351 MT Spring 1999

Here is an post-mortem anaylsis / "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: AB
Need to look at the geometry of the C atom and therefore the hybridisation. The amine CH3NH2 is sp3 and tetrahedral, so HCN angle is about 109.5o, the imine CH2NH is sp2 so HCN angle is about 120o and the nitrile HCN is sp so the HCN angle is 180o. Therefore the angle order is iii > ii > i.

Qu2: A
As these are all isomers we can look at the relative stability of the alkanes and so we need to use the fact that the more branched alkanes are more stable. These means that the stabiliity order is i > ii > iii and since the least stable material has the most exothermic heat of combustion, the order is i > ii > iii.

Qu3: B
The IHD or (pi + r) of the compounds are i: 5pi + 2r = 7, ii: 3r = 3 and iii: from the formula 0.5 x (2 x 6 + 2 - 4 -1 + 1) = 5, therefore i > iii > ii.

Qu4: C
Since all are C to C bonds we should look at the specific nature of the systems. ii is a single bond, iii is a double bond and so is shoter, but i is aromatic which means that due to the resonance it is halfway between C-C and C=C. Thus the order is ii > i > iii.

Qu5: E
Need to determine the hybridisation of each system (same as qu. 1). The amine CH3NH2 is sp3 = 25% s, the imine CH2NH is sp2 = 33% s and the nitrile HCN is sp = 50% s. Therefore the order is iii > i > ii.

Qu6: A
Need to look at the bond dipoles and think about how the vectors would add due to the 3D shape of the molecules. CH2F2 would have 2 very polar CF bonds whose vectors would add to give a high moleculat dipole. CF2Cl2 although the bonds are only polar there would be partial cancellation due to the tetrahedral geometry and in CF4 there would be total cancellation. Therefore i > ii > iii.

Qu7: C
Boiling points depend on overcoming the intermolecular forces. For an alkane these are the weak Van der Waals forces and these will be higher the more surface contact there is, so a linear hydrocarbon will have a higher boiling point than a more branched isomer. The diol will have hydrogen bonding, the strongest type of intermolecular force usually encountered in organic molecules and this significantly raises the boiling point. So the order is ii > i > iii

Qu8: C
The more highly substituted the alkene, the more stable it is due to the increased amount of hyperconjugation. Here we have isomeric di-, tri- and mono- substituted alkenes. Therefore ii > i > iii.

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: B
Since a liquid boils when the vapour pressure equals the applied pressure, boiling point decreases with decreasing pressure.

Qu10: B
A distillation flask should be heated slowly for maximum efficiency.

Qu11: A
Anhydrous salts such as magnesium sulphate and sodium sulphate are commonly used to remove trace amounts of water from organic solvents.

Qu12: B
If you did, this would probably lead to the solution boiling over and you would lose your sample ! It should be added to a cool solution.

Qu13: A
This flow of water will ensure that the condenser does its job.

Qu14: A
Saturated means that there is the maximum amount of material dissolved.

Qu15: B
Need to calculate the moles of each of the reagents and check the stoichiometry of the reaction (here 1:1), the limiting reagent will be the material that there is the least of, here that is the acetic anhydride.

Qu16: E
Need to calculate the (moles of crude product) / (moles of limiting reagent) which = 4.83 mmol / 4.6 mmol, so > 100% (!)

Qu17: AC
As it says that the crude product was "large off white crystals" and the pure material was "pure white" we know a coloured impurity was present and that the lumps were probably due to water holding the crystals together.

Qu18: C
95 % of 4.6 mmol = 4.37 mmol which corresponds to 4.37 mmol x 151g/mol = 0.66g

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

Qu19: A
C4 is attached to 3 electronegative atoms and 1 C, so +3

Qu20: C
Since the C6-C8 bond is a double bond it will be shorter than any of the others.

Qu21: E
C11 is sp3, all the others asked are sp2 as they are part of double bonds.

Qu22: C
N3 is sp2 so that it can be involved in resonance with the two adjacent carbonyls.

Qu23: B
Need to check the formal charges to see if they are reasonable based on the number of bonds and the octet rule at each N atom.

Qu24: A
Since N3 is between two carbonyls, deprotonation can give an anion that has two resonance structures in which the charge can be delocalised onto electronegative O atoms.

Qu25: E
Need to have got qu. 23 right or use the equation to get this one. (pi + r) = 5 + 2 = 7

AROMATICITY:
Best method is to work through the molecules and decide on the aromaticity based on the four criteria (cyclic, planar, conjugated pi system with 4n+2 pi electrons)

Qu26: AB
Since B is in group III, it only has 6 electrons in the valence shell and hence is sp2 with an empty p orbital available for conjugation. AE is 4pi, anti-aromatic but is a cation.

Qu27: A or AC
Pyridine, A, is obviously related to benzene, but remember that the lone pair on the N is in an sp2 orbital, perpendicular to the pi system, and that the N is already involved in the pi system in the C=N. The substituted furan AC is 6pi by virtue of one of the lone pairs on the sp2 O being in a p orbital.

Qu28: BC
If n=2, then we are after 10pi electrons which is satisfied by the outer loop in BC.

Qu29: B
Remember that conjugated only means interacting pi systems, so B is still conjugated, just not a cyclic conjugated system.

Qu30: D
Non-conjugated implies that the the pi systems are insulated from each other, probably by sp3 centers, such as those in D.

Qu31: B
Which are non-aromatic as drawn ? B, C, D, E, (not cyclic conjugated system) and AD. Removing a H- from B with give a 6pi aromatic cation (remember that carbocations are sp2 hybridised so there is an empty p orbital available as part of the pi system. On loss of the hydride, C would be 4pi and anti-aromatic, D would still not be a conjugated cyclic system and E would be 2+ve.

Qu32: AD
For this type of question, you probably need to look for a system that has an exocyclic pi bond, as in AD. The resonace of the carbonyl gives +C-O- which means we have a 2pi aromatic system.

Qu33: C or E
Similar to qu31, but now we remove a proton, H+ to get an anionic system with 2 more electrons. This would be good for C and E.

NOMENCLATURE:
Two types of questions here, however it is a good idea to use key descriptors like E, Z, cis or trans, numbering, functioanl 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.

Qu34: C
Trans 1,3-substitutuion pattern. Named as a disubstituted cycloalkane.

Qu35: C
E alkene (high priority groups are on opposite sides of C=C), a carboxylic acid.

Qu36: D
Longest chain is C6, disubstituted ketone.

Qu37: D
Two substituents on the N of the amide.

Qu38: B
Looking for an ester limits us to B or C. C cannot be described as E or Z. A and D are ethers, E a ketone.

Qu39: D
Nitro group is -NO2, -NH2 is amino and -CN is cyano or nitrile.

Qu40: B
Another ester, so must be B ! A and C are ethers, D is an 4-ethylbenzoic acid.

Qu41: B
Check the E/Z patterns and the postions of the 3 methyl groups.....

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