Chemie II Revision Tutorial J.D.Revell

Transcription

1 Chemie II Revision Tutorial J.D.Revell ydrolysis of the itrile Group: a) Acidic ydrolysis of a itrile (C) Group C Proton transfer is a ERY good leaving group! Final step is ERY fast - b) Basic (very slow) ydrolysis of a itrile (C) Group C a is a ERY poor leaving group! Final step is ERY slow -

2 Acetalisation Mechanism: Question: Draw the mechanism for the following acetal formation between cyclohexanone and ethylene glycol. + cat. acid Cyclohexanone (Electrophile) Ethylene glycol (di-nucleophile) Cyclic acetal (protecting group for ketones) Mechanism: Proton transfer cat. acid Protonated ketone more electrophilic! Tetrahedral geminal heteroatoms - Acid is re-formed = catalytic - Desired acetal product xonium Ion Intermediate Points to Remember: The ketone group of cyclohexanone is a good electrophile, but we can make it an even better electrophile if we protonate it! ow there is also the driving force to neutralise this new positive charged on oxygen and the reaction with nucleophiles will be faster. Ethylene glycol has hydroxyl groups which are both good nucleophiles. Since these hydroxyl groups are in the same molecule, you should see that if they both react with other molecule (an electrophile) this will form a ring (cyclic molecule). This is a big driving force for cyclic acetalisation reactions, and ethylene glycol is very often used to form stable cyclic acetals. ery importantly the reaction goes through an xonium ion intermediate (see above), you should always write this intermediate in the reaction mechanism! ALL of the steps of this mechanism are reversible, so if we want to make the reaction go to full conversion we must remove water to stop these backwards reactions. The acetal group is stable to most bases and is therefore used as a protecting group when we need to protect a carbonyl group during basic reactions (e.g. enolisation) on other parts of the same molecule. Since all the steps are reversible, we can also remove the acetal group simply by adding catalytic acid and water; now the reaction will run backwards and give us back the ketone and ethylene glycol.

3 Stereochemistry Question: Determine the configuration of the Carboxylic acid shown below. What to Remember: st we must find the stereocentre (chiral atom) in the molecule, remember that this will be a carbon atom with DIFFERET groups attached. nd we must give each group attached the chiral centre a priority (number of importance), remember that the EAIEST atom is the most important and is given the LWEST number (highest priority). rd we must (often but not always) re-draw the molecule so that the atom with the LWEST priority () goes to the back. th we must say if the other numbers, & count clockwise (Rectus) or anti-clockwise (Sinister). Ph C Chloride is the heaviest atom in the molecule and takes highest priority () ydrogen is the lightest atom in the molecule and takes lowest priority () Ph C ere we have carbon atoms which have the same priority, so we need to consider the EXT atom! In the Benzene ring this is again carbon, but in the Carboxylic acid the next atom is oxygen. Since oxygen is heavier than carbon, the Carboxylic acid takes the next highest priority Turn the molecule so that priority (LWEST) is to the back C Ph Count the Priorities starting with number then and. ere,, counts ATI-CLCKWISE and so the stereocentre is (S) The chiral centre of the Carboxylic acid has an (S) configuration.

4 Question: Determine the configuration of the compounds shown below. C F C(C ) C(C ) F C GRUP PRIRITY F > > C(C ) > C > > > eaviest atom > Lightest atom Turn the molecule to put number to the back Is the Priority from, & clockwise or anti-clockwise? Anti-clockwise count from to to = (S) (S) configuration. Ph C CBr C Priorities: C > Ph > C CBr > C Priority is already at the back so we only need to count the direction. Is the Priority from, & clockwise or anti-clockwise? Anti-clockwise count from to to = (S) (S) configuration. C C C C C C(C ) C C(C ) The compound is shown as a Fischer Projection. The lower half of the drawing can shortened to C(C ) because these groups are identical and therefore not stereogenic The Fischer Projection can now be re-drawn as above and priorities can now be given more easily Turn the molecule so that priority is at the back. ockwise or anti-clockwise? ockwise count from to to = (R) (R) configuration. C * * C C C Remember: The chiral centre has different groups attached. ere, the only chiral centre in the molecule is shown with a star* The carbon atom bearing atoms takes highest priority, the proton takes lowest priority. Priority is already at the back so we only need to count the direction. Anti-clockwise count from to to = (S) (S) configuration.

5 Question: Draw stereo diagrams of the active (S) enantiomers of Ibuprofen, methyldopa and Penicillamine shown below. Ibuprofen Methyldopa Penicillamine C C C C C C C C C S Remember: st find the stereocentre (chiral atom) in the molecule = C with DIFFERET groups attached. nd give priorities to each group on the chiral centre. EAIEST atom = LIGTEST atom = rd re-draw the molecule so that the atom with the LWEST priority () goes to the back. th if to to counts clockwise the chiral centre is (R) (Rectus) if to to counts anti-clockwise the chiral centre is (S) (Sinister). 5th if this drawing is not the enantiomer that you want ( to to counts the wrong way) change the positions of groups and (don't change the position of group, keep this at the back) this will give you the correct count direction. C C C C C C C C S C (S)-Ibuprofen (S)-Methyldopa (S)-Penicillamine

6 Question: Draw stereo diagrams of: (S)--methylbutanoic acid and (R)--(ethoxycarbonyl)propanoic acid. Et Me * Me C * Anti-clockwise count = (S) C C Et ockwise count = (R) Question: Assign R or S configurations to the chiral centre in each molecule below. C C C Br C Br Br Br Anti-clockwise count = (S) C ockwise count = (R) C Question: Convert the Fischer projections below into stereo diagrams and assign R or S configurations to each of them. C C C C Et C C Et Remember: Turn the molecules so that priority (lowest priority, e.g. ) is at the back! C C C C Et C C Et C C C C Et C C Et ockwise count (R) ockwise count (R) ockwise count (R) Question: Assign R or S configurations to the natural amino acids shown below. C C(C ) aline C C Serine C C S Cysteine Proline C (S)-aline, (S)-Serine, (R)-Cysteine (Sulfur is heavier than xgen!) and (S)-Proline.