Organic Chemistry 2510 Midterm 2 Osu

The bond strength of C-X decreases as

size of X increases

F>Cl>Br>I

bond dissociation

The bond between carbon and halogen is made up of an

sp3 orbital on carbon and a p orbital of the halogen

Short bonds are stronger than

longer bonds

Boiling points of haloalkanes are generally higher than

corresponding alkanes; due to dipole dipole interactions

Boiling points also rise with increasing

size of the halogen

polarizability

degree to which the electron cloud is able to deform

nucleophile

substances that contain an unshared electron pair

Nucleophilic substitution

reagent attacks the haloalkane and replaces the halide

Negatively charged nucleophile reacts with a haloalkane to yield

a neutral substitution product

An uncharged nucleophile yields

a positively charged substitution product( with the counteranion it becomes a salt)

substrate

starting organic material

Bimolecular Nucleophilic Subsitution

two reactants interact in one step

nucleophile attacks substrate with simultaneous expulsion of the leaving group

Sn2 is a

concerted reaction

backside displacement

the nucleophile approaches the carbon from the side opposite the leaving group

Inversion of Configuration

occurs with all Sn2 reactions

S goes to R, R goes to S

Stereospecific

a process whose mechanism requires that each stereoisomer of the starting material transform into a specific stereoisomer

As the nucleophile approaches the back lobe of the sp3 hybrid orbital used by the carbon to bind to the halogen, the rest of the molecule becomes

planar at the transition state by changing the hybridization to sp2.

Double inversion sequence of two Sn2 processes gives

desired product(if you want to maintain the same configuration); retention of configuration

Facility of Sn2 reactions depend on several factors

-nature of leaving group

-reactivity of nucleophile

-and the structure of the alkyl portion of the substrate

Leaving group ability is correlated with

its capacity to accommodate a negative charge

For halogens, leaving group ability increase

down the column

Sulfur Derivatives are also good

leaving groups

ROSO3- and RSO3-

Leaving group ability is inversely related to

base strength

Weak bases are the best

to accommodate negative charge and are the best leaving groups

Good leaving groups are the

conjugate bases of strong acids

Nucleophilicity depends on

charge,basicity,solvent,polarizability, and the nature of the substituents

increasing negative charge increases

nucleophilicity

Nucleophilicity decreases

to the right of the period table

The more basic the nucleophile

the more reactive

Increasing negative charge has a greater effect

than moving right to left on the periodic table for nucleophilicity

Nucleophilicity increases down

down a column

Solvents capable of hydrogen bonding are called

protic

Switching a protic solvent with an aprotic solvent

-reactivity of the nucleophile is raised

-base strength overrides solvation, thus F is better than I

sterically hindered nucleophiles are

poorer reagents

DMF is

aprotic

Branching at the reactive carbon

decreases the rate of the Sn2 reaction

Relative Sn2 displacement reactivity

methyl>primary>secondary>>tertiary

Branching on B-Carbon

retards subsititution

solvolysis

a substrate undergoes substitution by solvent molecules

For hydrolysis, the rates of reactions

increases with the substitution on the reacting carbon, thus tertiary>secondary>primary

Overview of Sn2 reaction

-has second order kinetics

-generates products stereospecifically with inversion of configuration

-Is the fastest with halomethanes and slower with primary and secondary halides

-takes place very slowly with tertiary substrates if at all

Overview of Solvolyses or Sn1

-follow first order rate law

-are not stereo-specific

-characterized by opposite order of reactivity

Uni molecular substitution

Sn1, only one molecule participates in the rate determining step

-the rate does not depend on the concentration of the nucleophile

Sn1 consists of three steps

1)formation of carbocation

2)attack by nucleophile

3)deprotonation

For solvolysis, a large excess of nucleophilic solvent ensures

complete solvolysis

To minimize electron repulsion, the positively charges carbon assumes

trigonal planar geometry,sp2 hybridization

Sn1 reactions obtain

racemic products

Sn1 rate increases as solvent

polarity increases

Protic solvents accelerate Sn1 because

it stabilizes the transition state by hydrogen bonding with the leaving group

Sn2 reactions are accelerated in

aprotic solvents

Sn1 speeds up with

better leaving group

Sn1 is not effected by the strength

of the nucleophile, but strengths may affect product distribution

Primary haloalkanes undergo only

bimolecular substitution

Relative stability of carbocations

primary

What is the reason for carbocation stability

hyperconjugation

The pathway of secondary haloalkanes depends on

the solvent, leaving group, and nucleophile

What makes Sn1 favorable for secondary substrates

substrate bearing good leaving group, poor nucleophile, and polar solvent

What makes Sn2 favorable for secondary substrates

high amounts of good nucleophile, reasonable leaving group, and aprotic solvents

Sn2 compared to Sn1 is

greener

Elimination

removal HX with the simultaneous generation of a double bond

E1

uni molecular eliminations

rate determining step is the formation of the carbocation

which hydrogens can participate in E1

any hydrogen posititions on any carbon next to the center bearing the leaving group

E2

bimolecular elimination

rate of alkene formation proportional to the concentrations of both the halide and the base

E2 reactions proceed in

a single step

three changes take place in E2 reactions

1)deprotonation by the base

2) departure of the leaving group

3) Rehybridization of the reacting carbon ceneter from Sp3 to Sp2 to furnish a double bond

How E2 and E1 differ

E2- base does not wait for carbocation formation because it is more aggressive

E1-Carbocation first, then base is protonated

If the leaving group is equatorial

all the corresponding hydrogons are not axial, making the elimination process slower

anti transition state is preferred

so the the base can extract a hydrogen the same time the leaving group leaves

Weakly basic nucleophiles give

substitution

Sn2-primary Secondary

Sn1-Tertiary

Weak nucleophiles such as water and alcohols react

at decent rates only with secondary and tertiary halides, substrates capable of Sn1, elimination is minor

Strongly basic nucleophiles give more

elimination as steric bulk increases

As steric bulk increases around carbon bearing leaving group,

substitution is retarded relative to elimination because an attack on carbon is subject to more steric hinderance relative to a attack on hydrogen

Branched primary substrates give

equal amounts of Sn2 and E2

Steric Bulk on the nucleophile

hinders attach at the electrophilic carbon, making elimination predominate

Good nucleophile weak base

subsitution more likely

Good nucleophile strong base

elimination increases

Sterically unhindered primary haloalkanes

substitution

Sterically hindered primary, secondary, and tertiary haloalkanes

elimination

Sterically unhindered nucleophiles that are strong bases

substitution

Sterically hindered nucleophiles that are strong bases

elimination

Primary Haloalkanes reactivity summary

-unhindered primary alkyl substrates will always yield Sn2 products except with sterically hindered nucleophiles then E2 becomes predominant

-However good nucleophiles will furnish Sn2

-Strong bases will give E2

-poor nucleophiles give NR

Secondary Haloalkanes reactivity summary

-Good nucleophiles favor Sn2

- Strong bases favor E2

-weakly nucleophilic polar media give E1 Sn1

Tertiary Reactivity haloalkanes summary

-strong bases(E2)

-non basic media gives E1 and Sn1

The name of the alcohol is based on the chain

containing the OH substiuent

When there is more than one hydroxyl group along the alkane stem

name is followed by diol, triol, etc

Alcohol boiling points are much higher than

their corresponding alkanes and haloalkanes

The high boiling points of alcohol are a result of

hydrogen bonding

Alkanes are

hydrophobic

Alcohols solubility in water

good; hydrophilic

The larger the alkyl part of an alcohol the

lower its solubility in water

The oxygen in alcohols hybridization is

sp3

Deprotonation of alcohols give

alkoxides

Protonation of alcohols gives

alkyloxonium ions

Why alcohols are acidic

the electronegativity of the O stabilizes the alkoxide molecule

Effects of branching on Alcohol acidity

methanol> primary>secondary>tertiary

Presence of halogens

increases acidity

Inductive effects

transmission of charge through sigma bonds in a chain of atoms; stabilizes the negative charge of the alkoxide Oxygen

Amphoteric

may be a base and acid

Synthesis gas

pressurized mixture of CO and H2 to make Methanol

usually consists of catalyst consisting of copper, zinc oxide, and chromium(III) oxide

changing catalyst to Rhodium leads to 1,2ethanediol(antifreeze)

Ethanol is prepared in large quantities by

fermentation of sugars or phosphoric acid catalyzed hydration of ethene