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chem3331-bean-06 - CHAPTER 6 Alky Halides Nucleophilic...

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Unformatted text preview: CHAPTER 6: Alky Halides: Nucleophilic Substitution and Elimination (No sections omitted) ‘ . -~. “."r*-‘:,’r-f‘!'i'-!: 27‘» . :"YT‘C": ms 311112-me11" MID! 1~ Types of Organohalogen Compounds. “ ‘ Emmi—11 my "121:: 17 11;): id R—X \ / c=c / \ X 0x Common Uses: section 3 F T i” I Cl—C—F F—c—Ic—H DDT | Cl F CI Nomenclature: No new rules, halogen has no priority. For name as substituent, drop "ide" add "0". Example: chloride becomes chloro new terms: vicinal dihalide - two halogens on adjacent carbons geminal dihalide- two halogens on the same carbon Structure: 1 °/ 2°/ 3° determined by the type of carbon the halogen is bonded to Example - In a 1° halide, the halide is bonded to a 1° carbon atom. Physical Properties: types of intermolecular attractions: 1 . 2. /\/\F /\/\Cl /\/\Br /\/\I Preparation: 1. Free radical halogenation: (Caution - mixtures of products) Best results when a _ single major product is produced. Rememberzbromine is more selective than chlorine. O Br2 /|ight /l\ Br2 / light /\/\/\/ Br2 / light 2. Allylic brominaton: O Br2 /|ight Reason for selectivity: Consider first step of propagation H H Better reagent: Nucleophilic Substitution Reactions H \ .. c—x H\‘“/ H General Reaction: 9 Nu 3 + R—X ——-—> R—Nu + X Two specific reactons: G A. :CN + CH30I —> CH3CN + Cl 9 e e B. :CN + CH3)3CC| ——> (CH3)3CCN + Cl Both reactions are nucleophilic substitution reactions, but A and B proceed through differentmechanisms. Before proposing a mechanism for each, we consider two kinds of experimental evidence: 1. The kinetics of the reaction. - Examine rate changes in relation to changes in the concentrations of the reactants. 2. The stereochemistry of the products. Reaction A 1. Kinetics: Observations - if cone. of Nu is doubled - if cone. of substrate is doubled - rate = The Mechanism: 2. Stereochemistry: Nucleophile (Nu) begins to form bond to electrophilic carbon while leaving roup (LG) is still bonded. Nu donates electron pair to the small back lobe of the sp hybrid that forms bond to LG. This is called "backside attack" and results in inversion of carbon's configuration (Walden Inversion). Carbon configuration is turned "inside out." \ 9 / A 3 Nu: OC<3E® ,. c—cr —CN—» NC—C H\\\\ H/H H H H e H \ :°C3H / .-- C——Br —L—> HO—C .,, CH3\\\\/ "IICH3 CH30H2 CH20H3 8N2 is a stereospecific reaction - a particular stereoisomer reacts to give one specific stereoisomer of the product, even though another is possible. 8N2 summary: 1. rate = k[Nu] [substrate] 2. bimolecular TS 3. proceeds with 100% inversion Reaction B G 9 :CN + (CH3)3CCI ———> (CH3)3CCN + Cl 1. Kinetics: Observations - if conc. of substrate is doubled - if conc. of Nu is doubled - rate = The Mechanism: 2. Stereochemistry: Nu attacks after LG has left. Nu attacks a planar carbocation and can attack from either side. Attack from the opposite side of the LG's original position results in inversion of the carbon's configuration. Attack from the same side as the LG's original position results in retention of the carbon's configuration. This result is known as racemization of stereochemistry. The 8N1 process is not stereospecific. 55% I 55% <°— ‘ CG ——'—> H C‘“ \ Racemization Example: CH3CHZCH2\ CH “w“ C Br H20/acetone 3 / heat CH30H2 SN1 summary: 1. rate = k[substrate] 2. unimolecular TS for RDS 3. proceeds with racemization (may not get a 50/50 mixture due to ion pair formation) We've looked at two reactions, two mechanisms for nucleophilic substitution of alkyl halides: 8N1 and 8N2. We are left with a BIG QUESTION: What determines which mechanism is taking place in the reaction of a given alkyl halide? Simple Answer: Rate - faster pathway wins Factors that Affect Rate: 1. Substrate Structure 2. Nature of the Leaving Group 3. Concentration and Reactivity (Strength) of the Nucleophile (3N2 only) 4. Solvent Substrate Structure vs Reaction Rate 3N2 Process: Relative Reaction Rates for Alkyl Halides: H H H// / H H H “n /H H/”'C/ (I? \H ll"c/ \ ’c 1 \ H F (I \“tC—X ( H \\.C—X H\ \\.-C—C4H H H .C—X H C—X \ \ \ \ \ I \\‘ H I u‘ C ‘ \\‘ H H‘ I C HI I C X HiiulC A H /: \ H /¢ \ / C H\ H H H\ H H H H: \H Substrate Structure vs Reaction Rate 3N1 Process: Relative Reaction Rates for Alkyi Halides: Allyl 0+ 2 + + HQC=CH——CH2 <——-——> HzC—CH=CH2 Benzyi C+ : + + QCHQ *——> QCHZ <—-> +GCH2 <—-> GCHQ + Order of 0+ stabiliy: 3° allyiic/benzyiic > 3° = 2° allyiic/benzylic > 2° = 1° allyiic/benzylic > 1° > methyl 8N1 ISN2 Competition: CH3X 1° 2° 3° Nature of the Leaving Group - Same influence on rate in 8N1 and 8N2 - only the timing of the departure is different — LG must be electron withdrawing - must polarize the C - X bond - must put 6+ charge on C - LG must form a stable species (a stable anion, a neutral molecule - a weak base) - The weaker the base formed - Order of Reactivity of Alky Halides (by 8N1 or 8N2): RI RBr RCI RF How is reaction rate affected by the LG? Table 6 - 4 gives other good LG Example: 0 || R——O——fi——<:::>F—mfi _ML, 0 Strong bases make bad leaving groups in 8N1 and 8N2 processes. Example: Nu R—OH —> But, under acidic conditions: R—OH ——> Concentration and Strength of Nucleophile a) concentration 8N1- 8N2- b) strength (nucleophilicity - affinity for the 6+ carbon) 8N1- 8N2- strong Nu - weak Nu - Io Trends in Nucleophilicig; (Affinity for a positive or 6+ carbon atom) 1. A negatively charged nucleophile is stronger than its conjugate acid. 2. Nucleophilicity decreases from left to right in the periodic table. Electronegative elements "hold on" to lone pairs more tightly; therefore lone pairs are less available to form new bonds. 3. N ucleophilicity increases down a group in the periodic table as size and polarizability increase. Larger atoms hold outer electrons more loosely so electrons move more freely toward a positive charge. This contributes to earlier and stronger bonding in the transition state. 4. Bulky groups on the nucleophile decrease nucleophilicity. H Strength of the Nucleophile and the SN1/SN2 Choice for 2° Substrate 8N2: Nu 05+ 8N1: N' 0* Strong Nu necessary Weak Nu OK Strong Nu favors 8N2 Weak Nu favors 8N1 CAUTION: Does not mean a 1° substrate with a weak Nu will proceed by 8N1 or a 3° substrate with a strong Nu will proceed by SN2l! Example: CH3 CH3 0' NaI 0' AL, ., —-—> ., heat [/H acetone [/H Solvent (influence on rate) 8N2 :1} Table 6-3 lists nucleophiles in order of strength in a particular type of solvent: The trends in nucleophilicity were also determined in this type of solvent. Example: I' > Br‘ > Cl' > F' Consider how the protic (hydroxylic) solvent interacts with Nu: "e °°|.‘e I2. Now consider nucleophilicity in a polar, aprotic (nonhydroxylic) solvent. Examples: 0 0 0 II g CH3 3 C / CHSCN ch/ \CH3 H/ \N “30/ \CH3 CH3 polar: aprotic: Using a polar, aprotic solvent greatly increases the rate of the 8N2 process. Switching from a hydroxylic solvent to an aprotic solvent can increase the rate as much as one million times! protic solvent: aprotic solvent: Example: Br 0/ KF/HgO ——> Br KF/ CH CN W Solvent effect on SW :> R—X ——> [R———X] ——> [[email protected]] x9 CH3 CHa—C—Cl —___, CH3 The more polar the solvent :1) Higher the dielectric constant, the more polar the solvent. dielectric const. relative rate of t-BuCI ionization H20 MeOH EtOH Acetone Ether Hexane Protic solvents I11) Solvent Summary: 8N2: Solvent affects strength of Nu. Aprotic solvents destabilize Nu (Nu is not solvated well), so increase the strength of Nu. Aprotic solvent is not required. SN1: Solvent facilitates ionization by stabilizing the TS, the carbocation and the LG. Protic solvent is required for significant ionizaton. Summary of SN1/SN2 choices: CH3X, 1° 30 2°, allyl, benzyl i’f Carbocation Rearrangement : Any that can rearrange, WILLI! E'sr CH CH OH/A (IDEt (I)Et CHS'CH'?H'CH3 ‘J—Zfi CHs-CH-ICH-CHs + CHs-CHg-(l3-CH3 CH3 CH3 CH3 CH3 Br CH3 OEt 0a CH3 CHg—Cll) —C|}H-CH3 W» CH3—C:3 ——(I3H-CH3 + CH3— —c —CH- c-H3 CH3 CH3 CH3 Il3r OEt CH-CH3 CH- -CH3 /5’ Elimination Reactions: H T I —§>—f— + Nu: ———» X Two pathways: (timing of bond breaking and forming different as in SN1/SN2) E1 : elimination, unimolecular - T8 of RDS involves a single molecule, the substrate rate = k[RX] Mechanism: l 'i I 'i step1: —c—c— ; —c—-c—— ' l + I X H I | \\ // step 2: —- C—C— EH C=C + l // \\ Consider the reaction and mechanism: 3”” CH3- C—CI 95%EtOH/H20 l 65°C CH3 lacs Order of reactivity by E1: 3° 2° 10 lb E2 : elimination, bimolecular - TS of RDS involves two molecules, the substrate and Nu rate = k[RX] [Nu] Mechanism: ,,, ill 4, H B I D 6 \ F—C-I'l/U + B R_S_> \ 0—0 ”II", ——> x \ \ Consider the reaction and mechanism: H3040 T _ + —-——> I \ H CH3CH20H Br H Order of reactivity by E2: 3° 2° 1° Orientation of Elimination and Alkene Stability: H CH3 H _ - + CH ON (CH ) CO Na CH3_('3H_b_éH2 3‘a[ l Br Saytzeff's Rule: in an elimination, the more highly substituted alkene (more stable) alkene predominates order of P\ R R H R\ H R R R\ H alkene 0:0 0:0 0:0 C=C C=C\ stability R R R R H H H H H H 17 )8 Stereochemistry of E2: A stereospecific reaction 1. All four substrate atoms involved in the reaction must be coplanar. Two conformations meet this requirement. 2. Preferred conformation: Exception to preferred conformation: Experimental Evidence for Anti-periplanar (Co-planar) Geometry in E2 Process: front carbon H\ \H Br H \\f——C\" P“ NaNH2 \ / + \“‘ etfianol ’ O: C Br Ph Br / \ Ph Ph back carbon A Anti-periplanar result: H H H Ph \ H Ph OR Br Ph \ Br Ph Br Br Syn-periplanar result: H Br ?' H OR ' Ph H Br Ph Br Ph Ph H I? Experimental Evidence for 1,2 - Diaxial Requirement in E2 Process on Cyclohexyl Substrates H30>O--HIICH(CH3)2 H3C -iu|CH(CH3)2 I I ’/ ”(3' CI A. neomenthyl chloride B. menthyl chloride H HsC—Q—CH(CH3>2 H CH(CH3)2 H EtO‘ Na+ + H30 ————’ C, EtOH H H3C><;>""”CH(CH3)2 A. neomenthyl chloride H CH3 H CH(CH3)2 ___ C, H Cl H30 H H H H EtO' Na+ H B. menthyl chloride ETOH H30>QMHICH(CH3)2 CH(CH3)2 Substitution vs Elimination (Wade summaries: sections 6-16 and 6-22) Nu = strong base ('OH, HO, 'NH2, RCEC:‘) Nu = weak base bulky bases particularly favor E2 (tBuO ') $fl_ SN2/SN1/E1/ETk -- CH3, 1°, 2°, allylic/benzylic —- 3°, 2°, allylic/benzylic --Strong Nu, but weak .. Weak Nu . . -' Ionizmg solvent base such as iodide, is necessary HS ', RS ‘ , or ' CN _. 8N1 usually majorprocess —- faster rate in aprotic high T increases E1 solvents *Exception: 1° halide with a strong, unhindered base - 8N2 gives major product Examples: 5+0“ 1. CH3CH28r + CH30H20'Na+‘m—) CH30HZOCHZCH3 + CH2=CH2 _ 5/070 aw m ‘70 (E2) (-5“ fl 3 2. CH3(CH2)150H2CHQBf + (CH3)3CO.K+ o ‘ CH3(CH2)15CHQCHQOC(CH3)3 + CH3(CH2)15CH:CH2 V52. (6N2) 5’52; (62) (5V l _ + E+OH 3. CHaCHCHa + CHscHQO Na fl CH3CH=CH2 + CH3$3HCH3 .2]? (5 Z) r ’ 77970 (E2) oéuzcgj d N ace} 4. CH30HCH3 + Na! __0n_€_) CHacHCHa (no? (<5 2) i ’” 0 l’ J. 9H3 CH3 / CHs-IC-OCHQCHg + CH2=C CH3 3 2, (5w) 7 77a (\ CH3 52) cH3 [CH3 CHg‘P'OCHQCHg + CH2=C\ CH3 CH3 8029611!) 9023 CE“ ...
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