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.

Qu1: AB
Catalytic hydrogenation reactivity depends on the strength of the p bonds.  This means that alkynes reduce more readily than alkenes since the second p bond is weaker, therefore iii > ii.  The extra stability of aromatic C=C means they are not as reactive as simple alkenes, so ii > i. Overall then iii > ii > i.

Qu2: D
Alkenes react via electrophilic addition where the rate determining step in the addition of the electrophile. Here the electrophile will is H+ and so the rate of reaction depends on the stability of the intermediate carbocation, the more stable the C+, the more readily (faster) it will form. Alkanes like i don't have electrons available to react with the H+ so they are essentially unreactive.  Cyclohexene ii will react to give a secondary C+ and the alkyne iii will give a vinyl cation which are only about as stable as primary cations.  Overall then ii > iii > i.

Qu3: C
If you know your pKas this is easy : terminal alkyne pKa = 26, amide ion -NH₂ about 35  and t-butoxide, (CH₃)₃CO- = 18.  Remember the higher the pKa the stronger the base, so ii > i > iii.   What if you don't remember your pKas ?  (why not ?) Do you know your reactions ?  Recall that terminal alkynes i.e. R-CºC-H can be deprotonated and the most common base for this is NaNH₂... so the amide ion  -NH₂ must be a stronger base than the acetylide ion R-CºC-  .  Alkoxides are weaker bases than amides and acetylides (or we would use it for this reaction !).  You can rationalise the N vs O system by thinking about electronegativity.

Qu4: E
Carbocation stability is dictated by the substituents at the C+ center.  i is a simple secondary cation, ii is a secondary vinylic cation (which is like a primary) due to the hybridisation at the C (note no resonance as the pi system is perpendicular = no overlap) and iii is secondary and allylic - the C=C means we should recognise the possibility of resonance stabilisation. Therefore we get iii > i > ii.

Qu5: E
i has a chirality center (the C with the Br attached) and so can be R or S.  ii is achiral. iii has a chirality center (the C with the Br attached) and the C=C can be either E or Z. So iii has 4 isomers, i has 2 and ii has only 1. Overall then we have iii > i > ii.

Qu6: C
Reaction is the hydroboration oxidation of an alkene. This gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Since no C+ forms, there is no carbocation rearrangement.  First draw 1-methylcyclopentene, then add the -OH to the least substituted end of the C=C.  So ii is the major product and ii is a minor product (the Markovnikov product). iii is only obtained is there has been a rearrangement. Overall then, ii > i > iii.

Qu7: A
Observed rotation is given by a = [a]_D.c.l    We are told the pathlength l is a constant.
The diagram in i is the S enantiomer (so +ve) and the concentration of the sample is 2/20 = 0.1 g/ml, so a = +23.1 x 0.1
The diagram in ii is also the S enantiomer (so +ve) and the concentration of the sample is 1.5/30 = 0.05 g/ml, so a = +23.1 x 0.05
In iii, we have more R than S, the observed rotation will be produced by 1g of R in 20ml i.e. a = -23.1 x 0.05
So i > ii > iii.

Qu8: D
The more stable the alkene, the less exothermic the heat of hydrogenation.  The more highly substituted the alkene, the more stable it is. i is tri-alkyl substituted, ii is mono- and iii is di, so i is the least exothermic and ii is the most, i.e. ii > iii > i.

Qu9: BC
Sucrose is the disaccharide of glucose and fructose, the pKa of HCl = -7 and acetals are formed by the reaction of aldehydes or ketones with alcohols (they are important elements of carbohydrate structure).

Qu10: ABC
Cystine acid is a very minor component of casein (did you see it on your chromatogram ?).  Less polar materials interact less with the paper (the stationary phase) of the chromatogram and so they run faster, a higher Rf value.

Qu11: ACE
Steam distillation relies on the fact that for immiscible liquids, the vapour pressure of the mixture is based on the sum of the vapour pressures of the components. In an extraction in a separatory funnel, the density of the solutions determines which layer is on top and which is at the bottom, for example water and dichloromethane, the water is typically the top layer whereas for ether and water, the water is usually the bottom layer.

Qu12: AD
Cross-linking makes the polymer less flexible and hence more rigid. Kevlar is a polyamide (like nylon). PETE is the polymer of terephthalic acid and ethylene glycol.

Qu13: CE
Buchner funnels are used for vacuum filtrations (i.e. for separation of solids from solutions or liquids). Boiling chips are used to provide a surface on which to initiate boiling and prevent super heating. Aqueous waste should be poured in the appropriate waste container and not washed away in a sink.

Qu14: E
Reaction is the hydroboration-oxidation (an addition reaction) of an alkene. This gives the anti-Markovnikov alcohol since the B adds as the electrophile and the H as a nucleophile (electronegativities : B = 2.0, H = 2.1). Hence the B atom adds at the least substituted end of the alkene. During the oxidation step the B is replaced by OH to give the alcohol. Since no C+ forms, there is no carbocation rearrangement.  A is an organoborane, which is formed initially but would react with the hydrogen peroxide to give the alcohol. B and C require a rearrangement product. D has the wrong regiochemistry, it is the Markovnikov alcohol.

Qu15: B
KOH / heat causes an elimination of an alkyl halide to give an alkene that is then subjected to ozonolysis with a oxidative work-up. Looking at the dicarbonyl product, we can work out the required alkene by "snapping" together the two C=O units to give the C=C. This gives us 1-methylcyclohexene (draw carefully, count carbon atoms). The fact that one end of the dicarbonyl product is a carboxylic acid and the other is a ketone tells us that the two ends of the alkene are different. A and D would not react with the KOH nor the ozone. C would only deprotonate with the base, no alkene would be formed. E would eliminate with KOH / heat to give 1-ethylcyclopentene with would give 5-oxoheptanoic acid, an isomer of the desired product (6-oxoheptanoic acid).

Qu16: C
CH₃CO₃H is a peracid which is used to convert an alkene to an epoxide.  H2SO4 / heat is therefore being used to form the alkene by the elimination (dehydration) of an alcohol. Recall that these are usually E1 reactions, giving the more stable (more highly substituted alkene) as the major product. Hence 1-methylcyclohexanol will dehydrate to give 1-methylcyclohexene and this in turn gives the epoxide C with the peracid. Note the epoxide unit defines the location of the C=C.

Qu17: B
Terminal alkynes are deprotonated with sodium amide, a strong base to give the acetylide ion, a good Nu. This reacts SN2 fashion with the secondary alkyl bromide to give Ph-C#C-CH(CH₃)₂.  Alkynes react with aq. H₂SO₄ / Hg²⁺ salts to under go hydration to an enol that will rapidly tautomerise to a carbonyl of a ketone. The hydration of the alkyne will proceed such that the water Nu attacks at the end of the alkyne that would give the more stable carbocation.... in this case this means at the end closer to the phenyl group (which could stabilise the C+ by resonance). Hence the ketone B is the major product rather than D.  C  would be formed from the alkene Ph-CH=CH-CH(CH₃)₂ and E would require Ph-(CH₂)₂C#CH.

Qu18: D
Radical bromination at the more highly substituted position followed by E2 elimination to give the alkene, probably the anti-Zaitsev, less highly substituted alkene due to the bulk of the base and then form the cyclopropane ring via the Simmons-Smith reaction of a carbenoid reaction (with :CH₂ as the reactive species). The position of the cyclopropane ring confirms the location of the C=C. A would give a bicyclic product, B would brominate in an allylic position (most stable radical) then eliminate to a diene.... so cyclopropanation would give either a bicyclic alkene or a tricyclic product. C D and E all have the right number of carbon atoms to be possible solutions but C would give a mixture of bromides and hence alkenes and would still have a t-butyl group in the product. E would result in the cyclopropane unit on C3 rather than C2.

Qu19: E
The trans alkene would react with the bromine to give 2,3-dibromobutane. Sodium amide is a strong base causing a double elimination of HBr to give the alkyne, 2-butyne.

Qu20: D
To get a geminal dibromide, the starting material needed to be an alkyne. A and B are alkenes which would react to give a monobromide. C, an alkane, would not react with HBr. E  would give a 50:50 mixture of 2,2- and 3,3-dibromohexane (equal stability of the intermediates).

Qu21: C
Reaction of an alkene with a hypohalous acid to give a 1,2-halohydrin followed by treatment with a base to allow an intramolecular SN2 reaction (also an example of a intramolecular Williamson ether synthesis) to give an epoxide (hence answer must be A, B or C).  HOBr reacts as HO- and Br+, the reaction will lead to the halohydrin via the cyclic bromonium ion, the regiochemistry will put the -OH at the more substituted position with anti stereochemistry. For the SN2 the nucleophilic HO must attack at 180 degrees to the C-Br bond, therefore the two methyl groups will still be cis in the product epoxide.  So A and B have the wrong regiochemistry and D and E are diols not epoxides.

Qu22: D
The product is a 1,2-dichloride, so this must come from a C=C rather than a CºC.  Cl₂ adds across C=C to add one Cl atom at each end of the p bond in an anti fashion.  Analysis of the Fisher diagram, remembering to redraw with the two Cl anti .... this means the two methyl groups need to be cis in the alkene. A would give a tetra-chloride if there were excess chlorine or a vinyldihalide.  B lacks a methyl group. E has the C=C in the wrong location to give the required product. So is it C or D. To sort them out, you need to pay attention to the stereochemistry of the reaction. You should redraw the product Fischer diagram first as a wedge-hash diagram then so that the two Cl are drawn anti as this is the way they are added in the reaction itself since the addition of  Cl is an anti addition process.  Once drawn this shows that that the original alkene is requires that the methyl and t-butyl groups are trans, i.e. D.  The alkene C where the two methyl groups are trans would give products that were stereoisomers of the required answer.

Qu23: BD
Dissolving metal reduction of an alkyne with Na in ammonia gives the trans alkene. Generation of a carbene (here loss of HCl gives :CCl₂) which then adds to the C=C to give a cyclopropane in a syn fashion so the stereochemistry of the C=C is preserved.  B and D are enantiomers so both will be produced in equal amounts.  A has the wrong alkene stereochemistry, C has the wrong carbene. E has the wrong reaction ! It's an epoxide not a cyclopropane system.

Qu24: C
Catalytic hydrogenation of the alkyne using Lindlar's catalyst gives the cis alkene.  Permanganate oxidation of an alkene gives the syn 1,2-diol, i.e. HO-C-C-OH. Overall anti addition to a trans alkene means a meso product.   A, D and E are not a diols. B has the wrong stereochemistry. Overall syn addition to a cis alkene means a meso product and B is not a meso compound (it's the S,S enantiomer).

Qu25: E
Reaction is the hydroboration-oxidation of an alkene. This 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. A and B have the wrong regiochemistry. C and D are from anti additions - consider the location of the -OH group - if the -OH and the methyl group are cis then the -OH and the -H that added must be trans (i.e. from an incorrect anti addition).

Qu26: C
Dissolving metal reduction of an alkyne with Na in ammonia gives the trans alkene.  Bromination of an alkene gives a 1,2-dibromide i.e. Br-C-C-Br with anti stereochemistry - thus just by looking at the added groups, it must be either C or D or E. Overall anti addition to a trans alkene means a meso product. Of the possible viable solutions then,  C is the (R,S) or meso isomer, D and E are the (S,S) and (R,R) enantiomeric pair, and so have the wrong stereochemistry.

Qu27: AE or CD
Need to have the same configurations.

Qu28: AB or BE
Need to have the opposite configurations and all centers.

Qu29: (A or B or E) with (C or D)
If they aren't enantiomers, they are diastereomers.... they need to have the same configuration at at least one chirality center.

Qu30: CD
Meso isomers have an internal mirror plane so are (R,S) configuration.

Qu31: AE
Fisher diagrams are defined based on an eclipsed conformation (so A). Eclipsed means zero degrees as a torsional angle, so E.

Qu32: E
The specific rotation [a]_D = a /c.l  where a = obs. rotation, 5.1^o, c = conc. g/ml and l = pathlength in dm. So [a]_D = +5.1^o / (1.5/10)(1) = +34^o

Qu33: C
Since A is (R,R) and B is (S,S), and there is more A than B, and the sample rotation has a +ve rotation, then the  [a]_D for (R,R) must be +ve. If we calculate the enantiomeric excess of A to B = (1.2 - 0.3)/(1.2 + 0.3) = 0.9/1.5 = 60% then based on optical purity, 60% [a]_D for (R,R) = +34^o therefore [a]_D for (R,R) = 56.67^o

Qu34: D
See answer to Qu 33. 

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