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CM1401  AY14/15  Suggested  Solutions    Question  1    (a)  2H+(aq)  +  2e-­‐  →  H2(g)    (b)  Q  =   !!!

!! !    (c)  E!  !"#  !!!! = E°− !"

!"ln !!!

!!  

E!"  !"#  !!!! = E°−RTnF ln

𝑝!!10!  

 ΔE = E!"  !"#  !!!! − E!  !"#  !!!!  

=𝑅𝑇𝑛𝐹 [ln

𝑝!!16 − ln  (

𝑝!!100)]  

=𝑅𝑇𝑛𝐹 ln

10016  

=0.0235  V      Question  2    (a)  𝛥𝐺! = −𝑅𝑇 ln𝐾 = +9.4  kJ  mol!! =  +9400  J  mol!!  

𝐾 = 𝑒!!"##

!.!"#×!"" = 0.104    (b)  K = !!

!!= !!!!

!!!!= !!

!!!!  since  xI  +  xB  =  1  

 (c)  K = !!

!!!!= 0.104  

 𝑥! = 0.09420  𝑥! = 1− 0.09420 = 0.9058    (d)     B  (g)   I  (g)  Initial  (mol)   6

154 = 0.03896  12154 = 0.07792  

Change  (mol)   +  x   –  x  Equilibrium    (mol)   0.03896  +  x   0.07792  –  x    0.07792− 𝑥0.03896+ 𝑥 =

0.094200.9058  

 x  =  0.06691  mol    At  equilibrium,    nB  =  0.1059  mol  nI  =  0.01101  mol  

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(e)  Q = !.!"!.!"

= 3.3  𝛥𝐺 = 𝛥𝐺°+ 𝑅𝑇 lnQ  = 9400+ (8.314)(500)(ln 3.3)  = +14.4  kJ  mol!!    Reaction  goes  backwards  towards  borneol.    Question  3    (a)  UV-­‐visible  spectroscopy    (b)  First  order    (c)    Given:  − ! !"#!!!

!"= 𝑘! HCrO!!  

−1𝑘!

1HCrO!!

 𝑑 HCrO!! = 𝑑𝑡!

!

!!

!!  

 

ln𝐶!𝐶!= −𝑘!𝑡

𝐶! = 𝐶!𝑒!!!!  

 Since  HCrO4-­‐  forms  Cr3+  (in  a  1:1  ratio),  concentration  of  Cr3+  is  given  by  C0  –  Ct  Therefore,  

Cr!! ! = 𝐶!(1− 𝑒!!!!)  

 Question  4    (a)   Here   we   assume   that   because   the   Ka1   of   H2SO3   is   much   higher   than   the   other   values,  dissociation  arising  from  HCN,  H2S  and  HSO3-­‐  are  negligible.  Hence,    H! HSO!!

H!SO!= 1.2×10!!  

Letting  𝑎  be  the   H! , then  𝑎!13− 𝑎

= 1.2×10!!  

 Solving  for  a  gives  𝑎 = 0.0575  Hence  pH = − lg 0.0575 = 1.24    ED   NOTE:   a   cannot   be   assumed   to   be   small   compared   to   [H2SO3]!   Working   out   a   on   that  assumption   gives   a=0.0632,   or   18%   of   [H2SO3].   A   has   to   be   less   than   5%   of   [HA]   for   the  assumption  to  be  justified.    (b)  Upon  addition  of  excess  OH-­‐,  the  ions  present  in  the  solution  in  (almost)  equal  proportions  are  CN-­‐,  SO32-­‐  and  S2-­‐;  Qualitatively,  the  weakest  base,  and  thus  the  least  dissociated,  would  be  S2-­‐,  thus  this  is  present  in  the  greatest  amount.  

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 We  confirm  this  by  calculating  their  Kb  values,  which  are:    

K! CN! =10!!"

4×10!!" = 2.5×10!!  

K! SO!!! =10!!"

6.2×10!! = 1.6×10!!  

K! S!! =10!!!

10!! = 10!!    Thus,  the  weakest  base  is  S2-­‐,  and  this  is  the  predominant  species.    Question  5    (a)  A  B:  H2,  Lindlar’s  catalyst  BC:  H2,  Pd/C  catalyst  CD:  3  equiv.  H2,  Rh/C,  ethanol  OR  Pt  with  high  pressure    (b)  Aliphatic  alkene  in  B  adopts  the  cis-­‐  configuration  if  Lindlar’s  catalyst  is  used,.    2  chiral  centres  present  on  C  where  double  bond  was  in  B;  Two  enantiomers  could  be  formed.  

   4  chiral  centers  present  on  D;  2  are  generated  in  the  carbons  originally  bonded  to  substituents  on   the   benzene   ring.   The   2   substituents   on   the   cyclohexane   ring   have   a   cis-­‐   conformation.  Depending  on  the  reaction  from  C,  4  enantiomers  could  form.                                          

(S)(S) (R)(R) (R)(R) (S)(S)

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Question  6  The   –SO3-­‐   end   of   the   prion   is   not   nucleophilic,   thus   the   nucleophilic   group   is   on   the   cell  membrane   end   and   hydrolyses   the   thioester   on   the   prion.   The   –SSH   group   is   more  nucleophilic  than  the  –NH2  group  thus  this  is  the  end  that  attacks:    

                 

O

S SO3-

H2N

SS H

O

HN

O

S

-O3S

H 2N

SHSH

O

HN

O

S

-O3S

H 2N

SS

O

HNH

S

-O3S

H 2N

SS

O

HNH

O

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Question  7    (a)    

   (b)  Secondary  amine  is  a  good  nucleophile,  and  bromide  is  a  good  leaving  group  on  a  primary  alkyl   halide.   The   solvent   is   ether,   which,   being   polar   aprotic,   prevents   association   of   the  nucleophile  with  cations.  Thus  SN2  is  favoured  and  inversion  of  stereochemistry  takes  place.  G  will  be  comprised  of  only  the  S-­‐  enantiomer  and  will  exhibit  optical  activity.  

Structure  of  G  is          Question  8  

 Nitrobenzene   is   chlorinated   first   because   it   is   inert   to   Friedel-­‐Craft   alkylation;   mesomeric  electron-­‐donating  effects  of   the  chlorine  atom  then  allows  alkylation.  A  hydride  shift  occurs  on  the  carbocation  to  produce  the  tert-­‐butyl  cation  in  step  2.    Question  9    (a)  1-­‐bromo-­‐4-­‐(1,1-­‐dimethylethyl)cyclohexanecarboxylic  acid  (1-­‐bromo-­‐4-­‐tert-­‐butylcyclohexanecarboxylic  acid  also  acceptable  as  in  IUPAC  convention)  (b)  tBuOK  is  bulky  and  poorly  nucleophilic,  thus  favours  elimination  of  an  alkyl  halide.  For  P  and  Q,   the   tert-­‐butyl   group   fixes   the   conformation   of   the   cyclohexane   such   that   it   is   in   the  equatorial  position.  P  eliminates  via  E2  with  any  of  the  two  hydrogen  atoms  anti-­‐periplanar  to  Br  to  give  the  same  product:  

 

NH

H

DN

NO2 NO2

Cl

NO2

Cl

Cl2FeCl3

(CH3)2CHCH2ClFe powerBr2

Br

CO2H

H

We assume excess base such that carboxylic acid is deprotonated

Either of these H can be removed because they are anti-periplanar

CO2H

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 Q  has  no  good  leaving  group  thus  will  not  eliminate  by  E2;  the  resultant  compound  is  simply  the  deprotonated  acid:  

       Question  10    Spectrum   (c)   is   S.   The   singlet   at   ~3.6   ppm   corresponds   to   the   highly   deshielded   –CO2CH3  protons.  The  quartet  at  ~2.4  ppm  corresponds  to  –OCOCH2–  protons,  slightly  deshielded  by  the  anisotropic  effect  of  C=O,  split  by  3  neighbouring  hydrogen  atoms,  which  is  the  adjacent  methyl  group.  This  methyl  group  is  also  split  by  these  two  protons  into  a  triplet  signal  at  ~1.1  ppm.    The  remaining  spectra  (a)  and  (b)  both  have  a  singlet  that  corresponds  to  the  –OCH3  group  on  R  or  the  methyl  ketone  on  T,  not  split  by  any  neighbours.  Deshielding  by  electronegativity  of  O  has  a  greater  effect  than  the  magnetic  anisotropy  of  C=O,  thus  the  more  deshielded  singlet  belongs   to   the   –OCH3   group.   This   is   the   signal   at   3.2   ppm   on   spectrum   (b),   thus   this  corresponds   to   R.   The     quartet   at   3.8   ppm   results   from   the   –OCH2–   protons   split   by   the  adjacent  methyl  group,  and  this  methyl  group  is  also  split  by  these  two  protons  into  a  triplet  at  1.2  ppm.  (a)  then  belongs  to  T;  the  singlet  at  ~2.1  ppm  belongs  to  the  methyl  ketone.    The  –OCH2CH3  protons  give  the  quartet  at  ~4.1  ppm  and  the  triplet  at  ~1.3  ppm.        Question  11    

     

OH

CO2

CO2CH3 1. LiAlH4

1. NaH2. CH3CH2I

2. H+/H2OOH

V

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       Question  12    (a)  Z  is  the  quaternary  amine,  

   X  is  less  nucleophilic  than  Y  because  the  lone  pair  of  N  is  delocalized  into  the  aromatic  system,  unlike  in  Y  where  it  is  more  available  in  an  sp3  orbital.  Upon  reaction  of  Y  with  1  equivalent  of  CH3I,  the  resultant  tertiary  amine  is  even  more  nucleophilic  because  of  the  addition  electron-­‐donating  methyl  group  and  thus  further  reacts  to  give  the  quaternary  amine.    (b)  Z  undergoes  Hoffmann  elimination  to  give  AA,  the  alkene.  

 

CO2CH3 CHO1. DIBAHToluene, -78ºC

1. MeMgBr2. H+/H2O

OH

1. NaH2. CH3I

W

2. H+/H2O

NI-

N