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a)
Interpretation:
Polar head and all hydrophobic tails in sphingomyelins need to be identified and the possibility of being major constituent of lipid bilayers need to be discussed.
Concept introduction:
Lipids are a group of compounds which occur naturally and are extracted from cells using a non-polar solvent. Lipids are further classified as simple and complex lipids. The long non-polar carbon chain present in the lipid is the hydrophobic tail. The hydrophilic group or the polar group present in the lipid is known as polar head.
To identify: the polar head and all hydrophobic tails.
(b)
Interpretation:
Polar head and all hydrophobic tails in sphingomyelins need to be identified and the possibility of being major constituent of lipid bilayers need to be discussed.
Concept introduction:
Lipid bilayers are the main fabric of cell membranes. The surface of the lipid bilayer is polar where polar heads assemble. The lipid bilayers are hence water soluble. The lipid bilayers has the phosphoglyceride self-assembled. The hydrophobic tails avoid contact with the water.
To predict: whether sphingomyelins can be a major constituent of lipid bilayers.
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Chapter 26 Solutions
Organic Chemistry, Binder Ready Version
- ASP.....arrow_forwardQuestion 7 (10 points) Identify the carboxylic acid present in each of the following items and draw their structures: Food Vinegar Oranges Yogurt Sour Milk Pickles Acid Structure Paragraph ✓ BI UAE 0118 + v Task: 1. Identify the carboxylic acid 2. Provide Name 3. Draw structure 4. Take a picture of your table and insert Add a File Record Audio Record Video 11.arrow_forwardCheck the box under each structure in the table that is an enantiomer of the molecule shown below. If none of them are, check the none of the above box under the table. Molecule 1 Molecule 2 IZ IN Molecule 4 Molecule 5 ZI none of the above ☐ Molecule 3 Х IN www Molecule 6 NH Garrow_forward
- Highlight each chiral center in the following molecule. If there are none, then check the box under the drawing area. There are no chiral centers. Cl Cl Highlightarrow_forwardA student proposes the following two-step synthesis of an ether from an alcohol A: 1. strong base A 2. R Is the student's proposed synthesis likely to work? If you said the proposed synthesis would work, enter the chemical formula or common abbreviation for an appropriate strong base to use in Step 1: If you said the synthesis would work, draw the structure of an alcohol A, and the structure of the additional reagent R needed in Step 2, in the drawing area below. If there's more than one reasonable choice for a good reaction yield, you can draw any of them. ☐ Click and drag to start drawing a structure. Yes No ロ→ロ 0|0 G Х D : ☐ பarrow_forwardटे Predict the major products of this organic reaction. Be sure to use wedge and dash bonds when necessary, for example to distinguish between different major products. ☐ ☐ : ☐ + NaOH HO 2 Click and drag to start drawing a structure.arrow_forward
- Shown below are five NMR spectra for five different C6H10O2 compounds. For each spectrum, draw the structure of the compound, and assign the spectrum by labeling H's in your structure (or in a second drawing of the structure) with the chemical shifts of the corresponding signals (which can be estimated to nearest 0.1 ppm). IR information is also provided. As a reminder, a peak near 1700 cm-1 is consistent with the presence of a carbonyl (C=O), and a peak near 3300 cm-1 is consistent with the presence of an O–H. Extra information: For C6H10O2 , there must be either 2 double bonds, or 1 triple bond, or two rings to account for the unsaturation. There is no two rings for this problem. A strong band was observed in the IR at 1717 cm-1arrow_forwardPredict the major products of the organic reaction below. : ☐ + Х ك OH 1. NaH 2. CH₂Br Click and drag to start drawing a structure.arrow_forwardNG NC 15Show all the steps you would use to synthesize the following products shown below using benzene and any organic reagent 4 carbons or less as your starting material in addition to any inorganic reagents that you have learned. NO 2 NC SO3H NO2 OHarrow_forward
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