Answer:The crystal structures of five 6-mercaptopurine derivatives, viz. 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(3-methoxyphenyl)ethan-1-one (1), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-methoxyphenyl)ethan-1-one (2), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-chlorophenyl)ethan-1-one (3), C15H11ClN4O2S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-bromophenyl)ethan-1-one (4), C15H11BrN4O2S, and 1-(3-methoxyphenyl)-2-[(9H-purin-6-yl)sulfanyl]ethan-1-one (5), C14H12N4O2S. Compounds (2), (3) and (4) are isomorphous and accordingly their molecular and supramolecular structures are similar. An analysis of the dihedral angles between the purine and exocyclic phenyl rings show that the molecules of (1) and (5) are essentially planar but that in the case of the three isomorphous compounds (2), (3) and (4), these rings are twisted by a dihedral angle of approximately 38°. With the exception of (1) all molecules are linked by weak C—H⋯O hydrogen bonds in their crystals. There is π–π stacking in all compounds. A Cambridge Structural Database search revealed the existence of 11 deposited compounds containing the 1-phenyl-2-sulfanylethanone scaffold; of these, only eight have a cyclic ring as substituent, the majority of these being heterocycles.
Keywords: crystal structure, mercaptopurines, supramolecular structure
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Chemical context
Purines, purine nucleosides and their analogs, are nitrogen-containing heterocycles ubiquitous in nature and present in biological systems like man, plants and marine organisms (Legraverend, 2008 ▸). These types of heterocycles take part of the core structure of guanine and adenine in nucleic acids (DNA and RNA) being involved in diverse in vivo catabolic and anabolic metabolic pathways.
6-Mercaptopurine is a water insoluble purine analogue, which attracted attention due to its antitumor and immunosuppressive properties. The drug is used, among others, in the treatment of rheumathologic disorders, cancer and prevent
Step-by-step explanation:
Answer:
What was the problem.
Step-by-step explanation:
Answer:
a) 0.023
b) 0.286
c) 10 students will be unable to complete the exam inthe allotted time.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 80 minutes
Standard Deviation, σ = 10 minutes
We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.
Formula:
a) P(completing the exam in one hour or less)
P(x < 60)
Calculation the value from standard normal z table, we have,
b) P(complete the exam in more than 60 minutes but less than 75 minutes)
c) P(completing the exam in more than 90 minutes)
P(x > 90)
Calculation the value from standard normal z table, we have,
15.87% of children of class will require more than 90 minutes to complete the test.
Number of children =
Approximately, 10 students of class will require more than 90 minutes to complete the test.
Answer:
(3,1)
Step-by-step explanation:
Multiplying the second equation by 3, we get
-12x-6y=-42.
Now, adding this to the first equation, we get
-13x=-39
Therefore, x=3
plugging in x=3 to the first equation, we get that
-3+6y=3
adding 3 to both sides,
6y=6
therefore, y=1
