Haloform reaction

Haloform reaction

NETT       REACTION :

Haloform Reaction is a type of organic reaction where haloform is produced by halogenation of methyl ketone in the presence of a base. It is a kind of nuclophilic substitution reaction. This reaction is used in quanitatve analysis to indicate the presence of methyl ketone.

If the hypohalite used in this reaction is sodium hypoiodite, the product is iodoform (CHI3). Such a haloform reaction is also known as iodoform test. As iodoform is a yellow crystalline solid with a characteristic odor and melting point it can be handled much more easily handled in the the laboratory.

                              


                          

• When methyl ketones are treated with the halogen in basic solution, polyhalogenaton followed by cleavage of the methyl group occurs.

• The products are the carboxylate and trihalomethane, otherwise known as haloform.

• The reaction proceeds via successively faster halogenations at the α-position until the 3 H have been replaced.

• The halogenations get faster since the halogen stablises the enolate negative charge and makes it easier to form.

• Then a nucleophilic acyl substitution by hydroxide displaces the anion CX₃ as a leaving group that rapidly protonates.

• This reaction is often performed using iodine and as a chemical test for identifying methyl ketones. Iodoform is yellow and precipitates under the reaction conditions.

MECHANISM OF THE HALOFORM REACTION OF METHYL KETONES
Step 1: First, an acid-base reaction. Hydroxide functions as a base and removes the acidic α-hydrogen giving the enolate.
Step 2: The nucleophilic enolate reacts with the iodine giving the halogenated ketone and an iodide ion.
Step 3: Steps 1 and 2 repeat twice more yielding the trihalogenated ketone.
Step 4: The hydroxide now reacts as a nucleophile at the electrophilic carbonyl carbon, with the C=O becoming a C-O single bond and the oxygen is now anionic.
Step 5: Reform the favourable C=O and displace a leaving group, the trihalomethyl system which is stabilised by the 3 halogens. This gives the carboxylic acid.
Step 6: An acid-base reaction. The trihalomethyl anion is protonated by the carboxylic acid, giving the carboxylate and the haloform (trihalomethane).

nucleophillic substitution examples

substitution reactions summary with examples

      
                   

For each reaction below, determine whether the primary reaction is S_N1, S_N2, E1, or E2, and then draw the product.

Note: Me = methyl (CH₃)

problem set 1

SOLUTION

a) NaCN is charged! (Na⁺ and CN^-), so it's S_N2 or E2. CN is not a strong base, so it's S_N2.

b) KOtBu (potassium tert-butoxide) is charged, so it's S_N2 or E2. ^-OtBu is a strong base, so if anything is more bulky than 1º it will go E2. ^-OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).

c) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. ^-OMe is 1º (actually, not even 1º), but the alkyl halide is 2º, so it will go E2.

d) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. But in this case there is no bulk whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º, so it will go S_N2.

e) Methanol (MeOH) is neutral so probably E1 or S_N1. Methanol is a weak base and there's no bulk, so S_N1. In general water and alcohol do a mixture of S_N1 and E1 with alkyl halides (mostly S_N1).

f) H₂SO₄ is acidic so probably E1 or S_N1. Can't be S_N1 though because there is no nucleophile in H₂SO₄. (HSO₄^- is a very weak nucleophile). An alcohol with H₂SO₄ or H₃PO₄ is a dehydration reaction- E1.

g) H₂SO₄ is acidic so probably E1 or S_N1. In this case we have a nucleophile- Cl^-, so it will go S_N1.

h) Amines are neutral but they don't so S_N1/E1- they tend to go S_N2/E2, because they are basic (an amine solution has a basic pH). This amine is really bulky so it will go E2.

problem set 2

Rank the following compounds in order of decreasing reactivity with water (solvolysis). (1 = most reactive)

This is an S_N1 reaction. We know this because none of the reagents have charges (H₂O is neutral; if it were HO^-, it would probably be S_N2 or E2). So a carbocation will be formed in this reaction, and the compound whose carbocation is the most stable will react the fastest. So the 3º bromide will react fastest with water.

Problem # 3

Rank the following compounds in order of decreasing reactivity with NaI in acetone. (1 = most reactive)

This is an S_N2 reaction. We know this because NaI is an S_N2 reagent- charges give it away; when we see charges (Na⁺ is positive and I^- is negative) it's probably an S_N2 reaction. So we want reactants that are less substituted: methyl is more reactive than 1º, which is more reactive than 2º, etc. So methyl bromide reacts the fastest with NaI.

Problem #4

Rank the following electrophiles in order of decreasing reactivity with NaN₃ in DMF. (1 = most reactive)

Good leaving groups are stable. Larger ions tend to be more stable than smaller atoms (due to a smaller charge:size ratio. See problem 288), so when going down the periodic table, stability increases. I^- is more stable than Br^-, which is more stable than Cl^-, etc. So I^- is the best leaving group, and 2-iodobutane will react the fastest with a nucleophile.

Problem # 5

Indicate the reagents necessary to carry out each transformation.

For a), the wedge remains a wedge, so we have to do two S_N2 reactions (wedge to dash to wedge again). So we use PBr₃ to turn the OH into a Br. (Bromination of an alcohol with PBr₃ is an S_N2 reaction and so inverts stereochemistry).

For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr₃ to make the OH a better leaving group, we use RSO₂Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.

NUCLEOPHILLIC SUBSTITUTIONS WITH EXAMPLES

SUBSTITUTION by making a new bond AT THE SAME TIME as breaking the old bond

• substitution requires a bond to be broken AND a new bond to be formed

• the LOWEST energy way of doing this (unless precluded by steric or other effects, see later) is to MAKE THE NEW BOND (getting some energy "back") at the same time as BREAKING THE OLD BOND, this is SN2

Example

This is fundamentally just a Lewis acid/base reaction of the kind we saw when we were learning about Lewis acid/base reactions, the Lewis base has the high energy chemically reactive electrons, which are used to make a new bond to the Lewis acid, and a stronger bond is formed (C-O in the example above) and a weaker bond is broken (C-Br above)

• HO– is the Lewis Base and Nucleophile

• the halide is the Lewis acid/electrophile

• the Br– anion is the Leaving Group

The halide AND the nucleophile (2 molecules) are involved in the rate determining step and so the reaction rate depends upon the concentration of them both, the reactions is kinetically SECOND (2nd) order

Examples of SN2 Reactions; Give the major organic product in the following reactions

• we understand these SN2 reactions a simple Lewis acid/base processes

• identify the Lewis base/NUCLEOPHILE as the reactant with the high energy electrons

• the Lewis acid/nucleophile must react with the Lewis acid/ELECTROPHILE

Factors Controlling SN2 Reactivity: Leaving Group Ability

Good leaving groups are:

• stable/less reactive as an anion

• which means generally weak bases (have weak X-H bonds)

• polarize the C–LG bond (and are polarizable to make strong partial bonds in transition state)

Recall

Similarly

• INCREASING leaving group ability going down the periodic

• anion stability increases I– > Br– > Cl–

• the bond strength increases C-I < C-Br < C-Cl, i.e. the iodide anion is stable because when it reacts it makes WEAK BONDS

2.2  Factors Controlling SN2 Reactivity: Solvent Effects

Polar APROTIC  (NON-hydrogen-bonding) Solvents: There are Several, you need to know these

• Polar protic solvents have WEAKER intermolecular ion-dipole forces with dissolved ions

• The ion-dipole interactions certainly solvent and stabilize ions, but the H-bonding effect particularly on anions is MISSING

• There is a usually a SMALLER energy difference between reactants and the transition state when SN2 reactions are performed in polar APROTIC solvents, SN2 reactions in polar APROTIC solvents are usually faster than in polar protic solvents 

• you NEED TO KNOW the polar APROTIC solvents, this is not easy because they are different structures, learn them by working with them

Factors Controlling SN2 Reactivity: Nucleophilicity

Summary: Nucleophilicity order is sometimes difficult to remember, but favored by.....

1. anion over neutral

2. less electronegative over more electronegative

3. small atomic size over large atomic size in aprotic solvents

4. large atomic size over small atomic size in protic solvents

• These factors are all quite predictable except GOING DOWN THE PERIODIC TABLE (again!)

Factors Controlling SN2 Reactivity:  Steric Effects

First Order Nucleophilic Substitution (SN1) Reaction

• here the solvent "helps" to break the C–Br bond, the reaction is a solvolysis reaction (lysis - bond breaking)

We need a new substitution MECHANISM to account for this: The SN1 Mechanism

• the SN1 reaction requires a polar protic solvent to stabilize the ionic (cation and halide) intermediates

• usually requires heat (energy) to break the C–X bond unimolecularly

• ONLY the halide (not the nucleophile) involved in the R.D.S., thus SN1 (1 means only 1 reactant in the R.D.S.)

• requires a stable intermediate cation, NO SN1 for methyl or primary halides

3.1  Stereochemistry of SN1 Reactions: Racemization (?)

Example

Distinguishing SN1 and SN2 Reactions

• NOTE: the factors above favor the reactions by making them go faster, e.g. SN2 is FASTER at a primary carbon, SN1 is faster at a tertiary carbon, SN1 is faster in polar protic solvents etc.

• However, weak nucleophiles do not favor SN1 because they make Sn1 reactions faster, they don't, but they do make competing SN2 reactions SLOWER

• SN2 reactions are not precluded by polar protic solvents, they are just faster in aprotic solvents

Examples:  assign the mechanism of the following reactions to SN1 or SN2

Example Problems: Give the major organic product of reactions

• polar aprotic solvent, strong nucleophile, SN2, Br- better leaving group

• 1 equivalent means exactly the same number of nucleophiles as organic reactants, which in this context means that there is only enough nucleophile to substitute one of the halide leaving groups

• polar protic solvent and heat, no strong nucleophile and allylic halide, must be SN1. 

• polar aprotic solvent, strong nucleophile, SN2, allylic position more reactive

• 1 EQUIVALENT will ONLY REACT at the carbon where SN2 will be fastest

Nucleophilic Aromatic Substitution

Study Notes

A nucleophilic aromatic substitution reaction is a reaction in which one of the substituents in an aromatic ring is replaced by a nucleophile.

A Meisenheimer complex is a negatively charged intermediate formed by the attack of a nucleophile upon one of the aromatic-ring carbons during the course of a nucleophilic aromatic substitution reaction. A typical Meisenheimer complex is shown in the reaction scheme below. Notice how this particular complex can be formed from two different starting materials by using a different nucleophile in each case.

Figure 16.2: The formation of a typical Meisenheimer complex

A Nucleophilic Aromatic Displacement Reactions of Aryl Halides

The carbon-halogen bonds of aryl halides are like those of alkenyl halides in being much stronger than those of alkyl halides (see Table 4-6). The simple aryl halides generally are resistant to attack by nucleophiles in either SN1 or Sn2 reactions (Table 14-6). However, this low reactivity can be changed dramatically by changes in the reaction conditions and the structure of the aryl halide. In fact, nucleophilic displacement becomes quite rapid

1. when the aryl halide is activated by substitution with strongly electron-attracting groups such as N02, and
2. when very strongly basic nucleophilic reagents are used.

Addition-Elimination Mechanism of Nucleophilic Substitution of Aryl Halides

Although the simple aryl halides are inert to the usual nucleophilic reagents, considerable activation is produced by strongly electron-attracting substituents provided these are located in either the ortho or para positions, or both. For example, the displacement of chloride ion from 1-chloro-2,4-dinitrobenzene by dimethylamine occurs readily in ethanol solution at room temperature. Under the same conditions chlorobenzene completely fails to react; thus the activating influence of the two nitro groups amounts to a factor of at least 108:

A related reaction is that of 2,4-dinitrofluorobenzene with the amino groups of peptides and proteins, and this reaction provides a means for analysis of the N-terminal amino acids in polypeptide chains. (See Section 25-7B.)

In general, the reactions of activated aryl halides closely resemble the SN2-displacement reactions of aliphatic halides. The same nucleophilic reagents are effective (e.g., CH3O⊖, HO⊖, and RNH2); the reactions are second order overall (first order in halide and first order in nucleophile); and for a given halide the more nucleophilic the attacking reagent, the faster the reaction. However, there must be more than a subtle difference in mechanism because an aryl halide is unable, to pass through the same type of transition state as an alkyl halide in SN2 displacements.

The generally accepted mechanism of nucleophilic aromatic substitution of aryl halides carrying activating groups involves two steps that are closely analogous to those briefly described in Section 14-4 for alkenyl and alkynyl halides. The first step involves attack of the nucleophile Y:⊖ at the carbon bearing the halogen substituent to form an intermediate carbanion 4 (Equation 14-3). The aromatic system is destroyed on forming the anion, and the carbon at the reaction site changes from planar (sp2 bonds) to tetrahedral (sp3 bonds).

In the second step, loss of an anion, X⊖ or Y⊖, regenerates an aromatic system, and, if X⊖ is lost, the overall reaction is nucleophilic displacement of X by Y (Equation 14-4).

In the case of a neutral nucleophilic reagent, Y or HY, the reaction sequence would be the same except for the necessary adjustments in the charge of the intermediate:

Why is this reaction pathway generally unfavorable for the simple aryl halides? The answer is that the intermediate 4, which we can express as a hybrid of the valence-bond structures 4a-4c, is too high in energy to be formed at any practical rate. Not only has 4 lost the aromatic stabilization of the benzene ring, but its formation results in transfer of negative charge to the ring carbons, which themselves are not very electronegative:

However, when strongly electron-attracting groups are located on the ring at the ortho-para positions, the intermediate anion is stabilized by delocalization of electrons from the ring carbons to more favorable locations on the substituent groups. As an example, consider the displacement of bromine by OCH3 in the reaction of 4-bromonitrobenzene and methoxide ion:

The anionic intermediate formed by addition of methoxide ion to the aryl halide can be described by the valence-bond structures 5a-5d. Of these structures 5d is especially important because in it the charge is transferred from the ring carbons to the oxygen of the nitro substituent:

Substituents in the meta positions have much less effect on the reactivity of an aryl halide because delocalization of electrons to the substituent is not possible. No formulas can be written analogous to 5c and 5d in which the negative charges are both on atoms next to positive nitrogen,  and 

In a few instances, stable compounds resembling the postulated reaction intermediate have been isolated. One classic example is the complex 7 (isolated by J. Meisenheimer), which is the product of the reaction of either the methyl aryl ether 6 with potassium ethoxide, or the ethyl aryl ether 8 and potassium methoxide:

Exercises

Questions

Q16.7.1

Propose a mechanism for the following reaction:

Solutions

S16.7.1