JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
Buna-S can be represented as:
- (a) [-CH₂-CH=CH-CH₂-CH(|)(C₆H₅)-CH₂-]ₙ
- (b) - [CH₂-CH=C(|(C₆H₅))-CH=CH-CH₂-]ₙ
- (c) [-CH₂-CH=CH-CH=C(C₆H₅; |)-CH₂-]ₙ
- (d) - [ CH=CH-CH=CH-CH(|)(C₆H₅)-CH₂ ] -ₙ
Answer and solution
Answer (a) [-CH₂-CH=CH-CH₂-CH(|)(C₆H₅)-CH₂-]ₙ
Solution
Chemistry ma’am · 3 steps
Each monomer spends one C=C to join the chain, so a Buna-S unit keeps only one double bond.
- The monomers. Buna-S is a copolymer of 1,3-butadiene, CH₂=CH-CH=CH₂, and styrene, C₆H₅-CH=CH₂. Bu stands for butadiene, Na for the sodium catalyst, and S for styrene.
- What each monomer becomes. Butadiene adds through its two ends (1,4-addition). It uses up its two terminal double bonds and forms a new one in the middle: -CH₂-CH=CH-CH₂⁻. Styrene opens its only C=C, so the phenyl group ends up on a saturated CH: -CH₂-CH(C₆H₅)⁻.
Option C=C in the unit Carbon carrying C₆H₅ H atoms (needed: 6 + 8 = 14) (a) one, in the butadiene part saturated CH 14 (b) two double-bonded C 12 (c) two double-bonded C 12 (d) two saturated CH 12 - So. Only (a) has one C=C, the phenyl group on a saturated carbon, and the full C₁₂H₁₄ formula of one butadiene plus one styrene. The answer is (a).
The idea to keep. In addition polymers, every monomer spends one C=C on joining the chain. Butadiene keeps one C=C in the middle and styrene keeps none. The repeat unit has exactly the atoms of the monomers added together.
Answer from the NTA final answer key. Solution by PrepWiser.