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Topic 6 – Alkyl halide and carbonyl compounds

Recap 1. Apply  your  knowledge  of  curly  arrows  to  work  out  structures  for  the  products  of  these  

steps.    

Organic compounds containing a halogen • Dense  liquids  and  solids  which  are  insoluble  in  water  .  • The  carbon-­‐halogen  bond  strength  decreases  in  the  order  C-­‐F  >  C-­‐Cl  >  C-­‐Br  >  C-­‐I.      • Alkyl  fluorides  tend  to  be  less  reactive  than  other  alkyl  halides,  mainly  due  to  the  higher  

strength  of  the  C-­‐F  bond.  • Halogen  atoms  are  electronegative,  so  C-­‐halogen  bonds  will  be  polarized  like  this:  

 Nucleophiles   could   therefore   attack   the   electron-­‐deficient   carbon   and   substitute   the   halogen  atom  directly  like  this:  

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Nucleophilic substitution                                                

2. Draw  the  organic  products  in  the  following  reactions:                        

3. Think  about   the  bulk  of   the  organohalide  –   compare  3∘   and  1∘.  Which  of   these  direct  substitutions  by  do  you  think  will  be  faster,  and  why?  

A wide range of charged and uncharged nucleophiles may be employed

Nucleophile + H3C I Product + I Product class

HO

RO

N C

R C C

H2N

R3N

H3C I

H3C I

H3C I

H3C I

H3C I

H3C I

H3C OH

H3C OR

H3C C

H3C C

H3C NH2

H3C NR3

N

C R

+ I

+ I

+ I

+ I

+ I

+ I

alcohol

ether

nitrile

alkyne

amine

tetraalkylammoniumsalt

+

+

+

+

+

+

δδ

Br OH

NH2

CN

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Mechanism – SN2 (2nd order)                    SN2  reaction  favoured  by  a  primary  alkyl  halide  as  the  starting  material.                      

4. Sketch   the  energy  profile   for   a   generic   SN2   reaction  below.  The   starting  materials   and  products   are   shown.   Add   in   a   transition   state   and   show   the   activation   energy   (Ea)  involved  in  the  rate  determining  step  (RDS).  

 Since   both   starting  materials   are   involved   in   the   transition   state,   and   since   there   is   only   one  step,  the  reaction  rate  is  proportional  to  the  concentrations  of  both  the  starting  materials.  

C Br

H3CH2C

CH3H

CH3O CCH3O

CH2CH3

CH3H

Br

CH2CH3

C

CH3H

H3CO Br~1/2 ~1/2

nucleophile electrophile transition stateleaving group

δδ

C X

H

HH

C X

Me

HH

C X

Me

HMe

C X

Me

MeMe

Nu

Nu

methyl halide2,000,000 (FAST)

primary alkyl halide40,000

secondary alkyl halide500

tertiary alkyl halide<1 (V. SLOW)

Nu

Nu

relative rate:

relative rate:

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Mechanism – SN1 (1st order)                                

 The  intermediate  in  an  SN1  reaction  is  a  carbocation.  

• The  C  atom  has  six  electrons  and  a  positive  charge.  • This  is  an  unstable,  highly  reactive  intermediate.  • Not  all  carbocations  have  the  same  relative  (un)stability  .  

                   SN1  reaction  more  likely  when  a  tertiary  carbocation  is  the  intermediate          

C Br

CH3

CH3H3C

OHCHO

CH3

CH3CH3

Br

CH3

C

CH3H3C

planarcarbocationintermediate

leaving group

slow

+

HOC OH

CH3

H3CH3C

fast

H3C CCH3CH3

carbocations have sp2

hybridised carbons (trigonlal planar) with a vacant p-orbital

fast

vacant p-orbital

Attack of nucleophile to either side of the planar carbocation intermediate is equally likely!

HCH3C

H

CH3CH3C

H3CHC

H3C

H3Ca tertiary alkyl

carbocationa secondary alkyl

carbocationa primary alkyl

carbocation

HCH

Ha methyl

carbocation

increasing stabilitymoststable

leaststable

C X

H

HH

C X

H3C

HH

C X

H3C

HH3C

C X

H3C

H3CH3C

methyl halide(V. SLOW) <1

primary alkyl halide1

secondary alkyl halide12

tertiary alkyl halide(FAST) 1,200,000

relative rate:

relative rate:

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5. Fill  in  the  missing  reagents.  

6. Draw  the  products  in  each  case,  assuming  the  reactions  proceed  by  the  SN2  mechanism.  

Summary SN2 reaction:

Substitution, nucleophilic, second order

The reaction rate is determined by the concentration of both nucleophile and electrophile

Involves the direct, single step substitution of an alkyl halide by a nucleophile with no intermediates

The SN2 reaction proceeds with inversion

The rate of the SN2 reaction of alkyl halides decreases in the order methyl > 1° > 2° >> 3°

Usually only methyl, 1°and 2° alkyl halides react by this pathway - less crowded

Summary SN1 reaction:

Substitution, nucleophilic, first order

The reaction rate is determined by the concentration the electrophile only

Involves the two step substitution of an alkyl halide by a nucleophile via a carbocation intermediate

The SN1 reaction of chiral compounds proceeds with racemisation

The rate of the SN1 reaction of alkyl halides decreases in the order 3° >> 2° > 1° > methyl

Usually only 3° alkyl halides react by this pathway

IOH

C

OCH2CH3

N(CH3)3 I

+ NaI+ NaI

+ NaI

CCH3

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Elimination of HX Dehydrohalogenation  –  require  H  and  X  on  adjacent  carbons.  

 

7. We’ve  seen  in  the  elimination  of  water  from  an  alcohol  reaction  that  if  we  lose  H+  from  a   carbocation  we   can   generate   a   double   bond.   Look   again   at   the   SN1   reaction   below.  There’s  a  carbocation  in  the  middle.  Can  you  draw  the  mechanism  for  what  happens  if  that  carbocation  loses  an  H+  (from  the  carbon  next  to  where  the  charge  is)?  

   

8. So   what   does   the   energy   profile   look   like?   The   start   is   exactly   the   same   as   the   SN1  reaction.  We  lose  a  group  to  give  a  carbocation.  But  instead  of  attack  by  a  nucleophile  (which  would  be  a  substitution  reaction)  we  lose  a  proton  to  give  an  alkene.  Sketch  this  on  the  profile  below.  

 

HCH

CH2Br

H3C

HC CH2

H3C

1-bromopropane

c. OH heat

+ H2O + Br

HCBr

CH2H

H3Cc. OH heat

2-bromopropane

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9. Draw  the  E1  reaction  mechanism  for  the  following  reaction.  The  H+  does  not  magically  leave  on   its  own.   It  will  be   carried  away  by  a  mild  base,   such  as  water,   that  might  be  present  in  the  reaction.  Try  to  include  this  in  the  second  step  of  the  mechanism.  

 Why   is   it   important   that   the   base   is   “mild”.   What   might   happen   if   you   added   in  something  that  was  stronger,  like  hydroxide?      

Zaitsev’s Rule  –  where  there  is  a  choice,  the  most  substituted  alkene  forms.    

   

 

Examples  of  eliminations                          

c. OH heat

+ H2O + Br

H3C CHH

HCCl

CH2

H

HC CH CH3H3C

HC CH2H2CH3C + H2O + Br

major product

C CH

H

H

HC C

R

H

H

H

C CR

H

R

H

C CR

R

H

H

C CR

R

R

HC C

R

R

R

R

unsubstituted ethylene

monosubstituted alkene

disubstituted alkene

trisubstituted alkene

tetrasubstituted alkene

Brhot, conc K OH

ethanol+ H2O + K Br

C OHH3C

H3C

H3Chot, conc H2SO4 C CH2

H3C

H3C+ H2O

hot, conc K OH

ethanolCl

+ H2O

+ K Clmajor

trisubstituted C=C bond3 carbon substituents

minordisubstituted C=C bond2 carbon substituents

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10. Fill  in  the  reagents  in  the  following  sequence  of  reactions.    

     

11. Give  products  for  the  following  reactions  involving  alcohols  and  amines.  

 

       

OHOH

Br

Br

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Carbonyl Compounds – aldehydes and ketones • The  double  bond  is  composed  of  one  σ  and  one  π  bond.  • The  electron  density  of  a  π  bond  is  above  and  below  the  plane  of  the  

double  bond.  • The  C=O  double  bond  is  polarised.  • Aldehydes  have  at  least  one  H  attached  to  the  carbonyl  group,  

ketones  have  two  carbon  groups  attached  to  the  carbonyl  group.  • Carbon  of  the  carbonyl  group  is  sp2  hybridised.    

                               

12. Think  about  the  electronegativity  of  carbon  vs.  oxygen  and  therefore  how  the  carbonyl  group  is  going  to  be  polarised.  How,  then,  might  the  group  react  with  a  nucleophile?    

 

Reactions  of  carbonyl  group  –  nucleophilic  addition    Dominated  by  weak  and  accessible  π-­‐bond  

• Polarity  of  C=O  directs  addition  reaction    • Reaction  conducted  in  two  stages  

§ Nucleophile  (eg  hydride  or  carbanion)  adds  to  C    § Then  O–  adds  to  H+  

         

 

overlap of electrons in sp2-hybridisedorbitals gives a σ-bond

oxygen has two lonepairs of electrons

overlap of electrons in p-orbitalsgive rise to a π-bond

in formaldehyde, overlap of electronsin an sp2-hybridised orbital from C and a 1s orbital from H gives a σ-bond

OCH

HC OH

H

C O

H

Hδδ

C O

H

Hδδ

Nustage 1

C O

H

H

Nu stage 2

H C O

H

H

Nu

H

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13. Let’s  draw  a  nucleophile  actually   reacting  with  a  carbonyl  group.  You’ll   initially   form  a  compound  with  a  negatively  charged  oxygen  atom.  

 

 

Let’s  pretend  we  treat  that  intermediate  compound  with  some  very  mild  acid  in  water,  to   give   a   neutral   organic   compound   that   we   could   isolate   and   purify.   Complete   the  mechanism  above.  

Notice   how   in   this   reaction   the   nucleophile   interacts   with   the   organic   compound  (containing   the   C=O)   first,   making   this   mechanism   different   from   the   electrophilic  addition  to  double  bonds  that  we  saw  previously.  

14. Complete   this  mechanism  where   hydride   reacts   with   a   ketone,   and   the   intermediate  product  is  worked  up  to  give  a  neutral  organic  compound.  

   This  is  an  example  of  a  nucleophilic  addition  reaction.  But  in  this  case  look  at  what  we’ve  achieved:  the  opposite  of  an  oxidation.  So  what  else  can  we  call  this  reaction,  when  we  use  hydride  as  the  nucleophile?          Notice   how   the   carbonyl   carbon   changes   from   being   sp2   hybridised   in   the   starting  material  to  being  sp3  hybridised  in  the  product.  This  is  OK,  and  has  happened  in  lots  of  other  reactions  you’ve  already  seen.  The  carbon  changes  its  hybridisation  as  part  of  the  reaction  pathway.  

 • To  generate  the  hydride  ion,  use  LiAlH4  or  NaBH4    • The  second  stage  of  the  reaction  is  always  addition  of  H+  • Aldehydes  give  primary  alcohols,  ketones  give  secondary  alcohols  

         

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15. Give  the  organic  products  for  the  following  nucleophilic  addition  reactions      

       

16. Here  are  some  toe-­‐curling  mechanism  gaffes.  What’s  wrong  in  each  case?  

17. This  reaction  gives  a  chiral  product.  Which  enantiomer  do  we  get?  

 

ONu

O

Nu

O

Nu

O

Nu

ONu

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Grignard reaction  Formation  of  a  Grignard  reagent  involves  the  reaction  of  an  alkyl-­‐  or  aryl-­‐halide  and  magnesium.                                

18. The  carbon  atom   in  a  Grignard  reagent   is  nucleophilic.   It’s  powerfully  nucleophilic.   It’s  reactive.  We  have  to  handle  Grignard  reagents  in  the  absence  of  water.  Why?    Draw  the  products  of  the  reaction.  

 So  why  doesn’t  the  following  reaction  work  as  shown?  

 This  is  a  very  versatile  reaction.  

• Can  use  most  organohalide  compounds  to  generate  Grignard  reagent.  • Grignard  reagent  reacts  with  a  carbonyl  grooup.  • Must  employ  anhydrous  conditions.  

 Examples:            

C nucleophile

use Grignard reagent: RMgX

R Mg X

R Mg X

HC O

H

formaldehyde

nucleophilicaddition

HC O

H

H3CH H

C O

H

H3CH

acid-base reaction

STEP 1 STEP 2

an alkoxideanion

a primary alcohol

methyl magnesium

bromide

MgBr

H3Cδ δ

MgBr

R X + Mg R Mg X X = Cl, Br, IR = alkyl, aromatic

δ δδδ

eg CH3Br + Mg CH3MgBr

I

+ Mg

Mg I

Cl+ Mg

MgCl

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19. A   very   neat   application   of   Grignard   reagents   is   in   the   synthesis   of   carboxylic   acids  (functional  groups  that  we’ll  talk  about  in  the  next  lecture).  What’s  the  Mystery  Reagent  below?  

     

20. Some  practice.  Give  the  products  of  the  following  reactions.  In  the  first  reaction,  will  we  see  an  optical  rotation  from  the  reaction  mixture?    

RC O

H

other aldehydes

RC O

H

H3CH R

C O

H

H3CH

secondary alcohol

methyl magnesium

bromide

MgBr

H3Cδ δ

MgBr

RC O

R

all ketones

RC O

R

H3CH R

C O

R

H3CH

tertiary alcohol

methyl magnesium

bromide

MgBr

H3Cδ δ

MgBr

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21. Draw  the  products  of  the  following  sequence  of  reactions.      

     

1. In  the  Grignard  reaction  to  form  3-­‐methylhexan-­‐3-­‐ol  there  are  three  possible  ketones  that  could  be  used  as  a  starting  material.    Identify  the  three  ketones  and  the  Grignard  reagent  they  need  to  react  with  to  give  the  desired  product.    

   

22. Supply   the  missing   details   for   this   sequence,  which   neatly   illustrates   how  we   can   use  carbon-­‐based  nucleophiles  to  build  up  complex  molecules  quickly.  

Cr2O72 / H

conc H2SO4heat

dil H2SO4

1. CH3CH2MgBr 2. H

conc KOHheat

Cl

2 products, use one in next step