Question 17N.2.hl.TZ0.8
| Date | November 2017 | Marks available | [Maximum mark: 8] | Reference code | 17N.2.hl.TZ0.8 |
| Level | hl | Paper | 2 | Time zone | TZ0 |
| Command term | Deduce, Draw, Explain, State | Question number | 8 | Adapted from | N/A |
The reactivity of organic compounds depends on the nature and positions of their functional groups.
The structural formulas of two organic compounds are shown below.
Deduce, giving a reason, which of the two compounds can show optical activity.
[1]
A AND it has a chiral centre/asymmetric carbon atom/carbon with 4 different substituents
Draw three-dimensional representations of the two enantiomers.
[1]
Accept structures without tapered bonds.
State the reagents used in the nitration of benzene.
[1]
concentrated HNO3 AND concentrated H2SO4
“concentrated” must occur at least once (with either acid).
State an equation for the formation of NO2+.
[1]
HNO3 + 2H2SO4 H3O+ + NO2+ + 2HSO4–
Accept: HNO3 + 2H2SO4 NO2+ + HSO4– + H2O.
Accept: HNO3 + 2H2SO4 H2NO3+ + HSO4–.
Accept single arrow instead of equilibrium sign.
Accept equivalent two step reactions in which sulfuric acid first behaves as strong acid and protonates nitric acid, before behaving as dehydrating agent removing water from it.
Explain the mechanism of the reaction between 2-bromo-2-methylpropane, (CH3)3CBr, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.
[4]
curly arrow showing Br– leaving
representation of tertiary carbocation
curly arrow going from lone pair/negative charge on O in –OH to C+
formation of (CH3)3COH AND Br–
Do not accept curly arrow originating from C of C–Br bond.
Do not accept arrow originating on H in –OH.
Accept Br– anywhere on product side in the reaction scheme.
Award [2 max] for an SN2 type mechanism.