Introduction to Organic Reactions Glycine by benbenzhou


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									Introduction to Organic Reactions/1
Electrophiles: H+(HBr, HI), Br-Br, NO2+, Br+, R+, R-CO+, +SO3H
Nucleophiles: OH-, RO-, CN-, H2O, ROH, RNH2, C6H5O-, RCOO-, Cl-, OSO2-
Nucleophilic substitution: HO- + R-X  R-OH + X- (X=I, a better leaving group)
Electrophilic substitution:                   +       Br+                 -Br   +       H+
                                                                  + |                         Br |
Electrophilic addition:         -C==C-   +    H-Br      --->     -C--C-H   +   Br-       --> -C--C-H
(Markownikoff addition)          | |                              | |                         | |
Nucleophilic addition: --C--        + CN-     --->     --C--CN
               | |       O                               O-
Elimination: --C--C-H       +   CH3ONa/CH3OH      --->      --C==C--   + CH3OH       +    NaBr
               H   Br                (solvent)               |   |
Reaction mechanism in details
SN1 mechanism: R--X     -- r.d.s.-->     R+       +    X-         Homolytic fission
               +        -
              R + OH        --fast--> R-OH        (OH- is the nucleophile attacking R-X)
SN1 mechanism (Monomolecular Nucleophilic Substitution):
(1) is a TWO-STEP reaction with a carbocation intermediate & 2 transition states
    │E        - T.S1
    │                   - T.S2
    │               -
    │         R+ +X- +OH-   ___ R-OH + X-
    │ R-X--
    └─────────────────────────────────────────── Reaction path
(2) is favored in tertiary alkyl halide due to electronic factors
    (electron-withdrawing alkyl groups help stabilize the tertiary carbocation
by I-effect, making SN1 mechanism important for tertiary alkyl halides)
     Stability of carbocation (carbonium ion) :
                    +            +
      -CH2+ > CH3-C-CH3 > CH3-CH > CH3-CH2+
                    CH3          CH3
(3) produces a racemic product mixture as the intermediate is a planar cation
(4) reacts faster in polar solvent as the carbocation intermediate formed in the
rate determining step is stabilized if a polar solvent is used.

SN2 mechanism (Bimolecular Nucleophilic Substitution):
                H     r.d.s       H   T.S              H
       HO       C::Br ---> HO...C...Br     --->   HO--C::H
              H CH3             H CH3      fast        CH3

(1) is a ONE-STEP reaction with a single transition state
    │E              - T.S
    │                     ___ R-OH + Br-
    │ R-Br + OH- --
    └───────────────────────────────────── Reaction path
(2) inversion of configuration occurs (optical activity varies if asymmetry                             of
molecule exists, that is there is a chiral carbon in the molecule)
Note: In the reactant, H, CH3 and Br groups are arranged in an anticlockwise way.
      In the product, H, CH3 and OH groups are arranged in a clockwise manner.
(3) follows second order kinetics (rate = k [OH-] [R-X])
    SN2 reaction becomes faster if a higher concentration of nucleophile is used.                      (4)
favored in less polar solvent such as propanone (no ionic intermediate)
Introduction to Organic Reactions/2
Electrophilic Substitution (An electrophile replaces a H-atom on benzene ring)

               +        Br+       --->                                                              -->                -Br           +   H+

The electrophiles are often generated in situ. FeBr3 +                                              Br-Br      --->    FeBr4- + Br+
conc. HNO3          +    conc. H2SO4             --->         HSO4-       +    H2O    +     NO2+
AlCl3    +    R-Cl --->            AlCl4-        +    R;  +
                                                                     AlCl3      +    R-CO-Cl         ---->         AlCl4-        +    R-CO+
Aromatic      compounds            will        not     undergo            electrophilic            addition        because           addition       of   an
electrophile will destroy aromatic stability (presence of 6 pi-electrons)
There are more than one H-atoms available for substitution and the reaction condition
must be controlled to avoid further substitution of the ring hydrogens.
The reactivity of monosubstituted aromatic compounds depends on the group present                                                                         -
CH3, -NH2, -OH are called activating groups because their presence makes the                                                                      aromatic
compounds more reactive than benzene itself.
-Br, -NO2, -CO-R, -COOH, -NH-COCH3 are called deactivating groups.

Activating          groups         are      often         termed          ortho-       or    para-directors            because              the     cation
intermediate formed is more stable if the coming electrophile attacks the ortho- or
para-positions. With the exception of -Br group, deactivating groups are often meta-
directors. (ortho-position refers to the carbon next to the group present in the
monosubstituted aromatic compound; para- = opposite)
Explanation                                                                                          -H+
         H3C--                +    Br       --->                    <->              <->            ---> H3C-                -Br

        HO2C--                +    NO2+ --->

When the 2nd electrophile is introduced in the correct position, the intermediate
cation will be stabilised by inductive/mesomeric effect.

        Br-                   +    Br+      --->

Electrophilic addition
(1) Markownikoff addition to unsymmetrical alkenes (                                             C==C      )
    The       electrophile               involved             is        H--Br@-.      The     alkene       substrate             reacts       with       the
electrophile            to        form      a        stable          carbocation            intermediate            (like            SN1)     which      is
subsequently attacked by Br- to form an addition product, bromoalkane.
                                                              +     H                                Br        H
             C==C        +        H--Br@-       -->           C--C--          + Br- ----->              C--C--
                    H                                               H                                 H
    The hydrogen of H-Br will be added to the less substituted carbon in                                                              C==C

(2) Bromination of alkenes (trans-addition - the electrophile used is                                                        Br--Br@-
                                       (bromonium ion)                                |     Br       trans-addition: Br-
             C==C             --->         --C--C--           +     Br-       ---> --C--C--           attacks bromonium ion
         Br--Br                                 Br                                     Br                from the opposite side

Note: Despite OH- is a bad leaving group, nucleophilic substitution of R-OH by X- is
  possible.         This          is     done        by       the       protonation         of     the     alkanol      so           that     the     good
leaving group, H2O is easily displaced by the weak nucelophile X-.
Introduction to Organic Reactions/3
Lucas test for alkanols                                                                                     80II2a
         CH3           c.HCl                  H3C             -H2O                     Cl-         turbidity         CH3CH2-C--
OH   +    H        ---->    CH3CH2-C-OH2     --->           CH3CH2-C(CH3)2    ---> CH3CH2C-(CH3)2
         CH3                                  CH3             r.d.s            +          ZnCl2            Cl

          E │          R-OH2+
               │                --                                       Ea >> Ea', 1st step is r.d.s.
               │                              -- T.S.                    Thus [Cl-] does not affect the
               │                      ----                               rate of Lucas reaction.
               │                       R+ +H2O +Cl-           ___                As 3o R+ is most stable, 3o alkanol
     │          ---                          R-Cl + H2O              shows fast Lucas reaction.
               │R-OH + H+
               └────────────────────────────────────────── Reaction path

Rate of appearance of turbidity: 3o R-OH > 2o R-OH                            > 1o R-OH
As [Cl-] has no effect on the rate of Lucas reaction, this confirms SN1 mechanism has
occurred. (SN2 rate is affected by [nucleophile])
1-phenylethanol            shows     turbidity      in       Lucas     test   within    a    few       seconds   because   the
carbocation C6H5-C+H-Ch3 is stabilized by delocalization of positive charges with                                           pi
electrons on the ring. 1-Cyclohexylethanol takes longer time to show positive Lucas test
because C6H11-C+H-CH3 is stabilized by electronic factors (+I effect) only, but not by
resonance (+M effect). 1-phenylethanol shows fast Lucas reaction
mainly as a result of mesomeric effect.

Elimination reaction - Dehydrohalogenation and dehydration
Such strong bases as CH3O- or OH- are used to abstract the hydrogen (alpha to the
bromine) in dehydrohalogenation.
E2 mechanism:          H                      H@+
                      --C--C--        --->   --C::C--                    ----->    --C==C--        +    CH3OH + Br-
         CH3O-               Br                     Br@- (T.S.)
E1 mechanism: H            Br                                 H           OCH3
                    --C--C         - r.d.s -> Br            --C--C+    -------> H-Br

Nucleophilic substitution versus Elimination
The structure of R-X is important in governing the mechanism for nucleophilic
substitution. As SN2 mechanism involves the formation of a five-coordinated T.S., it is
sensitive to steric hindrance. The rate of SN2 reaction decreases as the number of bulky
alkyl groups present in R-X increases. The reactivity towards SN2
reaction decreases in the order : primary R-X > secondary R-X > tertiary R-X.
As SN1 mechanism involves the formation of the carbocation intermediate, the presence of
electron-releasing alkyl groups in R-X makes SN1 reaction faster. Tertiary R-X undergoes
very fast SN1 reaction and shows very slow SN2 mechanism.
For tertiary R-X, elimination (E2 mechanism) also occurs at an appreciable rate. In
fact, secondary R-X undergoes slow SN2 reaction due to steric hindrance and SN1 reaction
competes with elimination for secondary R-X.
R-X undergoes both elimination and nucleophilic substitution, to different extent.
Elimination is favored if a very strong base is used in a polar solvent. Thus the use
of CH3ONa in CH3OH facilitates elimination. High temperature also promotes elimination.
A substitution product must be prepared at lower temperatures, using a nucleophile of
weaker basicity and in a less polar solvent. so as to minimize the alkene side product.
(Tertiary R-X is specially prone to elimination)
Introduction to Organic Reactions/4
    CH3         H      r.d.s.       H                                      (T.S.)                   H
CH3-C--O-                     C::Cl         ------> RO...C...Cl                    --fast--> RO--C           +    Cl-
      CH3                 H    H             SN2                  H    H                               H H
                CH3                             Transition state is not formed via the SN2 mechanism due
CH3O-            C::Cl        --->              to steric hindrance.
            H3C      CH3
As the attacking nucleophile CH3O- is a very strong base, capable of abstracting a proton
from the tertiary R-X, giving major elimination product.
      CH3                          E2 mechanism                        CH3
CH3--C--Cl                         ----------->                 CH3--C       Cl@-
                     -                                                        @+
     H2C--H            OCH3                                            CH2       H.OCH3

The     stable           +C(CH3)3          is       not    formed       as    intermediate        leading        to   substitution   product
because elimination is favored in the presence of CH3O- (strong base).
Q: Give the structure of the product if KCN reacts with 2-chloro-2-methylpropane.
Outline the mechanism of the reaction.

Q : Both CH3O-                     --C(CH3)2 and CH3O-                       -CH(CH3) hydrolyze via SN1 mechanism.
                                        I                                    I
      Outline a reaction mechanism for the hydrolysis.                                                                      88II7b
      Compare the rate of hydrolysis of the two compounds. Explain your answer.
                │E                 A1#
                │                      --
                                                               -- A2# (ROH2+)
                 │                                  ----
                │                               +
                                            R +H2O +I-             ___
                │         ---                                      R-OH + HI
                │R-I + H2O
                └────────────────────────────────────────── Reaction path
For the aromatic compound, the intermediate carbocation formed via SN1 mechanism can be
resonance stabilized by dispersion of the positive charge with the ring.
Hydrolysis of CH3O-C6H4-C(CH3)2-I is faster.
Q: 2-Iodopentane is dissolved in propanone and treated with KOH solution and it is
found that optical activity is lost gradually. Explain.
            A                          A              Interchanging any two of these groups, an optical
            X::D                       X::C           isomer (enantiomer) is obtained if X is carbon.
        B    C                     B    D
Q: Suggest how K could be prepared from J.                                                                                     84II4b

Q:    Draw        all      steroisomers                   of    2,3-Dibromobutanedioic            acid       using      a   three-dimensional
representation(compounds with 2 chiral carbon have less than 4 isomers)                                                     85II9b
     Hint:        As      free         rotation           of     C-C    bond       is   allowed   in    2,3-Dibromobutanedioic          acid,
geometric isomers are obtained as elimination products using alcoholic KOH.
Introduction to Organic Reactions/5
Q: 1) Using curly arrows to show electron pair displacement, outline a mechanism
  (SN2) for the reaction between 2-bromobutane with KCN.          93II8b(i)
   2) Is it possible to use hydrogen cyanide instead of aqueous potassium cyanide to
   carry out this preparation ? Explain.                            94II8b
   3) 2-Bromobutane can exist in isomeric forms. Draw a suitable representation for
  each of these isomers.                                           93II9c
Q: Show how you would change CH3CH=CH2 into CH3CH2CH2OCH3 (methoxypropane), giving the
structures of intermediate compounds and the reagents for each step.    out

Q: Explain the order of increasing stability of the following carbocations:
     C6H11+, C6H11-CH2+ and C6H5-CH2+.                                                                  95I3a

Q: Outline a mechanism for the reaction between HBr and HC==C(CH3)                                      95I3c

              C2H5ONa                 Br2
Q:       -Br ------->                ----->
   Dehydrohalogenation,   followd  by   electrophilic                            addition,      produces     1,2-Dibromo
cyclohexane from bromocyclohexane.                                                             80II1b(iv)

Q:       -CH-CH2 --> -->       -CH-CH2 - excess C2H5O-/C2H5OH ->    -C=CH
          H Br                  Br Br
   A dibromoalkane undergoes repeated dehydrohalogenation to give an alkyne                                            when
treated with excess strong base.                                 86II9a(iii)
Q: Give the structure of the major organic product(s) from the reaction
                               excess                      Markownikoff addition                    84II1a
     CH3CH2C==CCH2CH3         -- HBr -->                         -->                            81II1b
                         conc. H2SO4                        H+               H         H2O          H
Q:              -CH-CH2 ------------>                    -CH=CH2 ---->                       -C-CH3 ---->          -C-CH3
       H   OH        heat                                              +                       OH
     Dehydration,        followed       by   Markownikoff        addition        (acidic     hydrolysis)     produces    a
positional isomer (1-phenylethanol from 2-phenylethanol)                                     80II1b(ii)
Q: An aliphatic bromide has a molecular formula C7H15Br.
     Deduce possible structures for the compound.                                                   84II2a
Q: Outline the reaction mechanism for the reaction between C2H5OH and conc. HCl.

Q: Hydrogen bromide is added by an ionic pathway to alkene B. Show the structure of
the major product, giving a full mechanism for its formation, and explain why it                                        is
favored over any alternative product.                H3C               H
                                                              C6H5                H   (B)            85I3b(5%)
Q: CH3CH2CH-CH-CH3         - KOH(fused), 200oC ->
           Br Br                                             diene           +        alykne        92II9a
Q: Methylbenzene may be made to react with bromine to give either of two isomeric
products        A   or    B     (C7H7Br)     depending     upon        the       reaction      conditions,      with    the
formation of hydrogen bromide.
     1) Give the structures of A & B together with the necessary reaction condition
  for their formation.
   2) Give the mechanism for the reactions.                                                                     80II4d(6%)

Introduction to Organic Reactions/6
J, C4H6O, on oxidation gives K, C4H6O2. L, C5H12O, can exist as a pair of optical                                                isomers
(enantiomers), and reacts with PCl5 to give HCl.
Suggest possible structures and explain your deductions.
Oxidation of J involves no change in the number of H-atoms but addition of one O-atom. J
may be an aldehyde and K                       is an acid. As the degree of unsaturation in J is (8+2-6)/2
=2, J may be an unsaturated aldehyde and K is an unsaturated acid.                                                       The possible
structures of J are: cis or trans-isomer for H3CCH==CHCHO. K might be the cis or trans
isomer for H3CCH==CHCOOH.
L may contain -OH group as its degree of unsaturation is 0.                                         CH3
L is an saturated alkanol. The possible structures are :                                            C::OH
                                                                                               H        C3H7
Ozonolysis with subsequent hydrolysis causes the cleavage of C==C double bonds into 2
fragments, but gives one product for an alicyclic alkene.
M, C6H10, on ozonolysis gives N, C6H10O2. As M shows no change in the number of C-atoms on
ozonolysis. M is a ring compound. The degree of unsaturation is                                            =2
M is an unsaturated ring. Possible structures of M are:

The corresponding structures for N are:

The position of the C=C bond in the molecule may be established by ozonolysis.
Give the structures of the products which you would expect from the ozonolysis of
citronellal CH3C==CH-CH2CH2-CH-CH2-CH                           (3,7-dimethyloct-6-enal)
                       CH3                 H3C        O
The two fragments are: CH3CCH3, H-C-CH2CH2CH-CH2C=O
                                              O           O           CH3         H                             83II4a(iii)

Using E (CH3CH=CH2), give reagents and a mechanism to explain what is meant by:
1) A reaction obeying Markownikoff's rule,
   The reagent is HBr which is generated in situ by NaBr & dil. H2SO4.
   CH3             H                 H3C                  Br-          CH3
          C==C               ---->        +C--CH3     ----> H-C--CH3 (major product)                            91I1a(4%)
    H              H                      H                            Br
   In     the    addition           of    an      unsymmetrical             reagent   (H-Br)       to    an    unsymmetrical         alkene
(propene),       the         more        electropositive              atom/group(H)      becomes          attached       to    the    less
substituted C-atom of the C-C double bond. This is called Markownikoff's rule
2) A polymerisation reaction.
   The reagent is an organic peroxide. The mechanism is a free radical addition.
   Initiation: R-O--O-R                   ---> 2 RO.                                  H2C==CHCH3                 CH3     CH3
   Propagation: R-O.                 H2C==CHCH3           --> R-O-CH2-CHCH3. ---------> R-O-CH2CH-CH2CH.
                         --> --> R-O--(CH2-CH(CH3))n--O-R
   A reaction in which propene molecules join together to form large polypropene
   molecules is known as polymerization reaction.

Give the conditions needed for the reaction between bromine water and phenol.                                                    Outline
the mechanism involved.                                                      OH                           OH
           O-H                            O-H       -H+          OH     Br2(aq)           Br       2Br2(aq) Br            Br
              ---->           ---->          --->           ------->
              Br-Br         Br         Br            Br                   Br

Introduction to Organic Reactions/7
       H       H       SN2     H+ H       -HCl      H H
   H3C-N:     C::Cl ----> H3C-N..C..Cl ----> H3C-N-C       2o amine
      H     H    R              H H R                H R
   Order of reactivity of amines: 3o amine > 2o amine > 1o amine > amide
   The two alkyl groups on the N-atom in the 2o amine increases the electron density
on the N-atom. 2o amine is a stronger nucleophile than 1o amine toward SN2 reaction     2o
amine may further react with C2H5I to give a 3 amine or even a quaternary salt.          A
mixture of amines is formed.
   It is thus difficult to prevent further alkylation of amines and excess amine is
needed to minimize the yield of 3o amines.                  CH2R
   RCH2-NH-CH3 -- RCH2Cl --> R-CH2-N-CH3 -- RCH2Cl --> R-CH2-N+-CH3
                                   RCH2                        CH2R
   1o amine reacts with CH3COCl via nucleophilic substitution, referred to as
acetylation of (addition of CH3CO- group to) amine. Acetylation of 1o amine gives       N-
alkylamide which is a poor nucleophile with respect to 2o amine. The 2o amine       formed
after reduction of amide is thus of higher purity.
        O SOCl2        O    CH3NH2      O        LiAlH4/ether
   R-C    ----> R-C      ------> R-C          --------------> R-CH2NHCH3
        OH             Cl   - HCl       NHCH3     hydrolysis

Q: Why is an aliphatic amino acid a stronger acid than an aromatic amino acid?
   Due to conjugation of the lone-pair electron on N-atom, with the pi electrons of
the ring, the basic strength of the amino group is weaker so it cannot abstract a
proton from the -COOH group. It thus exists as C6H5-NH2 in the uncharged form.
   In H2N-CH2-COOH, the amino group is sufficiently basic to remove a proton from the
-COOH group to form the dipolar ion -OOC-CH2-NH3+.
   In a strongly alkaline solution (high pH) of glycine H2NCH2COOH, glycine exists as
the carboxylate anion H2NCH2COO- which migrates towards the anode when a potential
difference is maintained across two electrodes placed in the solution.
   On the other hand, it will exist as the cation H3N+-CH2COOH which migrates towards
the cathode in a strongly acidic solution.

A= 2-aminoethanoic acid (glycine), B= 2-amino-3-methylpentanoic acid (isoleucine) C= 2-
amino-3-phenylpropanoic acid (phenylalanine)
(1) Give the full structure of the dipeptide A-C.
(2) Show all the possible tripeptides which can be formed by linking together one
of each of the three amino acids A, B and C ? (Show as, e.g. A-B-C)
(3) Give the full structure of any one of these tripeptides.                 80II5a

Q: Discuss the reaction between 3-bromopropene and H2/PtO2 and KCN
                 H2/PtO2               KCN
   CH2=CHCH2-Br -------> CH3CH2CH2Br ----> CH3CH2CH2CN
                  CN-                H2/PtO2
   CH2=CHCH2-Br ----> CH2=CHCH2CN ------> CH3CH2CH2CH2NH2
   H2/PtO2 reduces both C=C and the -CN group if CN- group is introduced prior to
catalytic reduction. Catalytic hydrogenation is non-selective and it reduces  C==C
--> CH--CH and CN  CH2NH2

Q: Comment on the nucleophilic addition on carbonyl carbon of an aldehyde
         H CN-    NH3                   H+, heat
CH3CH2-C    ---> ----> ---> CH3CH2CH-CN ------> CH3CH2CH-COOH
         O                        NH2                  NH2
Glycine may act as a nucleophile attacking the 2-chloroethanoic acid, giving di- and
trichloro-derivatives. To minimize the side reactions, ammonia must be in      limited
quantities. Cl                         glycine
CH3-COOH -- Cl2 --> CH2-COOH -- NH3, SN2 --> H2N-CH2-COOH + HCl
Q: Outline the mechanism for the reaction between but-1-ene and hydrogen     bromide
Explain why 2-Bromobutane is the major product rather than 1-bromobutane……

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