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

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:

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

S N 1 Reactions General Mechanism: Step 1: Formation of carbocation is the RDS. Step 2: Attack of the nucleophile

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.

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. + +

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)

S N 1 Reactions Example: Draw the mechanism for the reaction of t-butyl bromide with methanol.

S N 1 Reactions

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.

S N 1 Reactions-Stereochemistry + Attack from top Attack from bottom - H +

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)

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

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

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

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.

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

S N 1 Reactions-Rearrangements Hydride Shift Mechanism: Step 1: Formation of carbocation and rearrangement: Br - ~H 2o2o 3o3o

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 + +

S N 1 Reactions-Rearrangements Example of a Methyl Shift (~CH 3 ):

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

S N 1 Reactions-Rearrangements Mechanism of ~CH 3 : Step 2: Attack of nucleophile and loss of proton (if needed)

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!)

S N 1 Reactions-Rearrangements Example: Propose a mechanism for the following reaction. CH 2 CH 3 OH  +

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

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 -

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

E1 Reactions E1 Mechanism: Step 1: Formation of carbocation (RDS) Step 2: Base abstracts proton

E1 Reactions E1 reactions almost always occur together with S N 1 reactions.

E1 Reactions CH 3 CH 2 -O-H H +

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

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

E2 Reactions In the presence of a strong base, elimination generally occurs in a concerted reaction via an E2 mechanism

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.

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.

E2 Reactions Many alkyl halides can eliminate in more than one way. Mixture of alkenes produced

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

E2 Reactions Example: Draw the structures for all possible products of the following reaction. Which one will predominate? NaOCH 2 CH 3 EtOH

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.

E2 Reactions

Example: Predict the structure of the elimination product formed by the following reaction. NaOCH 3 CH 3 OH

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