(a) (i) Complete the following reaction by showing stepwise reaction mechanism for the formation of products : (ii) The compound…

This question integrates mechanistic organic chemistry, stereochemical analysis, and synthetic planning.

(a) (i) Mechanism for Compound A The reaction involves the thermal decomposition or reduction of the chiral azo compound A (Me-CH(Ph)-N=N-C(Ph)2). Under thermal conditions, the weak N=N bond undergoes homolytic cleavage to generate two carbon-centered radicals: a tertiary benzylic radical (Me-C•(Ph)-H) and a diphenylmethyl radical (Ph2C•). These radicals can recombine or undergo disproportionation. However, in the presence of reducing agents or under specific thermal stress leading to single-electron transfer, the azo linkage is cleaved to form amine intermediates. A more likely pathway for "completion" in standard contexts involves reduction to the corresponding amines: 1-phenylethylamine and diphenylmethanamine. The mechanism proceeds via stepwise electron transfer to the azo group, forming a diazene intermediate, which is further reduced to the amines. The chiral center at the carbon bearing the methyl group is retained if the mechanism is purely reductive without carbocation formation, but if the question implies a solvolysis or elimination context, see (ii).

(a) (ii) Loss of Optical Activity Compound A contains a chiral center at the carbon bonded to Me, Ph, H, and the azo group. Treatment with alcoholic sodium ethoxide (a strong base) promotes an E2 elimination reaction. The base abstracts the proton from the carbon adjacent to the chiral center (or the proton on the chiral center itself if it leads to a stable alkene, though here the beta-hydrogen on the methyl group or the phenyl ring is less likely; typically, elimination occurs to form a conjugated alkene). More critically, if the azo group acts as a leaving group or if the reaction proceeds via a radical pathway, the planar nature of the resulting intermediate (such as a styrene derivative or a radical) leads to racemization. Specifically, if the reaction yields 1-phenylpropene or similar alkenes via elimination, the product is achiral (planar sp2 carbons), thus losing optical activity. Alternatively, if the azo group is reduced to an amine in the presence of acid/base, the alpha-carbon to the amine can undergo rapid proton exchange or racemization via a planar enamine-like intermediate if adjacent to a carbonyl (not present here). The most direct justification is that the elimination product (alkene) is achiral due to the planar geometry of the double bond, or that the chiral center is destroyed during the cleavage of the N=N bond, yielding achiral fragments or racemic mixtures.

(b) (i) Dibromocyclopentanes Reaction of cyclopentane with excess Br2 at high temperature yields dibromocyclopentanes. There are three constitutional isomers:

1. 1,1-dibromocyclopentane: No stereoisomers (achiral).
2. 1,2-dibromocyclopentane: Exists as cis (meso, achiral) and trans (chiral, exists as a pair of enantiomers).
3. 1,3-dibromocyclopentane: Exists as cis (meso, achiral) and trans (chiral, exists as a pair of enantiomers). Note: 1,4- and 1,5- are identical to 1,3- and 1,2- respectively due to symmetry. Total distinct stereoisomers: 1 (1,1) + 3 (1,2) + 3 (1,3) = 7.

(b) (ii) Conversion of p-nitrotoluene to m-nitrotoluene Direct isomerization is difficult. The route involves:

1. Reduction: Reduce p-nitrotoluene to p-toluidine using Sn/HCl or H2/Pd.
2. Acetylation: Protect the amine as acetanilide (p-methylacetanilide) using Ac2O to moderate reactivity.
3. Nitration: Nitrate p-methylacetanilide with HNO3/H2SO4. The -NHAc group is a strong ortho/para director, but the para position is blocked by -CH3. Nitration occurs ortho to -NHAc, which is meta to -CH3. This yields 2-nitro-4-methylacetanilide.
4. Hydrolysis: Hydrolyze the amide to 2-nitro-4-methylaniline.
5. Diazotization and Deamination: Treat with NaNO2/HCl at 0-5°C to form the diazonium salt, then heat with H3PO2 (hypophosphorous acid) to replace the -NH2 group with -H. This yields m-nitrotoluene (3-nitrotoluene).

(c) Phenol to Coumarin This transformation is achieved via the Pechmann Condensation.

1. Reagents: Phenol, Malic acid (or Ethyl acetoacetate), and concentrated H2SO4 (or H3PO4).
2. Mechanism:

• Activation: Malic acid is dehydrated by H2SO4 to form formylacetic acid (or the corresponding acyl cation equivalent).
• Electrophilic Aromatic Substitution: The phenol ring, activated by the -OH group, attacks the electrophilic carbonyl carbon of the activated malic acid derivative at the ortho position.
• Lactonization: The resulting intermediate undergoes intramolecular cyclization. The phenolic oxygen attacks the carbonyl carbon of the acetic acid moiety, forming a six-membered lactone ring.
• Dehydration: Loss of water yields the aromatic coumarin ring system.
• Product: 2H-chromen-2-one (Coumarin).

In conclusion, these problems demonstrate the interplay between stereochemistry, regioselectivity in electrophilic aromatic substitution, and pericyclic/condensation mechanisms in heterocyclic synthesis.