Polymers Test

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Polymers Test - 62

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Weekly Quiz Competition

Question 1
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The difference in the densities of low-density (LDP) and high-density polymers (HDP) is due to the fact that:

A
LDP are highly branched structures while HDP consists of closely packed linear molecules

B
LDP are linear chains while HDP are branched chains of polythene

C
both LDP and HDP are unbranched linear chains with different lengths

D
at high temperature, the density of polymer is reduced

Solution

LDP obtained through the free radical addition and H-atom abstraction has highly branched structure. Due to this branching, the density of LDP becomes low.
Whereas HDP is obtained when addition polymerisation of ethene takes place in a hydrocarbon solvent in the presence of a catalyst such as triethylaluminium and titanium tetrachloride. It consists of linear molecules and has a high density due to close packing.
Question 2
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Which one of the following is a condensation polymer?

A
PVC

B
Teflon

C
Dacron

D
Polystyrene

Solution

Dacron is a condensation polymer obtained from ethylene glycol and terephthalic acid.

Its properties include high tensile strength, high resistance to stretching, both wet and dry, and good resistance to degradation by chemical bleaches and to abrasion.

Hence, option C is correct.
Question 3
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Natural rubber or raw rubber consists of basic material latex which is a dispersion of isoprene. During the treatment, this isoprene forms a high molecular weight polymer of isoprene. Natural rubber can be obtained from five hundred different species of plants.
Which rubber is not a polydiene?

A
Polyisoprene

B
Polychloroprene

C
Thiokal rubber

D
Nitrile rubber

Solution

Thiokol rubber is a type of artificial elastomer, which is invariably a polymer.
it can be prepared by the condensation of 1,2-dichloroethane with sodium polysulphide. It does not contain any $$C=C$$ bond. So it is not a polydiene.

Question 4

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Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	Buna-N	(i)	Phthalic acid and ethylene glycol
(B)	Nylon-6,6	(ii)	Terephthalic acid and ethylene glycol
(C)	Dacron	(iii)	Hexamethylenediamine and adipic acid
(D)	Glyptal plastic	(iv)	Acrylonitrile and butadiene

(A) $$\rightarrow$$ (ii), (B) $$\rightarrow$$ (iii), (C) $$\rightarrow$$ (iv), (D) $$\rightarrow$$ (i)

(A) $$\rightarrow$$ (i), (B) $$\rightarrow$$ (ii), (C) $$\rightarrow$$ (iv), (D) $$\rightarrow$$ (i)

(A) $$\rightarrow$$ (iii), (B) $$\rightarrow$$ (iv), (C) $$\rightarrow$$ (i), (D) $$\rightarrow$$ (ii)

(A) $$\rightarrow$$ (iv), (B) $$\rightarrow$$ (iii), (C) $$\rightarrow$$ (ii), (D) $$\rightarrow$$ (i)

Solution

Buna$$-N$$	(iv) Acrylonitrile and butadiene
Nylon-6,6	(iii) Hexamethylenediamine and adipic acid
Dacron	(ii) Terephthalic acid and ethylene glycol
Glyptal plastic	(i) Phthalic acid and ethylene glycol

Condensation polymerisation

This type of polymerisation generally involves a repetitive condensation reaction between two bi-functional monomers. These polycondensation reactions may result in the loss of some simple molecules as water, alcohol, etc., and lead to the formation of high molecular mass condensation polymers

The correct option is [D].

Question 5
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The monomers of biodegradable polymer, nylon 2-nylon 6 are:

A
glycine + adipic acid

B
glycol + phthalic acid

C
phenol + urea

D
glycine + amino caproic acid

Solution

Nylon 2-nylon 6 is an alternating polyamide copolymer of glycine $$(H_2N–CH_2–COOH)$$ and aminocaproic acid $$[H_2N (CH_2)_5 COOH]$$.

Hence, the correct answer is option $$\text{D}$$.
Question 6
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Match the polymers given in column I with monomers in column II and mark the appropriate choice.

A
(A) $$\rightarrow$$ (iv), (B) $$\rightarrow$$ (ii), (C) $$\rightarrow$$ (i), (D) $$\rightarrow$$ (iii)

B
(A) $$\rightarrow$$ (i), (B) $$\rightarrow$$ (iii), (C) $$\rightarrow$$ (iv), (D) $$\rightarrow$$ (ii)

C
(A) $$\rightarrow$$ (iv), (B) $$\rightarrow$$ (i), (C) $$\rightarrow$$ (ii), (D) $$\rightarrow$$ (iii)

D
(A) $$\rightarrow$$ (ii), (B) $$\rightarrow$$ (iv), (C) $$\rightarrow$$ (iii), (D) $$\rightarrow$$ (i)

Solution

Melamine formaldehyde polymer- (iv) melamine+formaldehyde
Bakelite - (i) phenol + formaldehyde
Neoprene - (ii) 2-chloro 1,3 butadiene (Chloroprene)
Natural rubber - (iii) 2-methyl 1,3 butadiene (isoprene)
Formation of phenol-formaldehyde polymer-
These are obtained by the condensation reaction of phenol with formaldehyde in the presence of either an acid or a base catalyst. The reaction starts with the initial formation of o-and/or p-hydroxymethyl phenols derivatives, which further react with phenol to form compounds having rings joined to each other through –$$CH_2$$ groups. The initial product could be a linear product like novolac.
The correct option is $$[C]$$.
Question 7
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Nylon-$$6$$ is prepared by linear condensation homopolymerization reaction of:

A
caprolactum

B
hexamethylene diamine and adipic acid

C
$$6$$-aminohexanoic acid

D
acrylonitrile

Solution

Solution:- (A) Caprolactam
Nylon-$$6$$ is prepared by linear condensation homopolymerization reaction of caprolactam.
$$n \left( \text{Caprolactum} \right) \xrightarrow[{N}_{2}]{533 \; K} \text{Nylon}-6$$
Question 8
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Which monomer is responsible for the flexibility property in $$PHBV$$?

A
$$HO -\underset {\underset {CH_{3}}{|\,\,\,\,}}{CH} - CH_{3} - COOH$$

B
$$HO - \underset {\underset {\,\,\,\,\,\,\,\,CH_{2} - CH_{3}}{|\,\,\,\,}}{CH} - CH_{2} - COOH$$

C
$$HO - \underset {\underset {CH_{3}}{|\,\,\,\,\,}}{CH} - CH_{2} - CH_{2} - COOH$$

D
$$HO - \underset {\underset {\,\,\,\,\,\,\,CH_{2} -CH_{3}}{|\,\,\,\,\,\,}}{CH} - CH_{2} - CHO$$

Solution

PHVB is a co-polymer. The monomers, 3-hydroxybutanoic acid and 3-hydroxypentanoic acid, are joined by ester bonds; the back bone of the polymer is made up of carbon and oxygen atoms. The property of the PHBV depends upon the ratio of these two monomers in it. 3-hydroxybutanoic acid provides stiffness while 3-hydroxypentanoic acid promotes flexibility.
therefore option B is correct.
Question 9
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Identify X and Y in the given polymerization reactions :

A
X = Bakelite, Y = Novolac

B
X = Novolac, Y = Melamine

C
X = Bakelite, Y = Malamine

D
X = Novolac, Y = Bakelite

Solution

Option D is correct.
Question 10
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The monomer that undergo radical polymerisation most easily is?

A
$$CH_2=CH_2$$

B
$$C_6H_5CH=CH_2$$

C
$$CH_2=C{(Me)}_2$$

D
$$CH_3-CH=CH_2$$