Chapter8 - Organic Chemistry CHE 275 Chapter 8 Nucleophilic...

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Organic Chemistry CHE 275 Chapter 8 Nucleophilic Substitution
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Y : R X R + : X •nucleophile is a Lewis base (electron-pair donor) –often negatively charged and used as Na + or K + salt •substrate is usually an alkyl halide Nucleophilic Substitution +
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Substrate cannot be an a vinylic halide or an aryl halide, except under specific conditions to be discussed in Chapter 12. X C C X Nucleophilic Substitution
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+ R X Alkoxide ion as the nucleophile .. O : .. R' Examples of Nucleophilic Substitution gives an ether + : X R .. O .. R'
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(CH 3 ) 2 CHCH 2 ONa + CH 3 CH 2 Br Isobutyl alcohol (CH 3 ) 2 CHCH 2 OCH 2 CH 3 + NaBr Ethyl isobutyl ether (66%) Example
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+ R X Carboxylate ion as the nucleophile .. O : .. R'C O gives an ester + : X R .. O .. R'C O Examples of Nucleophilic Substitution
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OK + CH 3 (CH 2 ) 16 C CH 3 CH 2 I acetone, water O + KI O CH 2 CH 3 CH 3 (CH 2 ) 16 C Ethyl octadecanoate (95%) O Example
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+ R X Hydrogen sulfide ion as the nucleophile .. S : .. H gives a thiol + : X R .. S .. H Examples of Nucleophilic Substitution
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K S H + CH 3 CH(CH 2 ) 6 CH 3 Br ethanol, water + K Br 2-Nonanethiol (74%) CH 3 CH(CH 2 ) 6 CH 3 S H Example
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+ R X Cyanide ion as the nucleophile C N : : Examples of Nucleophilic Substitution gives a nitrile + : X R C N :
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DMSO Br NaCN + Cyclopentyl cyanide (70%) CN + NaBr Example
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Azide ion as the nucleophile .. .. N N N : : + + R X .. gives an alkyl azide + : X R .. N N N : + Examples of Nucleophilic Substitution
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Na N 3 + CH 3 CH 2 CH 2 CH 2 CH 2 I 2-Propanol-water CH 3 CH 2 CH 2 CH 2 CH 2 N 3 + Na I Pentyl azide (52%) Example
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+ R X Iodide ion as the nucleophile .. : I .. : gives an alkyl iodide + : X R .. : I .. Examples of Nucleophilic Substitution
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NaI is soluble in acetone; NaCl and NaBr are not soluble in acetone. acetone + NaI CH 3 CHCH 3 Br 63% + NaBr CH 3 CHCH 3 I Example
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Reactivity of Leaving Groups Reactivity of halide leaving groups in nucleophilic substitution is the same as for elimination. RI RBr RCl RF most reactive least reactive
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Br CH 2 CH 2 CH 2 Cl + Na CN A single organic product was obtained when 1-bromo-3-chloropropane was allowed to react with one molar equivalent of sodium cyanide in aqueous ethanol. What was this product? Br is a better leaving group than Cl Problem
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Br CH 2 CH 2 CH 2 Cl + Na CN A single organic product was obtained when 1-bromo-3-chloropropane was allowed to react with one molar equivalent of sodium cyanide in aqueous ethanol. What was this product? Problem CH 2 CH 2 CH 2 Cl + Na Br C N :
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• Many nucleophilic substitutions follow a second-order rate law. CH 3 Br + HO CH 3 OH + Br • rate = k [CH 3 Br][HO ] inference: rate-determining step is bimolecular Kinetics and the S N 2 Mechanism
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HO CH 3 Br + HOCH 3 Br + •one step HO CH 3 Br δ ± δ ± transition state Bimolecular Mechanism
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