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ChemHardAdv 2023 • Paper 2

Yield and Mass Calculation of Organic Product S

The reaction of 4-methyloct-1-ene (P\mathbf{P}, 2.52 g2.52\text{ g}) with HBr\text{HBr} in the presence of (C6H5CO)2O2(\text{C}_6\text{H}_5\text{CO})_2\text{O}_2 gives two isomeric bromides in a 9:19 : 1 ratio, with a combined yield of 50%50\%. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aq. K2CO3\text{K}_2\text{CO}_3 to give a non-ionic product S\mathbf{S} in 100%100\% yield.

The mass (in mg) of S\mathbf{S} obtained is _____.

[Use molar mass (in g mol−1\text{g mol}^{-1}): H=1\text{H} = 1, C=12\text{C} = 12, N=14\text{N} = 14, Br=80\text{Br} = 80]

Official Numerical Answer1791

Topics & Concepts

Haloalkanes and HaloarenesHaloalkanes and HaloarenesAmines

Step-by-Step Solution

To find the mass of the non-ionic product S\mathbf{S} in milligrams, we break down the reaction sequence step by step:

Step 1: Molar mass and initial moles of compound P

Compound P\mathbf{P} is 4-methyloct-1-ene.

  • Formula of P\mathbf{P}: C9H18\text{C}_9\text{H}_{18}

  • Molar mass of P\mathbf{P}: Molar mass of P=(9×12)+(18×1)=108+18=126 g mol−1\text{Molar mass of } \mathbf{P} = (9 \times 12) + (18 \times 1) = 108 + 18 = 126 \text{ g mol}^{-1}

  • Moles of P\mathbf{P} taken: nP=2.52 g126 g mol−1=0.02 moln_{\mathbf{P}} = \frac{2.52 \text{ g}}{126 \text{ g mol}^{-1}} = 0.02 \text{ mol}

Step 2: Hydrobromination of P in the presence of peroxide

The reaction of an alkene with HBr\text{HBr} in the presence of peroxide, (C6H5CO)2O2(\text{C}_6\text{H}_5\text{CO})_2\text{O}_2, proceeds via a free-radical mechanism (anti-Markovnikov addition).

  • Primary alkyl bromide (1-bromo-4-methyloctane) is formed as the major product (90% of the formed bromides).
  • Secondary alkyl bromide (2-bromo-4-methyloctane) is formed as the minor product (10% of the formed bromides).

Given:

  • Combined yield of bromides = 50%50\%
  • Ratio of primary to secondary alkyl bromide = 9:19 : 1

Therefore: Total moles of bromides formed=0.02 mol×0.50=0.01 mol\text{Total moles of bromides formed} = 0.02 \text{ mol} \times 0.50 = 0.01 \text{ mol}

Moles of primary alkyl bromide=0.01 mol×910=0.009 mol\text{Moles of primary alkyl bromide} = 0.01 \text{ mol} \times \frac{9}{10} = 0.009 \text{ mol}

Step 3: Formation of Product S

The primary alkyl bromide (R−Br\text{R}-\text{Br}, where R=C9H19\text{R} = \text{C}_9\text{H}_{19}) reacts with diethylamine, HN(C2H5)2\text{HN}(\text{C}_2\text{H}_5)_2, followed by treatment with aq. K2CO3\text{K}_2\text{CO}_3 to neutralize the acid formed, giving the tertiary amine S\mathbf{S}.

C9H19Br+HN(C2H5)2→aq. K2CO3C9H19N(C2H5)2+KBr+H2O+CO2\text{C}_9\text{H}_{19}\text{Br} + \text{HN}(\text{C}_2\text{H}_5)_2 \xrightarrow{\text{aq. K}_2\text{CO}_3} \text{C}_9\text{H}_{19}\text{N}(\text{C}_2\text{H}_5)_2 + \text{KBr} + \text{H}_2\text{O} + \text{CO}_2

  • Molecular formula of S\mathbf{S}: C13H29N\text{C}_{13}\text{H}_{29}\text{N}
  • Molar mass of product S\mathbf{S}: Molar mass of S=(13×12)+(29×1)+(14×1)=156+29+14=199 g mol−1\text{Molar mass of } \mathbf{S} = (13 \times 12) + (29 \times 1) + (14 \times 1) = 156 + 29 + 14 = 199 \text{ g mol}^{-1}

Given that the yield of S\mathbf{S} from the primary alkyl bromide is 100%100\%: nS=0.009 moln_{\mathbf{S}} = 0.009 \text{ mol}

Step 4: Mass calculation of S

Mass of S=0.009 mol×199 g mol−1=1.791 g\text{Mass of } \mathbf{S} = 0.009 \text{ mol} \times 199 \text{ g mol}^{-1} = 1.791 \text{ g}

Converting to milligrams (mg): Mass of S=1.791×1000 mg=1791 mg\text{Mass of } \mathbf{S} = 1.791 \times 1000 \text{ mg} = 1791 \text{ mg}

Final Answer: 1791

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