Here is an post-mortem analysis / "how to" for the MT. The questions are split by the sections. At the start of each section are a few suggestions of what to look for or how to tackle the question type.

Qu1: AB
The more highly branched the more stable the isomeric alkane is. This is due to the increased intramolecular forces between the branches within the molecule. Therefore the stability order is iii > ii > i.

Qu2: E
iii represents the most important structure as all atoms have a complete octet. In fact it represents the structure you should draw directly from the name.  ii has a resonance structure derived by using the C=C and so has charge separation. i can also be derived by pushing curly arrows from the C=C but now all the way out onto the O, putting the -ve charge on a more favourable electronegative atom. So the order is iii > i > ii.

Qu3: C
Solubility is controlled by intermolecular forces, where "like dissolves like". Here, the more the solute is polar like water, the more soluble it will be in water. So the short chain alcohol, ii, ethanol will be soluble in water. The longer chain alcohol will be less stable due to the increased contribution of the hydrophobic alkyl chain. However the hydrocarbon, pentane lacks any polar group and will be insoluble in water.  Therefore ii > i > iii.

Qu4: AB
Here we have the chair, boat (capsized !) and the half chair. The chair is the most stable conformation of cyclohexane (lowest energy) followed by the boat then the half chair which is the highest energy conformation of cyclohexane. Thus the order is iii > ii > i.

Qu5: AB
All about ring strain. For 6 membered rings and less, the smaller the ring, the higher the ring strain and the higher the energy of the system per -CH2- implying the more endothermic the heat of formation.  Therefore the order is iii > ii > i.

Qu6: B
Consider the location of the -ve charge in the conjugate base. In the conjugate base of  i there is resonance stabilisation of the  O anion with the charge being delocalised to a second O. In iii the charge is fixed on a single O, so i is more acidic than iii. In ii the charge would be fixed on N but N is less electronegative than O so the N stabilises the -ve charge less. Therefore i > iii > ii.

Qu7: AB
Conformational analysis. Ethane i has only a single eclipsed conformation. ii has two with the methyl group eclipsing either OH or H.  iii has 3 (methyl,methyl with methyl,-OH), (methyl,-OH with methyl,H) and (methyl,methyl with methyl,H). So the order is iii > ii > i

Qu8: D
All about geometry. The angles are about 104^o for water, 120^o for sp2 C in ethene and 109.5^o for sp3 C in methane. Therefore ii > iii > i.

Qu9: B
No ! Slowly to ensure accuracy between the thermometer and the metal block.

Qu10: B
Depends on the density of the solvent..

Qu11: A
Since boiling occurs when the vapour pressure equals atmospheric pressure, increasing the applied pressure will require that the vapour pressure be higher to boil and that will require a higher temperature.

Qu12: A

Qu13: A

Qu14: A
See distillation experiment.

Qu15: B
From extraction experiment... but the solvents can't mix they have to be immiscible.

Qu16: D
Work out the moles of each reagent : aminophenol = 2.18g /109g/mol = 0.02 moles and acetic anhydride = 1.08g/ml x 1.51ml /102g/mol = 0.016 moles and is therefore the limiting reagent.

Qu17: D
1.81g / 151g/mol = 0.012 moles.

Qu18: B
1.51g/151g/mol = 0.01 moles therefore yield = 0.01/0.016 = 62.5%

Qu19: C
C9 is attached to 1 electronegative atoms, 2C and 1H so oxidation state = -(-1 + 0 +1) = 0

Qu20: B
C13 is involved in 3 bonds to electronegative atoms and 1 C, so oxidation state = -(-3 + 0) = +3

Qu21: CE

Qu22: B
O7 is sp² since it is directly involved in a pi system.

Qu23: B
O12 is sp² so that it can be involved in resonance with the adjacent carbonyl pi system.

Qu24: ADE
recognise those functional groups !

Qu25: E
Need to count the pi bonds and the rings, each represents a unit of unsaturation since each requires that there be one H₂ less that the corresponding non-cyclic alkane, (pi + r) = 5 + 2 = 7

Qu26: A
Anti is best since not only does it imply the staggered conformation but also that the two methyl groups are at 180^o with respect to each other.

Qu27: D
Ankle ? Angle is a pair of connected bonds, ring is the extra strain associated with a cyclic structure and torsional is due to the repulsion between a pair of bonds.

Qu28: D
A is not a Newman projection the others all are but only D shows the anti conformation.

Qu29: D
Canonicals refers to resonance structures. Constitutional implies different bonding patterns (eg functional groups or branching). Configurational require bond breaking to interconvert. Conformational can interconvert by rotation about sigma bonds. Isotopes are atoms with the same number of protons but different numbers of neutrons in the nucleus.

Qu30: C
A is torsional strain, B is ring strain, D is torsional strain, E is angle strain.

Qu31: C
A is 1,4-, B is 1,3- trans but has the larger group in the less favourable axial position, D is 1,3-cis and E is 1,3-cis (more favoured than D).

Qu32: D
methyl groups are trans, and numbered 3,4 with respect to the ketone

Qu33: E
An amide based on C4 chain with two methyl groups at C2 and C3 with respect the important C=O

Qu34: C
An ethyl ester of a substituted C4 acid.

Qu35: B
The methoxy group is on C1 of the alkene with the complex alkyl substitutent.

Qu36: B
Only B and D are ketones, one is E, the other D is Z.

Qu37: E
Look for C6 with a single Cl on C4 with respect to the C=O of the acid.....

Qu38: A or B (the same : ( )
Looking for a C4 aldehyde with an amine with two methyl groups on the N and an ethyl group adjacent to the aldehyde CHO.

Qu39: B
The tough one ! A is monocyclic not bicyclic so forget that. The [2.2.0] indicates the number of C atoms in the sides of the cyclic units. C = [3.1.0], D = [2.2.0] but the relative position of the C=C is not 2,5 (need to start at a bridgehead C atom) and E is not a C6 system.

Return to Homepage