351 MT Fall 2004

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
Question about geometry goes back to hybridisation and geometries of some N systems. The only difference between N and C systems is that you have to remember to count the lone pair on N as it requires an orbital and therefore space - count it as a group when deducing the hybridisation. i is an amine, N = sp3, bond angles of about 109.5o, ii  is a nitrile, N = sp, bond angle is 180o, and iii  is an imine, N = sp2, bond angles of about 120o, therefore giving : ii > iii > i

Qu2:  AB
Relates to hybridisation and nomenclaturei is an alkane, C = sp3, therefore 25% s character, ii is an alkene  C = sp2, therefore 33% s character and iii is an alkyne, C = sp, therefore 50% s character.  Increasing the s character in the hybrid orbital makes it smaller and creates a stronger interaction with the H atom and so a stronger bond. Therefore  iii > ii > i

Qu3: E
Acidity .... Know your pKa's or work it out ?  If you need to work it out, then consider the general acidity equation HA <=> H+  A-.   Look at the factors that stabilise the conjugate base, A-.  First notice that we have two C-H systems and an S-H system.  i is a ketone (pKa = 19) where the conjugate base has the negative charge on a C atom but resonance can further stablise it to an electronegative O atom - remember spreading charge out (delocalisation) is a stabilising effect.  ii is an alkene (pKa = 40), this time the conjugate base will have the negative charge on the C atom - but the charge can not be delocalised so there is no further resonance stabilisation.  Finally iii , a thiol (pKa = 10) - the conjugate base puts negative charge on the S atom, S is in the same group of the periodic table as O, but one row lower. Remember that acidity increases as you go down a group since the larger atom can readily accommodate the extra charge.  So overall we have iii > i > ii

Qu4: B
Calculate the formal charges for the O atom = group number - number of bonds - lone pairs electrons or by relating to known structures : i = +1 based on 6 - 3 - 2, ii = -1 based on 6 - 1 - 6  and iii = 0 based on 6 - 2 - 4. Therefore i > iii > ii

Qu5: B
Physical properties relate to intermolecular forces... for alkanes this amounts to weak induced dipole interactions. The shape of the molecule affects the surface areas that are in contact. More branched structures are more spherical in shape and therefore have less surface area in contact. Larger surfaces in contact means greater interaction and hence a higher boiling point. There is NO relation between thermodynamic stability and physical properties. So  i > iii > ii

Qu6: A
Remember the rules for ranking resonance structures  : complete octets are most important. Here we focus on the C=C-O unit. Both i and ii have complete octets at both C atoms and O (despite the charges in ii) and both have the same number of bonds. iii has an incomplete octet on C and has one less bond than the others. So i > ii > iii

Qu7: B
Acidity.... which functional group is more acidic ? Why is this important ? The more acidic H will be removed first when reacted with a base. Here we have a phenol (pKa = 10) and a carboxylic acid (pKa = 5), so the H in the acid group will be removed first. Since only 1 equivalent of base has been added, i will be the major species in solution. Some of iii will also be formed but ii is impossible since the more acidic COOH group will protonate the adjacent phenoxide O- : i > iii > ii

Qu8: A
Oxidation states...count the bonds attached to the C, each H counts +1, C counts 0 and a bond to a more electronegative atom (e.g. O, N, X) counts -1. Total the count and then  since the atoms are neutral, just switch the sign since the oxidation state for the C plus the groups attached must equal 0.  In i the C is attached to 4 Cl  (count - 4) therefore oxidation state C = +4. In ii the C is attached to H (count +1), two bonds to O (count -1 per bond) and C (count +0) therefore oxidation state C = +2.   In iii the C has two bonds to C (count 0 per bond), a bond to H (count +1) and a bond to O (count -1 per bond) therefore oxidation state C = 0.  Therefore  i > ii > iii

Qu9: A
Basicity... the strongest base will create the most of the conjugate base. Compare to O systems first... O- is a stronger base than O, so ii  > iii.  Now compare N- and O- .  N is less electronegative than O so it is a better electron donor and hence a better base (Lewis definition) so > ii. Therefore overall  i > ii > iii

Qu10: D
Resonance stabilisation... look at how the -ve charge can be delocalised. In all 3 cases it can be delocalised into the ketone group on the left. In i this is the only thing that happens. In ii  and iii it can be delocalised due to the extra pi systems present, in ii  the -ve charge can be delocalised to an electronegative O atom (good !) and in iii  to a C atom (not as good), so overall ii > iii > i

LABORATORY:
Need to know the details of the various experiments performed so far this semester.

Qu11: B
From the extraction experiment. Seems a hard question needing complex math, but if you understand the principles, it's really pretty straight forward. Given that KD = 2 chloroform : water and the solvent volumes are in the ratio 1 : 1 so there will be half as much in the water layer as the extracting dichloromethane layer i.e. 1 : 2 ratio. This means that each dichloromethane extraction will extract 2/3 of the total.  So the first extraction takes out 67% than the second takes 2/3 of the remaining 33% = 22%, therefore in total 67 + 22 = 89 % will have been extracted.

Qu12: E
From the physical properties experiment. Remember boiling points increase with increasing pressure so they are higher at sea level than here in Calgary at almost 4000ft above sea level. Rough rule of thumb for Calgary is 1o for every 15o above 50o. Thus for 200o we are talking about a 10o increase.

Qu13: C
A yield calculation. Reaction stoichiometry 1:1.  1.38g of salicylic acid (138 g/mol) is 10 mmoles.  1.0g of acetic anhydride (102 g/mol) is 9.8 mmoles, so the anhydride is limiting. 1.2g of aspirin (180 g/mol) or 6.67 mmoles was obtained, so the yield is 6.67 / 9.80 = 68%.

Qu14: D
From the molecular models experiment.  First draw out 1,2-diethylbenzene (if you can't, then you need to work on your nomenclature). The number of types of C are shown below in blue on the left and the H are shown below in green on the right. The red line shows the symmetry due to the mirror plane that bisects the molecule.

Qu15: AB
From the molecular models experiment on IHD.  Either use the molecular structures and count pi + r or use the equation based on the molecular formula (see link). Benzene has 3 pi bonds and one ring so it has an IHD = 4.  Cubane is more difficult, but the IHD for C8H8 is 5.

Qu16: ACDE
Only B is false because magnesium sulfate is used as a drying agent to remove traces of water from organic solutions.

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

Qu17: C
Oxidation states
C2 has the following bonds : 1 x C (count 0) and 3 x O (count -3) so the sum = -3 therefore the oxidation state of C2 = +3.
C13 has the following bonds : 1 x N (count -1), 1 x C (count 0) and 2 x H (count +2) so the sum = +1 therefore the oxidation state of C13 = -1.

Qu18: D
The functional group in question is C-O-C so it's an ether.

Qu19: D
The functional group in question is -C(=O)-O-C so it's an ester.

Qu20: D
IHD....Unsaturation arises due to pi bonds or rings.  So count these up in the structure:  there are 6 C=C and 2 C=O plus 3 rings (2 x 6 membered and 1 x  7  membered), so 11 in total.

Qu21: E
O12 is part of a double bond, C=O, so it is sp2.  N10 is sp2 because the lone pair on the N is involved in resonance with the adjacent pi system, a C=O, it's an amide N.   N15 is sp3 because it has 3 C atoms attached, plus the N lone pair but there is not resonance since there is not adjacent pi system, it's an amine N.

Qu22: B
Hydrocarbon hybridisations.... C2 is part of a double bond, C=O, so it is sp2.  C5 is sp3 because it has 2 x C, 1 x O and 1 x  H atoms attached.  C17 is sp3 because it has 3 x H and 1 x O atoms attached.

Qu23: B
Apply the Cahn-Ingold-Prelog rules at the 2 chirality centers.  C5 priority order :  -OC > CS > C(=O) > H, the low priority H is already at the back, 1-2-3 order goes clockwise => R. C6 priority order :  -S > CO > CC > H, the low priority H is currently at the front, so take care. When H is at the back, the 1-2-3 order goes counter clockwise => S.

Qu24: D
Basicity by the Lewis definition is lone pair donation, so can look at the availability of the lone pairs. Neutral C atoms with 4 sigma bonds don't have lone pairs so they are not basic.  N is less electronegative than O so N is a better electron donor and hence a better base than O.  N10 is in an amide and N15 in an amine. The lone pair in the amide is also involved in resonance with the C=O so making it less available to react as a base. So the amine N, N15 is the most basic site.

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

Qu25: B
The conformation shown is an eclipsed conformation since the bonds on adjacent atoms are aligned. Here the two principle groups, the methyl groups are at 120o torsional angle. This conformation does not have a more specific name.  Anti means a 180o torsional angle and syn a 0o torsional angle.

Qu26: C
The conformation shown is a staggered conformation since the bonds on adjacent atoms have torsional angles of 60o.  Here the two principle groups, the methyl groups are at 180o torsional angle, so this is an anti conformation.  A syn conformation has a 0o torsional angle and gauche has them at 60o.  Trans is used to described substituents on cyclic structures or across double bonds.

Qu27: AE
The two terms that apply to cyclohexane locations are axial and equatorial. Substituents prefer to be equatorial to reduce the steric strain due to Van der Waals strain due to 1,3-diaxial interactions and torsional strain that occur when they are axial. The chair is the more stable conformation (staggered therefore less torsional strain) compared to the boat, so B and D must be incorrect.  A ring flip of A will have two axial methyl group so A is the more stable conformation.  C ring flips to give two equatorial methyl groups which will be lower energy and hence more stable.  E ring flips to give two axial and one equatorial methyl groups which will be higher energy and hence less stable than the conformation shown.

Qu28: CE
First draw out each structure or work out the molecular formula of each, if you can't do this, you should review your nomenclature. If they have the same molecular formula, then they are isomers.  2-butanone is a ketone = C4H8O and so has 1 unit of unsaturation.  2-butanol is a saturated alcohol = C4H10O. Diethyl ether is saturated, C4H10O. Butanal is an aldehyde = C4H8O and so has 1 unit of unsaturation.  Butanoic acid is a carboxylic acid = C4H8O2. Cyclopropylmethanol has a cyclic unit and is an alcohol = C4H8O.

Qu29: E
The molecule shown is 2-chlorobutane in both drawings, CH3CH(Cl)CH2CH3.  In the two drawings, the front C is a chiral center and they are drawn as mirror images, so the best answer is that the two structures are enantiomers (even though there has been a rotation of the C2-C3 bond).  Since both drawings are 2-chlorobutane, they can not be constitutional isomers.  They are not conformational isomers since the configuration at C2 is different in the two drawings.  They are a type of stereoisomer and configurational isomer, but the question asks for the best description (i.e. most accurate).

Qu30: A
The lowest energy conformation will be a staggered conformation with the lowest steric strain. B, C and E are all eclipsedA is more stable than D because A has only 2 methyl-methyl gauche interactions whereas D has 3 methyl-methyl gauche interactions.

Qu31: D
Torsional strain is due to the alignment of bonds on adjacent atoms.  A is angle strain, B and C are Van der Waals strain and E is ring strain.

Qu32: C
The chair is the most stable and the boat is the least stable of these conformations with the twist boat in between. The chair is all staggered whereas the boat is partially eclipsed. The twist boat is more stable than the boat because the slight twist relieves a little of the torsional strain.

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.

Qu33: C
Longest chain is C10, contains a C-C only so a decane.  This means it has to be C or E E is wrong because the methyl group in the complex branch is on C2 and not C1 as the name implies.

Qu34: C
A ketone group at C1 (priority functional group), then use first point of difference to get the methyl group at C3, ethyl at C5. Remember to alphabetise ignoring the "di" so ethyl before dimethyl.

Qu35: C
The compound is an ester, -C(=O)-O  not an ether - the carbonyl group makes a difference and the carbon of the carbonyl is C1.  So it can not be E.  The acid part of the ester (left of C=O as drawn) has 5C and contains a C=C starting at C2 so it's a pentenoate.... that rules out A and B. The alcohol part of the ester (right of the C=O as drawn) is from isobutanol. The double bond has E stereochemistry.  Hence isobutyl (E)-3-methyl-2-pentenoate.

Qu36: E
Alkene stereochemistry as described by E and Z. The longest chain including the two C=C is 4 carbons so we need a butene. This limits our choice to C, D or E. The double bond has E stereochemistry because the higher priority groups (the OCH3 and the CH3) are on the opposite sides of the double bond.  Use first point of difference to get the chloro group at C.  Hence we have (E)-1-chloro-2-methoxybutene = E.

Qu37: E
Meta implies we have a pair of 1,3-substituents so that restricts us to B, D or E. Nitro = -NO2 and phenol implies a  -OH substituent on the benzene ring.  A is ortho-nitrophenol, B does not have a nitro group, NO is a nitroso.  C does not have a nitro group or an -OH, D is meta-aminophenol.

Qu38: C
Use the descriptors trans- and (1,2) and look at the position of the two methyl groups.... A trans-(1,4)-, B trans-(1,3)-, C trans-(1,2)-, D cis-(1,2)-, and E is cis-(1,3).

Qu39: C
-2-one ending implies we need a ketone. This narrows the choices to A, C or D.   B and E are aldehydes. The name is a pentenone so we need a C5 chain including the C=O and a C=C. D is wrong because it's a C6 chain.  In terms of assigning configurations, the group order is -OH > C=O > C=C > H.  Remember to assign the sense of the rotation to the order of the groups when the H is away from you.... A is R, we need S.

Qu40: B
Did you look at the nomenclature of bicyclics ? Bicyclic means two fused rings. The [3.2.1] means that there are 3C, 2C and 1C in the links between the shared (bridgehead) C atoms. This gets rid of C, D and E which are [2.2.1], [2.2.2] and [2.2.2] respectively. Then we number from one of the shared C bridgehead atoms via the longest link first so as to give the hydroxyl group (the first point of difference) the lowest number.