Question 2 of 5: Predicting the Products of Five Organic Reactions

Question 2: Predicting the Products of Five Organic Reactions (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The major organic product of each reaction.

Approach

Reactions (i)–(iv) are all electrophilic additions across a C=C: identify the electrophile, add it across the double bond, and for the unsymmetrical alkenes in (iv) apply Markovnikov's rule (H to the carbon that already carries more H's, the other fragment to the more-substituted carbon, via the more stable carbocation). Reaction (v) instead substitutes onto the aromatic ring by the standard nitration mechanism (the ring's π system attacks the nitronium ion, NO₂⁺, generated in situ by protonation of HNO₃).

1. (i) Br₂ addition to 2-methylpropene. Bromine adds across the C=C of (CH₃)₂C=CH₂, one Br to each alkene carbon: $$\boxed{\text{1,2-dibromo-2-methylpropane, (CH}_3)_2\text{C(Br)CH}_2\text{Br}}.$$
2. (ii) HBr addition to cyclohexene. H and Br add across the ring's one C=C; because both alkene carbons of cyclohexene are equivalent (each bears one ring-H), there is no regiochemical choice to make: $$\boxed{\text{bromocyclohexane, C}_6\text{H}_{11}\text{Br}}.$$
3. (iii) Catalytic hydrogenation of methylenecyclohexane. H₂ adds across the exocyclic C=CH₂ on the Pd surface, converting the =CH₂ into a ring-attached –CH₃: $$\boxed{\text{methylcyclohexane, C}_7\text{H}_{14}}.$$
4. (iv) Markovnikov HCl addition to 3-methylpent-2-ene. The C2=C3 double bond is unsymmetrical: C3 already carries two carbon substituents (the ethyl chain and the methyl branch) while C2 carries one (a methyl) plus one H. Protonation at C2 (the carbon with more H) places the positive charge at C3, generating the more stable (tertiary-type, three-carbon- substituent) carbocation; chloride then attacks that carbocation: $$\boxed{\text{3-chloro-3-methylpentane, CH}_3\text{CH}_2\text{C(Cl)(CH}_3\text{)CH}_2\text{CH}_3}.$$
5. (v) Nitration of benzene. H₂SO₄ protonates HNO₃ to generate the electrophilic nitronium ion, NO₂⁺, which the aromatic ring attacks (electrophilic aromatic substitution); loss of H⁺ from the resulting arenium ion restores the aromatic sextet: $$\boxed{\text{nitrobenzene, C}_6\text{H}_5\text{NO}_2}\ (+\ \text{H}_2\text{O}).$$