OrganicFallChpt8

OrganicFallChpt8 - Chapter 8 HW problems • All in-chapter...

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    Chapter 8 HW problems All in-chapter problems End of chapter problems: 17-19, 22, 23, 29, 30, 37 Extra problems: Devise syntheses for both of the products here, starting with the same material Show a synthesis for the following. Describe any stereochemistry , if necessary N 3 OH CN HO H H Br H 3 C H H 3 C H SCH 3 H H 3 C H 3 C H
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    Chapter 8: Nucleophilic Substitution Transformation of one functional group by another via nucleophilic substitution Nucleophile A species with a lone pair of electrons • A negative charge helps, but a positive charge is detrimental • Common nucleophiles come in the form M- R – M is a metal , typically alkalai metal, like Li, Na, or K – R is an organic compound with a lone pair and negative charge, like O- • Nitrogen and oxygen can act as nucleophiles through virtue of their lone pairs alone, but the negative charge will increase reactivity – Makes sense-- more electron density, more reactivity
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    Electrophiles Zones of electron deficiency A carbocation is a good electrophile A carbon attached to a heteroatom, like oxygen, nitrogen, or a halogen, will be electron poor Learn to identify poor-electron zones. Often times these will be electrophiles Nucleophile-Electrophile interactions Simplest form More complicated form Br Br Nucleophile Electrophile Br OH Notice the overall charge hasn't changed OH Br
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    Generalized forms of these two reactions Sn1 reaction • Carbocation is formed Nucleophile reacts where the carbocation is formed Rate limiting step is the loss of leaving group from the molecule -- only one molecule is involved, so it is first order Sn2 reaction • Let’s see that again • Notice what is happening -- one group is displacing the other group • The leaving group, Br, gains two electrons as it is shot from the molecule. This gives it a negative charge. The hydroxide ion uses a lone pair to form a bond at that last carbon, giving up two electrons to form a bond and losing its negative charge There are two molecules reacting here: hydroxide and the bromobutane. This is a bimolecular reaction, so it is second order Br OH Notice the overall charge hasn't changed OH Br
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OrganicFallChpt8 - Chapter 8 HW problems • All in-chapter...

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