All the bonds indicated are CC bonds. We have C=C and 2 C-C. One C-C is sp2C-sp3C then other is sp3C-sp3C. The increased s character of sp2 makes that bond stronger (v1=A, v2=D)

Ranking resonance structures...we need to check for complete octets and maximised bonding (within the octet rule limitations) and then for charge separation in accord with electronegativity. (v1=AB, v2=AB)

Basicity...Think about the availability of the electrons in the bases, always a good idea to draw in the lone pairs because lone pairs are more available than bonded pairs. We have N systems. +ve N does not have a lone pair only bonded pairs, so it's not very basic. In the neutral N systems, one N is sp3 the other sp. A lone pair in an sp hybrid orbital has greater s character and is closer to the +ve nucleus which makes it less available than an sp3 N. (v1=C, v2=E)

Qu 4:

Radical stability of this set is affected by (i) degree of substituion and (ii) resonance. From most stable to least stable we have secondary allylic > primary allylic > primary (v1=A, v2=C)

Each different monochlorination product comes from a different type of H in pentane. We can do the calculations considering the number of H of each type and the relative rectivity of each position. For the 1-chloro isomer, it would be 6x1, for 2-chloro 4 x 3.9 and 3 chloro = 2 x 3.9 so % 2-chloropentane > % 3-chloropentane > % 1-chloropentane. (v1=D, v2=C)

Acidity...the structures are an amine NH, CH adjacent to a carbonyl in a ketone and a CH of a methy group on an internal alkyne.
The CH adjacent to the C=O in the ketone is made more acidic because the H are adjacent to the C=O so they give a conjugate base where the anion is resonance delocalised to allow the negative charge on to the electronegative oxygen atom (pKa ketone alpha-CH about 20). In the amine, the conjugate base will have -ve charge on an electronegative N atom, some resonance delocalisation to the adjacent aromatic system (pKa about 27 for aniline). The internal alkynes can be deprotonated to give a resonance stabilised carbanion (pKa = 45). So, in terms of acidity ketone > amine > internal alkyne (v1=C, v2=A)

MOLECULAR PROPERTIES:
No real method here, really just do you know various aspects of molecular structure and apply it to the molecule(s) in question.

Acidity...the most acidic functional group in the molecule is the phenol (typical pKa = 10) an -OH system with some resonance stabilistion of the conjugate base. (v1=A, v2=D)

O1 is part of a phenol where the O lone pairs interact with the adjacent pi system so the O is sp2 and C14 is part of a simple alkene so it is also sp2. (v1=D, v2=A)

N21 = is part of an amine, O8 is part of a ether (v1= AE, v2= BD)

The chirality center at C9 (where there are 4 different groups attached to the sp3 C) and it has the R configuration (O > C(OOC) > C(CCC) > H). Remember that the low priority group has to be away when determining the sense of rotation. The C23=C24 is terminal (=Ch2) and therefore does not have stereochemistry. (v1= E, v2= E).

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 as needed.

The IR shows a strong C=O near 1735 cm-1 (which would be highl for a simple ketone), no -OH above 3000 cm-1. Careful inspection suggests C-O at about 1250 and 1050 cm-1 supporting the ester. (v1=E, v2=C)

The IR does not shows a C=O near 1700 cm-1. The IR shows a medium double band near 3500-3350 cm-1 that could be an -NH₂. On closer inspection, the C-H stretches are > 3000 cm-1 indicating sp2 C-H only and there are possible aromatic C=C near 1600 and 1500cm-1.. This would be consistent with the aromatic primary amine. (v1=AE, v2=BC)

The IR shows a strong double C=O near 1810 and 1750cm-1 (which would be very high for a simple ketone, probably an anhydride), no -OH above 3000 cm-1. This would be consistent with acyl anhydride. (v1=C, v2=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.

The longest chain is C8 so an octane. There are two substituents, an ethyl group and the t-butyl or 1,1-dimethylethyl group. Since the alphabetical order is wrong for the t-butyl group, we need to need the complex version and alphabetise it based on the "d" (first letter in the brackets). The first point of difference does not resolve the numbering so it is dictated by the alphabetical order. Therefore for we 4-(1,1-dimethylethyl)-5-ethyloctane (v1=C, v2=C)

A substituted cyclohexene. Need to number to make the alkene C1 & C2. In this case, the first point of difference rule dictates needing to give the methyl group the lower number and then need to list the substituents alphabetically (ethyl before methyl) ignoring the di prefix. Hence 5-ethyl-1-methylcyclohex-1-ene. (v1=E, v2=E)

The longest chain including the principal functional group (a ketone) is C6, so we have a hexenone system. The C=O is the higher priority functional group and gets the lower number of C3 with the trans stereochemistry double bond at C4-C5 and a methyl group at C4. Therefore we have trans-4-methylhex-4-en-3-one (v1= E, v2= B)

Naming of simple alkyl groups : here we save an sec-butyl (v1 = D, v2 = C)

Substituted benzenes and ester naming. The C=O (acid) side of the ester is based on benzoic acid. The ortho methyl group needs to be on the ArC next to the C=O on the acid side of the ester and then an ethyl group on the alcohol side of the ester N (v1 = D, v2 = E)

R/S stereochemistry
Chirality center is marked with a *. Priority of groups there is O > C(C=C, CCH) > C(C, CHH) > H. Also need to remember to view from the back so that the low priority group is away when assigning R or S. (v1 = C, v2 = D)

The bicyclic system needs to have 8C (oct) and an alkene since it is an octene. The [3.2.1] indicates the number of C atoms in each link between the central C atom. Need to start to number from the bridgehead C then round the larger ring first with the alkene adjacent to the bridgehead. (v1 = D, v2 = C)