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
Go to:
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: There is sufficient evidence to reject the dealer's claim that the mean price is at least $20,500
Step-by-step explanation:
given that;
n = 14
mean Ж = 19,850
standard deviation S = 1,084
degree of freedom df = n - 1 = ( 14 -1 ) = 13
H₀ : ц ≥ 20,500
H₁ : ц < 20,500
Now the test statistics
t = (Ж - ц) / ( s/√n)
t = ( 19850 - 20500) / ( 1084/√14)
t = -2.244
we know that our degree of freedom df = 13
from the table, the area under the t-distribution of the left of (t=-2.244) and for (df=13) is 0.0215
so P = 0.0215
significance ∝ = 0.05
we can confidently say that since our p value is less than the significance level, we reject the null hypothesis ( H₀ : ц ≥ 20,500 )
There is sufficient evidence to reject the dealer's claim that the mean price is at least $20,500
Answer
Input= 5
Output= 120
Step-by-step explanation:
The input goes up by 1, so m=5
The output shows how many hours there are in that many days so blank=120
Answer:
(D)
Step-by-step explanation:
First, you have to graph the equation. Then you notice that it is a parabola facing downwards. This means that as the graph decreases on the left side, the x's become negative and the y's become negative. On the right side, the x's become positive and the y's become negative.
