NCEA Level 3 Chemistry 91391 (3.5) — page 4 of 8

SAMPLE ASSESSMENT SCHEDULE

Chemistry 91391 (3.5): Demonstrate understanding of the properties of organic compounds

Assessment Criteria

Achievement / Achievement with Merit / Achievement with Excellence
Demonstrate understanding involves naming using IUPAC conventions (no more than eight carbons in the longest chain), drawing structural formulae of organic compounds, and giving an account of their physical properties and / or reactivity. This requires the use of chemistry vocabulary, symbols, and conventions. / Demonstrate in-depth understanding involves making and explaining links between structure, functional groups, physical properties, and reactivity of organic compounds. This requires explanations that use chemistry vocabulary, symbols, and conventions. / Demonstrate comprehensive understanding involves elaborating, justifying, relating, evaluating or comparing and contrasting the links between the structure, functional groups, physical properties and / or reactivity of organic compounds. This requires the consistent use of chemistry vocabulary, symbols, and conventions.

Evidence Statement

One / Expected coverage / Achievement / Merit / Excellence
(a) / Ethanoyl chloride
2–chlorobutanoic acid
1-amino -methylpropane / TWO names correct.
(b) (i) / 3-D structure drawn, eg
/ ONE isomer drawn correctly with 3D arrangements of groups around chiral carbon
OR
in (b) (i), the isomers drawn are mirror images but an error in the way the groups are connected to chiral carbon. / Correct 3D arrangements drawn, which are mirror images
(b) (ii) / Structures of P and Q drawn, eg:

P has a chiral carbon, which contains a carbon with four different groups attached. / Structure of P or Q drawn.

/ Explanation of chiral carbon linked to structure of P.

(c) (i)
(c) (ii) /

Dipeptides contain the CO–NH (peptide / amide) linkage. This forms when the amine group of one molecule reacts with the carboxylic acid group of the other molecule. Two dipeptides are possible as either the COOH from glycine can react with the NH2 from serine, or the NH2 from glycine can react with the COOH from serine. / ONE dipeptide drawn correctly
OR

peptide link / An explanation of the TWO dipeptides formed, linked to the structures of the amino acids.
(d) / The amide link is hydrolysed in both acid and basic conditions.
In acid conditions the product is:

The acid will form a salt with the amine group, NH2.
In basic conditions the product is:
/ ONE product of hydrolysis drawn. (Products drawn but does not recognize salt formation.) / Product of hydrolysis drawn with acid OR base, with limited reasons. / Structures of BOTH hydrolysis products are justified in terms of the reaction of the molecule with acid and with base.
Not Achieved / NØ / No response; no relevant evidence.
N1 / Candidate provides some accurate statements without answering any question completely.
N2 / Candidate provides any ONE statement for Achievement.
Achievement / A3 / Candidate provides any TWO statements for Achievement.
A4 / Candidate provides any THREE statements for Achievement.
Merit / M5 / Candidate provides any ONE statement for Merit.
M6 / Candidate provides BOTH statements for Merit.
Excellence / E7 / Candidate provides statement for Excellence AND ONE Merit statement.
E8 / Candidate provides statement for Excellence AND TWO Merit statements.
Two / Expected coverage / Achievement / Merit / Excellence
(a) / Reagant: Alcoholic KOH.
Reaction and reason: This is an elimination reaction as a hydrogen atom and a chlorine atom on adjacent carbon atoms are removed, forming a carbon-to-carbon double bond. / KOH (alcoholic) used
OR
reason for elimination reaction given.
(b) (i) / Dilute solutions of Cr2O72-/H+ (or MnO4-/H+)
will not react.
CH3– CH2– CH2 –OH will react with Cr2O72-/H+ and will turn from orange to green (or MnO4-/H+ will turn from purple to colourless).
The product formed is:
/ Observation of ONE reaction with chosen reactant (out of TWO possible reactions given)
OR
two correct reagents (allow oxidation to aldehyde). / ONE pair of compounds distinguished (with chosen reactant and observations for BOTH compounds) with product.
(b) (ii) / Add water
will not react.
reacts vigorously with an exothermic reaction.
The product formed is CH3– COOH.
(c) /
Compound W
This compound is oxidised by Cr2O72-/H+ to an acid so it has to be a primary alcohol (or an aldehyde – rejected as too many H atoms). The OH has to be on an end C. It’s a branched chain molecule so there is a CH3 on the 2nd C atom.
Hence:

Compound X
This is formed when

is oxidised to an acid.
Hence:

Compound Y
A substitution reaction occurs with SOCl2; the OH in:

is replaced with Cl.
Hence:

Compound Z
A substitution reaction occurs with CH3NH2; the Cl is replaced with CH3NH2.
Hence:
/ The functional groups of TWO out of FOUR compounds given
OR
one reason for a functional group given. / TWO correct structures with reasons given. (Error in W may be carried through for these two structures.) / Structures correct with justification for the functional groups and reactions given.
Not Achieved / NØ / No response; no relevant evidence.
N1 / Candidate provides some accurate statements without answering any question completely.
N2 / Candidate provides any ONE statement for Achievement.
Achievement / A3 / Candidate provides any TWO statements for Achievement.
A4 / Candidate provides any THREE statements for Achievement.
Merit / M5 / Candidate provides any ONE statement for Merit.
M6 / Candidate provides BOTH statements for Merit.
Excellence / E7 / Minor errors (eg omission or inaccuracies) from the Excellence criteria.
E8 / Candidate provides ALL the evidence from the Excellence criteria.
Three / Expected coverage / Achievement / Merit / Excellence
(a) / / TWO products correct.
(b) / CH3– CH2– CO– CH3 with Cr2O72-/ H+
No reaction occurs.
CH3– CH2– CO– CH3 with NaBH4
Product is CH3– CH2– CHOH– CH3
CH3– CH2– CHOH– CH3 with concentrated H2SO4
Products are CH3– CH2– CH= CH2 (minor) and
CH3– CH= CH– CH3 (major) / TWO products correct.
(c) / Butanal /aldehydes can be oxidised using Cr2O72-/ H+ (Reactant 1) to form a carboxylic acid, butanoic acid. However, butanone/ketones cannot be oxidised.
Both butanal/aldehydes and butanone/ketones can be reduced using NaBH4 (Reactant 2) to form alcohols. Aldehydes form primary alcohols and ketones form secondary alcohols.
With concentrated H2SO4 alcohols form alkenes.
The primary alcohol CH3– CH2– CH2 – CH2OH forms only one product as the H and OH atoms on adjacent carbon atoms are replaced with a carbon-to-carbon double bond.
The secondary alcohol CH3– CH2– CHOH– CH3 forms a mixture of products as OH is on the 2nd carbon atom. Either the 1st H is removed, forming
CH3– CH2– CH= CH2 (minor) or the 3rd H is removed, forming CH3– CH= CH– CH3 (major). / ONE similarity OR ONE difference identified. / ONE similarity AND ONE difference explained in terms of the structures or functional groups. / Reaction scheme of butanal and butanone shows both similarities with NaBH4 and differences with both Cr2O72-/ H+ and concentrated H2SO4 including major and minor products.
Not Achieved / NØ / No response; no relevant evidence.
N1 / Candidate provides some accurate statements without answering any question completely.
N2 / Candidate provides any ONE statement for Achievement.
Achievement / A3 / Candidate provides any TWO statements for Achievement.
A4 / Candidate provides any THREE statements for Achievement.
Merit / M5 / Candidate provides ONE similarity OR ONE difference, explained in terms of the structures/functional groups.
M6 / Candidate provides ONE similarity AND ONE difference, explained in terms of the structures/functional groups.
Excellence / E7 / Minor errors (eg omission or inaccuracies) from the Excellence criteria.
E8 / Candidate provides ALL the evidence from the Excellence criteria.