2012 RI_H3 Chemistry Answer

Raffles Institution 2012 H3 Pharmaceutical Chemistry Prelim (Suggested Solutions)

1

(a) (i) Competitive inhibition is a form of enzyme inhibition where the inhibitor binds reversibly to the active site on the enzyme and prevents binding of the natural substrate to the enzyme and hence removes the enzymatic activity.[1] The effect of inhibition can be displaced by introducing higher concentrations of the natural substrate.[1]

(ii) The hydrophobic parts of valdecoxib can interact with the hydrophobic valine residue through van der Waals‟ forces of interaction[1]. (Show by drawing is acceptable) This occupation of the active site of COX-2 would inhibit the formation of prostanoids (e.g. prostaglandins) from arachidonic acid[1], which causes symptoms of inflammation and pain.

(b)

Isoxazole is less susceptible than furan to electrophilic aromatic substitutions (EAS). [1] Reason (either one) [1] The presence of an additional electronegative N atom makes the molecule less nucleophilic and the delocalized π electrons are less able to attack electrophiles. OR The nitrogen atom on isoxazole has a tendency to (act as a nucleophile instead) attack the Lewis acidic electrophiles. This places a positive charge on the ring, further deactivating it from electrophilic substitution Note that the last factor in the lecture notes “The lone pair of electron on N of isoxazole is perpendicular to the π system and is thus unable to stabilize the positively charged Wheland intermediate effectively.” is not an effective explanation for reduced susceptibility to EAS.

1

(c) (i) NH2OH [1] (actual reagent NH2OH.HCl) (ii)

structures [1]

*note that both isomers can be used to proceed with the synthesis. correct labels [1] (iii)

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[1] (iv) Intermediate C is

[1]

(c) (v) Nucleophilic addition [1]

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(c) (vi)

OR

Correct anti-periplanar arrangement in Newman projection [1] Correct arrow pushing [1]

(d) (i) Some vibrational modes do not result in a change in polarity. Or some peaks are too close together such that they become convoluted. [1] (ii) Accept 1610 – 1750 cm–1 [real value 1660 cm–1]. [1] Based on the formula, the stretching frequency is dependent on k (force constant) and μ (reduced mass). ~

1

  2 c

k



Since C=N has a double bond, and the reduced mass of C=N is between that of C=C and C=O, we should be looking at a stretching frequency between C=C and C=O.[1] (iii) Compound B‟s IR spectrum would have abroad O-H peak between 3230 and 3550 cm−1 which is absent in that of A. [1] Reject: using C=O stretch in A as B has a C=N stretch which would be similar.

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(a) Bacterial transpeptidase is a bacterial enzyme that cross-links the peptidoglycan chains to form rigid cell walls. [1] The cell wall weakens, and this leads to cell death by osmotic imbalance (pressure differences outside and inside of cell). [½] Hence penicillin is bactericidal.[½] (b) (i) Permanent covalent bond formation with the enzyme[1] (ii)

[3] (iii) The free carboxylic acid group exists as –COO–(carboxylate) ion at physiological pH of 7.4, which holds the drug molecule in the right orientation via ionic interaction[½]with the alkyl ammonium ion of lysine residue[½] on the active site of enzyme.

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2

(c)

(i)

[3]

(ii) Any structure with electron-withdrawing group (EWG) in the R group. [1] e.g.

(d) (i) Its extensive conjugated π system lowers energy gap between HOMO and LUMO such that photons in the UV region of electromagnetic spectrum can excite electrons from HOMO to LUMO. [1] (ii) Desired average concentration = (2+8)/2 = 5 mg dm–3 and given Mr of drug = 424 Page 5 of 16

Hence, desired average concentration =

5  103  1.18 x 10–5mol dm–3 [1] 424

(iii) Using Beer‟s Law, A = εcl=(3.0 x 104)(1.18 x 10–5)(1) = 0.354 [1] (e) (i) Minimum: 1.961 mg [1]; maximum: 501.961 mg [1] (See working below) For every 12-hour lapse, there are 8 half lives. Hence the amount (in mg) will have to be multiplied by (0.5)8 Hours 0 12 24 36 8am (forgot to take drug) 8pm (double dosage) 8am

Before Dose / mg 0 500 * (0.5)8= 1.953 501.953 * (0.5)8 = 1.961 501.961 * (0.5)8 = 1.961 1.961

After Dose / mg 500 501.953 501.961 501.961 1.961

1.961 * (0.5)8 = 0.00766

1000.00766

1000.00766 * (0.5)8=3.906

(ii) Minimum: 3.91 mg [1]; maximum: 1000 mg [1] (See working above) (iii) Continuing on the course of antibiotic would allow his body‟s natural defence to kill off totally even the penicillin-resistant bacteria.[1] Otherwise, the remaining bacteria may further mutate and develop greater resistance to the drug.[1] 3     

(a) Antibiotics may (any 3 modes of action) [3] Inhibit crosslinking of bacterial cell wall peptidoglycan by trans-peptidases, hence inhibiting cell wall synthesis and causing bacterial death by osmotic lysis Inhibit gyrase which uncoils the bacteria supercoil for nucleic acid transcription or replication, hence preventing bacterial replication. Block folic acid synthesis from 4-aminobenzoic acid by dihydropteroatesynthetase (distruption of cell metabolism) Block protein translation by inhibiting ribosomes (stated in question) Disrupt bacterial plasma membranes by acting as a hydrophilic channel throughwhich ions flow, allowing ions to enter the bacterium, upsetting ionic balance and causing cell lysis.

(b) Most unstable boat conformation:

[1] Show the two bulky substituents in the flagpole positions The two bulky substituents are in the flagpole positions and experiencing the largest steric hindrance[½]

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Most stable chair conformation:

[1] Show the lone pairs of electrons on N at axial positions & substitutions on N at the equatorial positions The two bulkiest substituents are in the equatorial positions, minimizing 1,3-diaxial interactions [½]

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(c) (i) The lone pair of electrons on N-1 and N-2 are not available for nucleophilic attack due to resonance with the C=S group. [1]

[1]

(ii) Nucleophilic addition-elimination(preferred) / Condensation[1] [1]

[1]

[1]

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Hence F is

[1]

3

(c) (iii)

Nucleophilic aromatic substitution [1]

[1]

[1]

–1 if no arrows all the way to Nitro group!

3

(c) (iv) Step 5: Acyl chloride (RCOCl), (pyridine solvent), room temperature [1]

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(d) (i)

As for the receptor site‟s protein side chains, do accept sensible alternatives

receptor site Show VDW forces, Hydrogen bond, ionic bond (any two) [2]

(ii) H is resistant to acidic hydrolysis. [1]  

4

Amide linkage is less susceptible(than ester) to hydrolysis H is made up of aromatic rings which are resistant to acidic hydrolysis

(a) (i) Compound L is the most polar.[½]It has the lowest Rf value.[½] Give full credit if the following is given: In the presence of ammonia, C will be deprotonatedto give a species which contains a full negative charge[½]which binds the strongest to the silica gel.

(ii) J binds more strongly to the hydroxyl groups of the silica gel, and hence having a lower Rf value, because it contains more potential sites for hydrogen bonding (two N with lone pair of electrons and two H attached to the N) as compared to K (one N with with lone pair of electrons and two H attached to the N). [1] (iii) Explanation (not required): All three compounds are now fully ionized (because of protonation of basic N in all three compounds by strong acid HCl) and bind very strongly to silica gel via ion-dipole interactions. They are not likely to move up the silica gel with the same composition of dichloromethane and methanol.

J K L

[1] Page 9 of 16

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(b) (i)

Note that the non-cyclic bonds are all eclipsed!

[1] (½ mark for each strain stated; no need to draw Newman)

(b) (ii) [1] – shows the lone pair of electrons (hence nucleophilic) [1] – shows the empty p orbital (hence electrophilic)

(b)

(iii)

[1] – correct arrow pushing [1][1] – correct products with correct stereochemistry label

(c) (i) Nucleus of proton spins and generates a magnetic moment. When a magnetic field is applied, the magnetic moment aligns with or against the applied magnetic field. [½]To switch from the lower energy spin state that aligns with the applied field to the higher energy state that aligns against the applied field, radio-frequency radiation is absorbed. [1] The exact frequency of absorption depends on the chemical environment of the proton. [½] Page 10 of 16

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(c) (ii) Each deduction [1] No Observations 1

(M+1)+ peak is 8.9% of M+ peak

Deductions for structure of W Using the formula n =

100  AM+1  100 (8.9%)  8   1.1  AM  1.1

the molecular formula of W is C8H10O4 2

integration data + fact that W contains only C, H and O + M+ = 170

Unsaturation index =

18  10 4 2

Cannot contain benzene ring as a cyclopropane ring is present.

3

None of 1H NMR signals disappears when D2O is used instead of CDCl3

no acidic/labile protons present W should contain 5-membered lactone.

4

The 1,4-diol moiety in M and N is achieved by treating W with LiAlH4.

ppm 0.88

integration 2

splitting doublet

1.23

3

triplet

2.07

1

multiplet

4.21

2

quartet

4.26

2

doublet

(Cannot be because missing aldehyde protons in 1H NMR)

Indicates –CH2CH3 group

[1]

Indicates –CH2CHCH2– group

MAX 4 reasoning marks Answer:

[2]

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[1]

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(c) (iii)

[1]

(iv) M has a more extended conjugation due to an additional  bond outside of the aromatic ring. [1] Hence the  → transition has a smaller energy gap for M which corresponds to a higher max. Hence N will have a shortermax.[1]

5

(a) (i) A stimulant is a drug that acts on the nerve synapses to „wake up‟ the central nervous system(CNS). A stimulant is a drug that „wakes up‟ the central nervous system(CNS), mimicking the physiological effect of the neurotransmitter noradrenaline.[1] (ii) An agonist is a drug that mimics the natural ligand for a receptor and will dock at a binding site of the receptor, causing it to change its shape and result in the same physiological effect as the binding of the natural ligand.[1] However while an antagonist is a drug that docks well at the binding sit, it does not cause the required change in the shape of the receptor or may distort it the wrong way and does not yield the desired physiological effect.[1]

(b) (i)

[1]

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(b) (ii)

[1] (b) (iii) Since the blood-brain barrier can be considered a hydrophobic barrier[1], neutral mazindol, instead of the positively charged keto-mazindol, will pass through it better. [1] BBB restricts the diffusion of microscopic objects (e.g. bacteria) and large or hydrophilic molecules into the cerebrospinal fluid (CSF), while allowing the diffusion of small hydrophobic molecules (O 2, CO2, hormones). (c) (i) Nucleophilic addition-elimination (no mark given here but it is always good to write it down)

[3]

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(c) (ii) The reaction completion can be monitored using moist red litmus paper which can detect that no more alkaline NH3 is evolved which marks the end of the reaction. [1]

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(iii) Solvent X is dry THF (or dry ether). [1] Reagent Y is

(Allow acid chloride) [1] (iv) No, the mazindol obtained here is optically impure/inactive. [½] Step 2b is nucleophilic addition. The nucleophile approaches the planar carbonyl compound from above and below the plane with equal chance to give a racemic mixture. [½]

5

(d) (i)

[2] Allow 7.76 for aromatic protons next to -Cl

(d)

(ii)

(doublet) Int = 2 highest ppm

(multiplet) Int = 3 highest ppm

Singlet Int = 1 variable ppm

Singlet Int = 4 ~3.5ppm

[2]

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3.79, singlet, 4 chemically equivalent protons. 7.83, m, 2 ortho-H, downfield, because they are closer to N atoms 7.52, m, 3 other aromatic H atoms on mono-substituted benzene

[1]

(d) (iii) Add D2O to a sample in CDCl3 and shake. The signal due to the labile proton (N-H) will be absent in the new spectrum.[1]

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(a) Any two from the following [2]: effect

aspirin

paracetamol

 causing gastrointestinal tract blood loss





 increasing tendency to bleed





More

Less

 potential for causing nausea and vomiting (b)

Caffeine might cause insomnia (increased adrenaline levels) and hence it should be removed from the Night capsule which people take before they sleep. [1] Any two from this list: increased respiration rate / increased heart rate / small increase in blood pressure / more blood flow to the peripheral&to heart & muscles, less to skin and inner organs / increased secretion of digestive juices / increased urine flow / increased blood sugar. [1]

6

(c) (i) Reason 1: NaH is a stronger base than KOH and hence can deprotonate the amide H+ and generate a stronger nucleophile.[1] Reason 2: As opposed to the protic solvent EtOH, DMSO is aprotic and it does not form favorable hydrogen bonds (or ion-dipole interactions) with the nucleophile (solvation). Hence the nucleophile would be more exposed for reaction.[1] OR KOH in Method I may undergo nucleophilicsubsitution with CH3I , resulting in less available CH3I for reaction with uracil. [1]

(ii) No. Reduction of secondary amide to secondary amines will occur instead. OR Yes. The amide is part of the aromatic ring and hence it could be resistant to the reducing properties of LiAlH4. LiAlH4 might act as a base instead and the reaction might proceed. [1]

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(iii)

[1] (iv)

Each correct step [1]X3 (v) Reduction of aromatic nitro [1] followed by nucleophilic addition-elimination [1]

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(d) (i) The movement of a liquid mobile phase carrying the solutes to be separated through a (solid) stationary phase, under high pressure, wherein the solute molecules partition themselves between the stationary phase and the moving mobile phase. [1] Separation is achieved due to their different adsorption/partition affinities of solutesto the solid stationary phase as a result of their different polarities. [½]The stronger the affinity, the slower the solute moves. [½] (ii) Decreasing polarity (binding affinity to polar silica gel) of compounds follows the order: theophylline, U and PhCH2I. PhCH2I is the least polar due to its inability to form any hydrogen bonds with hydroxyl functional group.[1]Since it has the highest retention time, reverse-phase HPLC must be used here. [1] Peak II corresponds to U [1] since Peak I with the shortest retention time should correspond to theophylline which is the most polar of all three compounds due to its ability to form the most number of hydrogen bonds with hydroxyl functional group (more than U which lost a H atom bonded to N in the reaction). (iii) Peak Area I : ½ x 1 x 10 = 5 units2[½] Peak Area II :½ x 10 x 14 = 70 units2[½] % yield =

70  100%  93.3% [1] 70  5 Page 16 of 16