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Concept explainers
(a)
Interpretation: The building blocks which represents the sphingosine residues in the given block diagram of sphingophospholipid have to be predicted.
Concept introduction: In sphingophospholipids, the platform molecule is sphingosine to which one fatty acid and one phosphate group are attached. An alcohol is attached to the phosphate group. There are two ester linkages and one amide linkage present in the structure of sphingophospholipids.
(b)
Interpretation: The building blocks which represents the alcohol residues in the given block diagram of sphingophospholipid have to be predicted.
Concept introduction: In sphingophospholipids, the platform molecule is sphingosine to which one fatty acid and one phosphate group are attached. An alcohol is attached to the phosphate group. There are two ester linkages and one amide linkage present in the structure of sphingophospholipids.
(c)
Interpretation: The linkages which represents ester linkages in the given block diagram of sphingophospholipid have to be predicted.
Concept introduction: In sphingophospholipids, the platform molecule is sphingosine to which one fatty acid and one phosphate group are attached. An alcohol is attached to the phosphate group. There are two ester linkages and one amide linkage present in the structure of sphingophospholipids.
(d)
Interpretation: The linkages which involve a phosphate residue in the given block diagram of sphingophospholipid have to be predicted.
Concept introduction: In sphingophospholipids, the platform molecule is sphingosine to which one fatty acid and one phosphate group are attached. An alcohol is attached to the phosphate group. There are two ester linkages and one amide linkage present in the structure of sphingophospholipids.
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Chapter 19 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- 13.84. Chlorine atoms react with methane, forming HCI and CH3. The rate constant for the reaction is 6.0 × 107 M¹ s¹ at 298 K. When the experiment was run at three other temperatures, the following data were collected: T (K) k (M-1 s-1) 303 6.5 × 107 308 7.0 × 107 313 7.5 x 107 a. Calculate the values of the activation energy and the frequency factor for the reaction. b. What is the value of the rate constant in the lower stratosphere, where T = 218 K?arrow_forwardMy Organic Chemistry textbook says about the formation of cyclic hemiacetals, "Such intramolecular reactions to form five- and six-membered rings are faster than the corresponding intermolecular reactions. The two reacting functional groups, in this case OH and C=O, are held in close proximity, increasing the probability of reaction."According to the book, the formation of cyclic hemiacetals occurs in acidic conditions. So my question is whether the carbonyl group in this reaction reacts first with the end alcohol on the same molecule or with the ethylene glycol. And, given the explanation in the book, if it reacts first with ethylene glycol before its own end alcohol, why would it? I don't need to know the final answer. I need to know WHY it would not undergo an intermolecular reaction prior to reacting with the ethylene glycol if that is the case. Please do not use an AI answer.arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
- Highlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. Cl N شیخ x Garrow_forwardQ4: Draw the mirror image of the following molecules. Are the molecules chiral? C/ F LL CI CH3 CI CH3 0 CI CH3 CI CH3 CH3arrow_forwardComplete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward
- 4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forwardNonearrow_forward3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forward
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