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:
When balls equals 7 nuts equal 18. eat my sack
Answer:
y=0.5x
Step-by-step explanation: Plot the points then draw the line to then use the slope formula y=mx+b m is slope which is 1/2 which is 0.5, you can use either, and b is the y-intercept which is (0,0)
Given :
A patient with high blood pressure who weighs 226 pounds is put on a diet to lose 26 pounds in 3 months. The patient loses 8 3/4 pounds the first month and 11 5/8 pounds the second month.
To Find :
How much weight must be lost the third month for the goal to be achieved.
Solution :
Let, weight lose in third month is x.
So, 8 3/4 + 11 5/8 + x = 26
Therefore, weight lose in third month is 5.625 pounds.
Hence, this is the required solution.
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