Some Answers to the 301X Final Examination, January 24, The two diastereomers are, of course, the cis and trans diols.

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1 Some Answers to the 30X Final Examination, January 4, 007. The two diastereomers are, of course, the cis and trans diols. 3 3 cis trans The simplest mechanism through which both diols would give and 3, is an S-like process going through a common tertiary carbocation. 3 3 cis trans (a) (b) (a) (b) 3 3

2 By contrast, the requirement for backside displacement in an intramolecular S reaction would lead to the formation of from the trans isomer and 3 from the cis isomer. 3 3 cis trans = 3 = 3 displace from rear then deprotonate 3. The first one is easy - protonate-deprotonate

3 The second is a bit harder, but still not really taxing. There are two possible initial protonations, but only one can lead to the product shown. goes nowhere 3 + / or 3 + The last one is another easy one - just protonate, then deprotonate: note extensive resonance stabilization

4 3. (a and b) (c) In order to test whether the second step is one-step or two we need some sort of a stereochemical label (round up the usual suspects!). As only one of very many possibilities, cis--butene would do fine: onestep cis close able to rotate around single bond close + trans

5 4. The first reasonable step is formation of either a bromonium ion or the corresponding open carbocation (in fact it s the open one - look at the resonance stabilization). or "forcing conditions" The reaction doesn t go on to add bromide - the structure of the product tells you that (and we didn t ask why). To make bromobenzene, a proton must be lost - it is just a gardenvariety E step most easily written from the open ion, but possible from both intermediates shown above. "forcing + conditions" Toluene can form three possible ions. Two are better stabilized by resonance than the third (note the tertiary carbocation in A and B, but not ), and the transition states leading to them will profit energetically from that stabilization. owever there are steric problems in B, and the major product is the,4-isomer (called para). A B 3 tertiary tertiary 3 3 3

6 5. (a) the IR data tell you that one of the product sis an aldehyde, so the workup must have been reductive = /Pd or (3)S. (b)the small coupling constant tells you that the alkene is cis (you knew the product was an alkene from the formula). Thus, the reagent must be /Pb/a3 (Lindlar catalyst). (c,d,e) Each is an alcohol (IR ~ 3400 cm - ). These are the three ways to hydrate an alkene: straight hydration (Markovnikov with possible rearrangements), oxymercuration (Markovnikov without rearrangements), and hydroboration/oxidation. So, the three products must be: (c) (d) (e) one triplet at delta 4.5 multiplet at delta 4.5 no signal at delta 4.5 The MR spectrum tells them apart, and so: Box (c) contains:. B3. / Box (d) contains:. g(ac)/. ab4/ and Box (e) contains: /3 + (f) Finally a really tough one. can add either way depending on whether the box contains peroxides or some innocuous solvent like l4. owever the methyl groups of the isopropyl group are enantiotopic if Markovnikov addition occurs (one signal) but diastereotopic if anti-markovnikov addition occurs (two signals). So the box must contain and peroxides. not

7 6. (a) there will be six Ms (b) s φb φa p x p y p z (c)the bonding M of interacts with s to give Ms and 3. The antibonding M of interacts with px to give Ms and 4. either the py nor the pz orbital interacts with either the bonding or antibonding M of (net zero, orthogonal interactions), and so are unchanged in the calculation. 3 φb ± s φa ± p x 4 (d) rder the Ms by counting the nodes. M and are net bonding, 3 and 4 are antibonding, and the residual p orbitals are nonbonding. 4 3 Energy p y pz

8 eutral will have seven electrons, two from the pair of hydrogens and five from nitrogen. The cation will have one fewer electron and the anion one more electron. ere are the electronic occupancies for, +, and Energy p y p z p y p z neutral - ESP "sees" an unpaired electron cation - two unpaired electrons- ESR will "see" them 4 3 Energy p y anion - no unpaired electrons

9 7. In Russia, with pure materials, we get Markovnikov addition. Ring closing gives the cyclopropane:.. metals In Sweden, where peroxides are present, anti-markovnikov addition occurs to give a pair of diastereomeric dibromides (you should write the mechanism for the radical chain addition). Eclipsed forms are drawn for clarity = meso 3 3 racemic Ring closing leads to cis- and trans-,3-dimethylcyclopentane. two dibromides, each metals 3 3 meso rotate race mic 3 ra cemic 7 4 racemic

10 8. If a bromonium ion were involved, a pair of enantiomeric products must be formed. Py 3 l opens here mirror images 3 Py 3 l So a bromonium ion cannot be exclusively involved - the resonance-stabilized open cation must be involved to some extent, as only it can add bromide to give both diastereomers A and B. The cyclic ion can only give anti/trans addition Py 3 l + A + B 3 not mirror images

11 eating in methanol leads to an S reaction (make sure to ionize the right bromine!) that gives diastereomers and D. 3 A + B D 3 3 Finally, the E elimination leads inevitably from to E and from D to F - remember the requirement for a 80 E in open-chain molecules. 3 3 E 3 3 D 3 3 F

12 9. (a) Two dibromides would be formed, the one shown and 3-bromopentane. (b) There would be a rearrangement to give the tertiary alcohol. The fix is to use oxymercuration. (g(ac)/,. ab4/ (c)ydroboration leads to the less substituted alcohol. The fix is to hydrate the alkene with 3 + /. (d) The product would be,-dibromobutane, not,3-dibromobutane. (e) The product would be the cis dideuterio compound not the trans compound shown. (f) The products are all wrong. The correct product would be: (g) ofmann elimination would lead to -butene. The fix is to change the leaving group to a Saytzeff LG, such as or Ts. (h) The product would be a mixture of cis- and trans--butene because the LG is a Saytzeff LG. The fix is to change to a ofmann LG such as ammonium, fluoride, or sulfonium. (i) Addition would be in the anti-markovnikov sense. The fix is to get rid of those peroxides.