Qu1: AB
Question about geometry goes back to hybridisation and geometries of cycloalkanes  i is cyclopropane with bond angles of 60^o, ii  is cyclopentane, bond angles are a little bit less than the tetrahedral optimum of  109^o, and iii  is cyclohexane where the bond angles are also close to those of a perfect tetrahedra, and are about 111^o, therefore giving : iii > ii > i

Qu2:  AB
All about hybridisation. Alkyne C have two attached groups and so are sp hybridised, a 1 : 1 mix of s and p, so 50% p character.  Alkene C have three attached groups and so are sp2 hybridised a 1 : 2 mix of s and p, so 66% p character.  Alkane C have four attached groups and so are sp3 hybridised a 1 : 3 mix of s and p, so 75% p character.  Therefore % p character is  iii > ii > i

Qu3: D
Acidity .... Know your pKa's or work it out ?  If you need to work it out, then consider the general acidity equation HA <=> H+  A-.   Look at the factors that stabilise the conjugate base, A-.  First notice that we have two O-H systems and a C-H system.  ii is a carboxylic acid (pKa = 5) where the conjugate base has the negative charge on the electronegative O atom  and resonance can further stablise it to the other O atom - remember spreading charge out (delocalisation) is a stabilising effect.  iii is an alcohol (pKa = 16), this time the conjugate base will also have the negative charge on the electronegative O atom - but the lack of a pi system means there is no further resonance stabilisation.  Finally i , a ketone (pKa = 19) - the initial conjugate base puts negative charge on a less electronegative C atom, but it can also be resonance stabilised to the O atom.  So overall we have ii > iii > i

Qu4: D
Calculate the formal charges for the O atom = group number - number of bonds - lone pairs electrons : i = -1 based on 6 - 1 - 6, ii = +1 based on 6 - 3 -2  and iii = 0 based on 6 - 2 - 4. Therefore ii > iii > i

Qu5: 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. Remember to ignore configurational isomers such as R / S or E / Z (cis- / trans) issues.  C₃H₆O is unsaturated, it could be propanal, propanone, 1-propen-1-ol,  1-propen-2-ol, 3-propen-1-ol, cyclopropanol, oxacyclobutane and methyloxacyclopropane. (7 isomers). C₄H₈ is unsaturated it could be 1-butene, 2-butene, 2-methylpropene, cyclobutane or methylcyclopropane (5 isomers). C₅H₁₂ could be pentane, 2-methylbutane, 2,2-dimethylpropane (3 isomers). Therefore  i > ii > iii

Qu6: E
Remember the rules for ranking resonance structures  : complete octets are most important.... iii has complete octets at C,N and O despite the charges, and note it has one more pi bond than the other two.  As for the other two, the negative charge on the more electronegative O atom is more important.  So iii > i > ii

Qu7: D
The more branched an alkane isomer is the more stable it is (maximises number of stronger primary CH bonds). Hence in terms of stability ii > iii > i

Qu8: C
Count the bonds attached to the C, each H counts +1, C counts 0 and a bond to a more electronegative atom (e.g. O or N) counts -1. Total the count and then  since the atoms are neutral, just switch the sign since the oxidation state for the C plus the groups attached must equal 0.  In i the C is attached to C (count 0) and H (count +1) therefore oxidation state C = -1. In ii the C is attached to N (count -1), two bonds to O (count -1 per bond) and H (count +1) therefore oxidation state C = +2.   In iii the C has two bonds to C (count 0 per bond) and H (count +1 per bond) therefore oxidation state C = -2.  Therefore  ii > i > iii

Qu9: A
Ring strain decreases the stability per -CH2- unit as the cyclic structure gets smaller for this series of cycloalkanes. So cyclobutane is the least stable and therefore has the most exothermic heat of combustion per CH2 unit. Hence least exothermic is cyclohexane then cyclopentane then cyclobutane or  i > ii > iii

Qu10: D
The chair is the most stable (lowest energy) conformation of cyclohexane, the boat is an energy maxima. The twist boat is more stable than the boat (some relief of eclipsing strain) so  ii > iii > i

Qu11: D
From the extraction experiment. Seems a hard question needing complex math, but if you understand the principles, it's really pretty straight forward. Given that K_D = 1 and the solvent volumes are in the ratio 2 : 1 so there will be twice as much in the water layer as the extracting dichloromethane layer i.e. 2 : 1 ratio. This means that each dichloromethane extraction will extract 1/3 of the total.  So the first extraction takes out 33% than the second takes 1/3 of the remaining 66% = 22%, therefore in total 55% will have been removed.

Qu12: D
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 140^o we are talking about a 6^o increase.

Qu13: AC
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.  A cooled receiver flask would prevent the loss of any volatile (low boiling) fractions.

Qu14: D
Filtrations are for removing solids from liquids.  Simple distillation will only work if there is a large difference in the boiling points of the liquids, fractional distillation is more likely to ensure a good separation.  Recrystallisation is for the purification of solids and evaporation is used to remove volatile liquids from solids.

Qu15: A
The halogenation experiment was based on the radical reaction of these systems and the reason the t-butyl system didn't react is that it doesn't have an H in the right place to form a stable radical.  If you didn't look this up or ask about it after not doing well on that laboratory report, you only have yourself to blame. Part of being at University is about being an independent learner and having the desire to locate and learn. 

Qu17: B
C1 is attached to 3 x C = 0 so the sum = 0, therefore C1 = 0

Qu18: C
The group in question is C-O-C so it's an ether.

Qu19: D
The group in question is -C(=O)-O-C so it's an ester.

Qu20: D
Unsaturation arises due to pi bonds or rings.  So count these up in the structure:  there are 4 C=C and 5 C=O,  4 rings (3 x  6 membered and 1 x  29  membered).

Qu21: A
Any atom is this structure that is involved in a double bond is sp2.  O18 and N19 are considered sp2 because the O or N is involved in resonance with the adjacent pi system.  C9 is sp3.

Qu22: A
Apply the Cahn-Ingold-Prelog rules at each chirality center.  C7 priority order :  -C=C > CH2C > CH3 > H.  C9 priority order :  -C=O > CH2C > CH3 > H.  C11 priority order :  -OCH3 > -C=O > CH(OH)C  > H.  C12 priority order : OH >  CH(OCH3)C > -C=C > H.  C15 priority order :  -C=O > -C=C > CH3 > H. 

Qu23: E
Apply the Cahn-Ingold-Prelog rules at each C=C.  In each of the 4 cases, the higher priority groups are on opposite sides of the C=C unit, so all the cases are E isomers.

Qu24: D
The two terms that apply to cyclohexane locations are axial and equatorial. In the structure shown the Cl is equatorial (around ring) and the ethyl group is axial (away). Chair is the name of the conformation shown.  Anti and gauche are used to describe torsional angles. 

Qu25: C
Eclipsed would require a zero torsional angle. Here we have a staggered conformation, torsional angle of the methyl systems = 60^o so gauche. Axial relates to  cyclohexanes.   Anti means a 180^o torsional angle and syn a 0^o torsional angle.

Qu26: AC
The most stable conformations of these cyclohexane systems will have the larger groups equatorial in a chair conformation.  A has both methyl groups equatorial whereas B has the unfavourable diaxial situation (ring flip gives A). C has the larger t-butyl group equatorial rather than the methyl group. D is a boat conformation and E has the larger group, the isopropyl group, in the less favourable axial position so a ring flip will give a more stable conformation.

Qu27: D
A and B are applied only to alkenes and cyclic structures.  C constitutional isomers have different branching patterns or functional groups .... that's not true here.  D is correct, you can "make" the right Newman projection by rotating the front C by 120 degrees.  E they have the same molecular formula so they are isomers.

Qu28: B
The small ring of cyclopropane means there is a lot of torsional strain due to the alignment of all the C-H bonds.

Qu30: D
Longest chain is C10, contains a C-C only so a decane.  This means it has to be C, D or E. C can't be right because you can't have a 1,1,1-trimethyl system (in fact any 1-alkyl group would imply a longer chain is possible). E is wrong because the methyl groups are wrongly located.  D is the answer.... IUPAC indicates that maximise the branches is the way to go as that leads to the simplest names for the substituents.

Qu31: C
Alcohol group at C1 (priority functional group), then use first point of difference to get the methyl group at C3, ethyl at C5. Remember to alphabetise ignoring the "di" so ethyl before methyl.

Qu32: C
The compound is an amide, -C(=O)-N  not an amine - the carbonyl group makes a difference. So it can only be C or D.  The group that is attached to the nitrogen is a benzyl group (C₆H₅CH₂-) NOT a phenyl group (C₆H₅-). It is indicated by the N-benzyl description.

Qu33: C
Alkene stereochemistry as described by E and Z. The longest chain including the two C=C is 6 carbons so we need a hexadiene. This limits our choice to C or D. Numbering to give the first C=C the lowest possible number we have an ethyl group at C2 and a methyl at C5. The C3-C4 double bond has Z stereochemistry because the higher priority groups (the C chains) are on the same side of the double bond.   Hence we have (Z)-2-ethyl-5-methyl-1,3-hexadiene = C.

Qu34: A
An ester.... narrows the choices to A or E.  B is an ether and C and D are incomplete names.  Since it's a methyl ester it must be A.  In terms of assigning configurations, the group order is -Br > CH2CO2CH3 > CH3 > H.  Remember to assign the sense of the rotation to the order of the groups when the H is away from you.... then it's counter-clockwise = S.

Qu35: B
Ortho implies we have a pair of 1,2-substituents so that restricts us to A or B. Methoxy = CH3O- so B.  A is ortho-bromomethylbenzene, C is meta-bromomethylbenzene, D is meta-bromomethoxybenzene and E is 3-bromo-2-methylphenol.

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

Qu37: A
The -al ending indicates an aldehyde (RCHO) which limits the choices to A, B and E.  C = alcohol and D = ketone. Since the question asks for a pent system we need C5, so that rules out E which is C6.  If we assign configurations, the group priority order is OH > C=C > CH2CHO > H ....the low priority H is already back in A so the sense of the priority groups is counter-clockwise = S.

Qu38: D
Did you look at the nomenclature of bicyclics ? You were supposed too ! Bicyclic means two fused rings. The [2.2.2] means that there are 2C, 2C and 2C in the links between the shared (bridgehead) C atoms. This gets rid of A, B and C which are [3.2.1], [3.2.1] and [2.2.1] respectively. Then we number from one of the shared C bridgehead atoms via the longest link first so as to give the methyl group (first point of difference) the lowest number.

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