Wednesday, 5 April 2017

Electrophilic substitution reactons of haloarenes

Electrophilic substitution reactons of haloarenes

The halogen atom is ortho and para 

directing and therefore, substitution takes 

place at ortho and para positions giving 

electrophilic substitution reactions at the benzene ring.
In the resonance structures of chlorobenzene the ortho and para positions get negatively charged, i.e. the electron density is relatively more at ortho and para positions. The incoming electrophile is more likely to attack these positions. But, because of steric hindrance at the ortho position, the para-product usually predominates over the ortho product. Also, the halogen atom has -I effect because it is a electron withdrawing group. As a result, it tends to deactivate the benzene ring. Therefore, the electrophilic substitution reactions of chlorobenzene occur slowly and under drastic conditions compared to benzene. Thus,
  • Haloarenes undergo electrophilic substitution reactions slowly as compared to benzene.
  • Halogen group is ortho and para directing (para-product usually predominates over the ortho product).


Haloarenes react with halogens in the presence of ferric salt as catalyst to give ortho para isomers.
Haloarenes react with halogens  to give ortho para isomers


Haloarenes with concentrated HNO3 in the presence of concentrated H2SO4.
nitration of haloarenes


Sulphonation occurs when haloarene is treated with concentrated H2SO4.
sulphonation of haloarene

Alkylation and acylation

The alkylation and acylation reaction, known as 'Friedel-Craft reaction', is carried by treating haloarene with alkyl chloride or acyl chloride in the presence of a catalyst like anhydrous aluminium chloride. For example,
alkylation of chlorobenzene
acylation of chlorobenzene

Friday, 31 March 2017

Haloform reaction

Haloform 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.

                              Haloform reaction scheme

                          Haloform Reaction

  • 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.


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.
                         Haloform Schritt 1.svg
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 Schritt 2.svg

nucleophillic substitution examples

substitution reactions summary with examples


For each reaction below, determine whether the primary reaction is SN1, SN2, E1, or E2, and then draw the product.
Note: Me = methyl (CH3)
problem set 1
a) NaCN is charged! (Na+ and CN-), so it's SN2 or E2. CN is not a strong base, so it's SN2.
b) KOtBu (potassium tert-butoxide) is charged, so it's SN2 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 SN2. 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 SN2. 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 SN2.

e) Methanol (MeOH) is neutral so probably E1 or SN1. Methanol is a weak base and there's no bulk, so SN1. In general water and alcohol do a mixture of SN1 and E1 with alkyl halides (mostly SN1).

f) H2SO4 is acidic so probably E1 or SN1. Can't be SN1 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 SN1. In this case we have a nucleophile- Cl-, so it will go SN1.

h) Amines are neutral but they don't so SN1/E1- they tend to go SN2/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 SN1 reaction. We know this because none of the reagents have charges (H2O is neutral; if it were HO-, it would probably be SN2 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 SN2 reaction. We know this because NaI is an SN2 reagent- charges give it away; when we see charges (Na+ is positive and I- is negative) it's probably an SN2 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 SN2 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 SN2 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.