3.5cm 3 of a 0.16 mol/l solution of s-butyllithium in toluene was added to a sol

Question

3.5cm3 of a 0.16 mol/l solution of s-butyllithium in toluene was added to a solution of 8.40 g styrene in 200 cm3. After complete conversion of the styrene, 28.00 g of isoprene was added. When the isoprene was completely polymerized, the reaction was terminated by the addition of 2.89 cm3 of 0.10mol/L dichloromethane (X-R-X) in toluene. Assume 8.40 g styrene = 8.40 cm3 and 28.00 g of isoprene =27.00 cm3.

Describe the block copolymer and give the Mns of the blocks. Please view the following polymerization and termination scheme:

Explanation / Answer

First we have a block of styrene followed by a block of isoprene. (note: the second doublebond of isoprene is less reactive and will not react. Following both blocks are linear.)

Block1:

n(BuLi) = c*V = 0.16 mol/L * 0.0035 L = 5.6*10-4mol

n(styrene) = m/M = 8.4g/ 104.15 g/mol = 0.08 mol

Degree of polymerization: DP(styrene) = n(styrene)/n(BuLi) = 0.08/5.6*10-4 = 142.9

Mn(Block1) = DP * M(styrene) = 142.9 * 104.15g/mol = 14878.6 g/mol

Block2:

n(BuLi) = c*V = 0.16 mol/L * 0.0035 L = 5.6*10-4mol

n(isoprene) = m/M = 28g/ 68.12 g/mol = 0.41 mol

Degree of polymerization: DP(styrene) = n(isoprene)/n(BuLi) = 0.41/5.6*10-4 = 732.1

Mn(Block1) = DP * M(isoprene) = 732.1 * 68.12g/mol = 49878.6 g/mol

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