In the following reaction sequence, major products X and Y are acyclic monomers. 500 mol of X completely reacts with 500 mol of Y to give 1 mol of a single biodegradable acyclic copolymer Z as the only product. The amount of Z formed in grams is ____. Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, Br : 80
A. A
Step 1: Identification of monomer X The given reaction sequence for X is: CH _3 I KCN CH _3 CN CH _3 CN H _3 O ^+, CH _3 COOH CH _3 COOH Red P, Br _2 BrCH _2 COOH (Hell-Volhard-Zelinsky reaction) BrCH _2 COOH NH _3 (excess) H _2 N - CH _2- COOH Thus, X is Glycine. Molar mass of X ( C _2 H _5 NO _2) = 2(12) + 5(1) + 14 + 2(16) = 75 g/mol . Step 2: Identification of monomer Y Caprolactam on acidic hydrolysis gives -aminocaproic acid: Caprolactam H _3 O ^+, H _2 N -( CH _2)_5- COOH Thus, Y is -aminocaproic acid. Molar mass of Y ( C _6 H _ 13 NO _2) = 6(12) + 13(1) + 14 + 2(16) = 131 g/mol . Step 3: Calculation of the mass of copolymer Z The copolymer Z is Nylon-2-nylon-6, which is a biodegradable polyamide formed by the condensation polymerization of Glycine and -aminocaproic acid. The problem states that 500 mol of X and 500 mol of Y completely react to form 1 mol of a single acyclic copolymer Z . This means each molecule of polymer Z contains 500 units of X and 500 units of Y , making a total of 1000 monomer units. In an acyclic chain of 1000 monomers, the number of peptide (amide) bonds formed is 1000 - 1 = 999. For each peptide bond formed, one molecule of water ( H _2 O ) is eliminated. Since 1 mol of polymer Z is formed, 999 mol of water are eliminated. Applying the law of conservation of mass: Mass of Z = (Mass of 500 mol of X ) + (Mass of 500 mol of Y ) - (Mass of 999 mol of H _2 O ) Mass of Z = (500 75) + (500 131) - (999 18) Mass of Z = 37500 + 65500 - 17982 Mass of Z = 103000 - 17982 = 85018 g Answer: 85018
Related: Chemistry — Biomolecules · All PYQ Banks