Question 19M.2.HL.TZ2.6
| Date | May 2019 | Marks available | [Maximum mark: 9] | Reference code | 19M.2.HL.TZ2.6 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate, Compare and contrast, Draw, Identify, Outline | Question number | 6 | Adapted from | N/A |
Phenylethene can be polymerized to form polyphenylethene (polystyrene, PS).
The major product of the reaction with hydrogen bromide is C6H5–CHBr–CH3 and the minor product is C6H5–CH2–CH2Br.
Draw the repeating unit of polyphenylethene.
[1]
[✔]
Note: Do not penalize the use of brackets and “n”.
Do not award the mark if the continuation bonds are missing.
Most candidates were able to draw the monomer correctly. Some candidates made careless mistakes writing C6H6.
Phenylethene is manufactured from benzene and ethene in a two-stage process. The overall reaction can be represented as follows with ΔGθ = +10.0 kJ mol−1 at 298 K.
Calculate the equilibrium constant for the overall conversion at 298 K, using section 1 of the data booklet.
[2]
ln k «= » = –4.04 [✔]
k = 0.0176 [✔]
Note: Award [2] for correct final answer.
Another calculation which most candidates were able to work out, though some failed to convert ΔG given value in kJ mol-1 to J mol-1 or forgot the negative sign. Some used an inappropriate expression of R.
The benzene ring of phenylethene reacts with the nitronium ion, NO2+, and the C=C double bond reacts with hydrogen bromide, HBr.
Compare and contrast these two reactions in terms of their reaction mechanisms.
Similarity:
Difference:
[2]
Similarity:
«both» involve an electrophile
OR
«both» electrophilic [✔]
Difference:
first/reaction of ring/with NO2+ is substitution/S«E» AND second/reaction of C=C/with HBr is addition/A«E» [✔]
Note: Answer must state which is substitution and which is addition for M2.
The strong candidates were generally able to see the similarity between the two reactions but unexpectedly some could not identify “electrophilic” as a similarity even if they referred to the differences as electrophilic substitution/addition, so probably were unable to understand what was being asked.
Outline why the major product, C6H5–CHBr–CH3, can exist in two forms and state the relationship between these forms.
Two forms:
Relationship:
[2]
Two forms:
chiral/asymmetric carbon
OR
carbon atom attached to 4 different groups [✔]
Relationship:
mirror images
OR
enantiomers/optical isomers [✔]
Note: Accept appropriate diagrams for either or both marking points.
Candidates were given the products of the addition reaction and asked about the major product. Perhaps they were put off by the term “forms” and thus failed to “see” the chiral C that allowed the existence of enantiomers. There was some confusion with the type of isomerism and some even suggested cis/trans isomers.
The minor product, C6H5–CH2–CH2Br, can exist in different conformational forms (isomers).
Outline what this means.
[1]
benzene ring «of the C6H5–CH2» and the bromine «on the CH2–Br» can take up different relative positions by rotating about the «C–C, σ–»bond [✔]
Note: Accept “different parts of the molecule can rotate relative to each other”.
Accept “rotation around σ–bond”.
If candidates seemed rather confused in the previous question, they seemed more so in this one. Most simply referred to isomers in general, not seeming to be slightly aware of what conformational isomerism is, even if it is in the curriculum.
The minor product, C6H5–CH2–CH2Br, can be directly converted to an intermediate compound, X, which can then be directly converted to the acid C6H5–CH2–COOH.
C6H5–CH2–CH2Br → X → C6H5–CH2–COOH
Identify X.
[1]
C6H5–CH2–CH2OH [✔]
Quite well answered though some candidates suggested an aldehyde rather than the alcohol, or forgot that C has two hydrogens apart from the -OH. In other cases, they left a Br there.