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Polymers question

2011 · Shift 2 · Q14
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Polymers question

2011 · Shift 2 · Q14

JEE AdvancedChemistryPolymersMultiple correct+4 / −2
The correct functional group X and the reagent/reaction condition Y in the following scheme are IIT-JEE 2011 Paper 2 Offline Chemistry - Polymers Question 5 English
  1. A
    X = COOCH3COOCH_3COOCH3​, Y = H2H_2H2​/Ni/heat
  2. B
    X = CONH2CONH_2CONH2​, Y = H2H_2H2​/Ni/heat
  3. C
    X = CONH2CONH_2CONH2​, Y = Br2Br_2Br2​/NaOHNaOHNaOH
  4. D
    X = CNCNCN, Y = H2H_2H2​/Ni/heat
View written solutionFree

Correct answer: B, C, D

Step-by-step Derivations

The problem asks to identify the correct functional group X and reagent/reaction condition Y to convert Benzene-1,4-dicarboxylic acid (terephthalic acid) into a monomer P, which can then be used to synthesize Nylon.

1. Understanding the Target Product

Nylon is a generic name for a family of synthetic polymers known as polyamides. Polyamides are characterized by repeating amide linkages (-CO-NH-). They are typically formed by the polycondensation of:

  • A diamine and a dicarboxylic acid.
  • An amino acid that self-condenses.
  • A lactam via ring-opening polymerization.

In this problem, the starting material is a derivative of terephthalic acid. The resulting monomer 'P' must be suitable for forming a polyamide. This means P is likely a diamine (to react with a dicarboxylic acid) or an amino acid (to self-polymerize). The structure of P will be based on the benzene-1,4-diyl skeleton.

The overall scheme is: HOOC−C6H4−COOHHOOC-C_6H_4-COOHHOOC−C6​H4​−COOH → X−C6H4−XX-C_6H_4-XX−C6​H4​−X → P → Nylon

We will evaluate each option to see if it produces a valid monomer P for Nylon synthesis.

2. Evaluation of Option A: X = COOCH3COOCH_3COOCH3​, Y = H2H_2H2​/Ni/heat

  • Step 1: Formation of the intermediate. Terephthalic acid is converted to its dimethyl ester. The functional group X is −COOCH3-COOCH_3−COOCH3​. HOOC−C6H4−COOH+2CH3OHightleftharpoonsCH3OOC−C6H4−COOCH3+2H2OHOOC-C_6H_4-COOH + 2 CH_3OH ightleftharpoons CH_3OOC-C_6H_4-COOCH_3 + 2 H_2OHOOC−C6​H4​−COOH+2CH3​OHightleftharpoonsCH3​OOC−C6​H4​−COOCH3​+2H2​O The intermediate is dimethyl terephthalate.
  • Step 2: Reaction with Y. The intermediate is treated with Y = H2H_2H2​/Ni/heat. This is a catalytic hydrogenation reaction, which reduces esters to alcohols. CH3OOC−C6H4−COOCH3+4H2ext−−(Ni/heat)ightarrowHOCH2−C6H4−CH2OH+2CH3OHCH_3OOC-C_6H_4-COOCH_3 + 4 H_2 ext{--(Ni/heat)} ightarrow HOCH_2-C_6H_4-CH_2OH + 2 CH_3OHCH3​OOC−C6​H4​−COOCH3​+4H2​ext−−(Ni/heat)ightarrowHOCH2​−C6​H4​−CH2​OH+2CH3​OH The product P is 1,4-bis(hydroxymethyl)benzene.
  • Step 3: Polymerization. P is a diol. Diols react with dicarboxylic acids to form polyesters, not polyamides (Nylon). For example, 1,4-bis(hydroxymethyl)benzene can react with terephthalic acid to form a polyester. Therefore, this route does not produce a monomer for Nylon.
  • Conclusion: Option A is incorrect.

3. Evaluation of Option B: X = CONH2CONH_2CONH2​, Y = H2H_2H2​/Ni/heat

  • Step 1: Formation of the intermediate. Terephthalic acid is converted to its diamide. The functional group X is −CONH2-CONH_2−CONH2​. HOOC−C6H4−COOHightarrowClOC−C6H4−COClext−−(NH3)ightarrowH2NOC−C6H4−CONH2HOOC-C_6H_4-COOH ightarrow ClOC-C_6H_4-COCl ext{--(NH_3)} ightarrow H_2NOC-C_6H_4-CONH_2HOOC−C6​H4​−COOHightarrowClOC−C6​H4​−COClext−−(NH3​)ightarrowH2​NOC−C6​H4​−CONH2​ The intermediate is terephthalamide.
  • Step 2: Reaction with Y. The intermediate is treated with Y = H2H_2H2​/Ni/heat. This reaction reduces amides to amines. H2NOC−C6H4−CONH2+4H2ext−−(Ni/heat)ightarrowH2NCH2−C6H4−CH2NH2+2H2OH_2NOC-C_6H_4-CONH_2 + 4 H_2 ext{--(Ni/heat)} ightarrow H_2NCH_2-C_6H_4-CH_2NH_2 + 2 H_2OH2​NOC−C6​H4​−CONH2​+4H2​ext−−(Ni/heat)ightarrowH2​NCH2​−C6​H4​−CH2​NH2​+2H2​O The product P is 1,4-bis(aminomethyl)benzene (or p-xylylenediamine).
  • Step 3: Polymerization. P is a diamine. Diamines are key monomers for synthesizing polyamides. For example, it can be reacted with adipic acid to form the polyamide Nylon MXD6.
  • Conclusion: Option B is correct.

4. Evaluation of Option C: X = CONH2CONH_2CONH2​, Y = Br2Br_2Br2​/NaOHNaOHNaOH

  • Step 1: Formation of the intermediate. The intermediate is the same as in option B: terephthalamide, H2NOC−C6H4−CONH2H_2NOC-C_6H_4-CONH_2H2​NOC−C6​H4​−CONH2​.
  • Step 2: Reaction with Y. The intermediate is treated with Y = Br2Br_2Br2​/NaOHNaOHNaOH. This is the Hofmann bromamide degradation reaction, which converts a primary amide to a primary amine with one less carbon atom. H2NOC−C6H4−CONH2+2Br2+8NaOHightarrowH2N−C6H4−NH2+2Na2CO3+4NaBr+4H2OH_2NOC-C_6H_4-CONH_2 + 2 Br_2 + 8 NaOH ightarrow H_2N-C_6H_4-NH_2 + 2 Na_2CO_3 + 4 NaBr + 4 H_2OH2​NOC−C6​H4​−CONH2​+2Br2​+8NaOHightarrowH2​N−C6​H4​−NH2​+2Na2​CO3​+4NaBr+4H2​O The product P is p-phenylenediamine (or benzene-1,4-diamine).
  • Step 3: Polymerization. P is an aromatic diamine. It can be reacted with a dicarboxylic acid (or diacyl chloride) to form an aromatic polyamide (aramid), which is a type of high-performance Nylon. For instance, its reaction with terephthaloyl chloride yields Kevlar.
  • Conclusion: Option C is correct.

5. Evaluation of Option D: X = CNCNCN, Y = H2H_2H2​/Ni/heat

  • Step 1: Formation of the intermediate. Terephthalic acid is converted to its dinitrile. The functional group X is -CN. This can be done by converting the acid to the amide, followed by dehydration. HOOC−C6H4−COOHightarrowH2NOC−C6H4−CONH2ext−−(dehydration)ightarrowNC−C6H4−CNHOOC-C_6H_4-COOH ightarrow H_2NOC-C_6H_4-CONH_2 ext{--(dehydration)} ightarrow NC-C_6H_4-CNHOOC−C6​H4​−COOHightarrowH2​NOC−C6​H4​−CONH2​ext−−(dehydration)ightarrowNC−C6​H4​−CN The intermediate is terephthalonitrile.
  • Step 2: Reaction with Y. The intermediate is treated with Y = H2H_2H2​/Ni/heat. This catalytic hydrogenation reduces nitriles to primary amines. NC−C6H4−CN+4H2ext−−(Ni/heat)ightarrowH2NCH2−C6H4−CH2NH2NC-C_6H_4-CN + 4 H_2 ext{--(Ni/heat)} ightarrow H_2NCH_2-C_6H_4-CH_2NH_2NC−C6​H4​−CN+4H2​ext−−(Ni/heat)ightarrowH2​NCH2​−C6​H4​−CH2​NH2​ The product P is 1,4-bis(aminomethyl)benzene, which is the same monomer obtained in option B.
  • Step 3: Polymerization. As established for option B, this diamine is a valid monomer for Nylon synthesis.
  • Conclusion: Option D is correct.

Final Summary

Options B, C, and D all describe valid chemical pathways to synthesize diamine monomers from terephthalic acid. These diamines can then be used to produce polyamides (Nylons). Option A produces a diol, which is a monomer for polyesters, not polyamides.

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