Chem 351 Final Spring 1999

Here is an post-mortem anaylsis / "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.

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.

Qu1: D
Small rings are highly strained, but there is also strain associated with larger rings. Cyclohexane is strain free as the bond angles are the optimum 109.5 degrees and the bonds are staggered. So ii > iii > i.

Qu2: B
As we are looking at hexane as the solvent, the soluble materials should be less polar hydrocarbons with less intermolecular forces. So i > iii > ii.

Qu3: B
Acidity.... Carboxylic acid (pKa = 5) is most acidic, OH system with resonance to other O atom. Both others are CH systems alpha to C=O, an aldehyde and an amide. The aldehyde is more acidic (pKa =17) because the resonce of the N lone pair opposes the delocalisation of the enolate in the amide (pKa = 30). So i > iii > ii.

Qu4: A
Diene stability. Alkene stability increases as degree of alkyl substitution increases. For dienes, the order is conjugated > isolated > cumulated. So together we get most substituted, conjugated is the most stable and the isolated diene is the least stable. The most stable compound with have the most exothermic heat of formation. So i > ii > iii.

Qu5: D
Need to use Hooke's Law eqn.or use the trend that the heavier the mass attached to C, the lower the frequency, so C-H > C-D > C-Cl, since H < D < Cl (masses) therefore ii > iii > i.

Qu6: B
Need to think of the deshielding. The aldehyde type H is deshielded by the proximity of the carbonyl (typical range 9-10ppm). When electronegative atoms are close by (eg O) we also see deshielding (-O-CH typically 3-4ppm). Hydrocarbon CH are generally shielded (0-2ppm). So i > iii > ii.

Qu7: E
Consider all stereogenic centers, R or S, and E or Z. So we get 2, 1 and 4 therefore iii > i > ii.

Qu8:D
General trends are due to alkyl stabilisation (hyperconjugation) and resonance effects. Recall that a phenyl cation is not resonance stabilised since the orbital is perpendicular to the pi system. Therefore ii > iii > i.

Qu9: B
To determine the number of types of C present we should look at the symmetry of the systems. Ethyl benzene has 6, p-dimethylbenzene has only 3 and biphenyl has 4 types. So i > iii > ii.

Qu10: AB
BAsed on counting bonds to electronegative atoms, so we get +2, +3 and +4, therfore iii > ii > i.

Qu11: A
IR frequency will increase with increasing bond strength, and bond strength can be related to the hybridisation of the bond atoms. Here we have CO that are sp2-sp2, sp2-sp3, sp3-sp3, therefore i > ii > iii.

Qu12: E
Electronegative atoms can stabilise negative charge due to inductive withdrawal of electron density due to the polariastion of the sigma bonds. Proximity and nuimber of electronegative groups is important here, so iii > i > ii.

Qu13: B
Apply the n+1 rule, but remember that equivalent H do not couple (so care in iii). We get 10 lines, 3 lines and 4 lines, i > iii > ii.

Qu14: AB
The least stable (highest energy) has the largest group axial. In the most stable, they are both equatorial. So iii > ii > i.

ACIDS and BASES:
Based on the Henderson-Hasselbalch equation and interpretation of pKa data. Best method is to determine the fully protonated form, then carry out a "mental" titration and remove the H stepswise in order of acidity.... A -> D -> AB -> AC

Qu15: D
Only carboxylate will be deprotonated at this pH.

Qu16: D
pH is still not basic enough to remove the first ammonium proton.

Qu17: AC
Now all 3 of the protons will be removed.

Qu18: AB
Must be one of the species produced in our mental titration, the only overall neutral one is AB.

Qu19: AC
Look for the conjugate base of the weakest acid as indicated by the large pKa = 40, so the amide ion.

Qu20: E
More acidic than the pKa by two units, so protonated / deprotonated = 100.

Qu21: B
Need to look at the acids, the only acid stronger that the alcohol is the thiol, so it would protonate the alkoxide.

Qu22: E
As for 21, given the pKa difference (4 units) the equilibrium constant = 104 favouring the alcohol and thiolate.

LABORATORY:
Based on the general principles covered in the laboratory. Need to know the principles and details of the steps in the experiments. Only comments about the statements that are false are listed below.

Qu23: A

Qu24: A

Qu25: A

Qu26: B
No casein is a protein that you hydrolysed of amino acids

Qu27: A

Qu28: B
No a Buchner funnel is used for vacuum filtrations.

Qu29: B
No quite, it is actally an acetyl amide, not an ester.

Qu30: B
No, it is not that exclusive.

Qu31: B
No, the trap is used to trap any condensation from the steam lines to prevent flooding the distillation flask.

Qu32: B
No, the position will depend on the relative density of the organic solvent used. Ether is usually the top layer but dichloromethane is heavier than water and will be the bottom layer.

Qu33: A

Qu34: B
No, it is the other way up..... obtained / maximum.

STEREOCHEMISTRY:
Best method is to work through the molecules assign their configurations and use these to answer the rest of the questions....
Here A is (R,R), B is (R,S) and meso, C is (S,S), D is (R,R) and E is (R,S) and meso. Recall that enantiomers have the opposite configurations at all chiral centers and diastereomers are any stereisomers that are not enantiomers, so they have the same configuration at at least one center.

Qu35: AD or BE

Qu36: AC or CD

Qu37: AB or AE or BC or BD or CE

Qu38: AE
Meso compounds have (R,S) configuration which is the same as (S,R). The simplest case in which this occurs is when we have two chiral centers, with the same 4 groups attached to each center such that are they mirror images of each other.

Qu39: AB
Gauche implies that the angles between the two bonds in the Newman projection would be 60o.

Qu40: E
Using the eqn. for specific rotation we get specific rotation = -5o / (5/20) = -20o

Qu41: A
Using the optical purity from the specific rotation of sample / specific rotation of pure enantiomer we get opt. purity = -20o / 25o = -80%. The negative sign in conjunction with the (S,S) rotation being +ve, indicates that the (R,R) enantiomer is present in excess. If we have an optical purity of 80% then the enatiomeric ratio is 90:10.

Qu42: E
Since A and D are the same, then we have the pure enantiomer, so the specific rotation of (R,R) = -25o.

KINETICS AND THERMODYNAMICS:

Need to understand the various aspects of kinetics and thermodynamics, no real general method here, just application of concepts.

Qu43: D
Bonds broken are C=C and H-Cl requiring 146 + 103 = 249 kcal/mol. Bonds made are C-H, C-Cl and C-C giving 99 + 81 + 83 = 263 kcal/mol. Using bonds broken - bonds made (which is the same as energy in - energy out), the calculated heat of reaction = 249 - 263 = - 14 kcal/mol.

Qu44: A
As heat of reaction is -ve, the reaction is exothermic.

Qu45: B
If we take the chloride off the product C we get cation B.

Qu46: B
Need to know the order of carbocation stability and the reason for that order.

Qu47: C
Need to know the order of carbocation stability and the reason for that order, here the effect resonance has in stabilising it.

Qu48: B
The question indicates that A is formed rapidly, therefore implying that it is the kinetic product.

Qu49: C
Need to know the relative order of stability of alkenes: more alkyl groups = more stable alkene.

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