aprotic, can be used (ex. H2O, ROH, DMSO, etc). The dipoles of the solvent molecules orient themselves so the electron-rich end of the dipole of the solvent molecule faces the positively charged carbon. This surrounds the electrondeficient with more electron density and helps stabilize the positive charge. H H H δ H O O δ- H δ-O - δ- H δ O - O H H H H Biggest drawback to the SN1: Use of carbocations that can rearrange!! Summary: SN2 versus SN1 Factor Primary Secondary Tertiary Leaving Groups Solvent Nuc SN 2 Yes Yes No SN 1 No Yes Yes same for both Polar, aprotic Strong Polar Weak Strong Nucleophiles • Usually anions with a full negative charge (easily recognizable by the presence of sodium, lithium or potassium counterions) • Participate in SN2-type substitutions Examples: NaOCH3 (any NaOR), LiCH3 (any RLi), NaOH or KOH, NaCN or KCN, NaC≡CR (acetylide anion), NaNH2, NaNHR, NaNR2, NaI, KBr, NaCl, KI, NaN3 Weak Nucleophiles • Typically neutral molecules • Participate in SN1-type substitutions Examples: H2O, ROH, H2S, RSH SN2 or SN1? Ex. NaI OTs Ex. OTs H2 O Ex. OTs NaBr Ex. OTs Ex. NaOCH3 Br Ex. CH3OH H2 O Br