Lecture6: 123.101

Unit One Part 6:analysing chemical reactions

at last...chemical reactions!

Unit OnePart6Balanced reactions (pg69-70)Reaction types (pg71-72)Reagents (pg72-77)

NUCLEOPHILEelectron richhigh electron

density

ELECTROPHILEelectron poorlow electron

density

Chemistry...

...it’s that simple

this slide sums up the majority of

chemical reactions (even the funkier ones

follow these principles)

NUCLEOPHILEelectron richhigh electron

density

ELECTROPHILEelectron poorlow electron

density

Chemistry...

...it’s that simple

...so that was a quick lecture...

lets look at a simple reaction...

lets just show its that simple (we do have 50 minutes to

kill after all)

OHH3C

H3CH3C + HCl

ZnCl2 (aq)

the Lucas test

water solubleH-bond

water solubleion-dipole

water insolubletwo layers

ClH3C

H3CH3C + H2O

the Lucas test is simple...you’ll do it lab this week...two water soluble compounds

react...

OHH3C

H3CH3C + HCl

ZnCl2 (aq)

the Lucas test

water solubleH-bond

water solubleion-dipole

water insolubletwo layers

ClH3C

H3CH3C + H2O

to give an insoluble compound. Why

insoluble? (no longer able to H-bond)

ZnCl2 (aq)Cl

H3C

H3CH3C + H2OOH

H3C

H3CH3C + HCl

the Lucas test} }reactants products

stoichiometry1 11 1 ::: simple

definitions

ZnCl2 (aq)Cl

H3C

H3CH3C + H2OOH

H3C

H3CH3C + HCl

the Lucas test} }reactants products

stoichiometry1 11 1 :::

ratio of compounds in

reaction

...simply count atoms (or electrons)...

sorting out the stoichiometry is

easy...just make sure you have the same number of atoms on both sides of

the equation and remember...

NOatoms (or electrons)

created or destroyed

we only move them (or change

the bonds)

unfortunately, chemists are lazy...

a quick word of warning...organic chemists are a little lazy and often

miss side products off their reaction schemes!

OMeH2SO4

OMe

HO3S

depiction of a standard reaction

...so, here is a standard reaction as shown in many text

books...

OMeH2SO4

OMe

HO3S


...problem is, it doesn’t show

everything...so lets have a closer look...

so what bonds are broken?

OMe


obviously lost OH

OMe

HO3SHO SO3H

redraw sulfuric acid and it becomes

clear that we will loose an OH...

OMe


obviously lost OH

OMe

HO3SHO SO3H

...we can’t just add it to the aryl ring or one C would have too many

electrons...

OMe OMe

HO3SH

HO SO3H


remember we don’t always draw H


which means we ‘forgot’ to add HOH

OMe OMe

HO3SHO

HH

HO SO3H

...overall we have lost H from

aromatic and OH from acid...or water


which means we ‘forgot’ to add HOH

OMe OMe

HO3SHO

HH

HO SO3H


O OMe

MeOH

HO

HO

Br

O OMe

MeO

O

O


all the bonds broken & formed

O OMe

MeO

HO

HO

Br

O OMe

MeO

O

O

H

...break O–H bonds...


all the bonds broken & formed

O OMe

MeO

HO

HO

Br

O OMe

MeO

O

O

H

...break C–Br bond...


...and the ‘mystery’ product

O OMe

MeO

HO

HO

Br

O OMe

MeO

O

O

H Br

H

...form C–O bond AND H–Br


don’t forget the stoichiometry

1 13 3

O OMe

MeO

HO

HO

Br

O OMe

MeO

O

O

H Br

H

what is happening in a reaction?

it is the movement of electrons

Lewis structures

OHH3C

H3CH3C + Cl + H Cl

H3C

H3CH3C + H

OH

as each bond is two electrons we have just moved electrons...

here is the Lucas reaction again...

C

C

C

CH

HH

HHH

HH

HO H

Cl

HC

C

C

CH

HH

HHH

HH

H

O HH

Cl

Lewis structures

OHH3C

H3CH3C + Cl + H Cl

H3C

H3CH3C + H

OH


...and here is the Lewis structures

C

C

C

CH

HH

HHH

HH

HO H

Cl

HC

C

C

CH

HH

HHH

HH

H

O HH

Cl

Lewis structures

OHH3C

H3CH3C + Cl + H Cl

H3C

H3CH3C + H

OH


...we take O with its complete octet of

electrons and share two of them with H

to...

C

C

C

CH

HH

HHH

HH

HO H

Cl

HC

C

C

CH

HH

HHH

HH

H

O HH

Cl

Lewis structures

OHH3C

H3CH3C + Cl + H Cl

H3C

H3CH3C + H

OH


...form water...

C

C

C

CH

HH

HHH

HH

HO H

Cl

HC

C

C

CH

HH

HHH

HH

H

O HH

Cl

Lewis structures

OHH3C

H3CH3C + Cl + H Cl

H3C

H3CH3C + H

OH


...then we share the electrons of the chloride octet to make sure the original C is still an a

happy octet...

C

C

C

CH

HH

HHH

HH

HO H

Cl

HC

C

C

CH

HH

HHH

HH

H

O HH

Cl

Lewis structures

OHH3C

H3CH3C + Cl + H Cl

H3C

H3CH3C + H

OH


so just swapped electrons around...

what types of reaction are there?

3basic reactions

substitution reactions

A B + C A C + Bdoes what it says...the reagent swaps with a

functional group in our molecule...

CH3NH2Br

OH

NHCH3

OH

H Br


here’s an example from the synthesis of prozac® where an amine substitutes / displaces /

exchanges with a bromide...

fluoxetineProzac®

F3C

Cl

NHCH3

ONa

NaCl

O

F3CNHCH3


here is another example finishing the synthesis of prozac®; an alkoxide (alcohol derivative) substitutes a

chloride...

addition reactions

A + Y A Ytwo molecules add

together or combine...all atoms in both starting

materials are found in the product

addition reactions

here hydrogen bromide adds across an alkene to give us a

new bromide

CH3 + H Br BrH

CH3

CH3 + H Br BrH

CH3

addition reactions

at some point you’ll have to learn why it adds the bromide

to the more hindered end of the alkene...but that’s someone

else’s job!

Ph

OH3C MgBrCH3 Ph

CH3H3C O MgBr

addition reactions

here a Grignard reagent is adding to a ketone to

give (eventually) an alcohol

elimination reactions

A X A + X

elimination is the opposite of addition...we rip a bit of

the molecule off.

CH3OH

H

CH3HO

H


this example shows the elimination of

water (dehydration) to form an alkene

H3C

CH3H3C

H

Br

H3C

CH3H3C H Br


or the elimination of hydrogen bromide

(hydrobromic acid) to form an alkene

what reagents are involved in these

reactions?

now we know what reactions we can perform...what reagents can we use?

nucleophileselectron rich molecules

donate2electrons

nucleophiles are electron rich compounds that

donate two electrons to form a new bond...examples include...

anionsH O ≡ OH

3 lone pairs

Br Br≡4 lone pairs

H3CPr C

H

H1 lone pair

≡

anionsH O ≡ OH

3 lone pairs

Br Br≡4 lone pairs

H3CPr C

H

H1 lone pair

≡

negatively charged compounds that possess a lone pair of electrons that

can form a new bond

HO H H

OH

anionsnucleophilic reaction

HO H H

OH


hydroxide donates a pair of electrons to form a new bond to

proton

HO H H

OH


note: both sides of equation have same

charge (both are neutral overall)

HO H H

OH


this is a good sign that you’ve got the reaction right!

note: both sides of equation have same

charge (both are neutral overall)

lone pairs

HO

Hwater

HN

HH

ammoniaH3C

SCH3

dimethyl sulfide (DMS)

lone pairs as nucleophile

lone pairs on neutral molecules are also good

nucleophiles...

HN

HH

H

Cl

lone pairsnucleophilic reaction

HN

HH

H Cl

HN

HH

H

Cl


HN

HH

H Cl

ammonia donates a pair of electrons to form a new bond to hydrogen chloride

HN

HH

H

Cl


HN

HH

H Cl

can’t have two bonds to H (4 electrons) so break H–

Cl bond with electrons flowing towards most

electronegative element

HN

HH

H

Cl


HN

HH

H Cloverall its been a substitution

HN

HH

H

Cl


HN

HH

H Cl

note: overall both sides are neutral (have same

charge)

bondsC Y

δ+δ–

nucleophilic site

Y=Li, MgHB

HHH

HC

HC

H

Hhigh electron density

nucleophile

RR = electron

donating group

reactive σ bond

bonds can also be a source of electrons...and these cause confussion!

electrophileselectron poor molecules

accept2electrons

the other kind of reagents are...electrophiles are electron

poor compounds that accept two electrons to form a new bond...

examples include...

H ≡ H

proton

empty 1s orbitalno electrons!

simplest electrophile...it has no electrons and

desperately wants some!

HO

H

HHO

HH

protonelectrophilic reagent

HO

H

HHO

HH

protonelectrophilic reagent

nucleophilic water donates two electrons to

proton, which accepts them to form new bond

Group 13F B

FF

empty 2p orbital

Cl AlClCl

empty 3p orbital

F BF

F

only have 6 valence electrons...need 8 to obey

octet rule so happily accept a pair of electrons

C Yδ+ δ–

electrophilic site

Y = Cl, Br, N, O

bondsC Y Y = O,

NRδ+ δ–

electrophilic site

bonds can be electrophiles if they are polarised and have a δ+

centre

H3C CH3

O OH

HHO

H H3C CH3

O

bondselectrophilic reagent

is it that easy?

nucleophile + electrophile product

yup, this equation sums up the majority of reactions...

...and no...

OH3C

H3CH3C

H

δ+ δ–Cl

H3C

H3CH3C

δ+ δ–

polarity in these two compounds looks the

same...so do they behave in the same way??

OH3C

H3CH3C

H

δ+ δ–Cl

H3C

H3CH3C

δ+ δ–

of course not!this is chemistry after all...

OH3C

H3CH3C

H

δ+ δ–

nucleophile

C ClH3C

H3CH3C

δ+ δ–

electrophile

alcohol uses lone pair to act as a

nucleophile

OH3C

H3CH3C

H

δ+ δ–

nucleophile

C ClH3C

H3CH3C

δ+ δ–

electrophilechloride is an electrophile

...you will learn to to identify which is which...

so what actually happens in a reaction?

H3C O

CH3H3C

H + H ClH3C Cl

CH3H3C +

HO

H

what order are the bonds made and broken?

so here’s the Lucas test again...overall

it’s a substitution of hydroxyl for chlorine

H3C O

CH3H3C

H + H ClH3C Cl

CH3H3C +

HO

H


OH

H3C

CH3H3C

HO

HH3C

CH3H3C

H

additionstepstep

one

CH3CCH3

CH3

O H H CH3CCH3

CH3

O HH

OH

H3C

CH3H3C

HO

HH3C

CH3H3C

H

additionstepstep

one

CH3CCH3

CH3

O H H CH3CCH3

CH3

O HH

nucleophilic alcohol donates 2 electrons to a proton from H–Cl in an addition reaction

OH

H3C

CH3H3C

HO

HH3C

CH3H3C

H

additionstepstep

one

CH3CCH3

CH3

O H H CH3CCH3

CH3

O HH

note: octet rule is obeyed and the

charge is the same on both sides

(positive)

C

C

C

CH

HH

HHH

HH

HO HH

C

C

C

CH

HH

HHH

HH

H

O HH

CH3

CH3H3C

HOH

OH

H3C

CH3H3C

H

eliminationsteptwo

C

C

C

CH

HH

HHH

HH

HO HH

C

C

C

CH

HH

HHH

HH

H

O HH

CH3

CH3H3C

HOH

OH

H3C

CH3H3C

H

eliminationsteptwo

elimination of water...molecule splits in two...electrons go with most electronegative

atom

C

C

C

CH

HH

HHH

HH

HO HH

C

C

C

CH

HH

HHH

HH

H

O HH

CH3

CH3H3C

HOH

OH

H3C

CH3H3C

H

eliminationsteptwo

charge is the sameon both sides (positive)...we now have a reactive

intermediate(charged species with 6

electrons)

C

C

C

CH

HH

HHH

HH

HCl C

C

C

CH

HH

HHH

HH

HCl

ClH3C

CH3H3C

CH3

CH3H3CCl

additionstepstep

three

C

C

C

CH

HH

HHH

HH

HCl C

C

C

CH

HH

HHH

HH

HCl

ClH3C

CH3H3C

CH3

CH3H3CCl

additionstepstep

three

finally, second addition step to complete octet

C

C

C

CH

HH

HHH

HH

HCl C

C

C

CH

HH

HHH

HH

HCl

ClH3C

CH3H3C

CH3

CH3H3CCl

additionstepstep

three

nucleophile donates two electrons to electrophile

that accepts them to form new bond

C

C

C

CH

HH

HHH

HH

HCl C

C

C

CH

HH

HHH

HH

HCl

ClH3C

CH3H3C

CH3

CH3H3CCl

additionstepstep

three

note: the charge is the same on both

sides (neutral)

3steps

1substitutionreaction

another representation...

we can follow the change of energy during a reaction...in a

reaction profile

ener

gy

reaction progress

OH + HCl

O + ClH

H

+ ClH2O

Cl + H2O

reaction profile

ener

gy

reaction progress

OH + HCl

O + ClH

H

+ ClH2O

Cl + H2O

reaction profileeach hill represents

one step in our reaction

ener

gy

reaction progress

OH + HCl

O + ClH

H

+ ClH2O

Cl + H2O

reaction profileeach dip

represents an intermediate we

can see

ener

gy

reaction progress

OH + HCl

O + ClH

H

+ ClH2O

Cl + H2O

reaction profilehigher the hill, the

harder the step

ener

gy

reaction progress

OH + HCl

O + ClH

H

+ ClH2O

Cl + H2O

reaction profile

the lower in energy any stage the more

stable it is

ener

gy

reaction progress

HOCH3Br

CH3OHBr

just one hill, so just one step

ener

gy

reaction progress

HOCH3Br

CH3OHBr

no dip so no intermediates

ener

gy

reaction progress

HOCH3Br

CH3OHBr

HO + CH3Br CH3OH + Br

direct substitution

would look like this

HBr

HH

H

HHHO Brδ– δ–

+ + BrH

NuH

H

transition state

12

12SLOW

RDSHO

ener

gy

reaction progress

HOCH3Br

CH3OHBr

once again, you will learn more about this but not

from me! Why don’t I get to teach the cool stuff?

what have....we learnt?

Picture: © Pittsburgh Supercomputing Center

•analyse reactions•classify reagents•react ion t ypes

readpart7

moreexamples


H BrHBr

H BrHBrH

H HH

what order are the bonds made and broken?First, remember that our

skeletal figures are simplifications...these are all the atoms around the site of the

reaction

substitutionstepstep

one

H BrH

Br

CH3C

CH

HH Br C

HH

HBr

H3CC

H3C

H3Cthe alkene is the nucleophile. Lots of

electrons between two carbon atoms


one

H BrH

Br

CH3C

CH

HH Br C

HH

HBr

H3CC

H3C

H3C

it participates in a substitution reaction,

replacing the bromide...


one

H BrH

Br

CH3C

CH

HH Br C

HH

HBr

H3CC

H3C

H3C

the alkene attacks the proton and not the

bromide. Why do the electrons go for the less electronegative atom?


one

H BrH

Br

CH3C

CH

HH Br C

HH

HBr

H3CC

H3C

H3CIf we attacked the bromide first then, to obey the octet rule, we

would have to give two electrons to H and it would leave as H–. The

hydride (H–) is not stable as it is not electronegative


one

H BrH

Br

CH3C

CH

HH Br C

HH

HBr

H3CC

H3C

H3C

electrons flow towards the most electronegative

element


one

H BrH

Br

CH3C

CH

HH Br C

HH

HBr

H3CC

H3C

H3C

reaction is not over as we have a C with only 6 electrons. It is not obeying the octet rule so is

very reactive

additionsteptwo

HBr

HBr

CH

HH

BrCH3C

H3CCH

HH

CH3CH3C

Br

additionsteptwo

HBr

HBr

CH

HH

BrCH3C

H3CCH

HH

CH3CH3C

Br

Second step is addition. The bromide shares 2 electrons with the carbon so that they

both obey the octet rule

2steps

1additionreaction

ener

gy

reaction progress

reaction profile

H Br

Br

Br

ener

gy

reaction progress

reaction profile

H Br

Br

Br

the cation (positive charge) is the intermediate


Br

HHOHOH Br

eliminationstepone

Br

HHOHOH Br

C HH

CH3CH3C

HOH

Br

OHH

CH

HC

H3C

H3CBr

eliminationstepone

Br

HHOHOH Br

C HH

CH3CH3C

HOH

Br

OHH

CH

HC

H3C

H3CBr

once again the electrons are flowing towards the most electronegative

element

eliminationstepone

Br

HHOHOH Br

C HH

CH3CH3C

HOH

Br

OHH

CH

HC

H3C

H3CBr

all atoms maintain the octet rule

1step

1eliminationreaction

ener

gy

reaction progress

reaction profile

Br

HHO

HOH

Br

Br

HHOδ–

δ–

transition state

no intermediate as the reaction occurs by a

single step

Lecture6: 123.101

Technology

Transcript of Lecture6: 123.101