Qu1: E
Boiling point is a physical property and is therefore controlled by intermolecular forces. The system with the strongest intermolecular forces will have the highest boiling point. Therefore since the ether iii has a polar group that can participate in dipole-dipole interactions, which are stronger than induced dipole - induced dipole intermolecular forces, iii will have the highest boiling point. If we look at i and ii then we see two isomeric alkanes, but they have different branching. More branching makes the alkane more spherical and decreases the surface areas in contact - this means less intermolecular forces and therefore a low boiling point. Hence iii > i > ii.

Qu2: AB
Stability of alkanes in terms of heats of formation. More branched alkanes are more stable so iii is the most stable and i is the least stable. The more stable the alkane the more exothermic the heat of formation. Therefore iii > ii > i

Qu3: C
Radical chlorination of alkanes is strongly influenced by the number of H within each subtype. The easiest way to evaluate this is by counting the number of H in each type and multiply that by the appropriate reactivity factor.... for i that means 2 x 3.9 = 7.8, ii 9 x 1 = 9 and iii 3 x 1 = 3, so the yield of ii > i > iii.

Qu4: D
Each type of C in the structure gives a peak in the 13C NMR so just count the number of types of C in each structure. 1,2-dichlorobenzene has 3, 1,2-bromochlorobenzene has 6 and biphenyl iii has 4. So ii > iii > i.

Qu5: C
Acidity.... First let's compare i and ii (because they are very similar, both are carboxylic acids) : the difference between them is the Cl atom... the presence of the electronegative Cl will stablise the conjugate base due to inductive electron withdrawal through the sigma bonds, so in terms of acidity, ii > i. Now let's compare iii and i, i is a carboxylic acid and iii is an ammonium salt derived from an amine (which are bases)... in terms of general principles, OH systems are more acidic than NH systems due to the electronegativity of the O compared to the N. Other things that would help answer this question are the solubility experiment and knowing the pKa's (carboxylic acid are about 5, ammonium salts about 10)... carboxylic acids protonate amines and therefore carboxylic acids are more acidic than ammonium salts. Therefore in terms of acidity, ii > i >iii.

Qu6: E
Basicity...either think about the availability of the electrons in the base or the stability of the bases. Here, they are all anions, but the nature of the atom bearing the charge is changing. First let's compare i and ii because O and S are in the same group of the periodic table. Since S is larger than O it is better able to accommodate the -ve charge, so the O- is more basic than the S-. Now let's compare i and iii because O and N are in the same row of the periodic table. The electrons on N- are more available than those on the O- because the O is more electronegative than the N. So in terms of basicity iii > i > ii 

Qu7: C
Assign the formal charges...i a carbon with 4 bonds so it is neutral, ii is an nitrogen with 4 bonds so it is +1 and iii is an oxygen with 1 bond and 3 lone pairs so it is -1, so ii > i > iii.

Qu8: B
Ranking resonance structures... ii is the least important because it has a C atom without a complete octet and has one less bond than the other two. iii is has charge separation and hence is less important than i. : i > iii > ii.

Qu9: B
i is an sp² CH, ii is a tertiary sp³ CH and iii is a primary sp³ CH. Due to the s character in the hybrids, sp³ CH are weaker than sp² CH. Primary CH bonds are stronger than tertiary CH bonds (based on data discussed in the context of either radical or alkane stability) so i > iii > ii.

Qu10: C
The index of hydrogen deficiency (IHD) is a count of the number of pi bonds and rings. i has a ring and an aldehyde which contains a C=O giving IHD = 2. ii is phenol, so 3 C=C and one ring has IHD = 4 units. iii C₂Cl₄ drawn out gives Cl₂C=CCl₂ so there is one C=C and an IHD = 1. So in terms of IHD ii > i > iii.

Qu11: E
The index of hydrogen deficiency (IHD) is a count of the number of pi bonds and rings. Here there are 2 rings and 6 pi bonds to give a total of 8.

Qu12: B
O17 is a simple alcohol oxygen and so it has 4 groups attached (1xC atom, 1xH atom and 2 lone pairs) and is sp³. O25 is part of an amide C=O and so it has 3 groups attached (1xC atoms and 2 lone pairs) and is sp².

Qu13: C
C10 has 4 groups attached, (2xC, 2xH atoms) so it is sp³ hybridised and therefore essentially tetrahedral with bond angles of close to 109.5 degrees.

Qu14: E
Given that we are looking at H attached to atoms all from the same row of the periodic table, then as a general trend, more acidic H are attached to more electronegative atoms, so the alcohol involving O30 is more acidic that the amide NH for N26.

Qu15: C
All the bonds are C-H bonds, but the hybridisation of the C atoms involved is changing. Less s character in the hybrid orbitals will mean a longer and weaker bond so sp3 C-H bonds are weaker than sp2 C-H bonds so we can rule out B and D. Comparing the three sp3 systems, we see that we have two that are secondary (A and C) and one that's primary (E)... primary C-H are stronger than secondary (the neighbouring alkyl groups weaken the secondary)... so the difference between A and C is that C is also next to a benzene ring (i.e. benzylic) and the resonance interaction further weakens the C-H bond (the resonance stabilises the resultant radical).

Qu16: D
All the bonds are C-C bonds, but the hybridisation and character of the C atoms involved is changing.... double bonds are shorter than single bonds sowe can rule out A and E which are both C-C. B and C both involve sp2 C in aromatic rings while D is an alkene. The resonance in the benzene rings mean those C=C are half way between C=C and C-C so they are longer than simple alkene C=C as in D.

Qu17: B
N26 is part of an amide so the N atom is sp² hybridised to allow the lone pair on the N to be in a p orbital so that it can be involved in a stablising resonance interaction with the adjacent C=O group.

Qu18: AC
Functional groups....

Qu19: B
C1 is attached to two other C atoms and is therefore classed as secondary.

SPECTROSCOPY:
Use the IR spectra provided to get the functional groups that are present in the structures, so look for the key groups : C=O (near 1700 cm^-1) , -OH or -NH (above 3000 cm^-1), aromatic C=C (two bands 1600-1400 cm^-1), C-O (near 1200 cm^-1) and triple bonds (near 2200 cm^-1).... use the tables and then apply simple logic...

Qu20: BC
The IR shows an OH stretch at just above 3300 cm^-1, but there is no C=O or C=C. This suggests the aliphatic alcohol BC.

Qu21: E
The IR shows an NH₂ stretch above 3100 cm^-1, and a C=O near 1660 cm^-1, and Ar C=C at 1626 and 1580 cm^-1, these suggest the aromatic amide E.

Qu22: B
The IR shows a broad OH stretch centered around 3000 cm^-1, and a C=O near 1700 cm^-1, and Ar C=C near 1600 and 1585 cm^-1, these suggest the aromatic carboxylic acid B.

Qu23: C
The IR shows a C=O near 1725 cm^-1, Ar C=C near 1600 cm^-1 and a C-O near 1300 and 1100 cm^-1 these suggest an ester C.

Qu24: D
The IR shows a broad OH stretch centered around 3200 cm^-1, and Ar C=C near 1600 cm^-1, but there is no C=O near 1700 cm^-1. These suggest the aromatic alcohol, i.e. phenol D.

Qu25: A
The IR shows a triple bond near 2230 cm^-1, but no alkyne C-H at 3300 cm^-1 so it is not the alkyne AB and has to be the nitrile A.

NOMENCLATURE:
Two types of questions here, however it is a good idea to use key descriptors like E and Z, R or S, cis or trans, numbering, functional group etc. to help screen the answers before getting too tied up in the words. For the first four, name the compound and then match it to the list, for the last four use the key descriptors and check each molecule in turn. So this section really depends on the nomenclature rules.

Qu26: C
The first point of difference rule requires that the system be numbered as a 1,2,4 system. Alphabetisation means that the ethyl group is named before the two methyl groups since the substituent multiplier is ignored when alphabetising in these simple systems.

Qu27: A
There is no first point of difference rule here, so the locants for the alkyl groups are decided based on alphabetical order.....which means that the ethyl group is named before the methyl group.

Qu28: A
The complex substituent = (1,2,2-trimethylpropyl) since the longest chain in the substituent starting at the attachment point is C3 with one methyl group on C1 (the attachment point) and two methyl groups are on the second carbon out from that point. The principle functional group here is the alkene so it is numbered as C1and C2 (need to number both alkene C as low as possible in this example). The first point of difference rule then requires that the substituent is at C3 by numbering counter clockwise.

Qu29: A
The longest chain is 7, so it's a hept system, the principle functional group is the alkene, and it has an amino substituent. To give the alkene the lowest number we number from right, giving a 4-aminohept-3-ene. To designate the C=C stereochemistry, use the Cahn-Ingold-Prelog rules : the CH₃CH₂ > H and the NH > alkyl chain so it's Z stereochemistry.

Qu30: D
Phenyl = C₆H₅- and isopropyl = (CH₃)₂CH- ... so D. Only A, B and D are ethers. A is benzyl isopropyl ether and B is t-butyl phenyl ether.

Qu31: D
Only A, C and D are esters. A and D are ethyl esters, B is a methyl ester. A is dichloro.

Qu32: A
Only A and B are 2-methyl.... -1-ols.... Use the Cahn-Ingold-Prelog rules to assign the configuration. The group priorities are -OH > C=C > C-C > H so A is and B is R.

Qu33: E
This is an example of a bridged hydrocarbon, all are 2.2.1 systems. Numbering starts at a bridgehead C then goes to an alkene (2-ene) around the longer branches first. A is 2,6-dimethyl, B is 2,3-dimethyl, C is 5,6-dimethyl and D is 1,6-dimethyl.

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