Lecture 2 Organic Reactions We will look at organic reactions in the context of:  hydrolysis

 Acid / Base 1. definition of acids and bases 2. nucleophilic and electrophilic compounds 3. nucleophilic substitution reactions 4. Elimination Reactions 5. acid and base catalyzed hydrolysis rates and life times in the environment 6. Hammett and Taft rate constant observations halomethanes DDT and DDE carboxylic acid esters carboxylic acid amides carbamates

1 Bronstead-Lowry acid - bases acid is defined as a species that can give up a proton, [H+] a base is a species that will accept a proton.

+ - HNO3---> H NO3 + - CH3COOH ---> H CH3OO conjugate bases + - C6H5OH ---> H +C6H5O

2 Lewis Acids and Bases (1923)

Acids have less than a full octet of electrons and are “electron deficient or electron poor” (electrophiles) Gould, Mechanism and Structure in Organic Chemistry, Holt, Rinehart and Winston, NY,1993 p 115, 1963

What the Lewis system does for us is allow species besides the hydrogen ion to display acidic behavior

page 116, Holt

3 Nucleophilic compounds:

From a reaction perspective nucleophilic species carry a negative charge and are usually polar in character.

They may have an electron rich bond or unbonded electron pair which is often the site of attack.

In the environment the majority of the nucleophiles that react with organics are inorganic

Table 12.5 page 359, Schwarzenbach

4 Because of its great abundance, water plays a pivotal role among the nucleophiles in the environment

A reaction in which water (or hydroxide) substitutes for another atom or group is called hydrolysis.

The resulting organic products of hydrolysis are typically more polar than their parent compounds and generally of less environmental concern

+ - CH3-Br + H2O ---> CH3-OH + H + Br

Chlorinated organics often find their way into the environment and hence their reaction with nucleophiles is of interest

5 Relative Nucleophilicty of inorganic Nucleophiles

Swain and Scot (see page363, Fig. 12.5) observed that for different nucleophiles, x, attacking different methyl halides that

- - - - X + CH3-Br  X--CH3--Br  CH3-X + Br

log (kx/kH2O) = s x n

Where n is an indicator of the attacking ability of x, the nucleophile, and s is the sensitivity of the organic to nucleophilic attack; n is really important because it represents the ability of the nucleophile to donate electrons

As a standard for methyl bromide, s is set equal to 1; we can now ask what does the concentration of a given nucleophile have to be to compete with water; ie when is its rate similar?

-n log (kx/kH2O) = 1 x n; so kH2O/kx = 10 6 To be similar, the nucleophilic rate must equal the water, d(CH3BrH2O /dt) or d CH3Brx/dt = d(CH3BrH2O) for the nucleophile CN- as an example,

) - d(CH3BrCN- /dt = kCN [CN ] and d CH3BrH2O /dt = kH2O [H2O ]

- so kCN [CN ]50% = kH2O [H2O ] and using the ratio of the rate constants above - -n [CN ]50% = [H2O ] X 10 using values for n (page 359) for different nucleophiles and [H2O ]= 55 mol/L

- see Table 12.6 page 364 of Schwarznebach for [x ]50% NO3- = 6 molar Br- = 7x10-3 2- -1 - -3 SO4 = 2x10 OH = 4X10 Cl- = 6x10-2 I- = 6X10-4

- -3 2- -4 - -6 In fresh water Cl = 10 ; SO4 = 2x10 ; OH =10 mol/L; so what do we conclude??

7 SN2 substitution (Substitution, nucleophilic bimolecular)

A nucleophile attacks a carbon from the opposite side of the leaving group. An intermediate is theorized in which the nucleophile is partially bonded to the molecule, while the leaving group is partially dissociated.

The nucleophile donates two electrons and the leaving group takes two electrons

blow up page 109 of Richard Larson’s book on Environmental Organic Mechanisms

8 The free energy of activation G‡ and the rate of reaction will depend on the:

nucelophlicity of Y: steric factors

Hydrolysis t1/2 of chloromethanes (years)

CH3Cl CH2Cl2 CHCl3 CCl4

0.93 704 3,500 7,000

9 Sterioisomers,dieldrin and endrin, are two examples of insecticides that contain epoxide moieties. Both hydrolyze - by SN2 reaction with H2O and OH , resulting in diols

Richard A. Larson and Eric J. Weber. Reaction Mechanisms in Environmental Organic Chemistry, Lewis Publishers, Ann Arbor, 1994

Carbon skeleton sterically impedes nucleophilic attack by - H2O and OH . As a result persistence in aquatic eco- systems are long and they have been banned in the US, but still used in other countries.

10 Epichlorlhydrin is used for the manufacture of glycerol and expoy resins. Its calculated half-life in distilled water at 20oC is 8 days

Page 122 Larson

11 The SN1 Mechanism

This mechanism (nucleophilic substitution, monomolecular) differs from the SN2 in that a dissociation of the organic molecule 1st takes place to form a carbonium ion (carbocation). The carbonium ions is then attacked by a nucleophile

 exhibits 1st order behavior  factors that stabilize the carbonium ion will increase reactivity, such as resonance or inductive effects

12 Figure 12.4 page 362

13 SN1 vs. SN2

For mono and di halomethanes, an increase in the # of halogen substituents on carbon increases the hydrolysis half-life. Why?

R-Cl+H2O --> R-OH + HCl

Cl Cl H-C-Cl Cl-C-Cl H Cl

SN2

By contrast, as the steric bulk in the form of methyl addition to the central carbon bearing the halogen occurs, a significant INCREASE in reactivity can be observed. Why?

Cl Cl CH3-C-CH3 CH3-C- CH3 H CH3

38 days 23 seconds

14 As halogen electronegativity decreases (F>Cl>Br) hydrolysis rates increase

Page 366 Table 12.7

15 Explaining mechanisms

Under neutral or basic conditions nucleophilic attack on the primary carbon occurs by SN2; the epoxide opens up and the deuterated oxygen appears at the primary carbon site

Under acidic conditions, the high conc. of [H]+ attacks the epoxide oxygen and water attacks the primary carbon.

16 Write analogous SN1 and SN2 mechanisms for the neutral hydrolysis of a substituted epoxide.

H O R2 C C

R2 R3

to be SN1 a bond breaks and water adds

to be SN2 water adds and the OH leaves

17 Elimination Reactions

 sterically hindered nucleophilic substitution  when acidic protons are present next to the carbon of the leaving group  presence of strong bases

-elimination -C-C------> C=C H X -HX example of such a reaction is the conversion of 1,1,2,2- tetrachlorethane to trichloroethylene. This can be viewed as an SN2 reaction followed by elimination

18 DDT conversion to the more environmentally stable DDE via elimination as a function of increasing pH or increasing strength of nucleophilic OH-

19 Carboxylic Acid Esters: Esters are important because they are derived from many organic acids and show up in lipids, plasticizers, pesticides, etc

X R1- C the ester bond; if x is oxygen, O-R2 its a typical called an acid ester; if If it is sulfur it is thioester

O R1-C R1COOH + R2OH-->R1COOR2 O-R2

20 Hydrolysis Rates

R1COOR2 + H2O -->R1COOH + R2OH Influence of pH

Basic hydrolysis seems to occur for all species; acid hydrolysis is important for only the slow reacting compounds

21 Acid Catalyzed Reactions of Carboxylic acid esters

OH O + + + H O R1 -C + H3O R1 -C 2 O-R2 O-R2 ester protonated-species

The equilibrium for the protonated species is

+ K’a = [ ester] [H ]/ [p-species]

The concentration of [p-species]

+ [p-species] = [ ester] [H ]/ K’a d[ester]/dt = k’a [p-species] [H2O]

substituting

+ d[ester]/dt = k’a / K’a [ester] [H2O] [H ] so the overall rate constant k’a

+ ka = k’a / K’a [H2O] [H ] so let’s look at what influences k’a / K’a looking at page 379 Figure 12.10 22 OH

+ + H2O R1 -C O-R R1 -C 2 k'A OH + O-R2 2 2nd step is the rate determining step with a rate constant of k’A what will electron with drawing groups do to the protonated species [p-species]??

OH

+ R1 -C ??? O-R2

It will make the original carbonyl carbon more + What will this do in terms of water attack on the positive carbonyl carbon and the rate constant k’A??

Going back to the 1st reaction in the acid catalyzed reaction, what will increased with- electron drawing R1 groups do to the equilibrium?? OH O + + + H O R1 -C + H3O R1 -C 2 O-R2 O-R2

23 We said the rate constant was

+ ka = k’a / K’a [H2O] [H ] and both k’a and K’a

Increase with increasing with drawing groups; these are approximately the same order of magnitude and hence qualitatively, ka does not change with induction ------

Base –catalyzed reactions (Figure 12.11page 381)

O- O kB1slow - R -C R1 -C + HO 1 O-R2 OH O-R2 kB2 fast [ester] tetrahedral intermediate [I]

O- kB3 fast to slow O - R -C + O-R2 1 O-R2 R1 -C OH kB4 slow OH

O O fast - + HO-R 2 + O-R2 R1 -C R1 -C - OH fast O

24 What is the rate for d [I]/dt ??

- d [I]/dt = 0 = kB1[ester] [OH ]- kB2 [I] -kB3 [I]

- solving for [I] = kB1[ester] [OH ]/{ kB3 + kB2}

If we assume that the rate of reaction of the tetrahedral intermediate is what determines the overall rate of reaction

O- kB3 fast to slow O - R -C + O-R2 1 O-R2 R1 -C OH kB4 slow OH d[ester]/dt = kB3 [I]

- d[ester]/dt = {kB1 kB3}/{kB2 + kB3}[ ester] [ OH] and kB = {kB1 kB3}/{kB2 + kB3}

O- O kB1slow - R -C R1 -C + HO 1 O-R2 OH O-R2 kB2 fast

If kB3 >> kB2, then kB = kB1; but not always!! depends on pKa ; inductive effects also reinforce kB1 and kB3

25 Going back The Taft Relationship

Attempts to extend Hammett type LFERs to aliphatic compounds.

‡ ‡ ‡ ‡ G = G ref + G i,electronic+G i,steric

k * * log     Es kref

  = polar effects Es= steric effects  and  are fitting parameters to a reference system

Taft chose the hydrolysis of carboxylic acid ester system because he could use different R groups with different steric and inductive effects

26 27 Steric effects are illustrated by the Taft relationship for dialkyl-substituted phthalates

k * * log     Es kref   = polar effects, Es= steric effects

* log kb/kb-CH3= 4.59  +1.52 Es

28 29 Hydrolysis rate constant as a function of pH

+ - kh= kA[H ] + kH2O[H2O] + kB[OH ]

-1 + (units kh= sec ; in kA[H ], kA is the second order rate constant for hydrolysis reaction of H+ with an electrophile ) if the system is dominated by acid catalysis

+ log kh= log kA+ log[H ]= log kA -pH for bases

- - + log kh= log kB+ log[OH ]; [OH ] = kw/[H ]

log kh= log kBkw + pH

1. Given the water environment [55.4 M], Is there a range where kH2O[H2O] or kN is important?

+ - the point at which kA[H ] = kB[OH ]

2pH= -log kBkw+logkA;

30 pH = 0.5 log (kA/ kBkw)= IAB

This pH we will call IAB which is defined as the point at which acid and basic hydrolysis are equal. 2. We now want to describe the pH at which hydrolysis via neutral water and [H] are equal.

+ kN= kA[H ]

kH2O[H2O] = kN

- + kw = [OH ][H ]

+ log[H ]= log(kN/kA);

pH= log (kA/kN) = IAN

3. for bases

- kN= kB[OH ]

- + kw = [OH ][H ]

pH= log {kN/(kBKw)} = INB

31 We will next look at hypothetical hydrolysis rate constants, kh as a function of pH and specifically the pH points of IAB, IAN, INB

32 What are the t1/2(s) of compounds that have a large range where kh= kN

33 An acid catalyst hydrolysis mechanism of organic esters page 379 Figure 12.10

If R1 or R2 are good electron withdrawing groups, what does this do?

34 Base catalyzed ester reactions

Page 381 Figure 12.11

- electron with drawing R1 and R2 groups will promote the addition of OH and HOH. Goes thru a tetrahedral intermediate (stable)

35

Organophosphorus Esters: Used as extensively as insecticides- biologically, acetylcholinesterase is inhibited. Balance between target bioactivity and hydrolysis stability is sought. X

R1 O-P-XR3

OR2 X = O, S

Hydrolysis may occur by either nucleophilic substitution a the central phosphorous or carbon atoms resulting in a cleavage of the C-O or C-S bonds.

Base-catalyzed hydrolysis favors attack at the phosphorous cite. Increasing the steric bulk at the at the o reactive P center increases t1/2 at 20 C

36 37 Carboxylic Acid Amides

This group of compounds are important because many simple amides are used as herbicides

O R2 R1- C-N R3

The -NR2R3group is less electronegative than the group of carboxylic acid esters.

The tendency of the -NR2R3group to leave is also much lower than the O-R group. for amines

- + R1R2NH  R1R2N H much less than ROH  RO- +H+

38 page 388 Table 12.12

4. Explain what IAB values are and then why are IABvalues for amides higher than for esters. Calculate the IAB values for IAN and INB for methyl amide, CH3-C(O)-R1R2 using the values in Table12.12

39 Carbamates

This class is an ester derivative of carbamic acid O HO-C-NH2 carbamic acid

O R-O-C-NR2R3 carbamic acid ester

They exhibit both ester and amide charter

These compounds are widely used as herbicides and insecticides

40