Qu1: B
Question about geometry goes back to hybridisation and geometries of some C systems. i is benzene, C = sp², bond angles of 120^o, ii  is cyclobutane, C = sp³, but due to the ring structure, the CCC bond angle about 90^o, and iii  is propane, a simple alkane, C = sp³, bond angles = 109.5^o, therefore giving : i > iii > ii

Qu2: E
Relates to hybridisation and nomenclature.  i and ii are both  alkanes, C = sp³, therefore 25% s character, while 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.
So how do we separate i and ii ? While i contains primary CH bonds whereas ii has secondary CH bonds.  Primary CH bonds are stronger (from alkane stability or radical laboratory expt). Therefore  iii > i > iI

Qu3: D
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 one C-H system and two O-H systems.
i is an ester, looking at the alkoxy (-OR group) where the conjugate base has the negative charge on a C atom next to an electronegative O atom.
ii is the protonated form on an ether (like H3O+ pKa anout -2), this time the conjugate base will be a neutral ether molecule.
Finally iii , an alcohol (pKa about 16) - the conjugate base puts negative charge on the O atom. So ii is more acidic than iii - neutral oxygen is favoured over negative oxygen, and i is the least acidic since a carbanion (C-) is less favourable than an oxyanion as in an alkoxide (O-). So overall we have ii > iii > i

Qu4: B
Calculate the formal charges for the N atom = group number - number of bonds - lone pairs electrons or by relating to known structures.
i = +1 based on 5 - 4 - 0, ii = -1 based on 5 - 2 - 4  and iii = 0 based on 5 - 3 - 2. Therefore i > iii > ii

Qu5: B
Isomers.... i has 6 (1-butene, trans-2-butene, cis-2-butene, 2-methylpropene, cyclobutane and methylcyclopropane).  ii has 3 (pentane, 2-methylbutane and 2,2-dimethylpropane) and iii has 5 (1,1-dichloropropane, (R)-1,2-dichloropropane, (S)-1,2-dichloropropane, 2,2-dichloropropane and 1,3-dichloropropane).
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 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: A
Acidity.... application to a simple amino acid.  But not a trivial question.  The carboxylic acid (typical pKa = 5) will protonate the amine (typical pKa about 10) so the amino acid exists as the zwitterion in solution..... (NH₃⁺)--CH₂--CO₂^- (i.e. ii) when one mole equivalent of NaOH (a base, pKa about 15) is added to this, the ammonium group will be deprotonated (since the pKa for the NaOH / H2O equilibirum is less acidic than that of the ammonium ion).  So the major species in solution will be i.  iii  is not "possible" since the carboxyic acid is a strong enough acid to protonate the basic amine (look at the pKas).  So  i > ii > iii

Qu8: E
Stability of alkanes in terms of heats of formation. More branched alkanes are more stable so ii is the most stable and iii is the least stable. The more stable the alkane the more exothermic the heat of formation.  Therefore  iii > i > ii

Qu9: E
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 i  > ii.  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 iii  > i. Therefore overall  iii > i > ii

Qu10: A
Resonance stabilisation... look at how the +ve charge can be stabilised. In i the +ve charge can be delocalised into the benzene ring. This is a benzyl cation. Now compare ii and iii. In iii  the nearby electronegative Cl atoms will destablise the +ve C due to indcutive effects. So overall i > ii > iii

Qu11: AE
Since water and dichloromethane are immiscible (think of the caffeine extraction experiment), a separatory funnel is an ideal method.

Qu12: C
Simply filter out the solid.  The larger apparatus it the best choice.

Qu13: D or BC or E
Check for melting point accuracy compared to literature values or use chromotography to look for impurities.

Qu14: C or AC
Simply filter out the crystals.

Qu15: B
Since water and ethanol are miscible, a distillation is needed.

Qu17: A
Oxidation states
C13 has the following bonds : 1 x C (count 0), 1 x N (count -1) and 2 x H (count +2) so the sum = +1 therefore the oxidation state of C13 = -1.
N26 has the following bonds : 1 x N (count 0) and 2 x C (count +2) so the sum = +2 therefore the oxidation state of N26 = -2.

Qu18: B
The functional group in question is C-N-C so it's an amine.

Qu19: D
The functional group in question is -C-O-C so it's an ether.

Qu20: E
IHD....Unsaturation arises due to pi bonds or rings.  So count these up in the structure:  there are 4 C=C, 2 C=N, 1 C=O and 2 S=O plus 4 rings (3 x 6 membered and 1 x 5 membered), so 13 in total.

Qu21: D
O10 is part of a double bond so it is sp2.  N14 is sp3 it's a simple amine N and N20 is part of a double bond C=N so it is sp2.

Qu22: AB
Hydrocarbon hybridisations.... C2 and C3 are part of aromatic double bonds, C=C, so they are sp2. N24 is sp2 because it is part of an amide and so the lone pair is delocalised in resonance with the adjacent C=O.

Qu23: A
There are no C atoms that are sp3 and have 4 different groups attached.

Qu24: D
If the ring exists as a chair then the two substituents on the N will the equatorial with the lone pairs axial. Hence there are 4 C atoms each with an axial H.

Qu25: B
The chair conformation of cyclohexane is a staggered conformation. Here the two principle groups, the cis-1,2-methyl groups are at 60^o torsional angle.

Qu26: B
The conformation shown is a staggered conformation since the bonds on adjacent atoms have torsional angles of 60^o.  Here the two principle groups, the chlorine atoms are on the same C atom (i.e. they are 1,1-dichloro) so no specific term can be used. An anti conformation requires 1,2-dichloro with a 180^o torsional angle .  A syn conformation requires 1,2-dichloro with a 0^o torsional angle and gauche has them at 60^o.  Trans is used to described substituents on cyclic structures or across double bonds.

Qu27: D
Look for the all staggered conformation of propane. A, B, C and E all some eclipsing.

Qu28: BCDE
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-ethoxypropene, CH2=CH(OCH2CH3)CH3, has an IHD = 1, and formula C₅H₁₀O.
2-pentanol is an alcohol = C₅H₁₂O and so has 0 unit of unsaturation.  2-methylbutanal is an aldehyde, one pi bond, C₅H₁₀O. 2-pentanone is a ketone, one pi bond, C₅H₁₀O. (Z)-1-ethoxy-1-propene is an ether with one C=C, C₅H₁₀O and cyclopentanol acid is an alcohol, with a cyclic unit, C₅H₁₀O. 

Qu31: C
Van der Waals strain is due to two groups being too close to each other.  A is angle strain, B and D are torsional strain, and E is ring strain.

Qu32: AB
The anti is the most stable (lowest energy) and the syn is the least stable (highest energy) of these conformations with the gauche in between.

Qu33: E
Longest chain is C9, contains a C-C only so we have a nonane with 3 methyl groups and a propyl group.  Numbering dictated by the first point of difference rule.  A is wrong because the alphabetisation is wrong, methyl should be before propyl. B is wrong because the substituent at C6 is not an isopropyl group. C is wrong because the substituent at C6 is not a butyl group.  D is wrong because the numbering is incorrect based ont the first point of difference rule. 

Qu34: E
The larger ring is C5 so a cyclopentane system with methyl and cyclopropyl substituents.  Alphabetise cyclopropyl before methyl and have to number based on this since the first point of difference rule does not resolve the numbering.  A is wrong because the numbering is wrong, B is wrong because the alphabetisation is wrong. C is wrong because the numbering is incorrect.  D is wrong because the cyclic nature of the propyl group has been omitted. 

Qu35: E
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.  The acid part of the ester (left of C=O as drawn) has 5C and contains a C=C starting at C3 so it's a pent-3-enoate.... that rules out B. The is a methyl substituent at C4... so that rules out B.  The alcohol part of the ester (right of the C=O as drawn) is from tert-butanol : that rules out A and B. The double bond has no stereochemistry so that rules out C and D.  Hence t-butyl 4-methyl-3-pentenoate.

Qu36: C
Alkene stereochemistry as described by E and Z. The longest chain including the C=C and the C=O is 5 carbons so we need a pentene. This limits our choice to C, D or E. The functional group is an aldehyde so need the suffix -al, this rules out D.  The double bond has Z stereochemistry because the higher priority groups (the CHO and the ethyl) are on the same side of the double bond.

Qu37: C
Para implies we have a pair of 1,4-substituents so that restricts us to C or D. The name indicates an ethyl ester so it must be C.  A is ethyl ortho-hydroxybenzoate, B is methyl meta-hydroxybenzoate.  D is the ether, para-ethoxyphenol. E is 2-ethyl-4-hydroxybenzoic acid.

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

Qu39:B
The name is a butanoic acid so we need a C4 chain including the C=O and one methyl substituent and one amino group. A is wrong because it's a C5 chain.  C is wrong because it has 2 methyl substitutents. This leaves B, D and E. In terms of assigning configurations, the group order is -NH2 > COOH > C(CH3)2 > H.  Remember to assign the sense of the rotation to the order of the groups when the H is away from you.... B is R, D and E are both S.

Qu40: A
Did you look at the nomenclature of polycyclics? Spiro means two rings linked at a single atom. The [3.4] means that there are 3C and 4C in the links between the commom C atom. This gets rid of C and D which are [2.5] and [2.4] respectively. Note that the octene means we have 8 C in the parent structure.  Then we number from the first C in the smallest ring being in mind the first point of difference rule. B is 2-chlorospiro[3.4]oct-6-ene and C is 1-chlorospiro[3.4]oct-5-ene.

Return to Homepage