Qu1: C
We need to compare C=C with C-C and C-H. If we look at the CC bonds first, a C=C will be shorter than a C-C since there is more bonding character in the C=C than the C-C. If we compare the C-C with the C-H then the C-H will be shorter because H is a smaller atom than C. We also need to know that the small size of the H atom is a more significant effect than the bond type change in the CC series. The actual values are C=C = 1.34 A, C-C = 1.53 A and C-H = 1.11 A. Hence ii > i > iii

Qu2: A
Relates to hybridisation. i is an vinyl CH with C = sp², therefore 33% s character, while ii and iii are both alkane CH, C = sp³, therefore 25% s character. Increasing the s character in the hybrid orbital makes the orbital smaller and creates a stronger interaction with the H atom and so a stronger bond so i is the strongest CH bond. Now how do we separate ii and iii? iii is a secondary CH bonds whereas ii is a primary CH bond. Primary CH bonds are stronger (from alkane stability or radical laboratory expt). Therefore i > ii > iii

Qu3: A
More branching means less surface area in contact and therefore less intramolecular forces and hence lower boiling point. Therefore i > ii > iii

Qu4: C
Basicity...either think about the availability of the electrons in the base or the stability of this base. If we look at the acids we have a carboxylic acid, an ammonium type system or HBr. We need to know that HBr is a very strong acid, a mineral acid, and these are stronger than carboxylic acids. The large Br atom means it can easily accept the -ve charge. Now for the ammounium compared to the carboxylic acid. The resonance stabilised carboxylate is more stabilised the amine if we compare the conjugate bases (this shows the importance of resonance). Or we can just use the pKa's of these acids : 5, 10 and -9 respectively). Therefore in terms of basicity, since the stongest acid has the waekest conjugate base, we have ii > i > iii

Qu5: C
The stronger the interaction of the orbital the stronger the bond and the lower the energy of the orbital. The CC s bond is the strongest, then the CC p bond. The CC p* is a high energy antibonding orbital, so ii > i > iii

Qu6: A
The trend is dominated by ring strain.The most stable is the cyclohexane then the cyclopentane then the cyclobutane. The ring strain effect is more significant than the branching. So i > ii > iii.

Qu7: E
Draw each of them out in their lowest energy conformations. i has an equatorial isopropyl group and an axial methyl group. ii has both groups equatorial and iii has 2 equatorial and 2 axial methyl groups.... therefore since substitutents generally prefer to be equatorial, ii is most stable and hence the lowest energy. Since iii has 2 axial groups it is less stable than i. Note that the axial interactions are more significant than the branching effects. (FYI : the calc. heats of formation are: -56.6, -60.8 and -47 kcal/mol for i-iii respectively). Hence in terms of relative energies iii > i > ii.

Qu8: D
Acidity....First let's consider dissolving the amino acid in water.... when that happens the carboxylic acid group will protonate the amine group. The pKa's tell you this or look at qu 4 above. This means iii forms when proline is dissolved in water. If we acid 1 equivalent of HCl (a strong acid to this), then the HCl will cause the carboxylate to protonate to give ii. Therefore ii > iii > i.

Qu9: A
Basicity...either think about the availability of the electrons in the base or the stability of the bases. The stronger the bases, then the more the reaction shown moves to the right. These are all oxygen systems, an alkoxide, a phenoxide and a carboxylate respectively. Really it all about the resonance. In the carboxylate iii, the -ve charge is delocalised to a second electronegative oxygen atom whereas in the phenoxide ii, the resonance can only delocalise the charge to carbon atoms and in the alkoxide i there is no resonance and the charge is fixed on the single oxygen. Therefore in terms of base strength and hence the amount of reaction:   i > ii > iii

Qu10: B
Oxidation states...count the bonds attached to each of the atoms to be considered. A bond to a more electronegative atoms counts -1, a bond to the same type of atom counts 0 and a bond to a less electronegative atom (e.g. H) counts +1. Total the count and then consider the formal charge on the central atom since the oxidation state for the central atom plus the groups attached must equal the atoms formal charge. In i the C is attached to 1 O (count - 1), 1H (count +1) and 2 Cl (count -2) therefore oxidation state C = +2. In ii the N is attached to C twice (count +2) and 1 bond to O (count -1) therefore oxidation state N = -1. In iii the C has 4 bonds to C (count 0 per bond) therefore the oxidation state C = 0. Therefore i > iii > ii

Qu11: CD
The theoretical yield in g will be changed by altering the amounts of the reagents used. The presence of solvent or other impurities will alter the actual yield. Changing temperature has no effect on the theoretical yield only the actual yield.

Qu12: A
The structure of a material does not change when it melts.

Qu13: C
A % yield calculation. Reaction stoichiometry is 1:1. Aminophenol has MW = 109 g/mol so 1.05 g = 9.6 mmol. The ethanoic anhydride has MW = 102 g/mol and we used 1mL. using the density of 1.08g/mL, this means we are using 1.08g and hence 10.59 mmol. Therefore the aminophenol is the limiting reagent. We get 1.05g of product, MW = 151, this is 6.95 mmol. Hence the yield is 6.95 mmol / 9.6 mmol = 72%

Qu14: D
Based on the molecular models experiment.... there are 5 types of H and 6 types of C in the acetaminophen structure. The benzene ring has a mirror plane through the substituted para carbons so there are 4 aromatic C plus the carbonyl and the methyl group. The 5H types are 2 aromatic H, OH, NH and the methyl CH.

Qu15: E
The most efficient method for purifying the solid product will be recrystallisation. TLC and melting point are analysis methods. However TLC can be used for small scale purifications (mg scale). Distillations are used for purifying liquids. The other items are simple procedures.

Qu16: E
A, B and C are false because the Norit (charcoal) was not added until the purification step. It would not be involved in the reaction itself in any manner. D is false : anhydrous ionic salts are the normal drying agents such as sodium or magnesium sulfate.

Qu17: E
Oxidation states
O36 has the following bonds : 1 x C (count +1), it has a formal charge = -1, therefore the oxidation state of O36 = -2.

Qu18: B
Oxidation states
C41 has the following bonds : 3 x C (count 0) and 1 x O (count -1), it is a formal charge = 0, therefore the oxidation state of C41 = +1.

Qu19: E
A primary amine has one alkyl group attached = R-NH2 : so that is N52. Note that N6 is a secondary amine. N10, 18 and 32 are amides

Qu20: A
Basicity...all N atoms so we need to look at lone pair availability. Amines are more basic than amides because the N lone pair of amides is involved in resonance with the C=O. So it's down to the two amines, N6 and N52. More alkyl groups tend to make the amine more basic so the secondary amine is more basic than the primary amine.

Qu21: D
Highest pKa means the weakest acid. Therefore the C-H systems are most likely since the C is less electronegative than N. If we look at C25 we can see that it is adjacent to a carbonyl so the conjugate base will be resonance stabilised which makes C25 more acidic. C50 is a simple alkane type CH2.  

Qu22: D
Hybridisation. N18 is part of an amide, therefore it is sp2 hybridised and therefore trigonal planar so the bond angle is about 120 degrees.

Qu23: D
Hybridisation. O36 is part of a carboxylate ion so there is resonance stabilisation of the -ve charge. This requires that the O be sp2. N52 is part of a simple primary amine and therefore has 4 groups attached including the lone pair and therefore it is sp3.

Qu24: C
Absolute configuration means chirality and assigning R or S at the chirality center. The 4 groups around C25 in priority order are N, C=O, C=C and H. With the low priority group away from you, the sense of the priority of 1-3 is clockwise = R.

Qu25: D
Absolute configuration means chirality and assigning R or S at the chirality center. The 4 groups around C51 in priority order are N, C-O, C-C and C-H. With the low priority group away from you, the sense of the priority of 1-3 is counterclockwise = S.

Qu26: E
The functional group in question is C-O-C so it's an ether.

Qu27: D
The functional group in question is O=C-O- so it's a carboxylate.

Qu28: C
The best description is gauche. Since in this staggered conformation - the two CH3-C C-Cl bonds have a 60 degree torsional angle. The term staggered is not the best term because there is another staggered conformation where the CH3-C and C-Cl torsional angle would be 180 degrees (the anti conformation).

Qu29: B
The torsional angle between the methyl groups is 60 degrees for a pair of adjacent equatorial substituents on a cyclohexane.

Qu30: B
The best description of the relative position is trans. Anti and gauche are used to describe conformation produced by torsional rotation (e.g. review butane). Chair is the most stable cyclohexane conformation but says nothing about the substituents. Axial (or equatorial) are cyclohexane substituent positions but you can have two axial groups that are cis.

Qu31: A
The highest energy (i.e. least stable) conformation will be eclipsed (A or D) rather than staggered (B, C, E). The least stable conformation will have the most eclipsing interactions. In D there are 2 x CH3 / H eclipsing interactions while in A there is one CH3 / H plus one CH3 / CH3. Since a CH3 is larger than an H, the CH3 / CH3 interaction is stronger than the CH3 / H and therefore A is less stable than D

Qu32: E
Cyclohexane itself is regarded as being strain free. Given that the two substituents here are equatorial, then the system is still strain free.

Qu32: ABD
Cyclopropanes are planar and have eclipsed bonds, so there is torsional strain. Due to the nature of the ring, there is also angle strain since the 60 degree angle of the CCC is a lot lower than the 109.5 degrees required for an sp3 hybridised atom. In cis-1,2-dimethylcyclopropane, the two methyl groups are syn to each other. This outs them in to close proximity and therefore there is also Van der Waals strain.

Qu34: D
In order for K to be 1, then the two conformations need to be of equal energy and that would require the equatorial and axial substituents to be very similar (or the same!).

Qu35: A
The ring is C6, contains a C-C only so we have a cyclohexane with 2 methyl groups and an ethyl group. B and D have incorrect alphabetisation : ethyl is before dimethyl. 
C and E have incorrect numbering based on the difference rule.

Qu36: A
Longest chain is C5, including a C=C so we have a pentene with a methyl group substituent.  Numbering dictated by the carboxylic acid as C1. Alkene stereochemistry as described by E and Z. Stereochemistry of the alkene is Z. B is wrong because the stereochemistry is wrong.  C and D are wrong because the longest chain is not C4, D also has the wrong stereochemistry.  E is wrong the double bond locant is incorrect.  

Qu37: B
The compound is named as a substituted alcohol (it's the priority group) - the chain including the -OH group is C3, and the OH is on C2. Hence it's a 2-propanol. The substituent is s cyclopentyl group attached to the C2 of the propane unit. A and E have the prop- and pent- parts of the names mixed up. C and D have the wrong cyclopentyl locant, plus C has the wrong suffix for an alcohol.

Qu38: A
This alkene can not be E or Z because the end of the alkene has two identical groups attached. The ketone carbonyl needs to be numbered as low as possible, C3. The longest chain is C6 and we have a C=C.... so we have a substituted 4-hexene-3-one. The methyl groups are at C2 and C5.

Qu39: D
Benzylamine implies that we have a C6H5CH2- group attached to a nitrogen in an amine. A and B are amides. C and E are phenyl NOT a benzyl systems.

Qu40: E
Use the descriptors cis- and substituent positions (1,3) and look at the position of the two alkoxy groups vs alkyl groups.... A trans-(1,3)-, B trans-(1,3)-, C cis-(1,3)-dialkyl and D is cis-(1,2).

Qu41: A
The name is a pentanone so its a ketone and we need a C5 chain including the C=O and an amine group. The amine needs a methyl substituent.... that rules out D and E.  In terms of assigning configurations, the group order is -N > C=O > CH(CH3)2 > H.  Remember to assign the sense of the rotation to the order of the groups when the H is away from you...B and C are both R.

Qu42: A
Did you look at the nomenclature of polycyclics? This is a bridged system. The [2.2.1] means that there are 2C and 1C in the links between the common C atoms. This gets rid of D and E which are [2.2.2]. Note that the heptene means we have 7C in the parent ring structure, the bicyclo means two rings, the ene means we need C=C.  Then we number from a bridgehead C bearing in mind the first point of difference rule. B is 8-methylbicyclo[2.2.1]hept-2-ene. C is 5-methylbicyclo[2.2.1]hept-2-ene and D is B is 2-methylbicyclo[2.2.1]hept-2-ene .

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