Here is an post-mortem analysis / "how to" for the FINAL. The questions are split by the sections. At the start of each section are a few suggestions of what to look for or how to tackle the question type.

Qu1: E
Carbocation stability....for simple alkyl cations, more alkyl groups means a  more stable carbocation but here we also need to consider the effects of resonance. i is a primary cation as drawn with a secondary contributor. ii is a simple secondary cation and iii is secondary, with two other resonance contributors, one primary and one secondary. So overall we have in terms of stability, iii > i > ii.

Qu2: AB
All about the nucleophilicity of these groups. There are all oxygen systems and all are negative. The possibility of resonance delocalisation in i and ii makes the electrons less available and therefore the phenolate iii is more nucleophilic than the other two which are carboxylates. Now compare the O in i and ii respectively.  The 3 F atoms in i inductively withdraw electrons from the carboxylate due to the F electronegativity (think about acidity in these systems). This spreads out the electron density more in i and means i is less nucleophilic.  Hence  iii > ii > i.

Qu3: C
An application of acidity and basicity. All the species are anions, two are in the same row (O,N) and two are in the same group(O,S). In the same row, basicity increases as an atom gets less electronegative, so N- is more basic than O-. Within a group, acidity increases as you go down the group, so basicity increases as you go up the group, so O- is more basic than S-. Or we could use the pKas... alcohol =15, amine = 35 and thiol = 10 : the lower the pKa the stronger the acid and the weaker the conjugate base. So we have in terms of basicity that ii > i > iii.

Qu4: AB
Another application of acidity and basicity. This time acidity.... The functional groups are i an alcohol, ii an ammonium salt and iii a carboxylic acid. Carboxylic acids are the most acidic since the carboxylate ion outs the negative charge on an electronegative O where there is resonance delocalisation. Carboxylic acids will protonate amines, but alcohols will not....so we know that the ammonium salt is less acidic than the acid but more acidic than an alcohol (e.g. hydroxide would deprotonate an ammonium salt). Or we could use the pKas... alcohol =15, ammonium salt = 10 and carboxylic acid = 5: the lower the pKa the stronger the acid and the weaker the conjugate base. So we have in terms of acidity iii > ii > i.

Qu5: A
Each type of carbon will give a single peak so it's counting types of carbon.  i has 5 types since the ring has a vertical plane of symmetry as drawn. ii has 4 types due to symmetry between the two methyl groups and iii has 3 types since the molecule can be divided in half through the C=O.  Hence  i > ii > iii.

Qu6: C
The reaction here is SN2 (note the strong nucleophile and the aprotic solvent) so it will depend on the degree of steric hindrance at the C-LG center and the nature of the leaving group. i is tertiary and has a good leaving group, ii is secondary with a good leaving group under these conditions and iii is an unreactive phenyl system. So ii > i > iii.

Qu7: E
The number of lines in H-NMR coupling patterns is given by the n+1 rule when n = number of equivalent neighbours. In i there is only 1 neighbouring H, OH don't normally couple => doublet (2 lines). In ii the H in question has no neighbours and therefore appears as a singlet, and iii the H has two neighbours and therefore appears as a triplet (3 lines). Hence iii > i > ii.

Qu8: D
Bonds lengths are affected by the nature of the atoms involved and the bond order of the bonds. Bonds to H atoms, i.e. i will be short since the H atom is small. Both the other bonds are N-C bonds. But in iii there will be some double bond character due to the resonance between the N and the C=O. Hence the C-N bond iii will be shorter than the C-N bond ii. Therefore, in terms of bond length ii > iii > i.

Qu9: D
The highest energy conformation is the least stable one... this will be ii because one of the methyl substituents must be axial. The most stable conformation will be i where both substituents are equatorial and they are far enough apart that there is no gauche interaction between them. So in terms of the energies ii > iii > i.

Qu10: C
Ranking resonance structures: ii is the most important because the C and N atoms obey the octet rule. iii is the least important because it requires that the octet rule be exceeded which is not viable for atoms in the first row of the periodic table.

Qu11: A
The reaction is the E2 elimination of an alkyl halide with a strong base. The smaller the base, the more likely the basre is to abstract the more hindered H which leads to the more stable (more substituted) alkene

Qu12: A
Really about alkene stability. Notice they are isomeric.  The more stable the alkene the less exothermic it's heat of combustion.  In terms of alkene stability, the general rule is the more alkyl groups on the C=C unit, the more stable it is. i has 3 alkyl groups, ii has 2 alkyl groups and iii has 1, so i is the most stable then ii then i is the least stable. So iii has the most exothermic heat of combustion.  Hence, in terms of the heats of combustion i > ii > iii.

Qu 13: C
Rf is the ratio of distance traveled by sample to distance traveled by solvent from the origin. Measure these distances with a ruler. Distance traveled by sample (middle of spot) = 24mm, distance traveled by solvent = 35mm therefore 24/35 = 0.68

Qu14: E
Since the sample shows as three spots, the sample contains at least 3 materials and hence is not a pure compound so a is true. In normal chromatography, the polarity of the stationary phase means that the more polar materials elute more slowly = lower Rf so b is false. If a material elutes more rapidly, then it travels further and has a higher Rf so c is true.

Qu15: AE
B is false, 1,3-dimethylbenzene has 5 types of C, 1 sp3 and 4 ArC. C is false, the structure has 5 types of H, 1 vinyl H, 1 tertiary H (at ring fusion) and 3 different types of secondary H. D is false as glycine does not have a chirality center it is superimposable on it's mirror image.

Qu16: ACE
The aq. ethanolic silver nitrate conditions are for SN1 of alkyl halides. The nitrate ion is a weak nucleophile (like sulfate) due to resonance delocalisation so B is false. Bromide is a better leaving group than chloride, so bromides are more reactive than chlorides (to both SN1 and SN2 reactions) so D is false.

Qu17: C
Boiling points are lower at altitude (lower pressure) so A is false. Silica gel is the statinary phase in TLC so B is false. Charcoal is used to remove coloured impurities so C is false. Filtering funnel are used to separate insoluble solids from liquids or solutions so E is false.

Qu18: E
An application of  acidity and basicity.  It's a question about relative acidity or pKa. The most acidic organic functional groups are carboxylic acids, E, pKa = 5.

Qu19: C
An application of nmr spectroscopy.  The most shielded hydrogen with have the lowest chemical shift... we should expect this to be a simple alkyl group, remote from electronegative atoms.

Qu20: C
An application of ir spectroscopy.  The lowest frequency carbonyl will be the amide C.

Qu21: C
Configurational isomers are a type of stereoisomers where the groups differ in space and can not be easily interconverted at room temperature. A is a conformational isomer, B, D and E are constitutional isomers.

Qu22: D
Conformational isomers are a type of stereoisomers where the groups differ in space but can be easily interconverted at room temperature by rotation about single bonds. A, B, C and E are constitutional isomers of the molecule in question.

Qu23: A
Chiral carbons will be sp3 hybridised with 4 different groups attached - there are no C atoms that fit this description.

Qu24: C
Each of the rings symmetry planes giving 4 + 4 + 3 types then there are 2 CH2 linking the N and the phenyl, the 2 C in the ethyl group plus the C=O C.

Qu25:E
O4 is part of a carbonyl so the O is sp2 (3 attached groups, 1C and 2 lone pairs). N10 is part of a simple alkyl amine (4 attached groups, 3C and the lone pair) so the N is sp3. C6 is part of a benzene ring (3 attached groups, 2C and H) so it's sp2.

Qu 26: B
N5 is part of an amide, so the N lone pair is in a p orbital to allow for the resonance interaction and stabilisation with the carbonyl pi system.

Qu 27: C
A classic ethyl group patter, triplet and quartet integrating 3:2 respectively, with the CH2 quartet next to the C=O with the greater chemical shift.

Qu 28: C
We should work forwards .... looking at the starting material it's an alcohol : this will undergo a nucleophilic substitution with thionyl chloride to give the alkyl chloride. Reaction of the alkyl chloride with a strong base like KOH and heat will cause an elimination, a dehydrohalogentation, in accord with Zaitsev's rule to give the more stable trans-2-pentene as the major product. 

Qu 29: D
We should work forwards .... looking at the starting material it's a tosylate - these react like alkyl halides. Reaction of the tosylate with a strong base like KOH and heat will cause an E2 elimination. In order to predict the product we need to look at the anti alignment of the H and the LG bonds. 

Qu30: C
Need to work backwards .... looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - looking at the structure of the ether, we should see that we need to identify the cyclic alcohol component. Since the alcohol is the Nu in this reaction, the required alcohol will have the same stereochemistry as the final ether, so the alcohol and the methyl group need to be 1,3- and trans. A gives the wrong configuration. B would give the ether where the methoxy and the methyl groups would be cis. Neither D nor E would react under the reaction conditions given.

Qu31: C
Need to work backwards .... looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - looking at the structure of the ether and the reaction step 3, we should be able to see that we need 1-phenylethanol which can be made via a nucleophilic substitution of an alkyl halide with water. The required alkyl halide can be synthesised via a radical halogenation of ethylbenzene, C.

Qu32: C
Need to work backwards .... looking at the product it's an alkyl iodide synthesis (nucleophilic substitution), the reaction conditions indicate a tosylate is being used so we need an alcohol. Since the iodide is introduced via an SN2 reaction, there will be an inversion, and this means the original alcohol requires the opposite stereochemistry i.e. methyl and OH need to be 1,3- and cis. A will not react. Since B is racemic, it would give racemic products and not the just the enantiomer shown. D has the wrong stereochemistry and E has the opposite configuration.

Qu33: E
We should work forwards .... looking at the product, it's an alkene - but it's the anti-Zaitsev product. Since the starting material is an alcohol, a simple dehydration with acid and heat is not going to work and those E1 reactions follow Zaitsev's rule. Instead, convert the -OH to a halogen and then do an E2 elimination. using a large base to favour the anti-Zaitsev product. A would give the wrong alkene. B and C would not cause an elimination reaction of the alcohol (poor LG). D would give the more stable Zaitsev alkene the same as A.

Qu34:E
We should work forwards .... looking at the starting material it's an alcohol : this will undergo to undergo a nucleophilic substitution to give the alkyl chloride. Of the reactions given, only A and E react with alcohols to give alkyl chlorides. In this case, SN1 would result in a carbocation rearrangement (via a 1,2-hydride shiftto give a tertiary carbocation)... so A can't be used (the acidic conditions favour SN1) so E is the correct answer (thionyl chloride favours SN2).

Qu35: B
Ethers can be made by either acid catalysed reaction of alcohols, good for symmetrical ethers or by the SN2 reaction in a Williamson ether synthesis. This corresponds to B or E. Note that the reagents given in E would not give the product shown, they would give ethyl methyl ether instead. In this case, the symmetrical diethyl ether could be made using the acid catalysed method... the acid protonates one of the alcohol molecules to give a better leaving group, and the second alcohol acts as the nucleophile.

Qu36: D
In this case we are trying to make an ester. We can do this using a substitution reaction by first converting the acid to a better nucleophile by reacting it with a base and then adding an alkyl halide to act as the electrophile : i.e. D.

Qu37: C
IR shows a carbonyl (i.e. C=O) at 1718 cm^-1 which is a typical ketone. The 13C-nmr peak at 209ppm supports the C=O and also suggests an aldehyde or a ketone.  This means it could be C, E, BD or BE. The H-nmr doesn't have a peak at 9-10ppm, so it is not an aldehyde. The H-nmr has 3 types of H and the C-nmr has 4 types of C so it has to be C. The H-nmr coupling shows an ethyl group and a methyl group singlet.

Qu38: AC
IR shows a carbonyl (i.e. C=O) at 1741 cm^-1 which is a little high for a ketone. The 13C-nmr peak at 175ppm supports the C=O being a carboxylic acid derivative.  This means it could be B, AB or AC. The IR does not show an -OH so it is not the carboxylic acid B. The H-nmr has 3 types of H and the C-nmr has only 4 types of C as do both esters AB and AC. The H-nmr coupling shows an ethyl group and a methyl group singlet, but it is the methyl that is deshielded and therefore attached to the oxygen.

Qu39: E
IR shows a carbonyl (i.e. C=O) at 1716 cm^-1 which is a typical ketone. The 13C-nmr peak at 212ppm supports the C=O and also suggests an aldehyde or a ketone.  This means it could be C, E, BD or BE. The H-nmr doesn't have a peak at 9-10ppm, so it is not an aldehyde. The H-nmr has 2 types of H and the C-nmr has 3 types of C so it has to be E. The H-nmr coupling shows an ethyl group. It's a symmetrical ketone.

Qu40: BC
The H-nmr has 3 types of H. The H-nmr peaks at 6.9 ppm suggests that it is an aromatic : AE, BC, BD or BE. Since the H-nmr has a peak at 4ppm, it suggests the ethers AE or BC rather than the ketones BD or BE. Since the C-nmr shows 3 types of ArC (114, 121 and 149 ppm) this implies the meta rather than para substitution pattern.

Qu41: AB
IR shows a carbonyl (i.e. C=O) at 1743 cm^-1 which is a little high for a ketone. The 13C-nmr peak at 171ppm supports the C=O being a carboxylic acid derivative.  This means it could be B, AB or AC. The IR does not show an -OH so it is not the carboxylic acid B. The H-nmr has 3 types of H and the C-nmr has only 4 types of C as do both esters AB and AC. The H-nmr coupling shows an ethyl group and a methyl group singlet, but it is the ethyl that is deshielded and therefore attached to the oxygen.

Qu42: B
IR does shows an -OH at 2700-3500cm^-1 and a band at 1712cm^-1 suggesting a C=O and a possible carboxylic acid.  The 13C-nmr peak at 181ppm supports the C=O being a carboxylic acid derivative. The H-nmr exchangeable peak at 11.5ppm supports the -OH and it being a carboxylic acid => B. The H-nmr shows an n-propyl group coupling pattern.

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