Hg(OAc)2, EtNH2 2) NaBH4, CCl4, Br2, CHCl3
The Correct Answer and Explanation is:
e.
Reactant: 2-methyl-2-butene
Reagents:
- Hg(OAc)₂, EtNH₂
- NaBH₄
This is an example of oxymercuration–reduction, but instead of water, an amine (EtNH₂) is used as the nucleophile. This results in amino-mercuration, where the nucleophile adds to the more substituted carbon of the alkene, following Markovnikov’s rule.
Product: The ethylamino group (EtNH) adds to the more substituted carbon (tertiary), and hydrogen goes to the less substituted carbon (secondary).
So the product is:
2-ethylamino-2-methylbutane
h.
Reactant: 1,2-dimethylbenzene (o-xylene)
Reagent: Br₂ in CCl₄
This is a radical bromination at the benzylic position, not an electrophilic aromatic substitution, because CCl₄ is a non-polar solvent favoring radical conditions. Since both benzylic positions (methyl groups) are equivalent, bromination occurs at one of them.
Product: Benzyl bromide derivative — one of the methyl groups becomes a –CH₂Br
So the product is:
1-bromomethyl-2-methylbenzene
l.
Reactant: 1,2-dihydronaphthalene
Reagent: Br₂ in CHCl₃
In alkenes, Br₂ in non-nucleophilic solvents (like CHCl₃) leads to anti-addition of bromine across the double bond, forming a vicinal dibromide.
Product: Bromines add anti to each other across the former double bond.
So the product is:
trans-1,2-dibromo-1,2-dihydronaphthalene
Summary of Products:
- e. 2-ethylamino-2-methylbutane
- h. 1-bromomethyl-2-methylbenzene
- l. trans-1,2-dibromo-1,2-dihydronaphthalene
Explanation
Each reaction represents a different classical organic transformation. In part e, the reaction uses oxymercuration–reduction to achieve Markovnikov-selective addition of an amine nucleophile across an alkene. The mercury complex adds to the less substituted carbon while the nucleophile (ethylamine) adds to the more substituted center. Subsequent reduction with NaBH₄ removes the mercury, leading to a stable amine-containing product.
In part h, benzylic bromination is shown. Under radical conditions (Br₂ in CCl₄), bromine atoms preferentially abstract hydrogen from benzylic positions due to the resonance stabilization of the resulting radical. This leads to selective monobromination of one methyl group on the aromatic ring, resulting in a benzylic bromide. This type of reaction is valuable for installing good leaving groups on aromatic rings for further substitution or elimination.
In part l, halogenation of an alkene is demonstrated. Br₂ in CHCl₃ adds across the alkene in a stereospecific anti manner via a bromonium ion intermediate. The product is a vicinal dibromide, with the two bromine atoms on opposite faces of the former double bond. This transformation is useful in organic synthesis to introduce halogen atoms that can later undergo substitution or elimination reactions.
These reactions showcase the versatility of alkenes and aromatic compounds in organic synthesis, demonstrating how different reagents and conditions direct selective functionalization.
