Question 22M.2.hl.TZ2.8e(i)
| Date | May 2022 | Marks available | [Maximum mark: 4] | Reference code | 22M.2.hl.TZ2.8e(i) |
| Level | hl | Paper | 2 | Time zone | TZ2 |
| Command term | Draw, Explain | Question number | e(i) | Adapted from | N/A |
Carbon forms many compounds.
Explain the mechanism of the reaction between 1-bromopropane, CH3CH2CH2Br, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.
[4]
curly arrow going from lone pair/negative charge on O in HO– to C ✔
curly arrow showing Br breaking ✔
representation of transition state showing negative charge, square brackets and partial bonds ✔
formation of organic product CH3CH2CH2OH AND Br– ✔
Do not allow curly arrow originating on H in HO–.
Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition
state.
Do not penalize if HO and Br are not at 180° to each other.
Award [3 max] for SN1 mechanism.
As usual, good to excellent candidates (47.5%) were able to get 3/4 marks for this mechanism, while most lost marks for carelessness in drawing arrows and bond connectivity, issues with the lone pair or negative charge on the nucleophile, no negative charge on transition state, or incorrect haloalkane. The average mark was thus 1.9/4.
