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:
<u>Part 1: C. $3,159.30</u>
<u>Part 2. C. –5; –135; –10,935</u>
Step-by-step explanation:
Part 1:
Price of the boat = $ 16,600
Depreciation rate = 14% = 0.14
Time of utilization of the boat = 11 years
Price of the boat after 11 years = Original price * (1 - Depreciation rate)^Time of utilization of the boat
Price of the boat after 11 years = 16,600 * (1 - 0.14)¹¹
Price of the boat after 11 years = 16,600 * 0.1903
<u>Price of the boat after 11 years = $ 3,159.30</u>
Part 2:
Let's find out the first term of the sequence given:
A(1) = -5 * 3¹⁻¹
A(1) = -5 * 1
A(1) = -5
Let's find out the fourth term of the sequence given:
A(4) = -5 * 3⁴⁻¹
A(4) = -5 * 3³
A(4) = -5 * 27
A(4) = -135
Let's find out the eighth term of the sequence given:
A(8) = -5 * 3⁸⁻¹
A(8) = -5 * 3⁷
A(8) = -5 * 2,187
A(8) = -10,935
Y = kx
80 = 16k
k = 80/16 = 5
direct variation
The first 2 choices are corect
Answer:
there will be about less times than the cube earns!
Step-by-step explanation:
