Polymers JEE 2019: Why B and D Are Correct

What are the correct options in the Polymers JEE 2019 question?

Select B and D in the Polymers JEE 2019 multiple-correct question from Chemistry, JEE Advanced 2019, Paper 2. The task is to decide which descriptions of polymer composition, preparation or bonding are valid. More than one option can be correct.

The complete question, paraphrased, asks you to choose the correct statements:

The correct selections are B and D. Every factual claim within an option must pass.

Why is option A wrong about cellulose?

Cellulose contains beta-D-glucose, not alpha-D-glucose. The reference to glycosidic bonds is true, but a correct bond type cannot rescue an incorrect building block. Check the polymer family, glucose configuration and linkage separately.

1. Step 1: Identify cellulose as a polysaccharide, a polymer built from sugar units.
2. Step 2: Its building blocks are beta-D-glucose units rather than alpha-D-glucose.
3. Step 3: These units have beta-1,4-glycosidic linkages. The numbers identify the linked carbon positions: carbon 1 of one glucose unit and carbon 4 of the next.

Replace alpha with beta to repair the statement. A is incorrect.

Why is option B correct about preparing Teflon?

Teflon is polytetrafluoroethene, or PTFE, made by polymerising tetrafluoroethene under the stated conditions. Both the monomer and preparation must match. The persulphate called a “catalyst” in the question functions as a free-radical initiator.

1. Step 1: Identify Teflon as polytetrafluoroethene.
2. Step 2: Identify its monomer as tetrafluoroethene: $\mathrm{C}{\mathrm{F}}_2=\mathrm{C}{\mathrm{F}}_2$

Its addition polymerisation is:

Addition across the monomer double bond produces the carbon-chain repeat unit while retaining the fluorine substituents. No small molecule is eliminated in this addition step.

3. Step 3: Confirm heating at high pressure in the presence of persulphate, which initiates free-radical polymerisation.

The monomer, repeat unit and stated preparation agree. B is correct.

Why does trans-polyisoprene not describe natural rubber?

Natural rubber is cis-1,4-polyisoprene, not trans-1,4-polyisoprene. Calling it polyisoprene gets only the polymer identity right. The configuration is a separate claim and must also be checked.

1. Step 1: Accept that natural rubber is polyisoprene.
2. Step 2: Identify its characteristic structure as cis-1,4-polyisoprene.
3. Step 3: Associate trans-1,4-polyisoprene with gutta-percha, not natural rubber.

Here, cis and trans describe the configuration about the double bonds remaining in the polymer chain. They do not identify different monomers. C is incorrect because its trans assignment is wrong.

Where is the amide linkage in nylon-6?

The amide connection joins the carbonyl end of one nylon-6 repeat to the nitrogen end of the next. Do not inspect only the middle of a written repeat unit. Read across the boundary between adjacent repeats.

1. Step 1: Identify caprolactam as the starting material.
2. Step 2: Classify nylon-6 as a polyamide, with repeat unit:

3. Step 3: Identify the amide connection formed where adjacent repeat units join: $-\mathrm{C}\mathrm{O}-\mathrm{N}\mathrm{H}-$

This connection lies in the polymer backbone. D is correct.

A false, B true, C false, D true; therefore select B and D.

Why is checking only “polyisoprene” a reasoning mistake?

It verifies the polymer identity but leaves the configuration unchecked. A faulty route to selecting C is: “Natural rubber is polyisoprene, so C must be correct.” That reasoning checks one claim and silently accepts the other.

Split C into its two claims:

• Polymer identity: Natural rubber is polyisoprene. True.
• Configuration: Its characteristic alkene geometry is trans. False.

A statement containing several factual claims is correct only when all are correct. Repair C by changing trans to cis.

Apply the same discipline to A: “glycosidic bonds” does not validate the false alpha-D-glucose assignment. Check family, building block, linkage and every qualifier actually asserted before accepting an option.

Can you solve three related polymer checks?

Use these checks to test glucose configuration, rubber stereochemistry and polymer linkages separately. These are original follow-up practice questions, not additional verified PYQs. Answer each before reading its worked answer.

1. A glucose polymer has beta-1,4-glycosidic bonds. Which polymer from the worked question fits, and what must replace alpha in its description?

Answer: Cellulose fits. Replace alpha-D-glucose with beta-D-glucose. The glycosidic bond description remains valid; the glucose configuration needs correction.

2. Two polyisoprene samples have cis-1,4 and trans-1,4 configurations respectively. Identify them.

Answer: They are natural rubber and gutta-percha, respectively. A shared isoprene monomer does not make their structures identical: their configurations about the remaining double bonds differ.

3. Match these repeat units to their polymer names and starting materials. Which contains an amide linkage? For a chain segment containing ten complete repeats of the first polymer, count its fluorine atoms.

Answer: The first is Teflon from tetrafluoroethene. The second is nylon-6 from caprolactam. Only nylon-6 contains the amide linkage: $-\mathrm{C}\mathrm{O}\mathrm{N}\mathrm{H}-$

Worked numerical check: Each Teflon repeat contains two carbon atoms bearing two fluorine atoms each. Ten complete repeats therefore contain:

Cover the answers and write the corrected versions of A and C from memory before moving to another question.

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