| Date | May 2017 | Marks available | 1 | Reference code | 17M.2.hl.TZ2.7 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Outline | Question number | 7 | Adapted from | N/A |
Question
A compound with a molecular formula C7H14O produced the following high resolution 1H NMR spectrum.
Deduce what information can be obtained from the 1H NMR spectrum.
Identify the functional group that shows stretching at 1710 cm–1 in the infrared spectrum of this compound using section 26 of the data booklet and the 1H NMR.
Suggest the structural formula of this compound.
Bromine was added to hexane, hex-1-ene and benzene. Identify the compound(s) which will react with bromine in a well-lit laboratory.
Deduce the structural formula of the main organic product when hex-1-ene reacts with hydrogen bromide.
State the reagents and the name of the mechanism for the nitration of benzene.
Outline, in terms of the bonding present, why the reaction conditions of halogenation are different for alkanes and benzene.
Below are two isomers, A and B, with the molecular formula C4H9Br.
Explain the mechanism of the nucleophilic substitution reaction with NaOH(aq) for the isomer that reacts almost exclusively by an SN2 mechanism using curly arrows to represent the movement of electron pairs.
Markscheme
Number of hydrogen environments: 3
Ratio of hydrogen environments: 2:3:9
Splitting patterns: «all» singlets
Accept any equivalent ratios such as 9:3:2.
Accept “no splitting”.
[3 marks]
carbonyl
OR
C=O
Accept “ketone” but not “aldehyde”.
[1 mark]
Accept (CH3)3CCH2COCH3.
Award [1] for any aldehyde or ketone with C7H14O structural formula.
[2 marks]
hexane AND hex-1-ene
Accept “benzene AND hexane AND hex-1-ene”.
[1 mark]
CH3CH2CH2CH2CHBrCH3
Accept displayed formula but not molecular formula.
[1 mark]
Reagents: «concentrated» sulfuric acid AND «concentrated» nitric acid
Name of mechanism: electrophilic substitution
[2 marks]
benzene has «delocalized» \(\pi \) bonds «that are susceptible to electrophile attack» AND alkanes do not
Do not accept “benzene has single and double bonds”.
[1 mark]
curly arrow going from lone pair/negative charge on O in –OH to C
curly arrow showing Br leaving
representation of transition state showing negative charge, square brackets and partial bonds
Accept OH– with or without the lone pair.
Do not allow curly arrows originating on H in OH–.
Accept curly arrows in the transition state.
Do not penalize if HO and Br are not at 180°.
Do not award M3 if OH–C bond is represented.
Award [2 max] if wrong isomer is used.
[3 marks]
