Chemistry

1. 1.

   A compound's infrared spectrum has a strong absorption near 1700 cm^-1. State a likely bond and explain why this alone does not identify the compound.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Infrared absorption identifies a vibration consistent with a bond type. Aldehydes, ketones and acids can all contain that bond, so one peak narrows possibilities but does not select one structure.

   Marking points

   • A carbonyl C=O bond is likely.
   • Several functional groups contain carbonyl bonds.
   • Other spectral/compositional evidence is needed for identity.

   Examiner tip: Do not infer an aldehyde specifically from a carbonyl peak alone.

2. 2.

   Explain why but-2-ene can have E/Z isomers but but-1-ene cannot.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Restricted rotation is necessary but insufficient. Distinct substituents on both carbons make alternative arrangements distinguishable; two identical hydrogens erase that distinction at one end.

   Marking points

   • Rotation about a C=C bond is restricted.
   • Each double-bond carbon in but-2-ene has two different substituents.
   • The terminal carbon in but-1-ene has two identical hydrogen substituents.

   Examiner tip: Check both alkene carbons, not just one.

3. 3.

   Ignoring OH splitting and assuming fast proton exchange, predict the proton NMR splitting of the CH3 and CH2 groups of ethanol and explain their relative integration.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Use the n+1 rule on adjacent nonequivalent proton sets. Integration counts the protons producing a signal, whereas splitting counts relevant neighbours, so the two numbers have different roles.

   Marking points

   • CH3 gives a triplet from two neighbouring CH2 protons.
   • CH2 gives a quartet from three neighbouring CH3 protons.
   • CH3:CH2 integration is 3:2.

   Examiner tip: Do not use the signal's own proton count for n+1.

4. 4.

   Two compounds have formula C3H6O. One gives a positive Tollens test and the other does not. Identify the aldehyde and ketone possibilities, and give an additional spectral distinction.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Both structures contain a carbonyl and share the formula, but only the aldehyde is oxidised by Tollens reagent under the test conditions. Proton NMR can independently detect the CHO proton.

   Marking points

   • The aldehyde possibility is propanal.
   • The ketone possibility is propanone.
   • Propanal has an aldehydic proton signal absent from propanone.

   Examiner tip: The formula permits other structures; these identifications are the requested aldehyde/ketone possibilities.

5. 5.

   A C4H8O2 compound shows an ester carbonyl, no broad O-H absorption, and proton NMR signals: singlet at 2.1 ppm (3H), quartet at 4.1 ppm (2H), triplet at 1.3 ppm (3H). Propose a structure and account for each signal.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The quartet at 4.1 ppm indicates OCH2 in an ethyl group, while the singlet at 2.1 ppm fits an acyl methyl; methyl propanoate would instead have an OCH3 singlet further downfield. Joining them through an ester satisfies both the formula and functional-group evidence.

   Marking points

   • A consistent structure is ethyl ethanoate, CH3COOCH2CH3.
   • CH3CO gives the 3H singlet with no adjacent carbon-bound hydrogens.
   • OCH2 gives the 2H quartet coupled to terminal CH3.
   • Terminal CH3 gives the 3H triplet coupled to CH2.

   Examiner tip: Use integration and splitting together; a triplet alone does not identify an ester.

6. 6.

   Explain why a racemic mixture of a chiral alcohol has no net optical rotation, and why that observation alone cannot prove a sample is racemic.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Zero rotation is a net measurement. It can arise from cancellation or from no optically active molecules in the first place, so it cannot distinguish those explanations unaided.

   Marking points

   • Enantiomers rotate plane-polarised light equally in opposite directions under the same conditions.
   • Equal quantities cancel their rotations.
   • An achiral substance can also show no rotation.
   • Identity/chirality evidence or enantiomer separation is required to establish a racemate.

   Examiner tip: A racemate is a mixture of enantiomers, not a single achiral molecule.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.