Fluorodeoxyglucose, namely the 18F isotopomer 2-deoxy-2-[18F]fluorodeoxyglucose (18F-FDG), is a comp — Analytical Chemistry Chemistry Question
Fluorinated radiotracers
Fluorodeoxyglucose, namely the 18F isotopomer 2-deoxy-2-[18F]fluorodeoxyglucose (18F-FDG), is a compound used in cancer diagnostics in a technique called positron emission tomography (PET). In this technique, the patient is treated with a radiotracer which is preferentially taken up by cancer cells. Upon radioactive decay, a positron is formed which rapidly annihilates with a nearby electron. A pair of γ-photons flying in opposite directions are produced and detected. This allows for the localization of the tumour with high sensitivity and spatial resolution.
The isotope 18F is produced by a proton bombardment technique. Which isotope of which element is used for the production of 18F?
Model Answer
Oxygen-18 (18O). 18-Fluorine is synthesized by the nuclear reaction: 18O + p → 18F + n.
What is the amount of 18F-FDG (in moles) present in one dose of 300 MBq (3×108 s−1)? The half-life of 18F is 109.771 min.
Model Answer
n = 4.73 × 10^-12 mol.
Calculation:
t_1/2 = 109.771 min = 6586 s
k = ln(2) / t_1/2 = 1.052 × 10^-4 s^-1
N = Activity / k = 3.00 × 10^8 s^-1 / (1.052 × 10^-4 s^-1) = 2.852 × 10^12 atoms
n = N / N_A = 2.852 × 10^12 / (6.022 × 10^23 mol^-1) = 4.73 × 10^-12 mol.
At what time point will the chemical energy of 18O-glucose, produced by the decay of 18F-FDG, be equal to the total energy of γ-photons not yet released from the remaining 18F-FDG? In other words, at which time point would decomposing all obtained 18O-glucose into CO2 and H2O produce the same amount of energy as the radioactive decay of all remaining 18F-FDG? The heat of combustion of glucose is 2 800 kJ mol−1.
Model Answer
t = 9.95 × 10^4 s = 27 h 38 min.
Calculation:
Chemical energy of one glucose molecule: E_c = 2 800 kJ / N_A = 4.650 × 10^-18 J
Energy of γ-photons per decaying 18F atom (due to positron annihilation): E_p = 2 * m_e * c^2 = 1.637 × 10^-13 J
At time t, total chemical energy of 18O-glucose equals total energy of unreleased γ-photons:
E_c * N_glucose = E_p * N_18F
E_c * [N_0(18F) - N(18F)] = E_p * N(18F)
E_c * [N_0(18F) * (1 - e^-kt)] = E_p * N_0(18F) * e^-kt
E_c * (1 - e^-kt) = E_p * e^-kt
E_c = (E_c + E_p) * e^-kt
e^-kt = E_c / (E_c + E_p)
t = ln((E_c + E_p) / E_c) / k = ln((4.650 × 10^-18 + 1.637 × 10^-13) / (4.650 × 10^-18)) / (1.052 × 10^-4 s^-1) = ln(35213) / (1.052 × 10^-4 s^-1) = 9.95 × 10^4 s.
Nevertheless, 18F-FDG is not the only fluorinated radiotracer in use. Compound 1 is a radiotracer used in the diagnostics of Parkinson’s disease (PD). Molecule 1 binds to the dopamine transporter (DAT), a membrane protein characteristic of dopaminergic neurons. Degeneration of this class of neurons is a symptom of PD. Therefore, targeted imaging of neural cells expressing DAT is advantageous in the diagnostics of the neurodegenerative disorder.
A freshly synthesized sample of K18F reacts with ditosylate A, producing monofluorinated precursor B. Molecule B further reacts with amine 2 to give the final radiotracer 1.
[VISUAL]
Model Answer
Structure of ditosylate A: TsO-CH2-CH2-OTs (ethane-1,2-diyl bis(4-methylbenzenesulfonate))
Structure of monofluorinated precursor B: 18F-CH2-CH2-OTs (2-[18F]fluoroethyl 4-methylbenzenesulfonate)
Additive X: Any K+ chelator, such as [2.2.2]cryptand (Kryptofix 222) or 18-crown-6 ether.
Amine 2 can be easily produced by a sequence of reactions starting from cocaine (3), a natural product from plants of the Erythroxylaceae family. The synthesis starts with acid-catalyzed hydrolysis of cocaine (3) leading to compound C (C9H15NO3). Subsequent elimination with POCl3 produces, after a methanolic workup, compound D. Addition of magnesium-containing reagent E to compound D provides, after subsequent aqueous workup, precursor 4.
[VISUAL]
Draw the structures of compounds C to E.
Model Answer
Structure of C: Ecgonine (hydrolysis product of cocaine, featuring a carboxylic acid group at C-2 and a hydroxyl group at C-3 in the tropane core).
Structure of D: Anhydroecgonine methyl ester (contains a double bond between C-2 and C-3 of the tropane ring, and a methyl ester group at C-2).
Structure of E: 4-chlorophenylmagnesium bromide (4-Cl-C6H4-MgBr).
Compound 4 is not the only stereoisomer which can be formed by the addition of E to D. Draw the structures of all the stereoisomers which are unwanted side-products of the transformation.
Model Answer
Unwanted side-products are the other three possible stereoisomers resulting from diastereoselective/non-selective additions across the double bond of D (varying the configuration of the ester group at C-2 and the 4-chlorophenyl group at C-3 on the tropane bicycle):
1) (2-beta, 3-alpha)-isomer (where the ester is beta and the aryl group is alpha, or trans to nitrogen bridge)
2) (2-alpha, 3-alpha)-isomer
3) (2-alpha, 3-beta)-isomer
The final step in the synthesis of secondary amine 2 involves demethylation with 1-chloroethyl chloroformate followed by workup with aqueous sodium carbonate. The mechanism of demethylation of 4 involves the formation of acylated intermediates F and G, and subsequent liberation of amine 2 with acidified hot methanol.
[VISUAL]
Draw the structures of intermediates F and G.
Model Answer
Structure of intermediate F: The quaternary ammonium salt formed by nucleophilic attack of the tropane nitrogen of 4 onto the carbonyl of 1-chloroethyl chloroformate ( Cl-).
Structure of intermediate G: The carbamate intermediate formed after thermal elimination of methyl chloride from F ().