353 MT Winter 2025

Here is an post-mortem analysis / "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.

RELATIVE PROPERTIES:
Identify the controlling feature, which is not always as obvious as it may appear. Look for two pairs of similar systems to compare that have minimal differences in structure. If a compound is named, draw it out. If a reaction is involved, identify the type of reaction and then what the controlling factors are.

Qu1: C
Carbocation stability.... Look at the C atom bearing the charge and what's attached to it. We have a secondary carbocation, a secondary carbocation with Br attached to the +ve C and a phenyl carbocation. The lone pairs on the Br can interact through resonance to delocalise and stabilise the +ve charge. Phenyl carbocations aren't delocalised (think about the orbital alignments) and are less stable than primary C+ (which are less stable than secondary C+)

Qu2: D
Alkenes undergo electrophilic addition and in many reactions, the first step is the attack of the electrophile, for example H+ which is the rate determining step. Hence the stronger the acid, the faster the reaction. In terms of acidity HCl > H3O+ > HCN (pKas = -7, 0 and 10 respectively).

Qu3: C
First identify the alpha positions, the positions adjacent to the carbonyl groups. C-H systems in the alpha positions are enolisable H. In the diagram below, the C-H in the alpha positions are shown in blue, 4 > 3 > 0

Qu4: E
The reaction is catalytic hydrogenation which reduces alkynes to alkenes and alkenes to alkanes. In contrast, the C=O in carbonyls are less reactive (as they are stronger bonds). This will mean that with H2 (which will naturally be present in excess, most of the alkene in the starting material (draw it out) will be reduced to the alkane but some C=O will be reduced to an alcohol as well (but C=C tends to reduce before C=O).

Qu5: D
Looking at the benzopquinone should allow you to recognise the Diels-Alder reaction of the conjugated dienes being reacted with a conjugated ketone (the dienophile). The diene in a Diels-Alder reaction is more reactive if it (1) has electron donating groups and (2) favours the reactive s-cis conformation. Cyclopentadiene is a very good diene because it is locked s-cis and effectively has alkyl groups (weak electron donors) at each end of the diene unit. Of the two acyclic dienes, the cis,cis isomer is less reactive because steric effects destablise the reactive s-cis conformation which means very little of the molecule is in the reactive conformation at any one time.

Qu6: A
Draw out the named structure... that alkene reacts with MCPBA (meta-chloroperbenzoic acid, a peracid) to give an epoxide. The epoxide ring opening is under basic conditions with a strong nucleophile which means the unsymmetrical epoxide is preferentially opened when the RO- nucleophile attacks the less substituted C (SN2 like fashion) to give the alkoxy alcohol.

Qu7: E
Acidity and pKas... if you know your pKas this is easy. Or think least acidic (high pKa) to most acidic (low pKa)... Ammonia has a pKa = 35, internal alkynes = 45 (it's terminal alkynes = 25) and alcohols = 15 (question asks for relative pKa). The pKa values are affected by the electronegativity of the atom to which the acidic H is attached and the hybridisation of that atom (think %s character and the effect of nuclear stabilisation).

Qu8: C
All about specific rotation. Looking at the three structures, we need to use the Cahn-Ingold-Prelog rules to assign the configurations. One (wedge hash diagram) is (R,R), one (Newman projection) is (S,S) and the other (Fisher projection) is (R,S), a meso compound and therefore is optically inactive and has by definition a specific rotation = zero.

Qu9: C
Draw out the named structure... the conjugated diene will undergo direct addition (1,2-addition = major product) of HCl at the lower temperature (i.e. kinetic control) forming the more stable tertiary allylic carbocation with the conjugate addition (1,4-addition) being the other product.

Qu10: A
Hydroboration / oxidation of alkenes tends to give the less substituted alcohol (which is the anti-Markovnikov product). Alkan-2-ol means R-CH(OH)-CH3. 2-Butene is C4 and symmetrical and can only give butan-2-ol, 2-pentene will give a fairly even mixture of pentan-2-ol and pentan-3-ol while 1-butene will give mainly 1-butanol.

STARTING MATERIALS, REAGENTS AND PRODUCTS:
If you are trying to find the product, then you should probably just work forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is probably the best way to go....
Basically depends on the need to know and identify the reactions, this is often triggered by looking at the functional groups in the molecules.

Qu11: D
Working forwards, terminal alkynes undergo hydration to give methyl ketones.

Qu12: E
Working forwards, dehydration of an alcohol to give the anti-Zaitsev alkene means we will need to first convert to a different levaing group (e.g. a halogen) and then heat with a bulky base.

Qu13: E
Working forwards... first note that we need to extend the C chain my 2 C atoms, likely using addition of an acetylide to an epoxide...

Qu14: A
Working backwards, the product is a primary alcohol made by the hydration of an alkene using the anti-Markovniov hydroboration / oxidation.

Qu15: A
Working forwards, the alkene is being converted into a primary chloride. This will require making the primary alcohol using the anti-Markovniov hydroboration / oxidation followed by an SN2 to convert the alcohol to the chloride. Note there not radical variant of the addition of HCl to alkenes.

Qu16: E
Working backwards. The product is a vicinal-dibromide being formed by the double radical hydrobromination of an alkyne adding HBr to the 2-alkyne

Qu17: E
Working forwards, deprotonation of the terminal alkyne (pKa about 25) using the amide base (pKa about 35) to form the nucleophilic acetylide ion followed by SN2 methylation using methyl bromide, adding 1 C atom.

Qu18: E
Working forwards, the alcohol undergoes acid catalysed dehydration of an alcohol with a rearrangement to give the tetrasubstituted alkene at the ring fusion followed by ozonolysis with a reductive work up giving the diketone (counting / labelling helps).

REGIO- and STEREOCHEMISTRY:
How well do you know your reagents ? Look at what has actually happened in terms of the reaction functional group transformation and then first look for any regiochemical issues then finally the stereochemistry last (it's the hardest to sort out).  In cases where more than one product is formed in equal amounts (e.g. the enantiomers), then both must be selected for full marks, part marks are given when only one of the pair is selected.

Advice : in each case draw the starting material in the conformation in which is reacts or the product in the conformation in which it is initially formed using wedge-hash diagrams. It is a good idea to draw the materials in such a way that the new bonds are in the plane of the page.  Once you have drawn the materials it may also be good for you to use model kits for these questions too.  Once you have drawn the materials in this way, you may need to consider rotations around sigma bonds to make your answer match the options.  An alternative approach could be to assign configurations to your drawn answer to compare them with the options - this can be slow and prone to error.

Qu19: B
Working forwards : Alkenes undergo hydration when reacted with BH3 then aq. H2O2 / NaOH via a syn addition with the -OH group at the less substituted C (reaction occurs via a concerted syn addition of H and B followed by oxidation with retention of configuration at the C-B center so the added -H and -OH are on the same face of the ring).

Qu20: E
Working forwards : The first reaction is catalytic hydrogenation which reduces the alkyne to the cis-alkene. This alkene reacts with the peracid via a syn-addition to give the epoxide with the same stereochemistry. The epoxide ring opening is under basic conditions with a strong nucleophile which means the unsymmetrical epoxide is preferentially opened when the RO- nucleophile attacks the less substituted C (SN2 like fashion) to give the alkoxy alcohol and then work out the required Fischer projection (remember to position the alkyl chain (vertical) and back into the page with all the horizontal bonds coming out towards you)

Qu21: B
Working backwards : addition of bromine to give a vicinal-dibromide implies we started from an alkene, counting 5 atoms tells us it was 2-pentene. Addition of bromine is an anti addition (redraw in that conformation) so we started from a trans-alkene.

Qu22: A
Working backwards : potassium permanganate reacts with alkenes to give cis-1,2-diols via a syn addition so rotate the product into the conformation in which it is formed to reveal the cis stereochemistry of the alkene that is required, hence use poisoned catalytic hydrogenation.

Qu23: B
Working forwards : Cyclopenta-1,3-diene is a "classic" Diels-Alder diene and will react with dienophiles in a Diels-Alder reaction. The stereochemistry of the dienophile is preserved in the product. This is then followed by a catalytic hydrogenation of the cyclohexene to the cyclohexane.

Qu24: E
Working forwards : The halohydrin reacting with Na₂CO₃ (a weak base) to make an epoxide via an SN2 reaction so we need to make sure we review the halohydrin in the reactive conformation with the O nucleophile undergoing a backside attack to the C-Cl bond so -OH and -Cl anti to each other... that reveals that the Ph- and CH3- are trans to each other... then just rotate to the whole molecule to get the groups lined up to ensure you get the right configurations.

Qu25: E
Working forwards : HBr will undergo the addition to the isolated diene as if it were an alkene, adding to the C=C unit that gives the more stable carbocation in accord with Markovnikov's rule.

PI SYSTEMS:
All about knowing properties of pi systems such as resonance, structure, bonding, shapes, and terminologies etc.

Qu26: CDE
Conjugation requires an extended pi system (at least 3 atoms in series) where the pi systems can interact (parallel orbitals, not perpendicular). The image below has highlighted the p orbitals (blue circles for top down view of vertical p orbitals.

Qu27: B
Resonance contributors are derived via the delocalisation of the pi electrons across the pi system and can be derived by pushing curly arrows (blue to the left contributor, black to the right contributor). Use the rules for recognising resonance structures.

Qu28: C
We are looking at a Diels-Alder reaction.1,3-butadiene is the diene component, so we need to use a dienophile. Electron withdrawing groups (e.g. -CO₂R, -C(=O)R) make better dienophiles.

Qu29: C
Draw the picture...The least exothermic heat of hydrogenation means we need to identify the most stable compound, hence the more highly substituted conjugated diene.

Qu30: B
s-trans (3E)-2,3-dimethylpenta-1,3-diene :

Qu31: CD
Nomenclature Z means zusammen where the two higher priority groups are on the same side of the alkene.

Qu32: D
The hydroboration step of the hydroboration oxidation reaction is a concerted addition where the electrophilic B goes to the less hindered end (electronics & sterics).

Qu33: E
H2 / Lindlars catalyst selectivly reduces alkynes to cis-alkenes.

Qu34: C
The shortest CC bond will be the triple bond between two sp C atoms.