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: D
The main factors for carbocation are degree of substitution (more alkyl groups = more stable due to hyperconjugation) and resonance. Here ii is 3^o and iii is 2^o, so ii > iii. i is the phenyl cation which is NOT stabilised by resonance (due to the spatial arrangement of the orbitals), in fact the phenyl cation is less stable than a 1o. So overall, ii > iii > i

Qu2: B
Acidity. Look at the functional groups: i = carboxylic acid, ii = alkyne, iii = alcohol.
Now think about the equation HA  =  H⁺  +  A^- and look for factors that stabilise A- (the conjugate base)
Comparing i and  iii, the acid is more acidic because in the conjugate base, the -ve charge can be shared by two electronegative oxygen atoms. If we now compare ii and iii, then the alkyne leads to a carbanion. Although this carbanion is reasonable as carbanions go, it is not as favourable as the alkoxide from iii.  pKas i = 5, ii = 26, iii = 16. So i > iii > ii

Qu3: D
Boiling points means intermolecular forces.  All these compounds have similar MW, so we need to look at the functional groups and the differences they cause.  ii is an alcohol and so will have hydrogen bonding, the strongest of the intermolecular forces and thus the highest bpt. iii is an ether, which will have the weaker dipole-dipole interactions. i is an alkane and so only has the weakest intermolecular forces, Van der Waals interactions. So ii > iii > i

Qu4: A
Draw them out first. 3-methylpentane has 4 different types, n-hexane has 3 types and cyclohexane has 1 type of H (think about how many different monochlorides you could get....). So i > ii > iii

Qu5: C
If you know the reaction for the preparation for the acetylide ion which uses the amide ion, NH₂^-, and that alkoxides are not string enough to do this, then you get the order ii > i > iii

Qu6: B
Remember pi + r. i has 4 double bonds, 1 ring so IHD = 5, ii has 1 double bond and 1 ring, so IHD = 2, iii has 2 double bonds and 1 triple bond, so IHD = 4.  Overall, i > iii > ii

Qu7: A
All the systems have the same "critical" atom, an oxygen, and all are anions.  For nucleophilicity, we need to look at lone pair availability. The more available the lone pair, the more nucleophilic the system. i is a simple alkoxide, where the charge is localised on a single O atom (most available) ii is a phenoxide where some resonance interaction with the ring system allows for the -ve charge to be distributed onto C atoms in the ring.  iii is a carboxylic acid, and the -ve charge is shared between 2 electronegative O atoms (least available). Therefore i > ii > iii

Qu8: C
Reaction conditions favour an SN2 reaction (CN^- = good nucleophile in polar aprotic solvent like DMSO) so steric factors and leaving groups.  The methyl iodide is the least hindered and has the best leaving group and so will be the most reactive. 2-chloropropane is next, more hindered, poorer LG and finally the hindered tertiary alcohol with the very poor LG, HO^-. Therefore ii > i > iii

Qu9: AB
High temperature and a strong base will favour elimination over substitution so the alkenes dominate, the more stable trans-alkene will be the major product.

Qu10: C
For radical chlorination there are two factors, the stability of the radical (reactivity) but also the number of each type of H (statistics).  Unlike radical brominations, the reaction does not always give the product from the more stable radical since the statistics can dictate the outcome. In this case, i is produced from 6 1^o H, ii from 4 2^o H, and iii from 1 3^o H.  Therefore ii > i > iii

Qu11: A
Good leaving groups are the conjugate bases of strong acids. So we need to look at HBr, H₂O and NH₃ or the stability of Br^-, HO^- and NH₂^-  : Acidity order is HBr > H₂O > NH₃ .  O is better able to stabilise charge than N due to the higher electronegativity of O. Br is able to accommodate -ve charge due to its larger size (think charge density).

Qu12: D
Stereoisomers are isomers that differ only in the arrangment of groups in space. i can be R or S at the chiral center , ii has 3 chiral centers, so has 2³ = 8 stereoisomers and iii can be either E or Z at both double bonds so 4 possible stereoisomers. Therefore, ii > iii > i

Qu13: C
The more alkyl group substituents on the C=C then the more stable the alkene (due to polarisability of alkyl groups)  i has 3, ii has 4 and iii has 2, so ii > i > iii.

Qu14: C
A tough one. The reaction with HCl and C=C is an addition via a carbocation, so carbocation stability is important.  ii leads to a stable tertiary carbocation, so will ii reacts the most rapidly. i leads to a secondary carbocation, whilel iii reacts to give a secondary carbocation too, however the Br has an impact. Although the Br can stabilise the carbocation due to resonance, its' inductive effect (electron withdrawal) tends to make the alkene less reactive. So ii > i > iii.

Qu15: AE
False statements: B, Rf = (distance solute spot moves)/(distance solvent front moves). C, not that exclusive. D, ninhydrin is used for amino acids.

Qu16: C
False statements: A, in an SN1 reaction, the loss of the leaving group to give the carbocation is the rate determining step. B, zinc chloride is a Lewis acid. D, proton transfers to O atoms tend to be very, very rapid. E, 3^o alcohols dehydrate easily since they give stable carbocations.

Qu17: D
False statements: A, bromine is used to test for alkenes. B, no, DNP tests for aldehydes and ketones. C, DNP does not react with a carboxylic acid. E, DNP does not react with alcohols.

Qu18: BCE
False statements: A, NaI / acetone favours SN2 (good nucleophile, weakly polar, aprotic solvent). D, SN2 reacts fastest for the primary bromide.

Qu19: ACE
False statements: B, nitrate is a poor nucleophile due to the effect of resonance. D, primary systems react very slowly under SN1 conditions.

Qu20: BD
Sulphuric acid / heat will cause dehydration of the alcohol (E1) to the more most stable alkene, 1-methylcyclopentene. The alkene will undergo an addition reaction with bromine in an anti fashion to give an equal mixture of the enantiomers B and D.

Qu21: AC
MCPBA is a peracid that reacts with alkenes to give an epoxide via an addition reaction. 1-methylcyclobutene will give enantiomeric products A and C depending which face of the alkene is attacked.

Qu22: C
Tosyl chloride with a base is used to convert an alcohol into a tosylate which is a better leaving group. The we do an SN2 with inversion to give C.

Qu23: D
Ozonolysis with a oxidative work-up of alkenes. As each end of each double bond has only one alkyl group, then the product will be the dicarboxylic acids due to the cyclic nature of the system. So break each double bond in half, make the carboxylic acids at each end.

Qu24: E
Reaction of an alcohol with a strong acid.... protonate the -OH to make the better leaving group, H₂O, loss of the LG to the carbocation.... watch for the rearrangement, then the good nucleophile, the bromide ion reacts to complete the SN1 reaction.

Qu25: B
NaNH₂ a strong base, removes the most acidic H in this system, the terminal alkyne H to make an acetylide ion that then undergoes an SN2 reaction with the ethyl iodide to extend the C chain by 2 C atoms.

Qu26: C
Need to pay attention to the stereochemistry. Addition of bromine occurs in a trans fashion, therefore to get the required product you must use the cis-alkene. The cis-alkene is formed by the reaction of the alkyne with the Lindlar type, poisoned catalyst.

Qu27: D
Alkynes can be made by elimination, so we are going to need a leaving group. This requires that we add Br₂ across the C=C then remove 2 HBr to get the triple bond. So Br₂ then NaNH₂.

Qu28: D
Alkenes can be made by elimination which requires a leaving group. Only chemistry of alkanes is radical halogenation, so use that to add a Br to make the tertiary bromide then eliminate with strong base and heat. i.e. Br₂ / hv then KOH /EtOH / heat

Qu29: D
To make an ether we need a alkyl halide and an alkoxide. We have the alkyl chloride already, so we need to get the oxygen in. This means an anti-Markovnikov addition to the C=C, so need to use the hydroboration / oxidation method: BH₃ then NaOH / H₂O₂  then NaI / acetone will promote the ring closure by substituting the Cl to the more reactive iodide.

Qu30: D
Need to add Br and HO across the double bond in an anti fashion. This can be achieved using Br₂ / H₂O.  Make the bromonium ion, then water attacks as the nucleophile at the more cationic end.

Qu31: E
Need to do an anti-Saytzeff elimination, so best to use a strong, bulky base system, KOtBu / DMSO / heat.  KOH / EtOH / heat will give the more stable 2-alkene.

Qu32: A
Radical bromination then nucleophilic substitution by EtOH. Counting carbons limits it to A, D or E.  D would add the Br at the Me group and E would give a dibromide....

Qu33: B
A tough one. D doesn't have enough carbon, so thats out. The changed carbon skeleton should suggest a carbocation rearrangement. Sulphuric acid will protoante the hydroxy group to make a better LG and generate the carbocation which can then rearrange.  As C is primary it is unlikely to generate a carbocation. Neither A nor E can lead to a cyclopentyl ring system, so we are left with B. Try drawing it out.... the ring expansion from a cyclobutane to cyclopentane will relieve some ring strain.

Qu34: C
Ozonolysis requires the C=C, so the sulphuric acid is probably causing a dehydration reaction of an alcohol, so either C or D. C would give 5-oxopentanal after ozonolysis.

Qu35: D
NaNH₂ will not react with an alkane or alkene, so A and B are out. C would eliminate to an alkene, 2-pentene. D and E would undergo double elimination to give alkynes, 1-pentyne and 2-pentyne respectively.  HgSO₄ / H₂SO₄ undergoes addition reactions to convert the alkyne to an enol which tautomerises to ketones. E would give a mixture 2-pentanone and 3 pentanone.

Qu36: A
Tosyl chloride is used to convert an alcohol -OH into a better leaving group (so rule out C). This will then undergo an E2 elimination with the strong, bulky base to give the alkene. E is out because it has too few carbons. The stereochemistry of the tosylate of B in the E2 elimination (antiperiplanar) would allow 1-methylcyclohexene to be formed rather that the 3-methylcyclohexene required. D is the enantiomer of B.

Qu37: E
Ozonolysis of an alkyne to a carboxylic acid (no oxidative work-up needed) followed by nucleophilic substitution on CH₃I to prepare the ester. E has the right number of carbons for this to happen.

Qu38: B
Since they have the opposite configurations at all centers they must be enantiomers.

Qu39: C
Since they have the same configuration at one of the two centers, they must be diastereomers.

Qu40: D
Since iii is (S,R) it is meso, so has no optical rotation, therefore 0^o.

Qu41: C
Since i and iv are enantiomers, the specific rotation of iv will be of the same magnitude but of opposite sign to that of i.

Qu42: A
Since i and v are the same, then using the equation: [a]_D = a / c.l  which rearranges to give a = [a]_D.c.l, inserting values, a = -60.0 x 0.15 x 1 = -9.0^o.

Qu43: A
Since i and v are the same compound, it would be enantiomerically pure, so 100%.

Qu44: D
In terms of group priorities, OH > R > CH₂OH > H (where R represents the rest of the chain), and need to recall that the horizontal bonds are forward from the plane in a Fischer projection.

Qu45: A
Since iv and v are enantiomers they have the same physical properites.

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