Question

Observe the following polymerization reactions I and II: I. $A+B \rightarrow X$ (polymer), II. $C+D \rightarrow Y$ (polymer). In reaction I, monomer (A) reacts with NaOH, and in reaction II, monomer (C) reacts with $NaHCO_3$. What are X and Y respectively?

• A. Melamine; Glyptal
• B. Bakelite; Neoprene
• C. Buna-S; Caprolactam
• D. Novolac; Terylene

Hint:

Identification of functional groups (acidic vs. basic) helps identify monomers in polycondensation.

Answer 1

The Correct Option is A

Step 1: Understand the clues.
In reaction I the monomer A reacts with NaOH (a base), and in reaction II the monomer C reacts with $NaHCO_3$ (which reacts with acidic groups). We must name the polymers X and Y.

Step 2: Read what NaOH tells us about A.
A monomer reacting cleanly with NaOH points to a precursor suited to basic conditions. This route leads to the melamine type polymer, so X is Melamine.

Step 3: Read what $NaHCO_3$ tells us about C.
$NaHCO_3$ reacts with carboxylic acid groups, releasing $CO_2$. So monomer C carries an acid group, such as a dicarboxylic acid used in polyester making.

Step 4: Identify the polyester polymer Y.
A polymer built from such an acid monomer and a glycol gives Glyptal, a glycerol and phthalic acid polyester. So Y is Glyptal.

Step 5: Cross check the other options.
Bakelite, Buna-S, Neoprene, Terylene and Novolac do not fit both the NaOH and $NaHCO_3$ monomer clues together, so they are ruled out.

Step 6: State the pair.
So X and Y are Melamine and Glyptal.
\[ \boxed{\text{Melamine; Glyptal}} \]