(a)
Interpretation:
The rings in pyrrole is saturated, unsaturated or fused has to be indicated with the help of figure 17-8.
Concept Introduction:
Saturated ring in
Unsaturated ring in organic chemistry means that there has to be at least one double bond or triple bond present in the compound that is under consideration.
Fused ring in organic chemistry means that there has to be at least two cyclic rings that share at least two common atoms between then.
(b)
Interpretation:
The rings in indole is saturated, unsaturated or fused has to be indicated with the help of figure 17-8.
Concept Introduction:
Amines are a class of organic compounds that contain amino functional group. Heterocyclic amine is also a class of organic compound in which the nitrogen atom of the amine group is present in the aromatic or nonaromatic ring.
Saturated ring in organic chemistry means there are no double bonds present between two carbon atoms in the compound that is under consideration.
Unsaturated ring in organic chemistry means that there has to be at least one double bond or triple bond present in the compound that is under consideration.
Fused ring in organic chemistry means that there has to be at least two cyclic rings that share at least two common atoms between then.
(c)
Interpretation:
The rings in pyrimidine is saturated, unsaturated or fused has to be indicated with the help of figure 17-8.
Concept Introduction:
Amines are a class of organic compounds that contain amino functional group. Heterocyclic amine is also a class of organic compound in which the nitrogen atom of the amine group is present in the aromatic or nonaromatic ring.
Saturated ring in organic chemistry means there are no double bonds present between two carbon atoms in the compound that is under consideration.
Unsaturated ring in organic chemistry means that there has to be at least one double bond or triple bond present in the compound that is under consideration.
Fused ring in organic chemistry means that there has to be at least two cyclic rings that share at least two common atoms between then.
(d)
Interpretation:
The rings in purine is saturated, unsaturated or fused has to be indicated with the help of figure 17-8.
Concept Introduction:
Amines are a class of organic compounds that contain amino functional group. Heterocyclic amine is also a class of organic compound in which the nitrogen atom of the amine group is present in the aromatic or nonaromatic ring.
Saturated ring in organic chemistry means there are no double bonds present between two carbon atoms in the compound that is under consideration.
Unsaturated ring in organic chemistry means that there has to be at least one double bond or triple bond present in the compound that is under consideration.
Fused ring in organic chemistry means that there has to be at least two cyclic rings that share at least two common atoms between then.
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Chapter 17 Solutions
Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card
- Nonearrow_forwardn Feb 3 A T + 4. (2 pts) Draw the structure of the major component of the Limonene isolated. Explain how you confirmed the structure. 5. (2 pts) Draw the fragment corresponding to the base peak in the Mass spectrum of Limonene. 6. (1 pts) Predict the 1H NMR spectral data of R-Limonene. Proton NMR: 5.3 pon multiplet (H Ringarrow_forwardPart VI. Ca H 10 O is the molecular formula of compound Tom and gives the in the table below. Give a possible structure for compound Tom. 13C Signals summarized C1 C2 C3 C4 C5 C6 C7 13C shift (ppm) 23.5 27.0 33.0 35.8 127 162 205 DEPT-90 + DEPT-135 + +arrow_forward
- 2. Using the following data to calculate the value of AvapH o of water at 298K. AvapH o of water at 373K is 40.7 kJ/mol; molar heat capacity of liquid water at constant pressure is 75.2J mol-1 K-1 and molar heat capacity of water vapor at constant pressure is 33.6 J mol-1 K-1.arrow_forwardPart VII. Below are the 'HNMR 13 3 C-NMR, COSY 2D- NMR, and HSQC 20-NMR (Similar with HETCOR but axes are reversed) spectra of an organic compound with molecular formula C6H13 O. Assign chemical shift values to the H and c atoms of the compound. Find the structure. Show complete solutions. Predicted 1H NMR Spectrum ли 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 f1 (ppm)arrow_forward3. Draw the expanded structural formula, the condensed structural formula, and the skeletal structural formula for 2-pentene. expanded structure: Condensed structure: Skeletal formula: 4. Draw the expanded structural formula, the condensed structural formula, and the skeletal structural formula for 2-methyl-3-heptene. expanded structure: Condensed structure: Skeletal formula: following structurearrow_forward
- Part IV. Propose a plausible Structure w/ the following descriptions: a) A 5-carbon hydrocarbon w/ a single peak in its proton decoupled the DEPT-135 Spectrum shows a negative peak C-NMR spectrum where b) what cyclohexane dione isomer gives the largest no. Of 13C NMR signals? c) C5H120 (5-carbon alcohol) w/ most deshielded carbon absent in any of its DEPT Spectivaarrow_forward13C NMR is good for: a) determining the molecular weight of the compound b) identifying certain functional groups. c) determining the carbon skeleton, for example methyl vs ethyl vs propyl groups d) determining how many different kinds of carbon are in the moleculearrow_forward6 D 2. (1 pt) Limonene can be isolated by performing steam distillation of orange peel. Could you have performed this experiment using hexane instead of water? Explain. 3. (2 pts) Using GCMS results, analyze and discuss the purity of the Limonene obtained from the steam distillation of orange peel.arrow_forward
- Part III. Arrange the following carbons (in blue) in order of increasing chemical shift. HO B NH 2 A CIarrow_forward6. Choose the compound that will produce the spectrum below and assign the signals as carbonyl, aryl, or alkyl. 100 ō (ppm) 50 0 7. 200 150 Assign all of the protons on the spectrum below. 8. A B 4 E C 3 ō (ppm) 2 1 0 Choose the compound that will produce the spectrum below and assign the signals to the corresponding protons. OH 6 OH 3 2 1 0 4 ō (ppm)arrow_forwardIn the Thermo Fisher application note about wine analysis (Lesson 3), the following chromatogram was collected of nine components of wine. If peak 3 has a retention time of 3.15 minutes and a peak width of 0.070 minutes, and peak 4 has a retention time of 3.24 minutes and a peak width of 0.075 minutes, what is the resolution factor between the two peaks? [Hint: it will help to review Lesson 2 for this question.] MAU 300 200 T 34 5 100- 1 2 CO 6 7 8 9 0 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 Minutes 3.22 0.62 1.04 O 1.24arrow_forward
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