Below are the IR spectra of two plastics (A and B); one is PETE, the other is low density polyethene (LDPE).

Deduce, giving your reasons, the identity and resin identification code (RIC) of A and B using sections 26 and 30 of the data booklet.

Describe the difference in their structures.

Explain why the difference in their structures affects their melting points.

A RIC: 1 AND B RIC: 4

ALTERNATIVE 1:
«only» PETE contains carbonyl/C=O/ester/COO groups
carbonyl groups absorb at 1700–1750 «cm^–1»

ALTERNATIVE 2:
LDPE contains more C–H bonds «than PETE»
C–H bonds absorb at 2850–3090 «cm^–1»

For either, accept specific frequencies in these ranges (eg 1735 «cm^–1» or 2900 «cm^–1»).

[3 marks]

HDPE less branched
OR
LDPE more branched

Accept “no branching in HDPE AND branching in LDPE”.

[1 mark]

HDPE «polymer» chains/molecules can pack together more closely «than LDPE chains»
OR
HDPE «polymer» chains/molecules have a higher contact surface area «than LDPE chains»

stronger intermolecular/dispersion/London/van der Waals’ forces in HDPE AND higher melting point

Accept converse arguments.

[2 marks]