351 MT Fall 2000

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.

Qu1: D
Calculate oxidation state, count 0 for bonds to C, +1 to H, -1 to electronegative atoms, sum, then change the sign (since it needs to add upto zero) : here we get in i  C=0, H=+1 so C= -1, ii  Cl = -1,  so C= +4, and iii  C=0, Cl = -1 so C= +1, therefore giving : ii > iii > i

Qu2:  AB
Since all are CH bonds we should look at the hybridisation of the C atom.... the greater the s character of the C hybrid used to form the sigma bond, the shorter that bond will be. Here we have an alkyne (sp), an alkene (sp2) and an alkane (sp3)... therefore iii > ii > i

Qu3: D
Based on the rules for ranking resonance structures.  ii  has complete octets at N,C and O unlike iii which has only 6 electrons at the C and i which has 6 at O.  iii is better than i since the charge distribution is in accord with electronegativity. Therefore, in order of importance we have ii > iii > i

Qu4: E
Calculate the  formal charges : i = neutral, ii = -1 and iii = +1... thus iii > i > ii

Qu5: C
Ring strain makes the heat of combustion more exothermic per -CH2-, so the smallest ring will have the most exothermic value, and thus (most endothermic = least exothermic) cyclohexane > cyclopentane > cyclobutane (most exothermic) or ii > i > iii

Qu6: A
More branched isomers are more stable and have more exothermic heats of formation. Therefore i > ii > iii

Qu7: E
Just have to draw them out, no quick way.  Constitutional isomers means different numbers and types of bonds giving different functional groups and branching patterns.
C3H6 could be propene or cyclopropane, C3H8  must be propane and C3H8O can be ethyl methyl ether, 1-propanol, or 2-propanol. So iii > i > ii

Qu8: AB
Geometry so either use VSEPR or hybridisation to sort out the shapes.  Ethane has sp3 C so tetrahedral = 109.5o , CH2O is methanal, where the C is sp2, trigonal planar at about 120o, then HCN is sp, linear and 180o. Therefore, iii > ii > i

LABORATORY:
Based on the general principles cover in the laboratory so far. Need to know the principles and details of the steps in the experiments.

Qu9: D
Solids can usually be purified by careful recrystalisation.

Qu10: AB
A melting point can be used to do this since contaminated materials have lower and less precise melting points.

Qu11: B
Mixtures of liquids can be purified by distillation, here, given the small difference, then fractional distillation is better.

Qu12: A
Simple filtration would easily do this.

Qu13: C
An extraction where you shake with an organic solvent in a separatory funnel would work here.

Qu14: A
Simple filtration would easily do this.

To do the rest you need to understand the principles of extraction. The calculation was simplified by making KD = 1, so that the concentrations in the two layers need to be the same and the partition will be based on the relative volumes of the solvent layers...

Qu15: C
Given that the concs. are the same and we have the same volume, then the sample will be evenly split between the two layers so 50% will be removed.

Qu16: D
Given that the concs. are the same and we have the same volume, then the sample will be evenly split between the two layers, given that we only have 50% of the original in the aqueous phase, a further  25% will be removed.

Qu17: B
Add the previous two to get 75%

Qu18: C
Given that the concentrations are the same, but we have double the volume of one layer, the partition will be in the ratio 2:1, so 67 % to 33 %

MOLECULAR PROPERTIES:
No real method here, really just do you know various aspects of molecular structure.

Qu19: B
Attached to 3 x O = -3 and 1 x C = 0, so the sum = -3, therefore C5 = +3

Qu20: C
Attached to 1 x N = -1, 2 x C = 0 and 1 x H = +1, sum = 0, therefore C18 = 0

Qu21: BCD
C1 is a methyl group so sp3. C5 is attached to 3 groups and is part of a pi bond, so sp2. O6 has two lone pairs and is attached to a C via a pi bond, so sp2. C14 is part of a benzene ring and is attached to 3 groups, 2 C and an H, so sp2 . C19 is attached to 4 groups, 2 x C and 2 x H, so sp3.

Qu22: C
N2 is attached to 3 C and has a lone pair, so 4 groups and no adjacent pi systems to worry about....so sp3.

Qu23: BE
If it is secondary, it must be attached to 2 other C atoms.  C1 is "methyl" , C3 is secondary, C4 is tertiary, C13 is tertiary and C20 is secondary.

Qu24: BE
The N2 is part of an amine and there are two ester functional groups.

Qu25: C
Ortho-, meta- and para- describe the positions of a benzene ring where meta- = 1,3- so C15.

CONFORMATIONAL ANALYSIS:
Understanding the terminology and the energies involved in conformational analysis.

Qu26: C
For butane, the 60o torsional angle of the two methyl groups is best described as a gauche conformation.

Qu27: B
In propane, we can only use staggered or eclipsed. Since the groups are aligned here, we have an eclipsed conformation.

Qu28: D
Isomers that can be interconverted by rotation about s bonds are conformational isomers.

Qu29: B
Isopropyl methyl ether and diethyl ether have different branching so they are constitutional isomers.

Qu30: E
A  is really an "angle" type strain between lone pairs (as used in VSEPR).  B is ring strain. C and D are both torsional strain. E is a steric interaction....

Qu31: D
A, C and E are all trans- not cis-.  Note that for 1,4-cis-  we need one axial and one equatorial substitutent.  B is cis- but has the larger tBu group in an axial position, whereas E is cis- and has the smaller Me group axial and is therefore preferred.


NOMENCLATURE:
Two types of questions here, however it is a good idea to use key descriptors like E and Z, R or S, cis or trans, numbering, functional group etc. to help screen the answers before getting too tied up in the words. For the first four, name the compound and then match it to the list, for the last four use the key descriptors and check each molecule in turn. So this section really depends on the nomenclature rules.

Qu32: E
A cycloalkene, C=C has priority, number to get both the alkene C numbered and give the Br the lowest substituent number.

Qu33: C
A cycloalkane. Based on the first point of difference, we need to give the two methyl groups number 1 then the ethyl group =3, remember to alphabetise, but the "di" does not count.

Qu34: A
An ester... an ethyl ester (look at the alcohol portion).... longest chain in the acid portion is C4 including the C=C....

Qu35: B
Complex substitutents in an alkane... locate longest chain, here C10, identify substituents, number based on first point of difference ( ie a 4- group rather than a 5- group). The complex substituent has a C3 chain (so propyl) with a methyl group at C2 so 2-methylpropyl..... Remember to alphabetise....

Qu36: B
Use the descriptors.... A is (2E,5S), B is (2E,5R), C cannot be E or Z (two groups the same on one end of  the C=C), nor does it have a 5-methyl group so cannot be R or S...., D is (2Z,5S), E is (2Z,5R).

Qu37: D
Phenol is an aromatic alcohol, need a 2- ethyl group... this gives us D.

Qu38: A
Use the descriptors.... first the (N,4)- means we have methyl groups on the N atom and at C4... so D and E must be wrong. For the rest, A is S and B and C are both R.

Qu39: C
All methyl esters... so need to use the descriptors... only B and C have methyl groups at 3,6,9.  In B C2=C3 is E and in C C2=C3 is Z....

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