Qu1:E
Acidity...i is a ketone, the most acidic H are those adjacent (a) to the C=O where they give a carbanion that is resonance delocalised to allow the negative charge on to the electronegative oxygen atom (pKa about 20). ii is an allylic system, the most acidic H will be those in the methyl group where again there will be resonance delocalisation of the carbanion, but only to other carbon atoms (pKa about 45) iii is an alcohol where the most acidic proton is the one attached to the electronegative oxygen atom (pKa about 15). Hence in terms of acidity iii > i > ii.

Qu2: A
Heats of formation.... more branched alkane isomers are more stable and more stable isomers have more negative heats of formation. Based on the branching, iii is the most stable and i is the least stable. Therefore i has the least exothermic heat of formation and iii the most exothermic thus: i > ii > iii

Qu3: D
Basicity...either think about the availability of the electrons in the base or the stability of the bases. Here, they are all carboxylate anions, but the number of Br atoms adjacent to the carboxylate group is changing. The electronegativity of the Br means it is capable of stablilising the -ve charge by inductive electron withdrawal, the more Br the better the stabilisation and the weaker the base so in terms of basicity, ii > iii > i 

Qu4: E
Increased intermolecular forces will raise the boiling points. iii is an alcohol so there will be hydrogen bonding, the strongest of the intermolecular forces and therefore the highest boiling point. Of the two isomeric alkanes, the more linear structure will have the most surface area in contact (hence more induced dipole interactions). So iii > i > ii.

Qu5: B
Ranking resonance structures... i is the most important because the C and both O atoms all have complete octets. iii is has charge separation but all the atoms still have complete octets. ii has charge separation against electronegativity and violates (exceeds) the octet rule on the negative O (= impossible) and hence is less important than iii so i > iii > ii.

Qu6: A
i is an sp CH, ii is an sp² CH (vinyl) while iii is a primary sp³ CH. Due to the s character in the hybrids, sp³ CH are weaker than sp² CH which in turn are weaker than sp CH, so in terms of bond strengths, i > ii > iii.

Qu7: D
Assign the formal charges...i an oxygen with 1 bond and 3 lone pair so it is -1, ii is an oxygen with 3 bonds and 1 lone pair so it is +1 and iii is an oxygen with 2 bonds and 2 lone pairs so it is neutral, so ii > iii > i.

Qu8: C
Radical chlorination of alkanes is strongly influenced by the number of H within each subtype. First draw the starting material, 2-methylbutane and then count the number of hydrogens of each type in the starting material. Then 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 1 x 5.2 = 5.2, ii 2 x 3.9 = 7.8 and iii 3 x 1 = 3, so the yield of ii > i > iii.

Qu9: AD
Oxidation states...count the bonds attached to the atom being 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. For C10 the C is attached to 2 x O (count - 1 each, total -2), N (count -1) and C (count 0) therefore total = -3 and therefore the oxidation state C = +3.

Qu10: ACDE
All the atoms are neutral, so you need to remember to consider the attached H atoms. Since C3 is the only C atom in the list that is involved in a pi bond, it is the only one that is not sp³ hybridised.

Qu11: B
Of the atoms listed, the lone pair on the N in the amine at N7 makes it the most basic. N12 is part of an amide and the resonance interaction of the lone pair with the adjacent C=O makes it less basic.

Qu12: E
C14-N15 is a triple bond hence it is the strongest bond.

Qu13: B
N7 is a simple amine, so we treat it as being sp³ hybridised. N12 is part of an amide (there is important resonance) so it is sp² hybridised.

Qu14: C
O11 is part of a ketone which is typially treated as being sp² hybridised with both of the lone pairs in sp² hybrids.

Qu15: B
Functional groups....N7 has two alkyl groups and H attached so it is a secondary amine.

Qu16: BE
C6 has two C groups attached (hence secondary) and it is adjacent to a benzene ring (hence benzylic).

Qu17: C
C13 is part of a cycloalkane type structure (all atoms sp3) where the bond angles are near 109.5 degrees.

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.

Qu18: B
The IR shows a strong and broad OH stretch at 2500-3500 cm^-1, and C=O near 1700 cm^-1 so this suggests a carboxylic acid (B or E) but there is no sign of C=C near 1600 cm^-1 so it's not an aromatic (i.e. not E)

Qu19: C
The IR does not show OH, NH, C=O or CN triple bond. On closer inspection, we can see sp² CH (about 3100 cm^-1) and sp³ CH (below 3000 cm^-1) and C=C near 1650 cm^-1. To have both sp² and sp³ CH it must C.

Qu20: AB
The IR does not show OH, NH or C=O. On closer inspection, we can see sp² CH (about 3100 cm^-1) but no sp³ CH (below 3000 cm^-1) and C=C near 1500 cm^-1. To have only sp² CH it must AB.

Qu21: BC
The IR shows an OH stretch around 3400 cm^-1, but no C=O near 1700cm^-1. Only AC and BC have OH without C=O. In the CH reagion, the IR also shows only sp³ CH (just to the right of 3000 cm^-1). Hence, we must be looking at BC.

Qu22: AD
The IR does not show OH, NH, C=O or CN triple bond (this only leaves the hydrocarbons C, AB and AD). In the CH reagion, the IR also shows only sp³ CH (just to the right of 3000 cm^-1). Hence, we must be looking at AD.

Qu23: D
The IR shows a C=O just above 1700 cm^-1, no OH or NH, no Ar C=C near 1600 cm^-1 or CN triple bond. This only leaves the ketone D. Closer inspection of the IR shows the sp3 C-H only (below 3000 cm^-1).

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.

Qu24: E
An alkane with the longest chain C8, so an octane. There are two possible C8 chains, so we need to pick the one with the simpler substituents, giving us two methyl groups and an ethyl group. These need to be numbered to 2,3,6 rather than 3,6,7 and 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.

Qu25: B
The principle functional group is a ketone, so the suffix "-one" is required. There is also an alkene, so also need "-ene", together an "-enone". Number to give the ketone the lowest number.

Qu26: A
The principle functional group is the alkene and those two C atoms need to be 1 and 2. Then the first point of difference rule requires that the system be numbered as a 1,6 dimethyl system rather than 2,3-.

Qu27: B
The longest chain is C9, with an alkene as the principle functional group so a nonene. Need to number to give the alkene the lowest number, i.e. -4-ene. Thus we have a 5-chloronon-4-ene. Now to designate the C=C stereochemistry, using the Cahn-Ingold-Prelog rules to assign E or Z to the alkene based on the priority rules, so Cl > CH2 and CH2C > CH3 so the higher priority groups in each pair are opposite making this the E isomer.

Qu28: B
We are looking for an ester based on a C4 alcohol portion and a C5 carboxylic acid portion i.e. A, B or C. The C4 alcohol portion requires a Br at C4. In A the Br is in the acid portion and in C it's at C1.

Qu29: A
We are looking for a carboxylic acid (which is defined to be at C1 on the benzene) with a Cl ortho (C2) and F para (C4). B and C have Cl and F in the wrong locations, D and E break the rules of valence.

Qu30: B
All the systems are cyclopentadienes, we need one with a substituent on C1 of the diene unit (B or E). Now to designate the R/S stereochemistry, using the Cahn-Ingold-Prelog rules to assign group priorities....Cl > C=C > CH3 > H. In B the sense of the 3 higher priority groups when the lowest (H) is at the back is clockwise = R.

Qu31: B or D (they are the same structure!)
This is an example of a spiro system. A, B, C and D are 3.4 systems. Numbering starts adjacent to the ring fusion (spiro center) C so A is 1-chloro...6-ene, C is 1-chloro... 5-ene.

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