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Concept explainers
(a)
Interpretation:
The presence or absence of glycerol, sphingosine, phosphate, ester, carboxyl group, hydroxyl group and steroid nucleus in prostaglandin needs to be determined.
Concept introduction:
Glycerol is a simple polyol compound with three carbons and with three carbons and hydroxyl groups in each carbon.
Sphingosine is an 18-carbon
Phosphate is a phosphorous atom bonded to four oxygen atoms. One oxygen is bonded to phosphorous with a double bond and others are bonded with a single bond and bear a negative charge.
Ester group is a derivative of a
Carboxyl group formed when a carbon atom is bonded to an oxygen atom with a double bond and to a hydroxyl group.
Hydroxyl group is formed when an oxygen is bonded to a hydrogen.
Steroid nucleus is found in all steroids. It is made up of three cyclohexanes and one cyclopentane bonded together. Steroids vary by the
(b)
Interpretation:
Presence or absence of glycerol, sphingosine, phosphate, ester, carboxyl group, hydroxyl group and steroid nucleus in androgen needs to be explained.
Concept introduction:
Glycerol is a simple polyol compound with three carbons and with three carbons and hydroxyl groups in each carbon
Sphingosine is an 18-carbon diol with an amino group and a double bond
Phosphate is a phosphorous atom bonded to four oxygen atoms. One oxygen is bonded to phosphorous with a double bond and others are bonded with a single bond and bear a negative charge.
Ester group is a derivative of a carboxylic acid group. The hydrogen attached to oxygen in carboxylic acid is replace with an organic fragment in esters.
Carboxyl group formed when a carbon atom is bonded to an oxygen atom with a double bond and to a hydroxyl group.
Hydroxyl group is formed when an oxygen is bonded to a hydrogen.
Steroid nucleus is found in all steroids. It is made up of three cyclohexanes and one cyclopentane bonded together. Steroids vary by the functional groups attached to this core and oxidation states of the ring.
(c)
Interpretation:
Presence or absence of glycerol, sphingosine, phosphate, ester, carboxyl group, hydroxyl group and steroid nucleus in triacylglycerol needs to be explained.
Concept introduction:
Glycerol is a simple polyol compound with three carbons and with three carbons and hydroxyl groups in each carbon
Sphingosine is an 18-carbon diol with an amino group and a double bond
Phosphate is a phosphorous atom bonded to four oxygen atoms. One oxygen is bonded to phosphorous with a double bond and others are bonded with a single bond and bear a negative charge.
Ester group is a derivative of a carboxylic acid group. The hydrogen attached to oxygen in carboxylic acid is replace with an organic fragment in esters.
Carboxyl group formed when a carbon atom is bonded to an oxygen atom with a double bond and to a hydroxyl group.
Hydroxyl group is formed when an oxygen is bonded to a hydrogen.
Steroid nucleus is found in all steroids. It is made up of three cyclohexanes and one cyclopentane bonded together. Steroids vary by the functional groups attached to this core and oxidation states of the ring.
(d)
Interpretation:
Presence or absence of glycerol, sphingosine, phosphate, ester, carboxyl group, hydroxyl group and steroid nucleus in wax needs to be explained.
Concept introduction:
Glycerol is a simple polyol compound with three carbons and with three carbons and hydroxyl groups in each carbon
Sphingosine is an 18-carbon diol with an amino group and a double bond
Phosphate is a phosphorous atom bonded to four oxygen atoms. One oxygen is bonded to phosphorous with a double bond and others are bonded with a single bond and bear a negative charge.
Ester group is a derivative of a carboxylic acid group. The hydrogen attached to oxygen in carboxylic acid is replace with an organic fragment in esters.
Carboxyl group formed when a carbon atom is bonded to an oxygen atom with a double bond and to a hydroxyl group.
Hydroxyl group is formed when an oxygen is bonded to a hydrogen.
Steroid nucleus is found in all steroids. It is made up of three cyclohexanes and one cyclopentane bonded together. Steroids vary by the functional groups attached to this core and oxidation states of the ring.
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Chapter 19 Solutions
EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM
- 3. 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_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
- 1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forward
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