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(a)
Interpretation:
The total number of carbon atoms present in an alkane that contains fourteen hydrogen atoms has to be identified.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Hydrocarbons are the organic compounds that contain only hydrogen and carbon atoms. Hydrocarbon derivatives are the one in which the compounds contain hydrogen and carbon atoms along with one or more additional elements. The additional elements that can be present in hydrocarbon derivatives are oxygen, nitrogen, sulphur, chlorine, bromine etc.
Hydrocarbons are further classified into two categories. They are saturated hydrocarbons and
(a)
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Answer to Problem 12.14EP
The total number of carbon atoms present is 6.
Explanation of Solution
Alkanes are saturated hydrocarbons that contain only single bonds between carbon atoms with no ring structure. The general molecular formula for alkane is
The total number of carbon atoms that will be present in the given alkane is found to be six.
The total number of carbon atoms present in the alkane containing fourteen hydrogen atoms is six.
(b)
Interpretation:
The total number of hydrogen atoms present in an alkane that contains six carbon atoms has to be identified.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Hydrocarbons are the organic compounds that contain only hydrogen and carbon atoms. Hydrocarbon derivatives are the one in which the compounds contain hydrogen and carbon atoms along with one or more additional elements. The additional elements that can be present in hydrocarbon derivatives are oxygen, nitrogen, sulphur, chlorine, bromine etc.
Hydrocarbons are further classified into two categories. They are saturated hydrocarbons and unsaturated hydrocarbons. The hydrocarbons that contain single bonds between carbon atoms in the entire molecule is known as saturated hydrocarbon. The hydrocarbons that contain atleast one double or triple bond between two carbon atoms in the entire molecule is known as unsaturated hydrocarbon.
Alkanes are a class of saturated hydrocarbons that do not contain a ring of carbon atoms but a chain of carbon atoms with carbon‑carbon single bonds. The general molecular formula for alkanes is
(b)
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Answer to Problem 12.14EP
The total number of hydrogen atoms present is 14.
Explanation of Solution
Alkanes are saturated hydrocarbons that contain only single bonds between carbon atoms with no ring structure. The general molecular formula for alkane is
The total number of hydrogen atoms that will be present in the given alkane is found to be fourteen.
The total number of hydrogen atoms present in the alkane containing six carbon atoms is fourteen.
(c)
Interpretation:
The total number of hydrogen atoms present in an alkane when the total atoms present in it was 32 has to be identified.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Hydrocarbons are the organic compounds that contain only hydrogen and carbon atoms. Hydrocarbon derivatives are the one in which the compounds contain hydrogen and carbon atoms along with one or more additional elements. The additional elements that can be present in hydrocarbon derivatives are oxygen, nitrogen, sulphur, chlorine, bromine etc.
Hydrocarbons are further classified into two categories. They are saturated hydrocarbons and unsaturated hydrocarbons. The hydrocarbons that contain single bonds between carbon atoms in the entire molecule is known as saturated hydrocarbon. The hydrocarbons that contain atleast one double or triple bond between two carbon atoms in the entire molecule is known as unsaturated hydrocarbon.
Alkanes are a class of saturated hydrocarbons that do not contain a ring of carbon atoms but a chain of carbon atoms with carbon‑carbon single bonds. The general molecular formula for alkanes is
(c)
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Answer to Problem 12.14EP
The total number of hydrogen atoms present is 22.
Explanation of Solution
Alkanes are saturated hydrocarbons that contain only single bonds between carbon atoms with no ring structure. The general molecular formula for alkane is
The total number of hydrogen atoms that will be present in the given alkane is found to be twenty two.
The total number of hydrogen atoms present in the alkane containing a total of 32 atoms is twenty two.
(d)
Interpretation:
The total number of covalent bonds present in the alkane that contains sixteen hydrogen atoms has to be identified.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Hydrocarbons are the organic compounds that contain only hydrogen and carbon atoms. Hydrocarbon derivatives are the one in which the compounds contain hydrogen and carbon atoms along with one or more additional elements. The additional elements that can be present in hydrocarbon derivatives are oxygen, nitrogen, sulphur, chlorine, bromine etc.
Hydrocarbons are further classified into two categories. They are saturated hydrocarbons and unsaturated hydrocarbons. The hydrocarbons that contain single bonds between carbon atoms in the entire molecule is known as saturated hydrocarbon. The hydrocarbons that contain atleast one double or triple bond between two carbon atoms in the entire molecule is known as unsaturated hydrocarbon.
Alkanes are a class of saturated hydrocarbons that do not contain a ring of carbon atoms but a chain of carbon atoms with carbon‑carbon single bonds. The general molecular formula for alkanes is
(d)
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Answer to Problem 12.14EP
The total number of covalent bonds present is 22.
Explanation of Solution
Alkanes are saturated hydrocarbons that contain only single bonds between carbon atoms with no ring structure. The general molecular formula for alkane is
The formula for the given alkane is
From the formula obtained for the alkane, it is clear that there are sixteen hydrogen atoms and six carbon atoms. Each hydrogen atom will be bonded to carbon atom through a covalent bond. Each carbon atom will be bonded to other carbon atom through a covalent bond. Hence, a total of 22 covalent bonds will be present in the alkane that contains seven carbon atoms.
The total number of covalent bonds present in the alkane that contains seven carbon atoms is 22.
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Chapter 12 Solutions
General, Organic, and Biological Chemistry Seventh Edition
- Curved arrows are used to illustrate the flow of electrons using the provided starting and product structures draw the curved electron pushing arrows for the following reaction or mechanistic steps Ether(solvent)arrow_forwardThis deals with synthetic organic chemistry. Please fill in the blanks appropriately.arrow_forwardUse the References to access important values if needed for this question. What is the IUPAC name of each of the the following? 0 CH3CHCNH₂ CH3 CH3CHCNHCH2CH3 CH3arrow_forward
- You have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you manipulated and exploited the acid-base chemistry of one or more of the compounds in your mixture to facilitate their separation into different phases. The key to understanding how liquid- liquid extractions work is by knowing which layer a compound is in, and in what protonation state. The following liquid-liquid extraction is different from the one you performed in Experiment 4, but it uses the same type of logic. Your task is to show how to separate apart Compound A and Compound B. . Complete the following flowchart of a liquid-liquid extraction. Handwritten work is encouraged. • Draw by hand (neatly) only the appropriate organic compound(s) in the boxes. . Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4 and 5. • Box 7a requires the solvent (name is fine). • Box 7b requires one inorganic compound. • You can neatly complete this assignment by hand and…arrow_forwardb) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forwardShow work with explanation. don't give Ai generated solutionarrow_forward
- Redraw the flowchartarrow_forwardredraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward
- • Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forwardNonearrow_forward
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