Question

Which of the following alkyl halide is treated with sodium metal to obtain 2,2,3,3-tetramethyl butane?

• A. tert-Butyl bromide
• B. n-Propyl bromide
• C. sec-Butyl bromide
• D. n-Butyl bromide

Hint:

To solve a Wurtz reaction backwards, count the total number of carbons in the symmetrical product (here, 8 carbons) and divide by 2. This tells you the starting alkyl halide must have exactly 4 carbons, immediately narrowing down your choices to the butyl families.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are to calculate the standard enthalpy of formation (ΔfH°) for one mole of NH₃ gas using provided bond enthalpy values.

Step 2: Key Formula or Approach:
ΔrH° = Σ(Bond Energies of Bonds Broken) – Σ(Bond Energies of Bonds Formed). Then, divide ΔrH° by the stoichiometric coefficient of NH₃.

Step 3: Detailed Explanation:
For N₂ + 3H₂ → 2NH₃, Bonds Broken = 1(N≡N) + 3(H–H) = 941 + 3(436) = 2249 kJ. Bonds Formed = 6(N–H) = 6(389) = 2334 kJ. ΔrH° = 2249 – 2334 = -85 kJ for 2 moles. ΔfH° for 1 mole is -85 / 2 = -42.5 kJ mol⁻¹.

Step 4: Final Answer:
The enthalpy of formation is -42.5 kJ, corresponding to option (C).