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: A
The main factors for carbocation are degree of substitution (more alkyl groups = more stable due to hyperconjugation) and resonance. Here i is a simple benzylic system with resonance stabilisation by the p bonds. ii is 2^o and iii is a 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 1^o. So overall, i > ii > iii

Qu2: AB
Nucleophilicity requires that we look a the availability of the electrons.  Here all 3 systems are negatively charged O atoms, butin i there is resonance that will delocalise the charge to the second electronegative O atom making the electrons less available.  In ii there is resonance again, but less effective, spreading the charge to C atoms. Finally the simple alkoxide is unstabilised and is the most reactive (note that the same trend would apply for basicity is this case), so iii > ii > i

Qu3: E
Acidity means looking at the stability of the conjugate bases, but first we should identify which atom is going to become negative by locating the most acidic H in each case. In i we have an alcohol OH, ii is an amine NH and iii is a thiol SH. So the negative charge will be on these heteroatoms. S and O are in the same group of the periodic table, but one row lower, this makes it larger which means that the anionic charge is more dispersed (and the SH bond is weaker) compared to the O system, so the thiol SH is more acidic. O is more electronegative than N meaning the O will stablise the anion more so the alkoxide is more stable than the amide ion. So iii > i > ii

Qu4: B
After drawing them out, count p bonds and rings. Benzene has an IHD of 4 = 1 ring + 3 C=C and cyanocyclobutane has IHD = 3 = 1 ring + 2 CN p bonds. For the formula, either carefully draw an example then count or treat the halogen like an H (well they both have one bond) and ignore the O (since it forms two bonds, it doesn't matter if it is there or not, the same number of H are needed) IHD = 1/2 ( 2 x 8 + 2 - 13 -1) = 2.....so i > iii > ii

Qu5: E
Each type of C will give a single line in the broadband decoupled C nmr... i has 2 types (CH₃ and the tertiary C), ii has only a single type of C and iii has 3 types (CH₃, ArC with the CH₃ and the ArC with H). So iii > i > ii

Qu6: A
Alcohols react with hydrogen halides to give alkyl halides in a nucleophilic substitution reaction via an SN1 type of reaction and so proceed via the carbocation intermediate.  Since i is a tertiary alcohol, it will react fastest, ii is secondary then iii is a phenol and will not react at all (the phenyl cation is not favourable, since Qu 1).  Overall, i > ii > iii

Qu7: AB
Coupling in H nmr is dictated by the number of neighbours of a different type.... In i ethane, the two methyl groups are equal, so no coupling is observed and we see a single line for the single type of H present.  In ii the methylene is split into 4 lines due to the 3 adjacent H and in iii the methine shows as 7 lines since there are 6 adjacent H. Therefore iii > ii > i

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

Qu9: AB
All are CC bonds but the hybridisation of the C atoms is changing and so is the bond order. The C=C will be shortest, followed by the single bond between the two sp² hybridised (greater s character implies a smaller orbital) C in ii and finally the single bond between the two sp³ C in iii.

Qu10: AB
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 3 x 1^o H, ii from 2 x 2^o H, and iii from 9 x 1^o H which will dominate....  Therefore we get iii > ii > i

Qu11: C
The Zaitsev product is the more stable alkene. Here we are looking at elimination reactions proceeding via the E2 mechanism (we have a strong base), all the bases are alkoxides of different sizes.... larger bases will tend to favour extracting the most accessible H which leads to the anti-Zaitsev products..... in order of size we have methoxide < ethoxide < t-butoxide, so the yield of the Zaitsev product will favour ii > i > iii

Qu12: B
Boiling points are controlled by intermolecular forces. The stronger these forces the more energy is required to separate the molecules, the higher the boiling point. i is an alcohol and will have hydrogen bonding, the strongest intermolecular force.  ii is more branched than iii which decreases the amount of surface contacts so reducing the intermolecular forces, so we have i > iii > ii

Qu13: D
The more alkyl group substituents on the C=C then the more stable the alkene (due to polarisability of alkyl groupsand hyperconjugation). The more stable the alkene, then the less exothermic its heat of hydrogenation will be.  i has 1 alkyl group, ii has 3 (most stable) and iii has 2, so ii > iii > i.

Qu14: AB
Good leaving groups are the conjugate bases of strong acids. So we need to look at H₂O, HCl and TsOH or the stability of HO^-, Cl^- and TsO^-  : HO^- is a very poor leaving group, whereas tosylate (TsO^-) is very good since the charge is delocalised to 2 other electronegative O atoms. So iii > ii > i.

Qu15: C
Calculate the yield from the question data. We have 10 mmol of aminophenol, and we get 7.5 mmol of product corresponding to 7.5 mmol x 151 g.....

Qu16: DE
False statements: A the Lucas test favours the SN1 reaction. B secondary alcohols oxidise to ketones and C sulphuric acid did not oxidise the alcohol, it caused elimination.

Qu17: C
False statements: A bromine reacts with the alkene. B bromine in chloroform is dark red / brown. D no there is no leaving group to be substitution, it is an addition. E where does the Na come from ?, no precipitate forms.

Qu18: A
False statements: B NaI is soluble in acetone (covalent character). C it is an SN2 reaction. D under SN2 reactions, primary systems react fastest. E Br is a better LG.

Qu19: ABDE
False statement: C the silver nitrate enhances the LG ability of the halogen by causing it to precipitate as the silver halide salt.

Qu20: E
Thionyl chloride, SOCl₂ will cause substitution of the alcohol to the alkyl chloride which will undergo an elimination reaction with the bulky t-butoxide via an E2 reaction to give the anti-Zaitsev alkene E.

Qu21: A
Tosyl chloride with a base (removes HCL by-product) is used to convert an alcohol into a tosylate which is a better leaving group. Then we do an SN2 with inversion to give A.

Qu22: C
Radically brominate the tertiary position giving the alkyl bromide then E2 elimination to the alkene by loss of the Br and an adjacent H giving C.

Qu23: B
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 loss of the proton to complete the E1 elimination giving the more stable alkene, B.

Qu24: B
In order to get an E2 reaction to give this product, the LG must have been trans to the methyl group in the reactive conformation. The LG was the tosylate prepared from the alcohol with the same stereochemistry (since the C-O bond is not broken during the formation of the tosylate).... this is B. A would give the enantiomeric alkene, C and D can both eliminate to give 1-methylcyclohexene.

Qu25: D
The product is a ether... so look for an intramolecular Williamson type reaction .... an alcohol and an alkyl halide, but pay attention to the stereochemistry of the SM , the SN2 inversion, and use the stereochemistry of the product (methyl groups trans) to help.....

Qu 26: D
Cyanide is introduced via an SN2 reaction of the bromide.... which was obtained by the substitution of the appropraite -OH (location of the functional group).

Qu 27: C
Another intramolecular Williamson... need an alcohol and an alkyl halide with 5 C atoms.... this can be obtained from C. First the better leaving Br is replaced by I then the I is replaced by the HO^- giving an alcohol and an alkyl halide. Under these basic conditions the cyclisation to give a favourable 6 membered ring can occur.
Note that since -OH is a poor leaving group, it will not react with iodide to create an alkyl halide, nor will it cyclise to give an ether.  In D, the fluoride is too poor a leaving group to be suitable.

Qu28: E
Need to do an anti-Zaitsev elimination, so best to use a strong, bulky base system, KOtBu / DMSO / heat after first converting the alcohol to the tosylate. Using TsCl then KOH or acid / heat will give the Zaitsev alkene.

Qu29: A
Need to do a Zaitsev elimination, this can be most readily achieved by using acid and heat.

Qu30: C
An ether synthesis.... convert the alcohol to the more reactive alkoxide with a base then add the alkyl halide. Rule out some of the other options by remembering that OH is a poor leaving group.  Option D will give the wrong stereochemistry since the tosylate will have the same stereochemistry as the starting alcohol (since the C-O bond is not changed.... review the mechanism) then the OTs is displaced via an SN2 reaction as so it inverted. Thus D gives the enantiomer of the desired product.

Qu31: D
To get the bromide in the right place, we must use a radical chlorination (there are 9 primary H to be replacerd vs 1 tertiary) then a nucleophilic substitution.

Qu32: D
To get the extra double bond we need to first introduce a LG then carry out the elimination.  So use radical bromination which will add the Br at the allylic position adjacent to the C=C.

Qu33: C
An SN2 reaction will be best.... SN1 with HBr would give a rearrangement (1,2-hydride shift) and lead to 2-bromo-2-methylbutane.

Qu34: AC
SN1 nucleophilic substitution reactions proceed via carbocations. SN2 do not have an intermediate.

Qu35: BC
A reaction where all the changes occur at the same time is a concerted reaction, otherwise it is a stepwise reaction.

Qu36: C
This is the product predicted by Zaitsev's rule.

Qu37: BD
This is the definition of a leaving group.

Qu38: CD
E2 reactions occur best when the LG and the H are at 180o to each other in the antiperiplanar geometry.

Qu39: CE
SN2 reactions are slow for tertiary substrates because the alkyl groups block the attack of the nucleophile (steric effects).

Qu40: AD
A species with an unpaired electron is a radical. Carbanions and carbocations have only paired electrons.

Qu41: A
Named after is discoverer, the Walden inversion.

Qu42: AC
The rate determining step of an E1 reaction is the loss of the leaving group to give the carbocations.

Qu43: DE
Unpaired electrons means radicals, and these are formed via homolytic processes.

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