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:
Alkenes undergo electrophilic addition with these reagents where the first step is protonation to give the carbocation intermediate and hence the rate of reaction depends on the acidity of the reagent: the greater the acidity, the faster the reaction HCl (pKa = -7) > HF (pKa 3) > H2O (pKa = 15) (the lower the pKa the stronger the acid).
HCl > HF > H2O
Qu2:
Use the pKas of the acids water OH= 15, terminal alkyne CH = 25, ammonia NH = 35 or think about the bases that are used to prepare an acetylide ion !
Remember that the hybridisation of the C (sp) makes it difficult to compare to the O and N systems where the electronegativity is a key factor.
Qu3:
The question is about the hydration reactions of hex-1-ene to give hexan-1-ol (which is the anti-Markovnikov product). H3O+ will give hexan-2-ol as the major product (Markovnikov product). The hydroboration / oxidation of alkenes adds the HO and H across the C=C in an anti-Markovnikov fashion and 9-BBN tends to show enhanced selectivity due to increased steric effects. Therefore is terms of yields of the hexan-1-ol: 9-BBN > BH3 > H3O+
Qu4:
The hydroboration / oxidation of alkenes adds the HO and H across the C=C in an anti-Markovnikov fashion. Here we are asked about 3,3-dimethylbut-1-ene. The major product is the anti-Markovnikov product, followed by the Markovnikov product but there will be none of the rearranged product since the reaction does not proceed via an carbocation.
Qu5:
When HCl reacts with a cyclohexene, the Cl will be at the more substituted position and this could (but does not have to) make a chirality center. Since the reaction proceeds via a carbocation, we need to consider attack from both faces.
Qu6:
All about specific rotation, optical rotation and optical purity. A 0.6g of (R,R) to 0.4g (S,S) mixture of the two enantiomers corresponds to an optical purity of 20% and hence a specific rotation that is 20% of the maximum value (which would be for a pure (R,R) enantiomer) = +2.54o. (2R,3S)-tartaric acid is a meso compound and therefore is optically inactive and has by definition a specific rotation = zero. An observed rotation of -0.635o calculates to have a specific rotation of -6.35o, so +2.54o > 0o > -6.35o.
Qu7:
The reactions of alkenes and alkynes with aq. acid are controlled by carbocation stability and the nature of groups attached to the pi unit. Alkynes are less reactive since they would require the much less favourable termolecular pathway to avoid the formation of the very unfavourable vinyl carbocations. Phenyl groups can stablise adjacent +ve charge via resonance (benzylic systems).
Qu8:
Carbocation
stability.... Look at the C atom bearing the charge and what's attached to it. Both (a) alkyl groups, which are weak electrons donors, and (b) resonance with Br lone pairs which adds stability due to charge delocalisation. Remember that vinyl carbocations are less stable than alkyl carbocations.
Qu9:
The relative stability of systems.... in this case we are looking at polyenes and the degree of alkene substitution. Conjugated dienes are more stable than isolated dienes, and more alkyl groups on a C=C makes it more stable. Conjugation is worth about 4 kcal/mol while
Qu10:
All the bonds are CC bonds. The factors involved are (1) the hybridisation of the C atoms involved and (2) the bond order. Double bonds are shorter than single bonds. Bonds involving sp2 C atoms are shorter than sp3 C atoms (due to higher s character).
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:
Working forwards... The hydration of an alkyne initially gives an enol which tautomerises to give a ketone. Since it is a terminal alkyne, the Markovnikov product, will be a methyl ketone.
Qu12:
Working backwards....The product is a 1,1- or geminal dibromide, formed using HBr, so the starting material needs to be an alkyne undergoing hydrohalogenation. To avoid a mixture of products (which is not indicated), the terminal alkyne is the best choice.
Qu13:
Working backwards...Step 2 is a peracid forming the epoxide from an alkene, step 1 the elimination of an alcohol to give the alkene. In order to get the methyl group and the alkene in the correct location, need to form a carbocation and then have a 1,2-alkyl shift.
Qu14:
Working backwards.....Sodium carbonate / methyl iodide is an SN2 reaction and has formed the methyl esters. Steps 1 & 2 look like alkene ozonolysis with an oxidative work-up which created carboxylic acids that led to esters in step 3. Therefore reconnect the carbonyl groups to reveal the appropriate alkene.
Qu15:
Working forwards... need to make a new CC bond to get from C3 to C7, using a terminal alkyne (C4). To do this, we need to keep the C=C but introduce a leaving group (Br) so use allylic radical substitution, then treat with the anion of 1-butyne and reduce to the trans C=C with Na/NH3.
Qu16:
Working backwards... the cyclic ether product suggests that the reactions look like intramolecular alkoxymercuration / demercuration of an alkene with an alcohol as the nucleophile. Counting / numbering C atoms, the position of the methyl group and the relative position of the alcohol and alkene will reveal the exact structure.
Qu17:
Working forwards... addition of HBr to the alkene giving the Markovnikov product, 2-bromo-2-methylpropane, followed by an conversion to the Grignard reagent and then reaction with carbon dioxide to give the carboxylic acid.
Qu18:
Working forwards.....alkene epoxidation using a peracid followed by acid catalysed ring opening by water will give a 1,2-diol.
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:
Anti-Markovnikov (Br at the less substituted end of the alkene) radical addition of HBr to the alkene followed by anti-Zaitsev E2 elimination using a bulky base to give the less highly substituted alkene. Last step is he hydroboration-oxidation of an alkene to give 3-methylbutan-1-ol. The reaction gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Note that the -OH is formed with retention of stereochemistry when the B atom is replaced.
Qu20:
The starting material is an alcohol reacting with acid and heat to undergo elimination of an alcohol to give the Zaitsev product, the more highly substituted alkene, followed by alkene ozonolysis with an oxidative work-up will give a the methyl ketone and carboxylic acid. Number the C atoms to confirm the chain length / functional group positions.
Qu21:
The catalytic hydrogenation using deuterium is a syn addition, giving the cis-alkene. This is followed by halohydration via an anti-addition and putting the HO- at the benzylic position. The answers are all Newman projections, so look for the ones with the -OH on the same C as the Ph and rotate about the middle C-C bond to put the HO- and -Br anti (the conformation in which the compound is formed) then look to see if the D atoms are gauche (60 degrees) and consistent with the cis-alkene.
Qu22:
Convert the alcohol to a bromide via an SN2 reaction then eliminate via an E2 reaction to form the cyclohexene system. This is followed by peracid epoxidation (syn) and epoxide ring opening with hydroxide (strong nucleophile, SN2 like) making the diol via what overall appears as an anti addition (i.e. trans-diol formed).
Qu23:
The trans-alkene is reacted with cold alkaline potassium permanganate to give a 1,2-diols via a syn addition. Remember how to decode Fischer projections vertical bonds away, horizontals towards you. The issue is that because we started with a trans-alkene (i.e. Ph and Me groups anti), we need to be careful. If you draw the diol in the conformation in which it is formed then rotate to create the conformation for the Fischer projection, this will reveal that the -OH groups need to be on opposite sides.
Qu24:
Rearrange the alkene to the more stable trisubstituted system via protonation / deprotonation and then cyclopropanation with dichlorocarbene to give a geminal dichlorocyclopropane:
Qu25:
The reaction of a 1,2-halohydrin with a base forms the epoxide via an SN2 reaction. This means that we need to set up the starting material in the conformation in which it reacts where the HO and the Br are anti to each other so that the Nu can attack the backside (at 180 degrees to the LG) and cause the normal SN2 inversion. The analysis shows that the Ph and CH3 end up cis and the configuration at the stereocenter with the O attached initially needs to remain S (watch that you don't pick the enantiomer, the initial material was a specific stereoisomer).
PI SYSTEMS:
All about knowing properties of pi systems such as resonance, structure, bonding, shapes, and terminologies etc.
Qu26:
Conjugation requires an extended pi system (at least 3 atoms) where the pi systems can interact
(parallel, not perpendicular). The image below has highlighted the p orbitals (blue circles for top down view of vertical p orbitals, and black for those that are perpendicular to the others).
Qu27:
Resonance contributors are derived via the delocalisation of the pi electrons with the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures.
Qu28:
Hydration of alkenes with aq. sulfuric acid is controlled by the rate of carbocation formation where the most stable (in these examples tertiary) carbocation forms fastest. The other alkenes here can only form either primary or secondary carbocations on initial protonation.
Qu29:
The key issues are (1) the pi bonds in an C=C are stronger that those in an C≡C and hence alkynes are less stable than dienes, (2) branching increases stability so less branched are less stable and (3) alkyl groups on pi bonds stabilise them, so a terminal alkyne is less stable. Therefore the least stable isomer is the unbranched terminal alkyne.
Qu30:
Qu31:
Tautomers (remember alkyne hydration ?) of ketones are the enols.
Qu32:
The reaction is a catalytic hydrogenation. The reaction is fastest for the weakest pi bond and that is found in an alkyne.
Qu33:
From the laboratory experiment, NaBH4 reduces carbonyls (C=O) and ketones are more reactive than esters.
Qu34:
Resonance contributors are derived via the delocalisation of the pi electrons with the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures.
