353 MT Winter 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
Anions are more nucleophilic than neutral species (high e density) so ii>i and nucleophilicity increases as you go down a group becuase the larger atoms are more polarisable and this facilitates the bond formation so iii>ii

Qu2: B
Good LG are the conjugate bases of strong acids so by considering the acids involved HBr, H2O and NH4+ we can see that i > iii > ii

Qu3: D
AgNO3 / aq EtOH indicates an SN1 reaction, so our first concern should be the stability of the carbocations formed. ii is the very favourable benzyl cation (resonance stabilised), iii is secondary and i is the very poor phenyl cation (no resonance due to perpendicular nature of the orbitals) so rate should be ii > iii > i

Qu4: B
Two things we can use to get the answer to this. First the trends from the periodic table. Basicity increases R to L so C- > N- > O-. But that is only based on comparing atoms in similar states. In this question the C is only sp hybridised and this makes it more stable and therefore less basic (inc. s character moves e closer to the +ve nucleus). SO how do we deal with that ? Need to recall how we prepare the acetylide anion..... CH3CCH + NaNH2 ---> CH3CC- Na+ and NH3
This tells us that -NH2 (pKa=33) is a stronger base than CH3CC- (pKa=25). If an alkoxide were strong enough then it would be the base of choice for the reaction... but it isn't. (pKa=16) So i > iii > ii

Qu5: AB
Radical stability follows the same order as for carbocations 3 > 2 > 1. SO in the question we have i = 2, ii = 3 but iii is 2 and benzylic so iii > ii > i

Qu6: D
Reaction is of cyanide (good nucleophile) in a polar aprotic solvent (DMSO) so SN2. Therefore we need to look at degree of substitution and the LG. They are all Cl, so we have i = 3, ii = Me and iii = 1 therefore rate is ii > iii > i since the nore hindered the system the slower the reaction.

Qu7: D
Reaction conditions tells us that we are dealing with an elimination of water under acidic condtions so probably E1 and therefore controlled by the stability of the carbocation. i = phenyl, ii = 2o and benzylic and iii = secondary. Therefore ii > iii > i. Infact the phenol (i) will not dehydrate at all under these types of conditions.

Qu8: AB
Nucleophilicity. Due to polarisability (see above) S systems are better Nu than O therefore iii > ii. Since steric bulk tends to inhibit Nu properties due to the difficulties in getting at the reactive center, ii > i. So iii > ii > i


LABORATORY:
Based on the reaction in the alcohols and substitution experiments. Need to know the tests and reactions involved: what is a positive result, what does that tell you, what are the SM and P in the tests etc. For each question work out what you are being told and fit it to the selcetion provided.

Qu9: A
Slow reaction with AgNO3 / aq EtOH (SN1) and fast with NaI/acetone (SN2) indicates an primary alkyl halide

Qu10: E
Colourless test with Br2 indicates an alkene or alkyne. The alcohol that gives the ketone BD on oxidation is 2-butanol which is AD. Dehydration of 2-butanol gives mainly trans-2-butene, E.

Qu11: AB
Colourless test with Br2 indicates an alkene or alkyne. The Lucas test tells us that the alcohol is tertiary so t-butanol, AE, which when dehydrated gives alkene AB.

Qu12: BD
Precipitate with 2,4-DNP indicates either an aldehyde or a ketone. The Lucas test tells us that the alcohol is secondary, so it has to be 2-butanol, AD, which when oxidised gives BD

Qu13: C
Slow reaction with NaI/acetone (SN2) but rapidly with AgNO3 / aq EtOH tells us that we are after an tertiary alkyl halide. The halide obtained from the Lucas test of t-butanol AE must be t-butyl chloride, C.

Qu14: BE
AC is a primary alcohol so it can be oxidised to either the aldehyde BC or the acid BE. The clear red colour with 2,4-DNP is not a positive test (a precipitate is required) so we must have the acid BE.


PRODUCTS:
Start with the SM and work through the reagents one at a time drawing the new product at each step to derive the answer. If don't know the reagent, look at the subsequent step to see what that may be doing as it should give you a hint in terms of type of SM for that step.

Qu15: C
The alcohol would react thionyl chloride (SN2) to give the alkyl chloride, which in turn reacts with the Mg to generate the Grignard reagent. Grignard reagents are good bases so any acidic H will just protonate them to generate the alkane (C).

Qu16: A
The alcohol reacts with HBr to give the bromide (SN2). React with Li gives the alkyl lithium which on treatment with Cu generates the lithium dialkyl cuprate (CH3CH2CH2)2CuLi. Unlike Grignards or alkyl lithiums, cuprates react with alkyl bromides to give what looks like an SN2 react and "couples" the alkyl groups, so here we get butane CH3CH2CH2CH3

Qu17: A
Radical bromination of pentane should give 2-bromopentane as the major product (2o 4H). Substitution with water gives 2-pentanol which is oxidised with the chromate (see Cr, think [O] ! ) to give the ketone, 2-pentanone, A.

Qu18: C
A Gabriel synthesis of primary amines. Phthalimide is treated with a base to remove the NH proton generating an N-ve nucleophile. This reacts in an SN2 reaction with the n-propyl bromide which is then reacted with hydrazine (NH2NH2) to release the primary amine, C.

Qu19: C
Treatment of an alcohol with conc. sulphuric acid should cause a dehydration reaction (elimination of water). Protonation of the -OH to make a better leaving group, then loss of water gives a 2o carbocation. A 1,2 methyl shift can then occur to form the 2,3-dimethylbutane skeleton and the more stable 3o carbocation. Loss of a proton gives the tetra-substituted alkene, 2,3-dimethyl-2-butene, C.

Qu20: B
Reaction of a terminal acetylene (alkyne) with the NaNH2 generates the anion (acid/base reaction). This good carbon nucleophile then undergoes an SN2 reaction with methyl iodide to produce 2-pentyne, B.


STARTING MATERIALS:
Tougher than the previous section because it requires a fuller appreciation of reactions, that dreaded "thinking backwards" Two tactics to use. First look at the first reagent to see if it is a give away and tells you where to start from (ie. certain reagents have a very limited range of substrates they are used on !). If this does not work, trying working backwards, by saying to get this product using this reagent, I must have started from what ? If you struggle, look at the previous reagent for a hint. Counting C can also help !

Qu21: C
The only molecule there that can react with NaOH other than to give a simple deprotonation is C. The hydroxide function as a Nu, attacks the less substituted end in SN2 fashion, opening the epoxide, which on protonation gives the 1,2-diol.

Qu22: B
If we are reacting with Mg, we must be preparing a Grignard reagent, so we must be starting from an alkyl halide, so it has to be B.

Qu23: C
A tricky one. TsCl is used to prepare tosylates from alcohols, but all choices are alcohols, R-OH. Once we have our R-OTs group we are substituting using an acetate, CH3COO-, to give an ester, CH3COO-R which is hydrolysed with aq. acid to give back the alcohol R-OH. Note how we have not changed the system by addding any atoms (C, H or O). However this final step gives an inversion of stereochemistry because it is an SN2 type process. Therefore we must have started form the enantiomer of the product, so C.

Qu24: D
PBr3 reacts with alcohols to give bromides, so we can narrow it to C or D. We know we must have a halide becasue the next step is the preparation of a Grignard reagent which is then reacted with H2C=O (formaldehyde) followed by an acidic work-up to give a primary alcohol, which on oxidation with Cr gives the carboxylic acid. Therefore, the starting material only contains 6C (the 7th in the product is from the aldehyde), so the SM = D.

Qu25: D
Product we are after is a cyclic ether. Final reagent is Na metal, which could be used to prepare an alkoxide from an alcohol as part of a Williamson ether synthesis, in this case an intramolecular example. Therefore we need to break open the cyclic ether to a 5-halo-pentan-1-ol (count C !) Working back from this NaI /acetone is SN2 method and would have put in a -I (to create a better LG for the Williamson step. The aq NaOH has probably been used to introduce the -OH group. Given the trend, the only reasonable conclusion is D. It has the right number of C, and means we can change the -Br to -OH, then the -Cl to the -I then use the Na to prepare the nuclophile for the ether formation.

Qu26: D
A Wittig reaction, given away by the PPh3 and the alkene in the product. By counting C, D has to be the answer. The sequence is the PPh3 reacts with ethyl bromide via SN2 to give a phophonium bromide. Deprotonation with a strong bsae (BuLi) gives the ylid, or Wittig reagent (Ph3P=CHCH3) which will react with the ketone to give the exocyclic, tri-substituted alkene.


APPLIED SPECTROSCOPY:
Best method is to work through the reaction sequence drawing the products of each step then thinking about the key features of the H-nmr of the compound, then match to the spectra. Remember that the reagents in the following step can also give you a hint about the functional groups in the product of the previous step.... If that just isn't happening because you don't know your reactions well enough, then the slower alternative is to try to determine what each spectra could be, and link that to you do know about the reactions.

Qu27: D
Reaction of the tolyl chloride (p-chlorotoluene) with Li generates the aryl lithium. This reacts with the epoxide, followed by an acid work-up to give 2-(p-tolyl)ethanol. H-nmr of this should be characterised by 3H singlet for the Ar-CH3 around 2.3ppm, 4 ArH (7-8ppm), 2 triplets for the two coupled -CH2- units (one deshielded due to -O-) and a singlet for the -OH. This is spectrum D.

Qu28: E
2-(p-tolyl)ethanol is just an alcohol that will be dehydrated by H2SO4 to give the alkene, p-tolylethene. The H-nmr of this will show a 3H singlet for the Ar-CH3 (@ 2.3ppm), 4 ArH (7-8ppm) and a 3H mulitplet pattern between 5-7ppm, characteristic of alkene H. This is spectrum E.

Qu29: A
Catalytic hydrogenation of the p-tolylethene will reduce the alkene to the alkane, p-tolylethane. The H-nmr of this will show a 3H singlet for the Ar-CH3 (@ 2.3ppm), 4 ArH (7-8ppm), then very characteristic 2H quartet and 3H triplet for the ethyl group. This is spectrum A.

Qu30: B
Radical chlorination of p-tolylethane will give 1-chloro-1-(p-tolyl)-ethane as the major product due to the stability of the secondary, benzylic radical. The H-nmr will be characterised by 3H singlet for the Ar-CH3 (@ 2.3ppm), 4 ArH (7-8ppm), a deshielded 1H triplet (above 4ppm due to Cl and Ar) and a 3H doublet, as shown by spectrum B.

Qu31: AE
Water will substitute the benzylic chloride to give 1-(p-tolyl)-ethanol via an SN1 reaction. The H-nmr will show 3H singlet for the Ar-CH3 (@ 2.3ppm), 4 ArH (7-8ppm), a deshielded 1H quartet (above 4ppm due to O and Ar), a 3H doublet (-CH3) and a 1H for the -OH. See AE.


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