Here is an post-mortem anaylsis / "how to" for the MT. 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: B
Since they are all CO bonds, we need to look at the bond strengths and remember that stronger bonds give higher frequencies. i contains a C=O and ii C-O. In iii resonance between the O and the pi system adds double bond character to the C-O so we get i > iii > ii.

Qu2: A
Resonance energy measures the extra stability of conjugated systems compared to the same number of isolated C=C.  Naphthalene (i) is like two benzene rings and so has a higher resonance energy than benzene (ii) itself (but not a great as two separated benzenes). Benzene is more aromatic and has a greater resonance energy than furan (iii). So i > ii > iii.

Qu3: AB
Need to look for the number of types of C, try looking for mirror planes or try one of the other methods we talked about in the 351 models lab. Benzene (i) has only 1, naphthalene (ii) has 3 and 1,2-dimethylbenzene has 4. So iii > ii > i.

Qu4: D
All about coupling, the lines = n+1 rule but remembering that equivalent H do not couple. So the Me group in ethane is 1 line (the two Me are equivalent), the Me in butane has 2 neighbours so a triplet (3 lines) and the Me groups in 2-methylpropane has only 1 neighbour and appears as a doublet (2 lines). So ii > iii > i.

Qu5: AB
i is a vinyl carbocation (quite are between primary and methyl cations is terms of stability).  ii is secondary and iii in a secondary benzylic system (so further resonance stability present), so iii > ii > i.

Qu6: C
The reaction is the addition of HCl to a diene under kinetic control which will favour 1,2 addtion over 1,4-addition. Thus ii > i. iii is not obsevered, try drawing out the mechanism. So ii > i > iii.

Qu7: D
Doesn't matter whether you look at C of H nmr shifts. i is TMS (reference compound, d = 0 ppm).  In the ether, the Me groups are deshielded by the adjacent O atom ( H d = 3.2 ppm,  C d = approx 60 ppm).  In the ketone, the deshielding is less since the O is further away from the methyl groups (( H d = 2.2 ppm,  C d = 31 ppm). Overall then, ii > iii > i.

Qu8: D
A Diels-Alder reaction and we are looking at the diene component. Benzene does not react like a diene very easily as is results in the loss of aromaticity.  Between the 2 substituted butadienes, the cis methyl groups in iii sterically destabilise the reactive s-cis conformation making it less favourable, so ii > iii. Overall, ii > iii > i.

Qu9: B
No, sucrose is glucose and fructose.

Qu10: A
Yes since the rate depends on the concentration of a single species.

Qu11: B
No Raoults Law applies to miscible liquids as in fractional distillation.

Qu12: A
Yes the anhydrous salt with absorb water.

Qu13: A
Yes, its the protein in milk you collected and hydrolysed to the component amino acids.

Qu14: B
No proteins have amide bonds.

Qu15: A
Lactose is the milk sugar.

Qu16: A
Yes this is true, from the detergent experiment.

Qu17: A
Yes in the soap experiment you hydrolysed the ester bond with NaOH to prepare the carboxylate salt.

Qu18: A
Yes is the soap experiment you hydrolysed the ester bond, creating to prepare the carboxylic acid (in its salt form) and the alcohol, glycerol.

Qu19: C
Furan, C, is a planar, cyclic,  conjugated, 6p system.  Of the other 6p systems, B is not cyclic and AD does not have a complete conjugated cyclic system. Both D and E are only 4p systems.

Qu20: BC
If n=3, then we need a 10p system..... got to be BC.

Qu21: B or AD
Either B or AD fit because neither have a complet cyclic conjugated system and are therefore non-aromatic.

Qu22: A
Only A has isolated C=C and is therefore non-conjugated. Remember than all the term conjugated means in general terms is that there is an extended p system.

Qu23: E
The only compound in the list that can become neutral when it loses a proton is E.  The loss of the proton creates pyrrole and releases the lone pair on N to become part of the 6p system.

Qu24: AE
Which compounds are non-aromatic as drawn ? A, B, D, E, AC, AD, AE.  Of these A, B and AC will not form cyclic conjugated systems when H+ is removed, so they are out. Loss of H+ from D or E will not give a carbanion. So we are left with AD or AE.  AD would give an anti-aromatic 8p system but AE would give an aromatic 6p system, so it must be AE.

Qu25: C or AB
Protonation of either furan, C, or indole, AB will still leave an aromatic system.  In furan the second lone pair on O is still part of the 6p system. In indole, protonation of the N lone pair removes the 10 p system but leaves the benzene 6p system in tact.

Qu26: C
Working through the sequence we get an alkene plus a diene, so a Diels-Alder to give cyclohexene. Allylic bromination then E2 elimination will yield the conjugated cyclohexadiene, C.

Qu27: C
Looks like the alcohol was created by a substitution reaction, of a bromide created via radical bromination at a benzylic position of ethylbenzene.  This would have been formed by the Freidel-Crafts reaction of EtCl to add the Et group, so we must have started with C.

Qu28: E
Addition of Br₂ to the alkene will gives trans-1,2-dibromocyclohexane. Strong base and heat will cause elimination (loss of HBr) to give 1,3-cyclohexadiene. Diels-Alder reaction with ethene and reduction of the resultant C=C will create a cyclohexane with a 2 carbon bridge as in E.

Qu29: D
Freidel-Crafts acylation of benzene will give the ketone without rearrangement of the alkyl chain. The Wolff-Kischener reduction converts the C=O to a -CH₂- and hence D.

Qu30: C
Freidel-Crafts alkylation of benzene with isobutyl chloride will result in rearrangement of the alkyl system to provide the more stable C+ and t-butylbenzene as the product.

Qu31: E
Looking for the diene of the Diels-Alder reaction. If you draw the curly arrows for the retro-Diels-Alder reaction , then the furan diene, E is revealed.

Qu32: E
Looking for the diene of the Diels-Alder reaction. If you draw the curly arrows for the retro-Diels-Alder reaction , then the dienophile is MeO₂CCH=CHCO₂Me... given that the product has the two ester groups trans-, the dienophile must have also been trans, E.

Qu33: A
Need to add 2 x Br, so an addition reaction of C=C with Br₂.

Qu34: AE
A Diels-Alder reaction to create the cyclohexene, so as we have an alkene, we need the diene, AE.

Qu35: AC
How do we add an ethyl group to ethylbenzene, use a Freidel-Crafts alkylation, AC using an alkyl halide and AlCl₃.

Qu36: BD
Alkene to alkane, a reduction using H₂ / Pd.

Qu37: D
How do we add Br to an aromatic ? Using Br₂ and FeBr₃. These conditions are more severe than those needed to add Br to alkene C=C.

Qu38: AB
Adding HBr to the C=C to give the anti-Markovnikov product, so need to use HBr / light.

Qu39: AD
Eliminating  HBr from the alkyl bromide to give C=Cso we need a strong base and heat, AD.

Qu40: E
Adding HBr to the C=C to give the Markovnikov product, so need to use HBr / dark (now goes via the more stable carbocation).

Qu41: C
To add the Br to the alkane at the benzylic position indicates a radical bromination using Br₂ / uv light.

Qu42: B
The H-nmr indicates a mono-substituted benzene (7.4-8 5H m) and an CH₃CH₂O- group (1.4 3H t and 4.4 2H, q).  The C-nmr indicates a C=O of an acid derivative (166 s), and a deshielded -CH₂- (61, t) . Only ethyl benzoate, B, has these features.

Qu43: A
The H-nmr indicates a mono-substituted benzene (7.3 5H m) and an isolated and quite deshielded -CH₂- (5.1 2H s)and an isolated -CH₃ (2.0 3H s).  The C-nmr indicates a C=O of an acid derivative (170 s), and a deshielded  -CH₂- (66 t). So we need an ester, in which the -CH₂- and -CH₃ are separated but not a methyl ester, so it has to be A.

Qu44: D
The H-nmr indicates a para-disubstituted benzene (6.9 2H d, 7.1 2H d) and two isolated -CH₃ (2.25 3H s and 2.3 3H s) neither of which are deshielded enough to be attached to an O atom.  The C-nmr indicates a C=O of an acid derivative (170 s).   So we need an ester, but not a methyl or ethyl ester, and it must contain 2 methyl groups, so it has to be D.

Qu45: E
The H-nmr indicates a mono-substituted benzene (7.4-9 5H m) and an isolated and deshielded -CH₂- (4.7 2H s)and an isolated and deshielded -CH₃ (3.5 3H s).  The C-nmr indicates a C=O of an aldehyde or ketone (196 s), and a deshielded  -CH₂- (75 t) and -CH₃ (59 q).  Since there is no aldehyde peak in the H-nmr (9-10 ppm), we need a ketone (E or AB or BC). Of these, only E is a mono-substituted aromatic, so it has to be E.

Qu46: BD
The H-nmr indicates a para-disubstituted benzene (7.3 2H d, 8.0 2H d), an CH₃CH₂- group (1.3 3H t and 2.7 2H, q), and a carboxylic acid, CO₂H (12.1 1H brd s).  Note that the -CH₂- is not deshielded enough to be attached to an O atom.  The C-nmr indicates a C=O of an acid derivative (172 s).  So we need a para-ethylbenzoic acid, BD.
 

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