Question

X₂/hv is used for free radical substitution reaction in alkanes R-H+X-X$\longrightarrow$R-X+H-X. The rate of halogens with alkanes follows as: F₂>Cl₂>Br₂>I₂. "Fluorination is too violent to be controlled" while Iodination is slow and reversible reaction. Relative rate of substitution of H-atoms of alkanes in free radical halogenation is: Cl₂$\Longrightarrow$3°H : 2°H : 1°H 5 : 3.8 : 1 Br₂$\Longrightarrow$3°H : 2°H : 1°H 1600 : 82 : 1 Isopentane reacts with Cl₂/hv to give a monochlorinated product P(major) and with Br₂/$\Delta$ to give a mono-brominated product Q(major).

alc. KOH (Q) $\longrightarrow$ (T) $\longrightarrow$ (U) $\Delta$ (Major) (Major) Q, T and U respectively are:

• A. 2-bromo-2-methylbutane, 2-methylbut-2-ene, 2-methyl-2-phenylbutane
• B. 1-bromo-2-methylbutane, 2-methylbut-1-ene, 2-methyl-1-phenylbutane
• C. 2-bromo-2-methylbutane, 2-methylbut-1-ene, 1-methyl-2-phenylbutane
• D. 1-bromo-2-methylbutane, 2-methylbut-2-ene, 1-methyl-1-phenylbutane

Answer 1

Answer: (A)

2-bromo-2-methylbutane, 2-methylbut-2-ene, 2-methyl-2-phenylbutane

Answer 2

Step-by-step Derivations:

1. Identify Isopentane: Isopentane is the common name for 2-methylbutane.

2. Determine Major Mono-brominated Product Q: Isopentane reacts with Br₂/Δ. This is a free radical substitution. The relative rates of substitution for Br₂ are given as 3°H : 2°H : 1°H = 1600 : 82 : 1.

   Comparing these values (9 : 164 : 1600), the highest statistical reactivity is for the tertiary (3°) H-atom. Therefore, the major mono-brominated product Q will be formed by substitution at the tertiary carbon. Q is 2-bromo-2-methylbutane.

3. Determine Major Product T: Q (2-bromo-2-methylbutane) undergoes reaction with alc. KOH/Δ. This is a dehydrohalogenation (E2 elimination) reaction. The base (alc. KOH) removes a proton from a carbon adjacent to the carbon bearing the bromine (β-carbon), forming an alkene. According to Zaitsev's rule, the major product is the most substituted (more stable) alkene.

   2-methylbut-2-ene is the more substituted and thus more stable alkene. Therefore, T (major) is 2-methylbut-2-ene.

4. Determine Major Product U: T (2-methylbut-2-ene) reacts with H⁺/Benzene. This is an acid-catalyzed Friedel-Crafts alkylation.

   The tertiary carbocation (2-methylbutan-2-yl cation) is more stable and will be the predominant intermediate. This carbocation then acts as an electrophile and attacks the benzene ring. The product U will be 2-methyl-2-phenylbutane (also known as tert-pentylbenzene).

5. Conclusion:

   • Q: 2-bromo-2-methylbutane
   • T: 2-methylbut-2-ene
   • U: 2-methyl-2-phenylbutane