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S N 1 Reactions t-Butyl bromide undergoes solvolysis when boiled in methanol: Solvolysis: “cleavage by solvent” nucleophilic substitution reaction in which the solvent serves as the nucleophile
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S N 1 Reactions The reaction between t-BuBr and methanol does NOT occur via an S N 2 mechanism because: t-BuBr: tertiary alkyl halide too hindered to be S N 2 substrate CH 3 OH: weak nucleophile Solvolysis reactions occur via an S N 1 mechanism:
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S N 1 Reactions substitution nucleophilic unimolecular Rate = k[R-X] 1 st order overall 1st order in [R-X] zero order in [Nuc] Only R-X is present in the transition state for the rate determining step Nucleophile is NOT present in RDS
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S N 1 Reactions General Mechanism: Step 1: Formation of carbocation is the RDS. Step 2: Attack of the nucleophile
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S N 1 Reactions Reaction Energy Diagram for S N 1 Reactions: Formation of a carbocation is highly endothermic. According to Hammond’s Postulate, the transition state most closely resembles the carbocation.
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S N 1 Reactions The reactivity of a substrate in an S N 1 reaction depends on the stability of the carbocation formed: 3 o > 2 o > 1 o > methyl Allylic and benzylic halides often undergo S N 1 reactions because the resulting carbocations are resonance stabilized. + +
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S N 1 Reactions S N 1 reactions involve: weak nucleophile H 2 O not OH - CH 3 OH not CH 3 O - Substrates that form stable carbocation intermediates: 3 o, benzylic, or allylic halide are most favored 2 o (sometimes)
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S N 1 Reactions Example: Draw the mechanism for the reaction of t-butyl bromide with methanol.
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S N 1 Reactions
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S N 1 Reactions-Stereochemistry The carbocation ion intermediate formed during an S N 1 reaction is sp 2 hybridized and planar. The nucleophile can attack from either side of the carbocation. A mixture of both possible enantiomers forms. S N 1 reactions occur with racemization: a process that gives both enantiomers (not necessarily in equal amounts) of the product Racemization occurs because both retention and inversion of configuration take place.
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S N 1 Reactions-Stereochemistry + Attack from top Attack from bottom - H +
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S N 1 Reactions-Stereochemistry When the nucleophile attacks from the side where the leaving group was originally, retention of configuration occurs. NaOCH 2 CH 3 Attack from top (R)
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S N 1 Reactions-Stereochemistry When the nucleophile attacks from the back side (opposite to the original leaving group), inversion of configuration occurs. (R) (S) NaOCH 2 CH 3 Attack from bottom
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S N 1 Reactions-Stereochemistry For most S N 1 reactions, the leaving group partially blocks the front side of the carbonium ion more inversion of configuration less retention of configuration
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S N 1 Reactions-Rearrangements Carbocations often undergo rearrangements, forming more stable cations. Structural changes resulting in a new bonding sequence within the molecule The driving force for a rearrangement is the formation of a more stable intermediate. 1 o or 2 o carbocation rearranges to a more stable 3 o carbocation or resonance-stabilized carbocation
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S N 1 Reactions-Rearrangements A mixture of products often forms as a result of rearrangements during S N 1 reactions. NOTE: Rearrangements cannot occur during S N 2 reactions since an intermediate is not formed. + CH 3 CH 2 OH Rearranged product Rearrangement occurs via hydride shift.
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S N 1 Reactions-Rearrangements Common rearrangements: Hydride shift (~H) the movement of a hydrogen atom and its bonding pair of electrons Methyl shift (~CH 3 ) the movement of a methyl group and its bonding pair of electrons Alkyl shift (~R) the movement of any alkyl group and its bonding pair of electrons
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S N 1 Reactions-Rearrangements Hydride Shift Mechanism: Step 1: Formation of carbocation and rearrangement: + + + Br - ~H 2o2o 3o3o
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S N 1 Reactions-Rearrangements Hydride Shift Mechanism: Step 2: Nucleophile attack and loss of proton (if needed) + CH 3 CH 2 OH + CH 3 CH 2 OH 2 + +
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S N 1 Reactions-Rearrangements Example of a Methyl Shift (~CH 3 ):
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S N 1 Reactions-Rearrangements Mechanism of ~CH 3 : Step 1: Simultaneous (because primary carbocation is unstable) shift of methyl group and loss of leaving group: 3 o carbocation formed preferentially
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S N 1 Reactions-Rearrangements Mechanism of ~CH 3 : Step 2: Attack of nucleophile and loss of proton (if needed)
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Important! Important! Important! How do you know if the carbocation forms first and then rearrangement occurs or if formation of carbocation and rearrangement occur simulataneously??? In general: Secondary (2 o ) halides form the carbocation first and then rearrangement occurs (i.e. 2 steps) In general: Primary halides undergo simultaneous formation of carbocation and rearrangement. (Primary carbocation is quite unstable!)
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S N 1 Reactions-Rearrangements Example: Propose a mechanism for the following reaction. CH 2 CH 3 OH +
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S N 1 vs. S N 2 SN2SN2 Strong nucleophile Primary or methyl halide Polar aprotic solvents (acetone, CH 3 CN, DMF) Inversion at chiral carbon No rearrangements Weak nucleophile (may also be solvent) Tertiary,allylic, benzylic halides Polar protic solvent (alcohols, water) Racemization of optically active compound Rearranged products SN1SN1
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E1 Reactions An elimination reaction involves the loss of two atoms or groups of atoms from a substrate, usually forming a new bond. Elimination reactions can occur via a first order (E1) or a second order (E2) process. Na + - OCH 3 CH 3 OH + Br -
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E1 Reactions E1 reactions: Elimination, unimolecular 1st order kinetics Rate = k[R-X] RDS transition state involves a single molecule General conditions: 3 o and 2 o halides weak bases
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E1 Reactions E1 Mechanism: Step 1: Formation of carbocation (RDS) Step 2: Base abstracts proton
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E1 Reactions E1 reactions almost always occur together with S N 1 reactions.
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E1 Reactions CH 3 CH 2 -O-H H +
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E1 Reactions Once formed, a carbonium ion can: recombine with the leaving group react with a nucleophile forming a substitution product (S N 1) lose a proton to form an alkene (E1) rearrange to form a more stable carbocation and then: react with nucleophile lose a proton to form an alkene
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E2 Reactions E2 reactions: Elimination, bimolecular 2nd order kinetics Rate = k[R-X][B - ] RDS transition state involves two molecules General conditions: 3 o and 2 o halides strong bases
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E2 Reactions In the presence of a strong base, elimination generally occurs in a concerted reaction via an E2 mechanism
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E2 Reactions S N 2 reactions require an unhindered methyl or 1 o halide steric hinderance prevents nucleophile from attacking 3 o halides and forming the substitution product E2 reactions generally involve the reaction between a 3 o and 2 o alkyl halides and a strong base.
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E2 Reactions The reaction of t-butyl bromide with methoxide ion gives only the elimination product. The base attacks the alkyl bromide much faster than the bromide can ionize.
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E2 Reactions Many alkyl halides can eliminate in more than one way. Mixture of alkenes produced
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E2 Reactions Saytzeff Rule: When two or more elimination products can be formed, the product with the most highly substituted double bond will usually predominate. R 2 C=CR 2 > R 2 C=CHR > RHC=CHR and R 2 C=CH 2 > RHC=CH 2
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E2 Reactions Example: Draw the structures for all possible products of the following reaction. Which one will predominate? NaOCH 2 CH 3 EtOH
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E2 Reactions E2 reactions follow a concerted mechanism: bonds breaking and forming simultaneously specific geometry required to allow overlap of orbitals of bonds being broken and bonds being formed E2 reactions commonly involve an anti-coplanar conformation.
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E2 Reactions
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Example: Predict the structure of the elimination product formed by the following reaction. NaOCH 3 CH 3 OH
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E1 vs E2 E1 Weak base 3 0 > 2 o Good ionizing solvent polar, protic (water, alcohols) Saytzeff product No required geometry Rearranged products possible Strong base required 3 o > 2 o Solvent polarity not important Saytzeff product Coplanar leaving groups (usually anti) No rearrangements E2
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