Qu1: A
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. ii is an alkyl aromatic 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. iii is benzene, the carbanion will not be resonance delocalised so there is no further stabilisation. Hence i > ii > iii.

Qu2: C
Heats of combustion.... more branched alkanes are more stable and more stable isomers have less negative heats of combustion. Based on the branching, ii is the most stable and iii is the least stable. Therefore ii > i > iii

Qu3: A
Counting types of hydrogen.... i has 4 types, ii has 3 and iii has only 1 type, so i > ii > iii.

Qu4: D
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 6 x 1 = 6, ii 4 x 3.9 = 15.6 and iii 2 x 3.9 = 7.8, so the yield of ii > iii > i.

Qu5: A
Acidity.... First let's compare i and ii (because they are very similar, both are carboxylic acids) : the difference between them is the halogen atom... the presence of the electronegative atom will stablise the conjugate base due to inductive electron withdrawal through the sigma bonds, so in terms of acidity, the more lectronegative F will stabilise more than the Cl so in terms of acidity, i > ii. Now let's compare iii, it's a thiol... In terms of general principles, -SH systems are more acidic than OH systems due to the size of the S compared to the O. But that would only explain an alcohol compared to a thiol. In the craboxylic acid, there is resonance delocalisation in the conjugate base.... carboxylic acids are stronger acids that thiols. Other things that would help answer this question are knowing the pKa's (carboxylic acid are about 5, thiol about10). Therefore in terms of acidity, i > ii >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 neutral, but the nature of the basic atom is changing. First let's compare i and iii because O and N both have lone pairs making then much better bases than the C in ii. Since are in the same row of the periodic table, electronegativity is a factor ..... since O is more electronegative than N, O holds on the it's electrons more than N so it is the N that is more basic. So in terms of basicity iii > i > ii 

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

Qu8: B
Ranking resonance structures... i is the most important because the C, N and O atoms all have a complete octet. iii is has charge separation but all the atoms still have complete octets. ii has charge separation in accord with electronegativity but it has one less bond....and hence is less important than iii. : i > iii > ii.

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

Qu10: B
The greater the s character in the orbital then the lower the energy of that orbital. sp3 is 25%, sp2 is 33% and 2p is 0% (remember than 2p is higher energy than 2s, so mixing s and p to make hybrids lowers the energy of those hybrids.... more s character will make them lower in energy). So therefore in terms of the energies, i > iii > ii

Qu11:ABCE
Functional groups....

Qu12: E
There is one pi bond in the carbonyl gorup, one in the alkene and two in the triple bond.

Qu13: D
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 O16 is more acidic than the CH systems. FYI... alcohols are about pKa = 15, C2 enolate about 20, C5 is a vinyl and will be quite high, C15 about 25 and C23 will be about 20 (resonance stabilised enolate).

Qu14: AB
sp2 hybridised atoms are usually in double bonds or involved in resonance. C15 is in an alkyne and sp hybridised (2 attached groups), O16 is sp3 (2 lone pairs and 2 attached groups) and C20 is sp3 (4 attached groups, 2 x C and 2 x H).

Qu15: C
All the bonds involve C. Single bonds are longer than double bonds. CC will be longer than similar CO because O is smaller than C. In the CC bonds, the hybridisation of the C atoms involved is changing. Less s character in the hybrid orbitals will mean a longer so sp3 to sp3 C-C bonds will be the longest and that is C7-C8.

Qu16: AD
Look for molecules where there are dipoles to to bonds involving electronegative atoms but then pay attention to the 3D geometry / shape to make sure that the dipoles don't cancel each other out.

Qu17: C
The only atom here without a complete shell is the B atom in C from group 3.

Qu18: C
The system is an cycloalkane type structure (all atoms sp3) where the bond angles are near 109.5 degrees.

Qu19: C
Counting types of C indicates that there are 3 types of C present (note the mirror plane horizontally as drawn that shows the symmetry of the top/bottom).

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... then use the supplemental data and apply simple logic...

Qu20: BC
The IR shows an OH stretch at just above 3300 cm^-1, there is no C=O or C=C but there is a band near 2100cm^-1 that could be an alkyne. There are 5 alcohols in the choices, but C contains a C=O so we can rule that out. Since there are no Ar C=C (about 1600 cm^-1) or ArC-H (just above 3000 cm^-1), then there is no aromatic which rules out A. Of D, BC and BD (the other alcohols), only BC has 5 H types.... in addition, closer inspection of the IR shows the terminal alkyne C-H superimposed on the broad OH at 3300 cm^-1, confirming BC.

Qu21: C
The IR shows an OH stretch around 3400 cm^-1, and a C=O at about 1700cm^-1. Only B, C and E have OH and C=O. The IR also suggests Ar C=C (about 1600 cm^-1). The supplemental information indicates a positive ferric chloride test, indicating a phenol and not a carboxylic acid (i.e. not B). Hence, we must be looking at C.

Qu22: E
The IR shows a very broad OH stretch centered around 3100 cm^-1, and a broad C=O near 1700 cm^-1. This together with the solubility data suggests a carboxylic acid. The absence of Ar C=C near 1600 and 1585 cm^-1, imply it is not the aromatic carboxylic acid B and it must be the aliphatic acid E.

Qu23: AB
The IR shows very little... no OH or NH (about 3300 cm^-1) no C=O near 1700 cm^-1, no Ar C=C near 1600 cm^-1. This only leaves the ethers AB and BE. Closer inpsection of the IR shows the sp3 C-H only (below 3000 cm^-1) and possible C-O near 1300 and 1100 cm^-1. Of the two ethers, only AB has two types of H (BE has 3 H types).

Qu24: BD
The IR shows a broad OH stretch centered around 3350 cm^-1, but no C=O near 1700 cm^-1, and no Ar C=C near 1600 cm^-1. As in qu 20 there are 5 alcohols, only BD has just 3 H types.

Qu25: AC
The IR shows an NH stretch above 3100 cm^-1. There are three amines to choose from, AC, AD and AE. The solubiluty data confirms that we are looking for a base (i.e. the amines). Since the IR shows two bands for the NH stretch, it indocates the primary amine, AC.

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 1,1,3 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
The longest chain is 5, so it's a pent system, has a ketone and an alcohol. The principal functional group at the end of the name needs to be given the lower number and the substituent at the start of the name numbered accordingly. Since the higher priority functional group is the ketone, it has to be given the lower number and the alcohol has to be a substituent. B has the wrong numbering, C and D have the wrong principal functional group (C also has incorrect numbering for that principal functional group). E has the wrong sufffix, no aldehyde here and incorrect numbering.

Qu28: A
We have a C6 ring with an alcohol, so we are looking at cyclohexan-1-ol systems. The alkyl group should be numbered to be at C2, and the alkyl group is an isobutyl group. If the isobutyl was named as a complex substituent it would be (2-methylpropyl). D has the wrong numbering, the alcohol as the principle functional group requires the lower number.

Qu29: C
The longest chain is 5, so it's a pent 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 2-aminopent-2-ene. To designate the C=C stereochemistry, use the Cahn-Ingold-Prelog rules : the CH₃CH₂ > CH₃ and the NH > alkyl chain so it's Z stereochemistry.

Qu30: A
Benzyl = C₆H₅CH₂- and sec-butyl = CH₃CHCH₂CH₃ where the group attaches at the CH...... so A. Only A, C and E are benzyl ethers. B and D are phenyl ethers. B and C are t-butyl, D is sec-butyl and E is isobutyl.

Qu31: A
Only A and C are esters and of those A is meta while C is ortho.

Qu32: E
Only B and E are 1-alkene-1- amine systems. The others would be 1-alkene-3-amines. In B the methyl group is a 3-methyl.

Qu33: E
This is an example of a bridged hydrocarbon. C, D and E are 3.2.1 systems. Numbering starts at a bridgehead C then goes around the longer branches first giving the C=O the lowest possible number. C is -5-one, and D is -2-one.

Return to Homepage