2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Which of the following is a thermosetting polymer?
A.
Buna-N
B.
PVC
C.
NyIon6
D.
Bakelite
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The correct match between Item-I and Item-II is :
| Item - I | Item - II | ||
|---|---|---|---|
| (a) | High density polythene | (I) | Peroxide catalyst |
| (b) | Polyacrylonitrile | (II) | Condensation at high temperature & pressure |
| (c) | Novolac | (III) | Ziegler-Natta Catalyst |
| (d) | Nylon 6 | (IV) | Acid or base catalyst |
A.
(a) $ \to $ (IV), (b) $ \to $ (II), (c) $ \to $ (I), (d) $ \to $ (III)
B.
(a) $ \to $ (III), (b) $ \to $ (I), (c) $ \to $ (IV), (d) $ \to $ (II)
C.
(a) $ \to $ (II), (b) $ \to $ (IV), (c) $ \to $ (I), (d) $ \to $ (III)
D.
(a) $ \to $ (III), (b) $ \to $ (I), (c) $ \to $ (II), (d) $ \to $ (IV)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
Which of the following is a condensation polymer?
A.
Neoprene
B.
Buna-S
C.
Nylon 6, 6
D.
Teflon
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Which of the following compounds is a
constituent of the polymer
A.
N-Methyl urea
B.
Ammonia
C.
Formaldehyde
D.
Methylamine
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The structure of Nylon-6 is :
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
The two monomers for the synthesis of Nylon 6, 6 are
A.
HOOC(CH2)6COOH, H2N(CH2)4NH2
B.
HOOC(CH2)4COOH, H2N(CH2)6NH2
C.
HOOC(CH2)4COOH, H2N(CH2)4NH2
D.
HOOC(CH2)6COOH, H2N(CH2)6NH2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
Poly $\beta $ –hydroxybutyrate-co-$\beta $-hydroxyvalerate (PHBV) is a copolymer of
A.
2-hydroxybutanoic acid and 3-hydroxypentanoic acid
B.
3-hydroxybutanoic acid and 2-hydroxypentanoic acid
C.
3-hydroxybutanoic acid and 3-hydroxypentanoic acid
D.
3-hydroxybutanoic acid and 4-hydroxypentanoic acid
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
The homopolymer formed from 4-hydroxy-butanoic acid is :
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
The polymer obtained from the following reaction is :
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
The major product of the following reaction is
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
Major product of the following reaction is :
A.
B.
C.
D.
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
Which of the following statements is not true ?
A.
Step growth polymerisation requires a bifunctional monomer.
B.
Nylon 6 is an example of step-growth polymerisation.
C.
Chain growth polymerisation includes both homopolymerisation and copylymerisation.
D.
Chain growth polymerisation involves homopolymerisation only.
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Morning Slot
The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is :
A.
B.
C.
D.
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 9th April Morning Slot
Which of the following is a biodegradable polymer ?
A.
B.
C.
D.
2017
JEE Mains
MCQ
JEE Main 2017 (Offline)
The formation of which of the following polymers involves hydrolysis reaction?
A.
Bakelite
B.
Nylon 6, 6
C.
Terylene
D.
Nylon 6
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
Which of the following polymers is synthesized using a free radical
polymerization technique ?
A.
Teflon
B.
Terylene
C.
Melamine polymer
D.
Nylon 6, 6
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 9th April Morning Slot
Assertion :
Rayon is a semisynthetic polymer whose properties are better than natural cotton.
Reason :
Mechanical and aesthetic properties of cellulose can be improved by acetylation.
Rayon is a semisynthetic polymer whose properties are better than natural cotton.
Reason :
Mechanical and aesthetic properties of cellulose can be improved by acetylation.
A.
Both assertion and reason are correct, and the reason is the correct explanation for the assertion.
B.
Both assertion and reason are correct, but the reason is not the correct explanation for the assertion.
C.
Assertion is incorrect statement, but the reason is correct.
D.
Both assertion and reason are incorrect.
2016
JEE Mains
MCQ
JEE Main 2016 (Offline)
Which of the following statements about low density polythene is FALSE ?
A.
Its synthesis requires high pressure.
B.
It is a poor conductor of electricity.
C.
Its synthesis requires dioxygen or a peroxide initiator as a catalyst.
D.
It is used in the manufacture of buckets, dust-bins etc.
2015
JEE Mains
MCQ
JEE Main 2015 (Offline)
Which polymer is used in the manufacture of paints and lacquers?
A.
Glyptal
B.
Polypropene
C.
Poly vinyl chloride
D.
Bakelite
2014
JEE Mains
MCQ
JEE Main 2014 (Offline)
Which one is classified as a condensation polymer?
A.
Teflon
B.
Acrylonitrile
C.
Dacron
D.
Neoprene
2012
JEE Mains
MCQ
AIEEE 2012
The species which can best serve as an initiator for the cationic polymerization is :
A.
LiAlH4
B.
HNO3
C.
AlCl3
D.
BuLi
2010
JEE Mains
MCQ
AIEEE 2010
The polymer containing strong intermolecular forces e.g. hydrogen bonding, is
A.
teflon
B.
nylon 6,6
C.
polystyrene
D.
natural rubber
2009
JEE Mains
MCQ
AIEEE 2009
Buna-$N$ synthetic rubber is a copolymer of :
A.
${H_2}C = CH - CH = C{H_2}$
and $\,\,\,\,\,$ ${H_5}{C_6} - CH = C{H_2}$
and $\,\,\,\,\,$ ${H_5}{C_6} - CH = C{H_2}$
B.
${H_2}C = CH - CN$
and $\,\,\,\,\,$ ${H_2}C = CH - CH = C{H_2}$
and $\,\,\,\,\,$ ${H_2}C = CH - CH = C{H_2}$
C.
D.
2008
JEE Mains
MCQ
AIEEE 2008
Bakelite is obtained from phenol by reacting with
A.
(CH2OH)2
B.
CH3CHO
C.
CH3COCH3
D.
HCHO
2005
JEE Mains
MCQ
AIEEE 2005
Which of the following is a polyamide?
A.
Teflon
B.
Nylon – 6, 6
C.
Terylene
D.
Bakelite
2005
JEE Mains
MCQ
AIEEE 2005
Which of the following is fully fluorinated polymer?
A.
Neoprene
B.
Teflon
C.
Thiokol
D.
PVC
2003
JEE Mains
MCQ
AIEEE 2003
Nylon threads are made of
A.
polyester polymer
B.
polyamide polymer
C.
polyethylene polymer
D.
polyvinyl polymer
2002
JEE Mains
MCQ
AIEEE 2002
Polymer formation from monomers starts by
A.
condensation reaction between monomers
B.
coordinate reaction between monomers
C.
conversion of monomer to monomer ions by protons
D.
hydrolysis of monomers
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Evening Shift
The Novolac polymer has mass of 963 g. The number of monomer units present in it are
Correct Answer: 9
Explanation:
Novolac is
Molar mass of monomer is $107 \mathrm{~g} / \mathrm{mol}$
$ \mathrm{n}=\frac{963}{107}=9 $
The number of monomer units present in it is 9.
Molar mass of monomer is $107 \mathrm{~g} / \mathrm{mol}$
$ \mathrm{n}=\frac{963}{107}=9 $
The number of monomer units present in it is 9.
2025
JEE Advanced
Numerical
JEE Advanced 2025 Paper 1 Online
The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ______.
Use: Atomic mass of N (in amu) = 14
Correct Answer: 280
Explanation:
Monomer $(X) \rightarrow$ positive carbylamine test
The monomer used in the synthesis of Nylon -6,6 are hexamethylene diamine and adipic acid.
Positive carbylamine test is given by primary amine compounds. Hexamethylene diamine is a primary amine compound and can give positive carbylamine test.
So, the monomer $X$ is hoxamethylene diamine.
$
\begin{aligned}
&\begin{aligned}
x \rightarrow \mathrm{H_2N}-\left(\mathrm{CH}_2\right)_6-\mathrm{NH}_2 \quad & \text { Molarmars } & \\
= & 6 \times 12 \mathrm{~g} / \mathrm{mol}+2 \times 14 \mathrm{~g} / \mathrm{mol} +16 \times 1 \mathrm{~g} / \mathrm{mol} & \\
= & 116 \mathrm{~g} / \mathrm{mol} &
\end{aligned}\\
&\text { Moles of } X \text { analyzed using Dumas method }=10 \mathrm{~mol}
\end{aligned}
$According to Dumas method, the nitrogen present in the compound is released as nitrogen gas $\left(\mathrm{N}_2\right)$,
The general chemical equation fol Dumas method can be written as$
\mathrm{C}_x \mathrm{H}_y \mathrm{~N}_z+\left(2 x+\frac{y}{2}\right) \mathrm{CuO}(\mathrm{~s}) \xrightarrow{}
$(organic
Compound)$
x \mathrm{CO}_2(g)+\frac{y}{2} H_2 \mathrm{O}(gold )+\frac{z}{2} \mathrm{~N}_2(g)+\left(2 x+\frac{y}{2}\right) \mathrm{Cu}(s)
$The compound used in Dumas method is hexamethylene diamine. Its formula is $\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2$$
\text { So, the equation can be written as, }
$$
\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2+20 \mathrm{CuO} \rightarrow
$$
6 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}+1 \mathrm{~N}_2+20 \mathrm{Cu}
$$
\text { So, } 1 \text { mole of } \mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2 \text { gives. } 1 \text { mole of } \mathrm{N}_2 \text {. }
$Given 10, moles of $\mathrm{C}_{6}\mathrm{H}_{16} \mathrm{~N}_2$ (hexamethylene diamine) gives 10 moles of $\mathrm{N}_2$ gas.Moles of $\mathrm{N}_2$ gas evolved $=10$ mol Mass can be calculated as, \begin{aligned}
&\text { Moles }=\frac{\text { mass }}{\text { molarmass }}\\
&\text { Moloumass of } \mathrm{N}_2=28 \mathrm{~g} / \mathrm{mol}
\end{aligned}mass $=$ moles $\times$ molar mass$
\begin{aligned}
&\begin{aligned}
& =10 \text { mol } x 28 \mathrm{~g} / \mathrm{mol} \\
& =280 \mathrm{~g}
\end{aligned}\\
&\text { Answer: } 280 \%
\end{aligned}
$
Positive carbylamine test is given by primary amine compounds. Hexamethylene diamine is a primary amine compound and can give positive carbylamine test.
So, the monomer $X$ is hoxamethylene diamine.
$
\begin{aligned}
&\begin{aligned}
x \rightarrow \mathrm{H_2N}-\left(\mathrm{CH}_2\right)_6-\mathrm{NH}_2 \quad & \text { Molarmars } & \\
= & 6 \times 12 \mathrm{~g} / \mathrm{mol}+2 \times 14 \mathrm{~g} / \mathrm{mol} +16 \times 1 \mathrm{~g} / \mathrm{mol} & \\
= & 116 \mathrm{~g} / \mathrm{mol} &
\end{aligned}\\
&\text { Moles of } X \text { analyzed using Dumas method }=10 \mathrm{~mol}
\end{aligned}
$According to Dumas method, the nitrogen present in the compound is released as nitrogen gas $\left(\mathrm{N}_2\right)$,
The general chemical equation fol Dumas method can be written as$
\mathrm{C}_x \mathrm{H}_y \mathrm{~N}_z+\left(2 x+\frac{y}{2}\right) \mathrm{CuO}(\mathrm{~s}) \xrightarrow{}
$(organic
Compound)$
x \mathrm{CO}_2(g)+\frac{y}{2} H_2 \mathrm{O}(gold )+\frac{z}{2} \mathrm{~N}_2(g)+\left(2 x+\frac{y}{2}\right) \mathrm{Cu}(s)
$The compound used in Dumas method is hexamethylene diamine. Its formula is $\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2$$
\text { So, the equation can be written as, }
$$
\mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2+20 \mathrm{CuO} \rightarrow
$$
6 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}+1 \mathrm{~N}_2+20 \mathrm{Cu}
$$
\text { So, } 1 \text { mole of } \mathrm{C}_6 \mathrm{H}_{16} \mathrm{~N}_2 \text { gives. } 1 \text { mole of } \mathrm{N}_2 \text {. }
$Given 10, moles of $\mathrm{C}_{6}\mathrm{H}_{16} \mathrm{~N}_2$ (hexamethylene diamine) gives 10 moles of $\mathrm{N}_2$ gas.Moles of $\mathrm{N}_2$ gas evolved $=10$ mol Mass can be calculated as, \begin{aligned}
&\text { Moles }=\frac{\text { mass }}{\text { molarmass }}\\
&\text { Moloumass of } \mathrm{N}_2=28 \mathrm{~g} / \mathrm{mol}
\end{aligned}mass $=$ moles $\times$ molar mass$
\begin{aligned}
&\begin{aligned}
& =10 \text { mol } x 28 \mathrm{~g} / \mathrm{mol} \\
& =280 \mathrm{~g}
\end{aligned}\\
&\text { Answer: } 280 \%
\end{aligned}
$
2022
JEE Advanced
MSQ
JEE Advanced 2022 Paper 2 Online
Among the following, the correct statement(s) about polymers is(are)
A.
The polymerization of chloroprene gives natural rubber.
B.
Teflon is prepared from tetrafluoroethene by heating it with persulphate catalyst at high pressures.
C.
PVC are thermoplastic polymers.
D.
Ethene at 350-570 K temperature and 1000-2000 atm pressure in the presence of a peroxide initiator yields high density polythene.
2019
JEE Advanced
MSQ
JEE Advanced 2019 Paper 2 Offline
Choose the correct option(s) from the following.
A.
Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high pressure.
B.
Natural rubber is polyisoprene containing trans alkene units
C.
Cellulose has only $\alpha $-D-glucose units that are joined by glycosidic linkages
D.
Nylon-6 has amide linkages
2015
JEE Advanced
MCQ
JEE Advanced 2015 Paper 2 Offline
Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are
A.
CH3SiCl3 and Si(CH3)4.
B.
(CH3)2SiCl2 and (CH3)3SiCl.
C.
(CH3)2SiCl2 and CH3SiCl3.
D.
SiCl4 and (CH3)3SiCl.
2011
JEE Advanced
MSQ
IIT-JEE 2011 Paper 2 Offline
The correct functional group X and the reagent/reaction condition Y in the following scheme are
A.
X = COOCH3, Y = H2/Ni/heat
B.
X = CONH2, Y = H2/Ni/heat
C.
X = CONH2, Y = Br2/NaOH
D.
X = CN, Y = H2/Ni/heat
2016
JEE Advanced
MCQ
JEE Advanced 2016 Paper 1 Offline
On complete hydrogenation, natural rubber produces
A.
ethylene-propylene copolymer.
B.
vulcanized rubber.
C.
polypropylene.
D.
polybutylene.
2009
JEE Advanced
MCQ
IIT-JEE 2009 Paper 1 Offline
Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is
A.
Nylon
B.
Poly(vinyl chloride)
C.
Cellulose
D.
Natural Rubber
2007
JEE Advanced
MCQ
IIT-JEE 2007 Paper 1 Offline
Match the chemical substances in Column I with type of polymers/type of bonds in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | cellulose | (P) | natural polymer |
| (B) | nylon-6, 6 | (Q) | synthetic polymer |
| (C) | protein | (R) | amide linkage |
| (D) | sucrose | (S) | glycoside linkage |
A.
A - (p, s); B - (q); C - (p, r); D - (s)
B.
A - (p, s); B - (q, r); C - (r); D - (s)
C.
A - (p, s); B - (q, r); C - (p, r); D - (s)
D.
A - (s); B - (q, r); C - (p, r); D - (s)
2005
JEE Advanced
Numerical
IIT-JEE 2005 Mains
Monomer A of a polymer on ozonolysis yields two moles of HCHO and one mole of CH$_3$COCHO.
(A) Deduce the structure of A.
(B) Write the structure of "all $cis$" forms of polymer of compound A.
Correct Answer: <img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1mh9kec77/1cc5949f-d9e0-406f-bbfe-4968a35d3026/e8384930-b36f-11f0-b034-2711d08df967/file-jaoe38c1mh9kec78.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1mh9kec77/1cc5949f-d9e0-406f-bbfe-4968a35d3026/e8384930-b36f-11f0-b034-2711d08df967/file-jaoe38c1mh9kec78.png" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="IIT-JEE 2005 Mains Chemistry - Polymers Question 2 English Answer">
Explanation:
(A) The monomer A on ozonolysis yields two molecules of HCHO and a molecule of CH$_3$COCHO. During ozonolysis, the double or triple bonds in the molecule are replaced by carbonyl groups.
Therefore, if we remove the oxygen atoms from 2 molecules of HCHO and a molecule of CH$_3$COCHO and replaced them with double bond, we can deduce the structure of monomer A.
The structure of monomer A we got is the structure of isoprene. Therefore, monomer A is isoprene.
(B) The isoprene is a monomer for natural rubber.
To write the “all $cis$” form of the polymer of monomer A that is isoprene, we need to write the structure of polymer such that the identical groups present across double bonds should lie in the same plane.
The “all $cis$” for of the polymer is shown below.
Final Answer
Hints:
The ozonolysis eliminates the double or triple bond from the molecule and replace it with carbonyl groups. The removal of oxygen atoms and replacing it with double bonds will give the structure of monomer A.