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: D
Question about geometry goes back to hybridisation and geometries of cycloalkanes  i is a cycloalkane with bond angles of 60^o, ii  is an alkyne so the central C is sp , i.e. linear, with bond angles of 180^o, and iii  is an alkene so the central C is sp², i.e. trigonal planar, with bond angles of 120^o, therefore giving : ii > iii > i

Qu2:  A
Just have to draw them out, no quick way.  Constitutional isomers means different numbers and types of bonds giving different functional groups and branching patterns. C₄H₁₀O is saturated, it could be 1-butanol, 2-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol, diethyl ether, methyl n-propyl ether, isopropyl methyl ether (7 isomers). C₅H₁₂ could be pentane, 2-methylbutane, 2,2-dimethylpropane (3 isomers). C₃H₆ is unsaturated it could be propene or cyclopropane (2 isomers), therefore i > ii > iii

Qu3: C
First you need to complete the formal charges, then apply the rules for ranking resonance structures.  In these three, the C, N and O all have complete octets  in all three structures.  ii is the major contributor since it has no formal charges.  In i the O is negative and N positive which matches the relative electronegativities whereas in iii the O is positive and N negative which opposes the relative electronegativities, so i is more important than iii. So overall we have ii > i > iii

Qu4: C
Calculate the formal charges for the C atom = group number - number of bonds - lone pairs electrons : i = neutral based on 4 - 3 - 1, ii = +1 based on 4 - 3  and iii = -1 based on 4 - 3 - 2. Therefore ii > i > iii

Qu5: AB
Acidity .... consider the general acidity equation HA <=> H+  A-.   Look at the factors that stabilise the conjugate base, A-.  All C-H bonds, but the hybridisation of the C atom changes as we go alkane to alkene to alkyne.  Think about how the lone pair of electrons that results in the conjugate base is stabilised by being in sp³ vs sp² vs sp hybrid orbital.  The critical factor is the % s character (25%, 33%, 50% respectively). More s character means a less diffuse orbital and puts the electrons closer to the positive nucleus and hence means there is more stabilisation. The greater the stabilisation of the conjugate base. So we have iii > ii > i

Qu6: AB
The easiest way to count up the valence electrons here is to use the molecular formulae of each compound and use the valence electron count for each element (H = 1, C =  4, N = 5, O = 6, F = 7) then remember to adjust the overall count to match the systems overall charge. This way you get i = 23 +1 = 24, ii = 26 -1 = 25, and iii = 26. Therefore  iii > ii > i

Qu7: B
Cycloalkane stability so need to think of ring strains.  Cyclohexane is the most stable system (lowest energy) so its DH_fwill be the most exothermic. Cyclopropane is highly strained and is the least stable, so its DHfwill be the least exothermic. This means we have  i > iii > ii

Qu8: A
Boiling point is controlled by intermolecular forces. For non-polar alkanes this is the weak induced dipole interactions. More branched structures tend to be more spherical in shape and hence have less surface areas in contact with other molecules and so less energy is required to separate them. Hence they have lower boiling points.  (Remember physical properties are not connected to thermodynamic stability which is controlled by the forces that hold the atoms in the molecules together).  Therefore here we have  i > ii > iii

Qu9: A
Either use the equation that relates energy of a photon to frequency E = h n and c = l n or just use the fact the frequency, n, is directly proportional to E, or use the fact that i corresponds to visible light, ii is a "classic" IR frequency for a C=O stretch and 60 MHz is the frequency of a low end H NMR in the radio frequency region. Converting to frequencies, 500 nm = 6 x 10¹⁴ Hz, 1715 cm^-1 corresponds to a wavelength = 5831 nm = 5.15 x 10¹³ Hz and 60 MHz = 60 x 10⁶ Hz.  Any of the methods should lead you to the energies being  i > ii > iii

Qu10: D
Look at the relative positions of the two alkyl group substituents. In i it is staggered and they are anti (180^o) , in ii it is eclipsed and they are syn (0^o)  and in iii it is staggered and the groups are gauche (60^o). Staggered conformations are more stable than eclipsed and strain is lowest when the biggest substituents are well separated. Hence  ii > iii > i

Qu11: B
From the extraction experiment. Seems a hard question needing math, but if you understand the principles, it's pretty straight forward. Given that K_D = 1 and equal volumes of solvents are used it means the relative amounts in the two layers are the same, so the amount extracted each time is half. Do two extractions means you extract 50% then 50% of 50% i.e. 25%, so the total amount extracted is 75%.  See, that wasn't too difficult was it ?

Qu12: B
From the physical properties experiment. Remember boiling points increase with increasing pressure so they are higher at sea level than here in Calgary at almost 4000ft above sea level. Rough rule of thumb for Calgary is 1^o for every 15^o above 50^o. Thus for 200^o we are talking about a 10^o increase.

Qu13: BC
The distillation experiment. Slow patient heating ensures a good thermal equilibrium is set up in the fractionating column to facilitate the separation process.  A longer column means there are more evaporate - condense "cycles" or theoretical plates meaning improved separation.  There is no cooling water in the fractionating column and cooling the column would not allow any separation as the vapours would reflux rather than reach the top of the fractionating column and enter the condenser.

Qu14: E
Simple or vacuum filtration may not remove the impurity, it would need to be soluble while the desired compound would need to be insoluble. Distillations are most commonly used for miscible liquids. Recrystallisation would mean that the impurity would remain in solution. Evaporation would not get rid of a non-volatile impurity.

Qu15: E
From the molecular models experiment.  1,3-dimethylbenzene the number of types of C are shown below in blue on the left and the H are shown below in green on the right. The red line shows the symmetry due to the mirror plane that bisects the molecule.

Qu17: C
C13 is attached to 1 x O = -1, 1 x H = +1 and 2 x C = 0 so the sum = 0, therefore C13 = 0

Qu18: CDE
First fill in the lone pairs on the heteroatoms one on each N and two on each O.  N1 has 2 attached groups and a lone pair = sp², C5 has 3 attached groups = sp², O14 has 2 attached groups and two lone pairs = sp³, N16 has 3 attached groups and a lone pair = sp³, and C21 has 4 attached groups (don't forget the 2H atoms) = sp³.

Qu19: CD
A tertiary C will have 3 other C atoms attached to it.  C2 has only 1 C attached,  C12 has no C attached,  C19 has 3 C attached,  C20 has 3 C attached and  C22 has 2 C attached.

Qu20: BD
Ortho positions are those adjacent to the substituent.

Qu21: AD
Alcohol, ROH at O14, no amides RCONH₂ or esters RCO₂R. Ethers, ROR at C7-O11-C12, and no phenol, ArOH.

Qu22: D
Count the rings and the p bonds to get 4 rings plus 6 p bonds = 10

Qu23: E
Since all are CC bonds we need to consider the hybridisation of the atoms involved and the type of bond. A double bond will be shorter than a single bond due to the increased bonding interaction. C8-C9 is sp² to sp² (benzene so between C-C and C=C in character), C4-C13 is a single bond sp² to sp³, C13-C15 is a single bond sp³ to sp³, C20-C23 is a single bond sp³ to sp², C23-C24 is a double bond sp² to sp². Therefore C23-C24 will be the shortest.

Qu24: D
All the other options are examples of stereoisomers that have the same groups and bonds and differ only on how they are connected spatially.

Qu25: C
A and D are both cis.  B is trans-1,2-. So C or E ? Put the larger tBu group in the more favourable equatorial position rather than the axial position.

Qu26: A or B
Credit was given for either A or B as with hindsight, the question is potentially ambiguous depending on whether you decide to look at the worst single 1,3-diaxial interaction  (B) or worst total 1,3-diaxial interactions (A). Larger substituents have a strong perference to be equatorial because of unfavoutable Van der Waals strain with the other axial groups when they are axial themselves.

Qu27: A
It shows a staggered conformation where the two methyl groups are at 180^o, so anti is the best term to use here. B syn means the two methyl groups are at 0^o, C gauche means the two methyl groups are at 60^o, D  eclipsed is the more general description when the three bonds off the front carbon are aligned with the three off the rear carbon, E cis and AB trans are used to describe a pair of substituents on a C=C or in a cyclic system.

Qu28: C
See the images in the etext

Qu29: C
There are 10 C atoms in the decalin system, each with one axial and one equatorial substitutent. The two F atoms in this molecule are both axial and there are no other axial substituents other than H atoms. So 10 - 2 = 8.

Qu30: E
Longest chain is C6, contains a C=C so a hexene. If alkene is the main group, then its a 2-ene. Dimethyl substituted amino group hence N,N-dimethylamino, located at C4.  Given that the C substituents on the C=C out rank the H atoms, the higher priority groups are on opposite sides of the alkene so it is an E configuration.

Qu31: C
Alcohol group at C1, then use first point of difference to get the methyl group at C3, ethyl at C4. Remember to alphabetise.

Qu32: B
An ester not an ether, so it can't be A or C. The carbonyl defines the acid "half" of an ester, so here, where it is attached to the benzene ring, we are looking at a benzoate ester.  In this case the straight chain in the other half, the alcohol "half", means we have the n-butyl ester. Thus overall we have n-butyl benzoate.

Qu33: A
The alkene has an E configuration (C > H and CºC > CH₃). The longest chain is C6, so a hexenyne (note order alkene - alkyne). Since the alkyne is terminal it is numbered as "1" so we have a hex-3-en-1-yne... now just locate the two methyl groups at C3 and C5.

Qu34: B
Complex substituents in an alkane... locate longest chain, here C11, identify substituents, number based on first point of difference (i.e. 5,6- rather than just 6,7- diethyl groups). The complex substituent has a C2 chain (so ethyl) with two methyl group at C1 so 1,1-dimethylethyl located at C8 on the main chain.... Remember to alphabetise....prefixes such as "di-" don't count normally, but they do if they are part of the bracketed term as in complex substitutents, hence "dimethyl" preceeds "ethyl" and then "methyl" so B is correct and D has an error in the alphabetisation. We could consider naming as a tert-butyl (e.g. as per C) but this name is incorrect based on the first point of difference rule for the locants (it would have needed to be 8-tert-butyl-5,6-diethyl-4-methylundecane).

Qu35: D
Meta implies we have 1,3-substituents so that restricts us to C, D or E. A and C are amines, B and D ares amides and E has a nitro group.

Qu36: B
Use the descriptors trans- and (1,3) and look at the position of the two methyl groups.... A cis-(1,3)-, B trans-(1,3)-, C trans-(1,2)-, D trans-(1,4)-, and E is trans-(1,4).

Qu37: B
C and D are not carboxylic acids. The others are stereoisomers.  The relative priority of the groups (Cahn-Ingold-Prelog rules) at the chirality center (4 different groups attached) is : OH > C=C > CH₂ > H.  A and E where the sense of rotation of the three higher priority groups is counter clockwise are therefore both S.  B is R (remember to put the lost priority group, the H, at the back when determining the sense of rotation).

Qu38: A
Did you look at the nomenclature of bicyclics ? Bicyclic means two fused rings. The [3.2.1] means that there are 3C, 2C and 1C in the links between the shared (bridgehead) C atoms. This gets rid of C, D, and E which are [2.2.1], [2.2.2] and [2.2.2] respectively. We number from one of the shared C atoms via the longest link first so as to give the alkene the lowest number.

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