Haloform Reaction

Haloform reaction From Wikipedia, the free encyclopedia

Jump to: navigation, search The haloform reaction is a chemical reaction where a haloform (CHX3, where X is a halogen) is produced by the exhaustive halogenation of a methyl ketone (a molecule containing the R-COCH3 group) in the presence of a base.[1] R may be H, alkyl or aryl. The reaction can be used to produce CHCl3, CHBr3 or CHI3.

Contents      

1 Scope 2 Mechanism 3 Uses 4 Iodoform test 5 History 6 References

[edit] Scope Substrates that successfully undergo the haloform reaction are methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol. The only primary alcohol and aldehyde to undergo this reaction are ethanol and ethanal, respectively. 1,3-Diketones such as acetylacetone also give the haloform reaction. Beta-ketoacids such as acetoacetic acid will also give the test upon heating. The halogen used may be chlorine, bromine, or iodine. Fluoroform (CHF3) cannot be prepared from a methyl ketone by the haloform reaction due to the instability of hypofluorite, but compounds of the type RCOCF3 do cleave with base to produce fluoroform (CHF3); this is equivalent to the second and third steps in the process shown above.

[edit] Mechanism In the first step, the halogen disproportionates in the presence of hydroxide to give the halide and hypohalite:

X2 + OH−→ XO−+ X− + H+ (X = Cl, Br, I) (Unbalanced) If a secondary alcohol is present, it is oxidized to a ketone by the hypohalite (hydroxide depicted below should be hypohalite and the water≡H2O should be hypohalic acid≡HXO):

If a methyl ketone is present, it reacts with the hypohalite in a three-step process: (1) R-CO-CH3 + 3 OX- → R-CO-CX3 + 3 OH− (2) R-CO-CX3 + OH− → RCOOH + −CX3 (3) RCOOH + −CX3 → RCOO− + CHX3 The detailed reaction mechanism is as follows: Under basic conditions, the ketone undergoes keto-enol tautomerization. The enolate undergoes electrophilic attack by the hypohalite (containing a halogen with a formal +1 charge). When the α position has been exhaustively halogenated, the molecule undergoes a nucleophilic acyl substitution by hydroxide, with −CX3 being the leaving group stabilized by three electronwithdrawing groups. The −CX3 anion abstracts a proton from either the carboxylic acid formed or the solvent, and forms the haloform.

[edit] Uses This reaction was traditionally used to determine the presence of a methyl ketone, or a secondary alcohol oxidizable to a methyl ketone through the iodoform test. Nowadays, spectroscopic techniques such as NMR and infrared spectroscopy are preferred because they require small samples, may be nondestructive (for NMR) and are easy and quick to perform. It was formerly used to produce iodoform and bromoform and even chloroform in industry.[citation needed] In organic chemistry, this reaction may be used to convert a terminal methyl ketone into the appropriate carboxylic acid.

[edit] Iodoform test

negative and positive iodoform test When iodine and sodium hydroxide are used as the reagents, a positive reaction gives iodoform. Iodoform (CHI3) is a pale-yellow substance. Due to its high molar mass caused by the three iodine atoms, it is solid at room temperature (cf. chloroform and bromoform). It is insoluble in water and has an antiseptic smell. A visible precipitate of this compound will form from a sample only when either a methyl ketone, ethanal, ethanol, or a methyl secondary alcohol is present.

[edit] History The haloform reaction is one of the oldest organic reactions known.[2] In 1822, Serullas reacted ethanol with iodine and sodium hydroxide in water to form sodium formate and iodoform, called in the language of that time hydroiodide of carbon. In 1831, Justus von Liebig reported the reaction of chloral with calcium hydroxide to chloroform and calcium formate. The reaction was rediscovered by Adolf Lieben in 1870. The iodoform test is also called the Lieben haloform reaction. A review of the Haloform reaction with a history section was published in 1934.[3]

[edit] References 1. ^ Chakrabartty, in Trahanovsky, Oxidation in Organic Chemistry, pp 343-370, Academic Press, New York, 1978 2. ^ László Kürti and Barbara Czakó (2005). Strategic Applications of Named Reactions in Organic Synthesis. Amsterdam: Elsevier. ISBN 0-12-429785-4. 3. ^ Reynold C. Fuson and Benton A. Bull (1934). "The Haloform Reaction". Chemical Reviews 15 (3): 275–309

Haloform Reaction

This reaction has been used in qualitative analysis to indicate the presence of a methyl ketone. The product iodoform is yellow and has a characteristic odour. The reaction has some synthetic utility in the oxidative demethylation of methyl ketones if the other substituent on the carbonyl groups bears no enolizable α-protons.

Mechanism of the Haloform Reaction The reaction readily proceeds to completion because of the acidifying effect of the halogen substituents.

Chapter 18: Enols and Enolates

The Haloform reaction

Reaction type : Nucleophilic substitution

Summary      

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.

A yellow precipitate indicates a positive result in the iodoform test (centre tube).

QUESTION Could a methyl aldehyde undergo this reaction ? ANSWER

Related Reactions  

α-halogenation under acidic conditions α-halogenation under basic 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). Haloform reaction

Compounds which have one of the following structural units, on heating with a halogen ( Cl 2 , Br 2 or I 2 ) in the presence of a base like NaOH or KOH give rise to a haloform (chloroform, Iodoform).

As in

Acetaldehyde, Acetone

Ethanol, 2-Propanol

Chloroform It is prepared by heating bleaching powder [Cl 2 + Ca(OH) 2 ] and ethanol. It is Chloroform is formed as a colorless liquid. It is generally preserved in the presence of small amounts of ethanol (negative catalyst) to prevent the formation of carbonyl chloride that is phosgene which is highly poisonous. Chloroform was used as an anesthetic during surgery; it is no longer used now because of severe side effects like liver and cardiac toxicity. An anesthetic used at present is Halothane. ( F 3 CCHBrCl).

Net reaction using NaOH as the base

Steps

Note:

1. First step is oxidation of alcohol by the halogen.

Aldehydes and ketones with a methyl group next to the carbonyl group, will also give rise to haloform. The difference is the oxidation step will not be there.

Substitution of the three α-hydrogens by halogen atoms take place in the second step. The hydrogens replaced by halogens are all from the same side of the carbonyl group, in compounds like CH3COCH3. Once hydrogen is replaced by halogen the other hydrogen atoms at that carbon become more acidic causing further substitution at the same place.

Attack of the base on the trisubstituted compound resulting in haloform.

Haloform Reaction of Methyl Ketones

The haloform reaction is a variation of α-halogenation. Under basic conditions for enolate formation of a methyl ketone, halogenation of continues until its supply of hydrogens is exhausted. The resulting trihalo-derivative is unstable, undergoing an acyl type substitution res

formation of carboxylate and haloform. The iodoform version of this reaction is used as a qualitative test for methyl ketones. Yellow iodo convincingly indicates upon reaction with iodine the presence of a methyl ketone.

Haloform test is ideally positive for methyl ketones; the C=O group attached to a methyl group increases the overall acidity of methyl protons due to tautomerization to enol form. So the X you wrote should be a strong electron withdrawing substituent, or at least a group with a highly electronegative atom. However, you should take into account that the reactant is alkali iodine, or hypoiodite; so any reactant capable of reducing hypoiodite would consume it, causing a positive error. It would be wise to include a bit excess of hypoiodite. Try to find the mechanism responsible for iodoform formation, the mechanism probably goes from an enol intermediate. I am referring to the reaction of a given compound with KOH/I2 which we call the Iodoform Test. The mechanism as I know it involves first the iodination of the Me group attached to the carbonyl carbon to a CI3 group followed by "base removal of H+ ions" with a I- nucleophile successively attacking the intermediate to finally form iodoform (precipitate).

So it now does appear that having a MeCO- group is a necessary condition....