Lecture6: 123.101

Unit One Part 6:analysing chemical reactions at last...chemical reactions!

6Unit OnePart reactions (pg69-70)BalancedReaction types (pg71-72)Reagents (pg72-77)

Chemistry... NUCLEOPHILE ELECTROPHILE electron rich electron poor high electron low electron density density this slide sums up the majority of chemical reactions (even the funkier ones follow these ...it’s that simple principles)

Chemistry... NUCLEOPHILE ELECTROPHILE electron rich electron poor high electron low electron density density ...so that was a quick lecture... ...it’s that simple

lets look at a simple reaction... lets just show itsthat simple (we dohave 50 minutes to kill after all)

the Lucas test H3C H3C ZnCl2 (aq)H3C OH + HCl H3C Cl + H2O H3C H3C the Lucas test is simple...you’ll do it lab this week...two water soluble compounds react...water soluble water soluble water insoluble H-bond ion-dipole two layers

the Lucas test H3C H3C ZnCl2 (aq)H3C OH + HCl H3C Cl + H2O H3C H3C to give an insoluble compound. Why insoluble? (no longer able to H-bond)water soluble water soluble water insoluble H-bond ion-dipole two layers

the Lucas test H3C H3C ZnCl2 (aq)H3C OH + HCl H3C Cl + H2O H3C H3C} } reactants products stoichiometry 1 : 1 : simple definitions 1 : 1

the Lucas test H3C H3C ZnCl2 (aq)H3C OH + HCl H3C Cl + H2O H3C H3C ratio of} } compounds in reaction reactants products stoichiometry 1 : 1 : 1 : 1

...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) we only move them (or change the bonds)created or destroyed

unfortunately,chemists are lazy... a quick word of warning...organic chemists are a little lazy and often miss side products off their reaction schemes!

depiction of a standard reaction OMe OMe H2SO4 HO3S ...so, here is a standard reaction as shown in many text books...

depiction of a standard reaction OMe OMe H2SO4 HO3S ...problem is, it doesn’t show everything...so lets have a closer look...

so what bonds arebroken?

depiction of a standard reaction redraw sulfuric acid and it becomes clear that we will loose an OH... OMe OMe HO SO3H HO3S obviously lost OH

depiction of a standard reaction OMe OMe HO SO3H HO3S ...we can’t just add it to the aryl ring or one C would have too many electrons... obviously lost OH

depiction of a standard reaction OMe OMe HO SO3HH HO3Sremember we don’t always draw H

depiction of a standard reaction OMe OMe HO SO3H O H HH HO3S ...overall we have lost H from aromatic and OH from acid...or waterwhich means we ‘forgot’ to add HOH

depiction of a standard reaction OMe OMe HO SO3H O H HH HO3Swhich means we ‘forgot’ to add HOH

depiction of a standard reaction OH HO Me Br O O MeHO O OMe O O OMe

depiction of a standard reaction H O HO Me Br O O MeHO O OMe ...break O–H bonds... O O OMe all the bonds broken & formed

depiction of a standard reaction H O HO Me Br O O MeHO O OMe O O ...break C–Br bond... OMe all the bonds broken & formed

depiction of a standard reaction ...form C–O bond AND H–Br H O HO Me Br O H Br O MeHO O OMe O O OMe ...and the ‘mystery’ product

depiction of a standard reaction H O HO Me Br O H Br O MeHO O OMe O O OMe 1 3 1 3 don’t forget the stoichiometry

what is happening in a reaction?

it is the movementof electrons

Lewis structures here is the Lucas reaction again... H3C H3CH3C OH + + H3C Cl + O Cl H H H H3C H3Cas each bond is two electronswe have just moved electrons...

Lewis structures H H H C H H C H H H H ClH C C O H H C C Cl O H HH H HH C H C H H H H3C H3CH3C OH + + H3C Cl + O Cl H H H ...and here is the H3C H3C Lewis structuresas each bond is two electronswe have just moved electrons...

Lewis structures ...we take O with its complete octet of electrons and share H two of them with H H to... H C H H C H H H H ClH C C O H H C C Cl O H HH H HH C H C H H H H3C H3CH3C OH + + H3C Cl + O Cl H H H H3C H3Cas each bond is two electronswe have just moved electrons...

Lewis structures H H ...form water... H C H H C H H H H ClH C C O H H C C Cl O H HH H HH C H C H H H H3C H3CH3C OH + + H3C Cl + O Cl H H H H3C H3Cas each bond is two electronswe have just moved electrons...

Lewis structures ...then we share the electrons of the chloride octet to make sure the original C is still an a happy octet... H H H C H H C H H H H ClH C C O H H C C Cl O H HH H HH C H C H H H H3C H3CH3C OH + + H3C Cl + O Cl H H H H3C H3Cas each bond is two electronswe have just moved electrons...

Lewis structures H H H C H H C H H H H ClH C C O H H C C Cl O H HH H HH C H C H H H H3C H3CH3C OH + + H3C Cl + O Cl H H H so just swapped H3C electronsC H3 around...as each bond is two electronswe have just moved electrons...

what types ofreaction are there?

3basicreactions

substitutionreactionsA B + C A C + B does what it says...the reagent swaps with a functional group in our molecule...

substitutionreactions OH OH Br CH3NH2 NHCH3 H Br here’s an example from the synthesis of prozac® where an amine substitutes / displaces / exchanges with a bromide...

substitutionreactions Na O Cl O NHCH3 NaCl NHCH3 F3CF3C fluoxetine here is another example finishing the Prozac® synthesis of prozac®; an alkoxide (alcohol derivative) substitutes a chloride...

A + Y A Y two molecules add together or combine...all atoms in both startingaddition materials are found in the productreactions

CH3 CH3 + H Br Br H here hydrogen bromide adds across an alkene to give us aaddition new bromidereactions

at some point you’ll have to learn why it adds the bromide to the more hindered end of the CH3 CH3alkene...but that’s someone + else’s job! H Br Br Hadditionreactions

O H3C O MgBr H3C MgBr CH3 CH3Ph Ph here a Grignard reagent is adding to a ketone to give (eventually) anaddition alcoholreactions

eliminationreactionsA X A + X elimination is the opposite of addition...we rip a bit of the molecule off.

eliminationreactions OH CH3 CH3 O H H H this example shows the elimination of water (dehydration) to form an alkene

eliminationreactions CH3 BrH3C CH3 H3CH3C H Br H3C H 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?

nucleophiles 2electron rich molecules nucleophiles are electron rich compounds that donate two electrons to form a new bond... examples include... donateelectrons

anionsH O ≡ H O Br ≡ Br 3 lone pairs 4 lone pairs H H3C ≡ Pr C H 1 lone pair

anionsH O ≡ H O Br ≡ Br negatively charged compounds that possess a 3 lone pairs lone pair of electrons that 4 lone pairs can form a new bond H H3C ≡ Pr C H 1 lone pair

anionsnucleophilic reaction O H O H H H

anionsnucleophilic reaction hydroxide donates a pair of electrons to form a new bond to proton O H O H H H

anionsnucleophilic reaction O H O H H H note: both sides of equation have same charge (both are neutral overall)

anionsnucleophilic reaction O H O H H H note: both sides of equation have same charge (both are this is a good sign that you’ve neutral overall) got the reaction right!

lone pairs lone pairs as nucleophile O N S H H H H H3C CH3 H water ammonia dimethyl sulfide (DMS) lone pairs on neutral molecules are also good nucleophiles...

lone pairsnucleophilic reaction H N H Cl N ClH H H H H H

lone pairsnucleophilic reaction ammonia donates a pair of electrons to form a new bond to hydrogen chloride H N H Cl N ClH H H H H H

lone pairsnucleophilic reaction can’t have two bonds to H (4 electrons) so break H– Cl bond with electrons flowing towards most electronegative element H N H Cl N ClH H H H H H

lone pairsnucleophilic reaction H overall its been a N H Cl substitution N ClH H H H H H

lone pairsnucleophilic reaction H N H Cl N ClH H H H H H note: overall both sides are neutral (have same charge)

bonds bonds can also be a source of electrons...and these cause confussion! δ– δ+ H C Y Y=Li, Mg B H H nucleophilic H site reactive σ bond H H C C H H R high electron density R = electron nucleophile donating group

electrophiles 2electron poor molecules the other kind of reagents are...electrophiles are electron poor compounds that accept two electrons to form a new bond... examples include... acceptelectrons

proton H ≡ H empty 1s orbital no electrons! simplest electrophile...it has no electrons and desperately wants some!

protonelectrophilic reagent H H H O H O H H

protonelectrophilic reagent H H H O H O H H nucleophilic water donates two electrons to proton, which accepts them to form new bond

Group 13 F F Cl F B Cl Al F B F Cl F empty 2p empty 3p orbital orbital have 6 valence only electrons...need 8 to obey octet rule so happily accept a pair of electrons

bonds δ+ δ– Y = Cl, δ+ δ– C Y Y = O, Br, N, O C Y NRelectrophilic electrophilic site site bonds can be electrophiles if they are polarised and have a δ+ centre

bondselectrophilic reagent H H O O H O O H H3C CH3 H3C CH3

is it that easy?

nucleophile + electrophile product yup, this equation sums up the majority of reactions...

...and no...

H3C δ+ δ– H3C δ+ δ–H3C O H3C Cl H3C H H3C polarity in these two compounds looks the same...so do they behave in the same way??

H3C δ+ δ– H3C δ+ δ–H3C O H3C Cl H3C H H3C of course not! this is chemistry after all...

H3C δ+ δ– H3C δ+ δ–H3C O H3C C Cl alcohol uses lone H3C H H3C pair to act as a nucleophilenucleophile electrophile

H3C δ+ δ– H3C δ+ δ– H3C O H3C C Clchloride isC H3 an H H3C electrophile nucleophile electrophile

...you will learn to toidentify which is which...

so what actuallyhappens in a reaction?

H3C O H3C Cl H H + H Cl + OH3C H3C CH3 CH3 H so here’s the Lucas test again...overall it’s a substitution of hydroxyl for chlorine what order are the bonds made and broken?

H3C O H3C Cl H H + H Cl + OH3C H3C CH3 CH3 H what order are the bonds made and broken?

H H3C O H3C O H H H H3C H3C CH3 CH3 CH3 CH3 H H3C C O H H H3C C O H CH3 CH3 addition stepstepone

H H3C O H3C O H H H H3C H3C CH3 CH3 nucleophilic alcohol CH3 donates 2 electrons to CH3 H a proton from H–Cl in H3C C O H H anH3C Creaction addition O H CH3 CH3 addition stepstepone

H H3C O H3C O H H H H3C H3C CH3 CH3 note: octet rule is obeyed and the charge is the same CH3 on both sides CH3 H (positive) H3C C O H H H3C C O H CH3 CH3 addition stepstepone

H CH3 H H3C O H O H3C H3C CH3 CH3 H H H H C H H C H H H H HH C C O H H C C O H HH HH C H C H H Hsteptwo elimination

H CH3 H H3C O H O H3C H3C CH3 CH3 H H H elimination of H C H H water...molecule splits in C H H H H two...electrons go with H most electronegativeH C C O H H C C atom O H HH HH C H C H H Hsteptwo elimination

H CH3 H H3C O H O H3C H3C CH3 CH3 H charge is the same H on both sides (positive)... H we now have a reactive H C H intermediate H C H H H H H (charged species with 6H C C O H electrons) H C C O H HH HH C H C H H Hsteptwo elimination

CH3 H3C Cl Cl H3C H3C CH3 CH3 H H H C H H C H H H H C C Cl H C C Cl HH HH C H C H H H addition stepstepthree

CH3 H3C Cl Cl H3C H3C CH3 CH3 H H H C H H C finally, second H H H addition step to H C C Cl H C complete octet C Cl HH HH C H C H H H addition stepstepthree

CH3 H3C Cl Cl H3C H3C CH3 CH3 H H H C H H C H H H H C C Cl H C C Cl HH HH C H C H H H nucleophile donates two electrons to electrophile addition step that accepts them tostep form new bondthree

CH3 H3C Cl Cl H3C H3C CH3 CH3 H H H C H H C H H H H C C Cl H C C Cl HH note: the charge is HH C H the same on both C H sides (neutral) H H addition stepstepthree

3steps1substitutionreaction

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

reaction proﬁle energy Cl + H2O H O + Cl H OH + HCl reaction progress Cl + H2O

reaction proﬁle each hill represents one step in our reaction energy Cl + H2O H O + Cl H OH + HCl reaction progress Cl + H2O

reaction proﬁle each dip represents an intermediate we can see energy Cl + H2O H O + Cl H OH + HCl reaction progress Cl + H2O

reaction proﬁle higher the hill, the harder the step energy Cl + H2O H O + Cl H OH + HCl reaction progress Cl + H2O

reaction proﬁle the lower in energy any stage the more stable it is energy Cl + H2O H O + Cl H OH + HCl reaction progress Cl + H2O

just one hill, so just one step energy HO CH3Br Br reaction CH3OH progress

no dip so no intermediatesenergy HO CH3Br Br reaction CH3OH progress

direct energy substitution would look like this HO CH3Br Br reaction CH3OH progressHO + CH3Br CH3OH + Br

once again, you will learn energy more about this but not from me! Why don’t I get HO to teach the cool stuff? CH3Br Br reaction CH3OH progress H 1 H 1 H SLOW 2 δ– 2 δ–HO + Br HO Br Nu + Br H RDS H H H H H transition state

what have ....we learnt? • analyse reactions • classify reagents •r e a c t i o n t y p e sPicture: © Pittsburgh Supercomputing Center

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moreexamples

Br H H Brwhat order are thebonds made and broken?

H Br H H Br H H Hwhat order are thebonds made and broken? First, remember that our skeletal figures are simplifications...these are all the atoms around the site of the reaction

H H Br Br H3C H H3C H C C H Br C C H Br H3C H the alkene is the H3C nucleophile. Lots ofH electrons between two carbon atoms stepstepone substitution

H H Br Br it participates in a substitution reaction, H H3C H H3C replacing the bromide... C C H Br C C H Br H3C H H3C H stepstepone substitution

H H Br Br the alkene attacks the proton and not the H3C H bromide. H do the H3C Why electrons go for the less C C H Br electronegativeH C C atom? Br H3C H H3C H stepstepone substitution

H H Br Br H3C H H3C H C C H Br C C H Br H3C H H3C H If we attacked the bromide first then, to obey the octet rule, we would have to give two electrons to step H and it would leave as H–. Thestep substitution hydride (H–) is not stable as it is not electronegativeone

H H Br Br electrons flow towards H3C H the most electronegative H3C H element C C H Br C C H Br H3C H H3C H stepstepone substitution

H H Br Br reaction is not over as we have a C with only 6 electrons. It is not obeying the octet rule so is very reactive H3C H H3C H C C H Br C C H Br H3C H H3C H stepstepone substitution

H Br H Br H3C H Br H C C H Br H3C C C H H3C H H3C Hsteptwo addition

H Br H Br H3C H Br H C C H Br H3C C C H H3C H H3C H Second step is addition. The bromide shares 2 electrons with the carbon so that theystep addition both obey the octet ruletwo

2steps1additionreaction

reaction proﬁle energy Br H Br reaction progress Br

reaction proﬁle the cation (positive charge) is the intermediate energy Br H Br reaction progress Br

HO H HOH Br Brwhat order are thebonds made and broken?

HO H HOH Br BrH O H H H3C H H3C C C H H O C C Br H3C Br H H3C Hstepone elimination

HO H HOH Br BrH O H H H3C H H3C C C H H O C C Br H3C Br H H3C H once again the electrons are flowing towards thestep elimination most electronegative elementone

HO H HOH Br BrH O H H H3C H H3C C C H H O C C Br H3C Br H H3C H elimination all atoms maintain thestep octet ruleone

1step1eliminationreaction

reaction proﬁle δ– HO H δ– Br no intermediate as the transition state reaction occurs by a single step energy H HO Br reaction progress HOH Br

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Lecture6: 123.101

by Gareth Rowlands on Mar 08, 2009

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Lecture6: 123.101 — Presentation Transcript