Fluoro-deoxyglucose and PET imaging are explored in this problem [1, 2]. It covers the stereochemica — Physical Chemistry — Kinetics Chemistry Question
Fluoro-deoxyglucose and PET imaging
Fluoro-deoxyglucose and PET imaging are explored in this problem [1, 2]. It covers the stereochemical relationships of [18F]-FDG anomers [2], the nuclear decay kinetics of fluorine-18 [3], the synthesis of the mannose sulfonate precursor [4], and the chemical details of the radiofluorination reaction using Crypt-222 as a phase transfer catalyst [5, 6].
The endocyclic bond cleavage path makes use of the equilibrium between two acyclic forms that can recyclize to yield both compounds [18F]-α-FDG and [18F]-β-FDG. Draw the two conformers of the acyclic form that lead to the formation of [18F]-α-FDG and [18F]-β-FDG, respectively. [VISUAL]
Model Answer
The solution shows the conformers of the acyclic form that recyclize to form [18F]-α-FDG and [18F]-β-FDG [7].
Choose the correct structural relationship(s) between [18F]-α-FDG and [18F]-β-FDG.
Model Answer
[18F]-α-FDG and [18F]-β-FDG are epimers. [18F]-α-FDG and [18F]-β-FDG are diastereoisomers [7, 8].
[Question text not fully available in source excerpts]
[Question text not fully available in source excerpts]
[Question text not fully available in source excerpts]
Establish the equation of the radioactive decay as a function of time and determine the value of the radioactive constant (λ, that corresponds to the kinetic constant associated with the radioactive disintegration reaction).
Model Answer
N(t) = N(0)×e−λt, hence λ = ln(2) / t1/2 = 0.0063 min−1 [8]
Given that one injection to a human requires 370 MBq for imaging purposes and that the patient needs to rest for one hour before imaging is processed, calculate the remaining activity at the time of imaging. Similarly, calculate the remaining activity after 4 hours.
Model Answer
(i) One hour (60 min) is required between injection and imaging, and N(0) = 370 MBq so the remaining activity at the time of imaging is N(60) = N(0)×exp(−0.0063 × 60) = 254 MBq. (ii) Similarly after 4 h (240 min), N(240) = 370×exp(−0.0063 × 240) = 81.5 MBq [8, 9].
The synthesis of the required peracetylated mannose sulfonate E is described below. E is the precursor for the radiofluorination reaction. Choose the mechanism of the nucleophilic substitution:
- first-order nucleophilic substitution
- second-order nucleophilic substitution
- elimination-addition mechanism
Model Answer
second-order nucleophilic substitution [9]
The first step is a peracetylation of D-mannose yielding A. Represent the tetrahedral intermediate involved in this reaction. [VISUAL]
Model Answer
The solution represents the tetrahedral intermediate of the peracetylation of D-mannose [9].
[Question text not fully available in source excerpts]
Choose the reactivity of the fluoride ion in this step:
- nucleophile
- electrophile
- base
Model Answer
nucleophile [9]
In the same transformation, choose the reactivity of 2,4,6-collidine among the following:
- nucleophilic catalyst
- base
- co-solvent
Model Answer
base [9]
One of the intermediates in the transformation from C to D is C’. Draw the structure of the ionic reaction intermediate between C and C’. [VISUAL]
Model Answer
Mesomeric structures featuring the positive charge on both oxygens are also acceptable [9].
Identify which of these three compounds F, G or H is the most reactive toward the nucleophilic substitution leading to I. [VISUAL]
Model Answer
F [9]
Crypt-222 has the following structure: Why is the use of crypt-222 a way to increase fluoride nucleophilicity? Choose the correct answer:
- Crypt-222 provides a chelation of potassium ion enhancing the fluoride nucleophilicity. [VISUAL]
Model Answer
Crypt-222 provides a chelation of potassium ion enhancing the fluoride nucleophilicity [9].
In the process of [18F]-FDG production, the first wash (or elution) is carried out with a slightly acidic aqueous phase (pH around 3) followed by the second wash with acetonitrile. Choose the correct statement among the following (a list of pKa values is given at the end of the problem):
- Crypt-222 is eluted in first place, then [18F]-FDG. [VISUAL]
Model Answer
Crypt-222 is eluted in first place, then [18F]-FDG [9].
[Question text regarding the toxicity and injection of glucose with FDG]
Model Answer
Glucose is not a toxic compound and can be injected together with FDG [10].
Calculate the minimal molar amount of 18F− required at the beginning of the process to allow one injection to human for imaging purposes.
Model Answer
370×10^6 Bq are required for one injection, the transformation and process from E to J take 30 min.
So N(0) = N(t)×e^(λt) with t = 30 min and λ = 0.0063 min−1, thus N(0) = 447×10^6 Bq.
Chemical yield of 75%: n = N(0) / 0.75 = 447×10^6 / 0.75 = 596×10^6 Bq are initially required.
The determination of the amount of 18F− is obtained by considering the specific activity of 18F. SA = 6.336×10^19 Bq mol−1:
n = 596×10^6 / 6.336×10^19 = 9.41×10−12 mol (9.41 pmol) [10, 11].