Qu1: A
i is a ketone, ii is an amide and iii also looks like a type of amide. First compare i and ii.... in an amide, the lone pair on the N is involved in resonance with the C=O, that resonance contributor adds single bond character to the C=O and therefore weakens it compare to the C=O in the ketone. So in iii there is more opportunities for resonance because of the second N atom and therefore the effect is greater. So i > ii > iii.

Qu2: D
Radical bromination of alkanes is controlled by radical stability so the the Br tends to add to the more highly substituted position. In i the situation is impossible, since the C has too many bonds, therefore none of this product is formed. ii is from the substitution of a tertiary C-H bond and iii is from a secondary C-H bond, hence ii > iii > i.

Qu3: E
Boiling point is a physical property and is therefore controlled by intermolecular forces. The system with the strongest intermolecular forces will have the highest boiling point. Therefore since the alcohol iii has a polar group that can participate in hydrogen bonding, which are the strongest intermolecular forces, iii will have the highest boiling point. If we look at i and ii then we see two isomeric alkanes, but they have different branching. More branching makes the alkane more spherical and decreases the surface areas in contact - this means less intermolecular forces and therefore a low boiling point. Hence iii > i > ii.

Qu4: B
First, draw the structures for each of the names and then count the number of types of C in each structure. Pentane has 3, cyclohexane has 1 and 2,2-dimethylpropane has 2. So i > iii > ii.

Qu5: A
In normal TLC, more polar materials with more run more slowly (and therefore have lower Rf values) because they interact with the polar coating on the TLC plate more strongly and this slows down their journey up the plate.  In the three molecules, we have substituted benzenes. i is methyl benzene and in non polar and can’t hydrogen bond so it will travel near the solvent front with an Rf close to 1. ii is an ester and has a polar C=O group. iii has two polar groups, the hydroxyl and the carboxylic acid group… both can also hydrogen bond, this means iii will travel quite slowly compared to i and ii, so iii has a low Rf value.  Therefore, in terms of Rf values, i > ii > iii.

Qu6: AB
H NMR chemical shifts… i is a CH₃ group next to a CH₂ and therefore we should expect to be the most shielded (i.e. lowest chemical shift).  ii is a CH₂ next to a CH₃ and a C=O. The C=O will cause some deshielding due to the pi bond and the electronegative O atom, typically to about 2.1 ppm. iii has a CH₃ that is attached directly to an electronegative O atom which strongly deshields those H to about 3.5 to 4 ppm. So in terms of chemical shifts, iii > ii > i. 

Qu7: C
Assign the formal charges...i a carbon radical is neutral, ii the hydronium ion has a +1 oxygen and iii a chloride ion is -1, so ii > i > iii.

Qu8: E
Ranking resonance structures... ii is the least important because it is impossible because it exceeds the octet rule on the central N atom. iii is has the least charges and hence is the most important : iii > i > ii.

Qu9: C
i is a CC in a benzene ring so the resonance in the aromatic system means that it is half way between a C-C and C=C. ii is a C-C between 2 sp² C atoms. iii is a C=C. C=C are shorter than C-C because of the more bonding interaction (i.e. higher bond order), so overall, in terms of lengths, ii > i > iii.

Qu10: C
The index of hydrogen deficiency (IHD) is a count of the number of pi bonds and rings. The key here is the effect of the substituents since each system has 4 IHD units due to the benzene rings alone. i has a carboxylic acid which contains a C=O and 1 IHD unit giving 5 in total. ii has a nitrile group which has a C≡N so has 2 IHD units giving 6 in total. iii has a CF3 and hence only single bonds are no additional IHD units. So in terms of IHD ii > i > iii.

Qu11: D
The index of hydrogen deficiency (IHD) is a count of the number of pi bonds and rings. Here there are 3 rings and 4 pi bonds to give a total of 7.

Qu12: B
O9 is a simple ether type oxygen and so it has 4 groups attached (2xC atoms and 2 lone pairs) and is sp³. N12 is part of a C=N and so it has 3 groups attached (2xC atoms and 1 lone pairs) and is sp².

Qu13: C
C7 has 4 groups attached, (1xC, 1xO and 2 lone pairs) so it is sp³ hybridised and therefore essentially tetrahedral with bond angles of close to 109.5 degrees.

Qu14: B
Higher infra-red stretching frequencies tend to involve H atoms, so that narrows it to B, C or E. O-H tend to be about 3500 and sp³ C-H about 2900 amd sp² C-H about 3100 cm^-1.

Qu15: C
While talking about alkane stability, we saw that 1^o C-H are stronger that 2^o C-H which in turn are stronger than 3^o C-H. So since A,B and D are 1^o and C and E are 2^o, we can focus of C and E. The difference here is that C is also allylic while E is not. The allylic character further weakens the C-H bonds due to the electron rich nature of the adjacent C=C (we also looked at bond strengths of these types of systems while looking at radical substitutions).

Qu16: A
C-C are longer than C-H (since C is a larger atom than H) and single bonds between sp³ C atoms are longer than those involving sp² C due to the lower s character of the sp³ orbitals.

Qu17: D
O21 is part of an ester so one of the lone pairs on the O is in a p orbital due to the resonance interaction with the C=O. The other lone pair is in an sp² hybrid orbital.

Qu18: CD
Functional groups....

Qu19: AD
C20 is only attached to one other C atom and is therefore classed as primary. Since it is also adjacent to a C=C of an alkene, it is also described as allylic.

SPECTROSCOPY:
Use any IR information to get the functional groups. Use the 13C NMR to get the number of types of C and what types of C are present. H-NMR to get the number of types of H, how many of each type from the integral and what they are next to from the coupling patterns. Chemical shifts should tell you if the group is near -O- or maybe C=O groups etc.

Qu20: BE
IR shows 2 C=O stretches that are quite high frequency, characteristic of an acid anhydride. The 13C confirms that the C=O is a carboxylic acid derivative. The H NMR only shows an ethyl group coupling pattern. All point to BE as the only possible answer.

Qu21: CE
IR shows a C=O stretch at a typical C=O value for a simple ketone. The 13C confirms that the C=O is a either an aldehyde or ketone and the lack of a peak at 9-10ppm in the H NMR indicates that it is a ketone. The H NMR only shows an ethyl group coupling pattern. This points to CE as the only possible answer.

Qu22: BC
IR shows a C=O stretch at 1650, a low C=O frequency. The IR also shows two bands at 3363 and 3192 that are higher than most C-H and probably indicate an NH₂ group (OH tend to be broad). The 13C indicates that the C=O is a carboxylic acid derivative so maybe an amide. The H NMR shows 3 types of H, including an ethyl group coupling pattern and 2 H that match to the an NH₂ group. This points to BC as the answer.

Qu23: C
IR shows a possible C-O stretch at 1266. The 13C indicates 3 types of Ar C and 2 other C, the one at 65 possibly deshielded by O. The H NMR shows 3 types of H, including an ethyl group coupling pattern and 2 H consistent with benzene ring. The H NMR integrals indicate a disubstituted aromatic with two substituents that have ethyl groups. The fact that there are 3 ArC indicates the ortho substitution pattern. C as the answer.

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.

Qu24: A
The first point of difference rule requires that the system be numbered as a 1,2,4 system. Alphabetisation means that the ethyl groups are named before the methyl group.

Qu25: E
The principle functional group here is the alcohol so it is numbered as C1. The longest chain including the OH is C6. Counting from the C1, the alkene is between C4 and C5, so we have a 4-ene system. To designate the C=C stereochemistry, use the Cahn-Ingold-Prelog rules : the CH₃ > H and the alkyl chain > CH₃ so it's E stereochemistry.

Qu26: B
The longest chain is C8, so it's an octane, and it has an ethyl substituent and a complex substituent = (1,1-dimethylethyl).

Qu27: A
The longest chain is C6, so it's a hex system, and since it has a C=C it's an alkene.... a hexene. The alkene get the lowest numbers, a 2-ene and this means the Cl are at 3 and 5. To designate the C=C stereochemistry, use the Cahn-Ingold-Prelog rules : the CH₃ > H and the Cl > alkyl chain so it's Z stereochemistry.

Qu28: B
Benzyl = C₆H₅CH₂- and t-butyl = (CH₃)₃C- ... so B. A and D are phenyl ethers not a benzyl. C is an isobutyl, D and E are sec-butyl and not a t-butyl ethers

Qu29: A
Only A, B and D are esters. B is ortho and D is dichloro.

Qu30: C
Only C, D and E are amides. E is an N-methyl not an N-ethyl amide. Use the Cahn-Ingold-Prelog rules to assign the configuration. D is R.

Qu31: C
This is an example of a bridged hydrocarbon. D and E are 2.2.2 systems, not the required 2.2.1. Numbering starts at a bridgehead C and then goes to an alkene (2-ene). A is 1,6-dimethyl and B is 2,3-dimethyl.

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