Aromatic Chemistry

ELECTROPHILIC AROMATIC SUBSTITUTION

Solution to Subjective Problems 

LEVEL – I

2. In the acidic medium aniline is protonated to give

which is m-directing.

4. I) Aromatic II) Anti-ariomatic

III) Aromatic IV) Aromatic

1. V) Anti-aromatic VI) Non-aromatic

8. The use of a strong base and a dihalogen derivative suggests the possibility of the formation of a methylene intermediate (compare with CH2Cl2 ⎯→ CHCl). This is supported by the formation of a three – membered ring compound by reaction  with unsaturated compound . Hence, a possible path is.

PhCHCl2

The product is triphenylcyclopropenium bromide.

LEVEL – II

1. A trace of AlCl3 in the acetophenone bromine reaction serves as a catalyst for enolization.

3. The initially formed primary carbocation rearranges by methyl migration to form a more stable tertiary carbocation which acts as a electrophile and attack the benzene to give the major product.

(A) is iso propyl benzene or cumene and electrophile involved is CH3 – CH⊕ – CH3 (Isopropyl Carbocation). Rearrangements are also possible in Friedel Crafts, reactions when benzene is treated with n-butyl chloride and Lewis acid the product obtained is isobutyl benzene.

In aniline, benzene ring is directly attached –NH2 which releases electrons by resonance effect, where as the activation of –NH2 group can be decreased by converting it to anilide, and in amide, ring is attached to electron withdrawing group which deactivates the ring.

Tertiary butyl carbocation is Electrophile, which further attacks benzene ring. Many other reagents that can generate carbocations can be used for Friedel Craft’s reactions. Some of the reagents are

1. Alkenes in presence of acids like BF3 or Mineral acid (HF).
2. Alcohols in presence of acids like BF3 or Mineral acid (HF)
3. The product of acylation co-ordinates with the catalyst and removes the latter from the reactant side and thereby stops further acylation. For this reason, Friedel-Crafts acylation requires an excess of the catalyst. The product of alkylation does not co-ordinate with the catalyst. So, the catalyst can form complex with the alkylating agent and the catalyst propagates the reaction.
4. i) – o/p; (ii) –m
5. i) One chlorine atom has been replaced by a carboxyl group. Hence, controlled conditions have to be used, and if we work backwards, two immediate possibilities are: (a) CuCN/H2O; (b) Grignard and carbon dioxide. 

LEVEL – III

There is no availability of lone pair of electrons on the nitrogen atom in case of acid (A), hence the ring is deactivated. In sodium salt of the acid i.e. (B) the lone pair are available hence the ring can be easily acetylated.

Two molecules of aluminimum chloride must be used, since the acylating species is acetic anhydride. The formation of the p-product is probably due to steric effects.

The cyano-group directly attached to the benzene ring has a very strong –R effect and so is m-orienting. When ‘isolated’ from the benzene ring, it exerts only a –I effect on the ring. This effect must be weak, otherwise m-substitution would have occurred. At the same time, the CH2 group exerts a hyperconjugative, which favours o/p substitution. Presumably the steric effect largely decides p-substitution rather than o–.

Both alkyl groups have a +I effect, Me2CH > Me. Both also have a hyperconugative effect, Me> Me2CH. Since these effects are in opposition, it is not easy to predict the orientation. However, steric effects due to Me2CH are far larger than those due to Me, and this decides (presumably) in favour of substitution ortho to Me. 

The explanation is similar to that given for (iii).

1. v) Ph(CH2)3Me PhCO2H

NO matter how long the side – chain is, oxidation with strong oxidising reagents results in benzoic acid.

1. vi) PhNO2 + EtCl no reaction
2. Here, the student either known the answer or looks up a source of information. Possible oxidising agents are:

The presence of a –I group ortho to the side-chain usually require the use of alkaline permanganate.

6. i) Since the final product is an aldehyde, one precursor could be the 1,1-dichloride; these compounds are readily hydrolysed to carbonyl compounds.

There is also an alternative route.

1. ii)   Ph2C=CPh2

The use of two molecules of a 1,1-dichloride and copper suggests the formation of an alkene. This is an extension of the Wurtz reaction.

The use of the reagents given clearly indicates that a Grignard reaction is involved. However, aryl chlorides do not form Grignard reagents in ether; THF is necessary as solvent. On the other hand, aryl bromides (and iodides) readily react in ether solution.

The use of aqueous ammonium chlorides is a safety precaution to prevent the possibility of dehydration of the alcohol (which might  occur if acid is used).

The use of iron indicates nuclear substitution. Me is o/p – orienting, and since there is always the possibility of a steric effect at the o-position, as we can suggest is that the 4-Br product will perdominate. Experimental work would have to be done to find out what are the actual results. From the literature it appears that the 4-Br compound is formed exclusively.

7. The use of a strong base and a dihalogen derivative suggests the possibility of the formation of a methylene intermediate (compare with CH2Cl2 ⎯→ CHCl). This is supported by the formation of a three – membered ring compound by reaction  with unsaturated compound . Hence, a possible path is.

PhCHCl2

The product is triphenylcyclopropenium bromide.

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