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: C
The main factors for carbocation are degree of substitution (more alkyl groups = more stable due to hyperconjugation) and resonance. Here i is a simple 1^o but allylic system with resonance stabilisation by the p bonds. ii is also an allylic system, 1o as drawn but with a 2^o resonance contributor. iii is the very unstable methyl cation. So overall, ii > i > iii

Qu2: B
Nucleophilicity requires that we look a the availability of the electrons.  Here, 2 systems are O atoms, but i it is negative making the electrons more available.  In iii we have a N system in ammonia. Since N is less electronegative than O, the N system is a better electron donor and so a better nucleophile ie. ammonia > water. The simple alkoxide is unstabilised and is the most reactive (note that the same trend would apply for basicity is this case), so i > iii > ii

Qu3: C
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 also an alcohol, but with electronegative F atoms present and iii is an amine NH. So the negative charge will be on these heteroatoms. O is more electronegative than N meaning the O will stablise the anion more, so the alkoxide (from ROH) is more stable than the amide ion (from RNH₂).  But what about i vs ii ? The F atoms in ii will stabilise the conjugate base due to the inductive effect of the electronegative F atoms.  So for the acidity,  ii > i > iii

Qu4: AB
s bonding overlap is stronger than p bonding overlap the s CH will be the lowest energy orbital.  Anti-bonding orbitals are higher energy than bonding so p^* CC is highest energy, so we have iii > ii > i

Qu5: C
Each type of C will give a single line in the broadband decoupled C nmr... i has 4 types (CH₃ and 3 Ar C), ii has 5 types of C (CH₃ and 4 Ar C) and iii has 3 types (CH₃, and 2 ArC). So ii > i > iii

Qu6: E
We have a nucleophilic substitution reaction via an SN2 type of reaction (iodide = good Nu, polar aprotic solvent enhances nucleophilicity). SN2 is influenced strongly by the degree of substitution at the electrophilic center (steric effects hinder Nu approach). Since iii  is a primary alkyl halide, it will react fastest, i is secondary alkyl halide and then iii is a secondary alcohol. Under these conditions, -Cl is a better leaving group than -OH, so overall, iii > i > ii

Qu7: A
Coupling in H nmr is dictated by the number of neighbours of a different type.... For the H on C3 in pentane i , there are two methylene groups so a total of 4 neighbours and we see a quintet.  In ii the 2 methylene are equivalent and so don't couple. The CH₂ is split into 4 lines due to the 3 adjacent H of the methyl group and in iii the methyl group has a single neighbour and so shows as a doublet. Therefore i > ii > iii

Qu8: B
Reaction conditions favour an E1 reaction (alcohol plus strong acid and heat) so carbocation stability is the key factor. More stable C+ implies it is easier to form and reaction will be quicker. i is a tertiary alcohol, ii is primary and iii is secondary. Therefore i > iii > ii

Qu9: D
All are H-X bonds with X being a different atom. Recall that a longer bond tends to be a weaker bond.  Iodine in i is the largest atom, and F is smaller than O... so we have in terms of bond strength, HF > HOH > HI or, ii > iii > i

Qu10: A
(note this exact question was also on the midterm.)
Based on the rules for ranking resonance structures, look at the octets and then charge separation.  i has complete octets at C, N and O and the negative
charge on the more electronegative O atom. ii also has complete octets at C, N and O but now the negative charge on the less electronegative C atom. In iii
there N has violated the octet rule by having 10 electrons around it, which it can't do. Therefore, in order of importance we have i > ii > iii

Qu11: E
The location of the C=O stretch in IR is influenced by the attached groups and can be appreciated in terms of Hooke's Law and by thinking about how the attached groups affect the balance between the single and double bond resonance contributors.  The chlorine in the acyl halide of iii will be withdrawing electrons from the C (due to its electronegativity) so has more double bond character and therefore higher frequency. i and ii are both esters, but ii is a conjugated ester as has extra resonance contributors with single bond character (therefore lower frequency). So iii > i > ii

Qu12: D
Alcohols react with HBr via via an SN1 type of reaction (good LG formed via protonation of the -OH by the strong acid) and SN1 are controlled by C+ stability since the C+ is formed in the rate determining step. The more stable the C+ the easier it is to form and the faster the reaction will be.  i is a primary ROH, ii is a tertiary ROH and iii is a secondary ROH. so we have ii > iii > i

Qu13: A
The more alkyl group substituents on the C=C then the more stable the alkene (due to polarisability of alkyl groups and hyperconjugation). The more stable the alkene, then the less exothermic its heat of hydrogenation will be.  i has 4 alkyl group (most stable), ii has 3 and iii has 2 (least stable), so i > ii > iii.

Qu14: C
Good leaving groups are the conjugate bases of strong acids. So we need to look at H₂O, HBr and NH₃ or the stability of HO^-, Br^- and NH₂^- . In terms of acidity, HBr > H₂O > NH₃  since the large Br^- can accommodate the extra electrons more readily compared to O or N. We should know that HO^- is a very poor leaving group but since O is more electronegative than N,  NH₂^- is even worse !  So ii > i > iii.

Qu15: D
Calculate the yield from the question data. We have 10 mmol salicylic acid (MW=138) and use 1.59 mmol of acetic anhydride (MW=102), salicylic acid is the limiting reagent. We get 9 mmol of aspirin product (MW=180) corresponding to 9/10 = 90%.

Qu16: BDE
False statements: A tertiary alcohols are not oxidised at all. C since reaction (Lucas test) is SN1, 3^o > 2^o > 1^o.

Qu17: ABC
False statements: D chloroform is CHCl₃. E cyclobutane has that formula, but would not react with bromine in chloroform as there is no C=C...

Qu18: BE
False statements: A  acetone is a polar, aprotic solvent (no H to donate into a hydrogen bond since only CH bonds). C it is an SN2 reaction. D the precipitate is the sodium halide salt.

Qu19: A
False statements: B if the EtOH reacted as the nucleophile, the product would be an ether. C the silver nitrate enhances the LG ability of the halogen by causing it to precipitate as the silver halide salt. D as it is an SN1 reaction, the systems with stable C+ react fastest. E the intermediate is a carbocation not an anion.

Qu20: C
Phosphorous tribromide, PBr₃ will cause substitution of the alcohol to the alkyl bromide via an SN2 type process.  The resulting alkyl bromide will then undergo a 1,2-elimination reaction with the potassium hydroxide via an E2 reaction to give the alkene C. Since there are no C+ involved, there will be no rearrangements. E is not possible as the C has 5 bonds !

Qu21: C
Tosyl chloride with a base is used to convert an alcohol into a tosylate which is a better leaving group. Then we do an SN2 with inversion with water as the nucleophile to give C.

Qu22: E
Alcohols react with sulfuric acid via an E1 type of 1,2-elimination to the more highly substituted alkene E.

Qu23: C
Reaction of the alkyl bromide with NaI in acetone is an SN2 reaction so the product will be the iodide with inversion, the cis isomer C. B and E have the wrong stereochemistry (trans).

Qu24: A
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 bromide and since we don't do anything to the center where the methyl group is attached. it must be the same, so the starting material has to be either A or B. A is the trans isomer.

Qu25: A
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 starting material, the SN2 inversion, and use the stereochemistry of the product (methyl groups cis) to help..... Remember to draw the starting materials with the OH system at 180 degrees to the C-Br bond (as required for an SN2). B doesn't have a good leaving group (no halide), C doesn't have enough methyl groups, D and E will give the trans isomer.

Qu 26: B
The triple bond is introduced via an SN2 reaction of the acetylide ion with the chloride.... which was obtained by the substitution of the appropriate -OH (location of the functional group)... (count C atoms ?)

Qu 27: B
The product is an ester.  The second step uses an an alkyl bromide to give the ester via a nucleophilic substitution reaction.  The nucleophile needed is a carboxylate ion which is obtained from the parent acid using a base (Na2CO3).  The prepare the carboxylic acid, oxidise a primary ROH to a carboxylic acid (laboratory expt). Make sure you have enough C atoms..... A is 1 C short, C has 1 extra C. Neither D or E will an ester.

Qu28: D
Need to do a substitution to replace the -OH with a -CN. But to make an alcohol react, we need to get a better leaving group first, the tosylate being the viable option here. A will fails (no good LG). B and C do not give the CN unit. E will give the alkene which will not react with NaCN.

Qu29: C
Need to do an elimination (this can be most readily achieved by using acid and heat) then add hydrogen to convert the alkene to the alkane. A will just give the alkene, B and D will not do anything since the hydroxide will not cause the alcohol to eliminate. E will not do anything either....

Qu30: C or D
An ether synthesis....Williamson would suggest we 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.  Non of A, B or E will do anything : they don't create a good leaving group so no substitution can occur....

Qu31: D
To get the fluoride in via a substitution we will need to make the alcohol a better leaving group and use polar aprotic conditions to get the fluoride to work as a nucleophile. A and C don't provide a good LG, B lacks a good Nu since the F will be very strongly solvated and E has no F at all (DMF is the polar aprotic solvent dimethyl formamide).

Qu32: A
Simple E1 elimination of an alcohol so use an acid and heat. B and C will not eliminate due to poor LG. D will give the rearranged tertiary bromide then the Hofmann alkene, 2-ethyl-1-butene. E will give the tosylate and then the Hofmann alkene, 3-methyl-1-pentene.

Qu33: B
Substitution, but rearranged so C+ involved implying an SN1 reaction.... SN1 with HBr would give a rearrangement (1,2-hydride shift) and lead to 2-bromo-2-methylbutane. A will not react (reacts with C=C), C is SN2 so no rearrangement, D and E no reaction (only poor LG).

Qu34: AD
Move the larger groups further apart increases the stability.

Qu35: C
The halide is lost with the bonding electrons = leaving group

Qu36: AC
A single step reaction is concerted, multistep is stepwise.

Qu37: DE
This is the Zaitsev rule.

Qu38: D
E1 reactions via a C+ due to the initial loss of the LG.....

Qu39: BC
C+ are affected by the hyperconjugation effects of alkyl groups due to the aligned C-H bonds with the empty p orbital.

Qu40: CE
This is the basis of the Williamson ether synthesis.

Qu41: D
Carbocations in SN1 and E1 reactions can rearrange to give a more stable C+.

Qu42: AD
Size of a nucleophile can inhibit its reactions due to steric effects.

Qu43: BD
Larger atom allows it to accommodate the extra electrons in the conjugate base.....

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