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 an polarisability) and resonance. Here i is a simple 1^o, ii is 3^o also doubly benzylic system, and iii is a simple 3^o  t-butyl cation. So overall, ii > iii > i

Qu2: E
Nucleophilicity requires that we look a the availability of the electrons.  The systems here are all halide ions... increasing size increase the polarisability of the Nu and this enhances its reactivity so iii > i > ii

Qu3: A
pKa real means acidity and therefore looking at the stability of the conjugate bases. In i we have a C-H system, ii is also a C-H but it is stabilised by resonance creating an O- (an enolate pKa about 20), and iii is an alcohol -OH so the negative charge will be on O.  The acidity is iii > ii > i so the pKa is  i > ii > iii

Qu4: AB
Coupling is given by l = n+1, so count neighbours..... i  has 4 neighbours => quintet, ii has 6 neighbours => septet and iii has eight neighbours => 9 line pattern. So we have iii > ii > i

Qu5: A
Each type of C will give a single line in the broadband decoupled C nmr... i has 5 types (1 x CH₃ and 4 Ar C), ii has 4 types of C (1 x CH₃ and 2 sp3 C and 1 sp2 C in the ring) and iii has 3 types (1 x CH₃, and 2 ArC). So i > ii > iii

Qu6: AB
Heats of combustion can be deduced from alkene stability which is determined by the degree of substitution (i.e. number of attached alkyl groups) and the stereochemistry. i is monosubstituted, ii and iii are disubstituted but ii is the less stable cis isomer and iii is the more stable trans isomer. The more stable the isomer, the less exothermic the heat of combustion will be, hence overall, iii > ii > i

Qu7: D
Bond strength can be related to bond length. All 3 are H-X bonds. iis the longest due to the large Br atom. ii and iii are both C-H bonds, but ii is methyl and iii is tertiary.  From stability and radical  reactions we should know the the tertiary bonds are weaker, hence overall we have ii > iii > i

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: A
Boiling point is determined by intermolecular forces.  iis an alcohol and so has hydrogen bonding (strongest), ii is an alkyl fluoride where there are dipole-dipole interactions and iii is an alkane where there are only induced dipole-induced dipole forces (weakest). The stronger the intermolecular forces the more energy (and hence higher temperature) is needed to separate them . Thus the boiling points are i > ii > iii

Qu10: B
Draw out the conformations of the cyclohexanes and look at the positions of the substituents. i must have one axial and one equatorial, ii has 1,4- groups so they are well separated and both can be in the preferred equatorial positions and iii has 1,2- groups again both in the preferred equatorial positions. So ii is the most stable then iii and finally i due to the unavoidable, unfavourable axial substituent.  The more stable the isomer, the more exothermic its heat of formation. So, therefore, we have i > iii > ii

Qu11: C
The reaction is a dehydrohalogenation (removal of HX), an example of a 1,2-elimination, via the E2 mechanism (strong base).  For each option look at the H on the C next to the C with the Br attached and see how many different alkenes you can get, remember cis and trans isomers. i has 2 : 2-methyl-1-butene and 2-methyl-2-butene. ii has 3 : 1-butene, cis-2-butene and trans-2-butene and iii has just 2-methylpropene (note different types of H will lead to different alkene products here). So ii > i > iii

Qu12: A
IR stretching frequencies relate to bond strengths.  These are all C-H bonds, the factor that is changing is the hybridisation of the C. Increasing the s character of the C hybrid means a shorter and therefore stronger C-H bond. i  is an alkyne, sp  ii is benzene so sp2 and iii is an alkane so sp3. Therefore we have i > ii > iii

Qu13: A
Either use the formula or draw an example and then count rings and p bonds. i has 5, ii has 4 and iii has 3 so i > ii > iii.

Qu14: B
Good leaving groups are the conjugate bases of strong acids. i  is a tosylate a very good leaving group. The other two are halides, Br is a better leaving group than F (think of HBr and HF in terms of acidity, HBr is the stronger acid) but not as good as the tosylate. So i > iii > ii.

Qu15: D
Calculate the yield from the question data. We have 10 mmol aminophenol (MW=109) and used an excess of acetic anhydride (MW=102, 1.12g/1.02g/mol), so aminophenol is the limiting reagent. We get 6 mmol of acetaminophen product (MW=151) corresponding to 6/10 = 60%.

Qu16: ACD
False statements: B tertiary alcohols are not oxidised at all. E the insoluble layer is the alkyl chloride.

Qu17: D
False statements: A the DNP reacts with the carbonyl group in aldehydes and ketones. B the reagent itself is red. C DNP doesn't react with the carbonyl group is acid derivatives. E the oxidant was a chromate...

Qu18: BDE
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.

Qu19: AC
False statements: B the fractionating column is the vertical one, if it were cooled there would be not distillation. D charcoal can be used to remove coloured organic impurities (acetaminophen expt.). E a separatory funnel is no good if the liquids are mixed... they need to exist as layers i.e. be immiscible.

Qu20: A
Phosphorous trichloride, PCl₃ will cause substitution of the alcohol to the alkyl chloride via an SN2 type process (inverts).  The resulting alkyl chloride will then undergo a nucleophilic substitution reaction with the sodium methoxide / methanol via an SN2 reaction (inverts again) to give an ether where the alkoxy group has the same stereochemistry as the original -OH group A.... this is an example of a Williamson ether synthesis.

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 the cyanide in a polar aprotic solvent to give C.

Qu22: D
Alcohols react with HBr via an SN1 type of nucleophilic substitution.... the secondary C+ intermediate will rearrange to give the tertiary C+ by a 1,2-hydride shift so the tertiary bromide D will result.

Qu23: C
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 two methyl groups in the starting material, the SN2 inversion, and use the stereochemistry of the starting material to help..... Remember to draw the starting materials with the nucleophilic  -OH system at 180 degrees to the C-Br bond (as required for an SN2), this puts the methyl groups trans leading to C.

Qu24: C
In order to get an E2 reaction to give this product, the LG (chloride) must have been trans to the methyl group in the reactive conformation. The chloride was introduced via an SN2 reaction of the alcohol, so taking into account the inversion, the -OH must have been cis to the methyl group. So we need to look for the cis-1,2-methylcyclohexanol : A and B are trans, D is cis, but has the wrong stereochemistry at the methyl group and E is not an alcohol.

Qu25: D
The product is a ether, the reagents suggest a Williamson ether synthesis : an alcohol and an alkyl halide. We need to pay attention to the stereochemistry of the starting material... the ether is cis, so the 1,3-methylcyclohexanol use to make the ether must also have been cis.  The reagents imply the alcohol was made by the SN2 (inversion) of an alkyl halide. Therefore we need the trans-1-bromo-3-methylcyclohexane. C and E are cis-1,3- , A is cis-1,4- and B has the wrong stereochemistry at the methyl group (it's the enantiomer of the answer).

Qu 26: C
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 2 C short, B is 1 C short. Neither D or E will allow the preparation of an ester.

Qu 27: D
The product is a just an ether... so look for an intramolecular Williamson type reaction .... an alcohol and an alkyl halide. Count C atoms ! A has 5 C and would not react in the sequence (-OH is poor LG), B has 4 C but like A would not react in the sequence, C has 4 C but would yield a cyclic system with O atoms at 1 and 3.  D the I replaces the Br (SN2) and in turn is replaced by -OH, under the basic conditions the alkoxide would form and allow the favourable 6 membered ring to form with O-C-C-O-C-C-O sequence.... E has only 3 Cl.

Qu28: C
Need to do an anti-Zaitsev elimination via an E2 pathway of an alkyl halide to get the less highly substituted alkene - so use a strong bulky base KOtBu / DMSO / heat.

Qu29: A
Need to do a Zaitsev elimination to the more highly substituted alkene from an alcohol - this can be most readily achieved by using acid and heat - conc. H₂SO₄ / heat.

Qu30: D
Need to remove -OH to -H... can't do this directly so dehydrate the -OH to C=C (acid / heat) the reduce that alkane with H₂ / catalyst.

Qu31: D
A simple ether synthesis from an alcohol.... use a base then add the alkyl halide with the right number of C atoms.

Qu32: C
A substitution of an alcohol to alkyl bromide but we need to ensure SN2 to avoid C+ rearrangements during an SN1. Br₂ reacts with alkenes, HBr will be SN1, KBr will not react (only a poor LG) and NaBr / H₂SO₄ will be SN1 too.

Qu33: B
A substitution of an alcohol to an alkyl chloride but rearranged so C+ involved implying an SN1 reaction.... SN1 with HCl would give a rearrangement (1,2-alkyl shift) and lead to 2-bromo-2,3-dimethylbutane. A will not react (reacts with C=C), C and E are SN2 so no rearrangement, D no reaction (only a poor LG).

Qu34: AD
Only two types of H, no coupling in a 3:1 ratio. Only A, E, AB and AD have two types of H and of those only AD is in a 3:1 ratio.

Qu35: A
Three types of H, no coupling in a 3:2 ratio. Only A, E, AB and AD have two types of H : so we can rule out AD. Chemical shifts can help us...the 2H signal is more deshielded than the 3H signal and is close to 4ppm, suggesting that the CH2 is attached to an O atom.

Qu36: E
IR implies a C=O, must be E, AB or AC. Two types of H means it can't be AC. Since the 3H chemical shift is at 2.1ppm, it is more likely next to C=O than O.

Qu37: B
Only B, C, and AC have three types of H. They are in a 3:6:1 ratio which means it is B and the coupling patterns tell us that we have a CH3CH system as in B.

Qu38: C
IR implies an -OH - so it must be C or D. But the number of types of H or the coupling patterns tell us it must be C.

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