A question about stability carbocations. Remember that the order of stability is tertiary > secondary > primary > methyl and that allylic carbocations are further stabilised by resonance...but vinyl cations are less stable than primary. Here we have a tertiary allylic > secondary allylic > vinyl

Here we have 3 ionic compunds, where the heteroatom in the organic part is -ve. We have the conjugate bases of an alcohol (alkoxide), a carboxylic acid (carboxylate) and a thiol (thiolate). If we compare the alkoxide and the carboxylate (both O-) then the resonance delocalisation in the carboxylate stabilises the -ve charge and makes it less nucleophilic. If we compare the alkoxide and the thiolate (O- vs S-) then the large S atom will be more polarisable and more nucleophilic.

First identify the reaction... SN2....(good Nu in polar aprotic solvent) which will tend to be controlled by the degree of substitution where the general trend is that primary > secondary > tertiary. We don't get simple SN2 at vinyl centers.

Qu 4:

First draw out the structures from the names and the identify the reaction, and the reaction conditions imply alcohol dehydration.... which will tend to be E1 and therefore the stability of the intermediate carbocation. Recall that for C+ stability tertiary > secondary > primary (due to e donating properties of alkyl groups). Alkyl bromides need base to eliminate.

Qu 5:

Radical chlorination of the cycloalkane (draw it) will be controlled by (i) the relative reactivity of each location, Ri and (ii) the number of H at each position, Hi

For chlorination, the reaction is controlled by a combination of the relative reactivity of the radical formed and the number of H that yield that radical. The relative reactivity of tertiary : secondary : primary systems are 5.2 : 3.9 : 1. The required calculations are outlined on the figure on the starting material.

Qu 6:

The number of lines = multiplicity (i.e. coupling) of the signals for each of the positions indicated is determined by the number of neighbours (n) that are of a different type and the n+1 rule. Here we see 7 lines (6 neighbours), a singlet and a doublet (1 neighbours). You need to remember that one doesn't observe coupling between H of the same type.

Qu 7:

The IR stretching frequency of a C=O is governed by the nature of the attached groups in the specific functional group. A ketone is the base / typical value at 1715 cm-1. Esters are a little higher (about 1735 cm-1) due to the increased double bond character due to the electronegativity of the O, while the anhydride in much higher (1810 & 1760) due to the electron withdrawing character of a carbonyl group.

We should work backwards.... Looking at the reagents used and the structure of the product, the methoxy ether has been formed by a nucleophilic substitution (SN2) of a tosylate which would have been formed from an alcohol. The SN2 will cause an inversion of stereochemistry, so we need the alcohol with the opposite configuration at the -OH center (same configuration at the isopropyl group). (v1 = B, v2 = C)

We should work forwards and review the transformation by comparing starting material and product. Since we are starting with an alkane, we will need to do a radical halogenation to get a reactive functional group. The halide is then been trasformed into an alcohol (SN1) and there has been a change in the skeleton which means we must have had a carbocation rearrangement during an SN1 reaction. Radical chlorination is going to tend to give a large % of primary chlorides, while bromination will give a high yield of a secondary bromide. The bromide will undergo an SN1 with aq. AgNO₃ (recall the SN lab expt) to give the required rearranged alcohol. (v1 = E, v2 = A)

We should work forwards... the first reaction is an SN2 involving the conversion of the alcohol to the alkyl halide using thionyl chloride via a nucleophilic substitution and then an anti-Zaitsev E2 elimination (due to the use of the strong but bulky base) with the loss of the better leaving group (the chloride) to give the less highly substituted alkene. (v1 = E, v2 = A)

We should work forwards... LDA (lithium diisopropyl amide, i.e. LiN(iPr)₂ ) is a strong base and will remove the most acidic H from the keto-ester starting material, that's the H in the "active methylene" between the two C=O groups. This forms a resonance stabilised "enolate" which is just a C nucleophile that then undergoes an SN2 reaction with the alkyl iodide to add the methyl group to the central C. (v1 = C, v2 = D)

We should work backwards... we are trying to make an ether using AgNO₃ (recall the SN lab expt) via an SN1 reaction. Ethers can be made by the reaction of an alcohol and an alkyl halide. Since the AgNO₃ reacts with an alkyl halide to make a carbocation we need an alkyl halide and it is best if it yields a more stable carbocation. (v1 = C, v2 = D)

We should work forwards...we have an alcohol being converted into an alkene : a dehydration, an elimination. It the more highly substituted alkene (Zaitsev's rule) so that all means simple acid catalysed dehydration is all that is required. (v1 = A v2 = C)

We can work forwards... the alcohol is being reacted with the acid HBr, which will convert the alcohol via a nucleophilic substitution to an alkyl bromide (SN1) which means a carbocation will be formed and we should consider the possibility of a rearrangement. In this case, the protonation of the R-OH will form the better LG that then leaves to give the secondary C+ which undergoes a 1,2-hydride shift to the more stable tertiary C+ and hence gives the tertiary bromide. (v1 = E, v2 = D)

IR has no carbonyl (i.e. C=O) near 1715 cm^-1 but shows C=C near 1600 cm^-1. The C-NMR shows 5 C types including 4 ArC 110-170 ppm that suggests a substituted benzene. The H-NMR has at least 2 types of H (note that all the ArH are showing as a multiplet). The 3H singlet at 2.3 ppm, is consistent with a -CH₃ group attached to the benzene ring. Checking the integration shows a 3:2 ratio which implies there needs to be 2 methyl groups to create a disubstituted benzene (i.e. 6:4). Given the number of ArC, it must be meta substituted, i.e. 1,3-dimethylbenzene.

IR shows a carbonyl (i.e. C=O) at 1720 cm^-1 which is typical for a ketone as is the C-NMR peak at 212 ppm. The H-NMR has 2 types of H. The 3H triplet at 1.1 ppm, suggests a -CH₃ group next to a CH₂, and the 2H quartet at 2.4ppm suggests a CH₂ that has 3H neighbours, i.e. an ethyl group. Remember that integration only gives empirical ratios. These pieces of data are consistent with pentan-3-one.

IR shows a carbonyl (i.e. C=O) at 1717 cm^-1 which is typical for a ketone as is the C-NMR peak at 209 ppm. The C-NMR shows 5 C types including a peak for the C=O at 209 ppm. The H-NMR has 4 types of H. The 3H triplet at 0.9 ppm, suggests a -CH₃ group next to a CH₂. The 2H sextet at 1.6 ppm, suggests a -CH₂ group with 5 H neighbours, the 3H singlet at 2.1 ppm, suggests a -CH₃ group with 0 neighbours (slightly deshielded) and the 2H triplet at 2.4 ppm suggests a slightly deshielded CH₂ that has 2H neighbours, together this suggests i.e. a methyl group and a propyl group. These pieces of data are consistent with the ketone, pentan-2-one.

IR shows a carbonyl (i.e. C=O) at 1715 cm^-1 which is typical for a ketone as is the C-NMR peak at 212 ppm. The C-NMR shows 5 C types including a peak for the C=O at 212 ppm. The H-NMR has 3 types of H. The 1H pentet at 1.7 ppm, suggests a -CH group next to 4H, the 2H pentet at 1.9 ppm, suggests a -CH₂ group with 4 H neighbours, and the 2H triplet at 2.3 ppm suggests a slightly deshielded CH₂ that has 2H neighbours. These pieces of data are consistent with the ketone, cyclohexanone.