Pentanedioyl dichloride, ClOC(CH2)3COCl, is reacted with butane-1,4-diamine, H2N(CH2)4NH2, to form a polyamide. Per repeat unit of the polymer formed, how many molecules of hydrogen chloride are eliminated, and how many amide bonds are formed?

Answer: B. Step 1: Count the reactive groups on each monomer Pentanedioyl dichloride, ClOC(CH2)3COCl, has two acyl chloride groups, -COCl, one at each end. Butane-1,4-diamine, H2N(CH2)4NH2, has two amine groups, -NH2, one at each end. Both monomers a…

A2 Chemistry 9701: HCl Elimination in Polyamide Formation from a Diacyl Chloride

Step 1: Count the reactive groups on each monomer

Pentanedioyl dichloride, $\text{ClOC}(\text{CH}_2)_3\text{COCl}$ , has two acyl chloride groups, $-\text{COCl}$ , one at each end. Butane-1,4-diamine, $\text{H}_2\text{N}(\text{CH}_2)_4\text{NH}_2$ , has two amine groups, $-\text{NH}_2$ , one at each end. Both monomers are difunctional.

Step 2: Form the amide bonds

Each $-\text{COCl}$ group reacts with an $-\text{NH}_2$ group, forming an amide bond and eliminating one molecule of hydrogen chloride:

$-\text{COCl} + \text{H}_2\text{N}- \;\longrightarrow\; -\text{CO-NH}- + \text{HCl}$

Because a repeat unit is built from one diacyl chloride molecule joined to one diamine molecule, this reaction happens twice per repeat unit, once at each junction, so two amide bonds form, and two molecules of $\text{HCl}$ are eliminated per repeat unit.

Why the other options are wrong

A and D only account for one of the two reactive-group pairs, missing that both ends of each monomer react.
C correctly counts two HCl molecules eliminated but incorrectly counts only one amide bond. Each HCl eliminated corresponds to exactly one amide bond formed, so the two counts must match.

Final answer

B, 2 molecules of HCl eliminated; 2 amide bonds formed, giving repeat unit $\left(-\text{NH}(\text{CH}_2)_4\text{NH-CO}(\text{CH}_2)_3\text{CO}-\right)_n$ .