
Concept explainers
(a)
Interpretation:
The molecular formula for the given compound 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)

Answer to Problem 13.33EP
Molecular formula of the compound is
Explanation of Solution
Given structure is,
Carbon atoms are present at the intersection and at the end points. The above structure has six intersections and two end points. Therefore, there is a total of eight carbon atoms. The given compound is found to have one double bond in it. The molecular formula for the given compound can be found by substituting in the general molecular formula of alkene that contain a single double bond as shown below,
The molecular formula of the given compound is identified as
The molecular formula for the given structure is identified.
(b)
Interpretation:
The molecular formula for the given compound 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
Alkenes and cycloalkenes are a class of unsaturated hydrocarbons that contain at least one double bond in its structure. The general molecular formula for alkene with one double bond is
(b)

Answer to Problem 13.33EP
Molecular formula of the compound is
Explanation of Solution
Given structure is,
Carbon atoms are present at the intersection and at the end points. The above structure has six intersections and two end points. Therefore, there is a total of eight carbon atoms. The given compound is found to have one double bond in it. The molecular formula for the given compound can be found by substituting in the general molecular formula of alkene that contain a single double bond as shown below,
The molecular formula of the given compound is identified as
The molecular formula for the given structure is identified.
(c)
Interpretation:
The molecular formula for the given compound 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
Alkenes and cycloalkenes are a class of unsaturated hydrocarbons that contain at least one double bond in its structure. The general molecular formula for alkene with one double bond is
(c)

Answer to Problem 13.33EP
Molecular formula of the compound is
Explanation of Solution
Given structure is,
Carbon atoms are present at the intersection and at the end points. The above structure has six intersections and two end points. Therefore, there is a total of eight carbon atoms. The given compound is found to have two double bonds in it. The molecular formula for the given compound can be found by substituting in the general molecular formula of alkene that contain two double bonds as shown below,
The molecular formula of the given compound is identified as
The molecular formula for the given structure is identified.
(d)
Interpretation:
The molecular formula for the given compound 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
Alkenes and cycloalkenes are a class of unsaturated hydrocarbons that contain at least one double bond in its structure. The general molecular formula for alkene with one double bond is
(d)

Answer to Problem 13.33EP
Molecular formula of the compound is
Explanation of Solution
Given structure is,
Carbon atoms are present at the intersection and at the end points. The above structure has six intersections and four end points. Therefore, there is a total of ten carbon atoms. The given compound is found to have two double bonds in it. The molecular formula for the given compound can be found by substituting in the general molecular formula of alkene that contain two double bonds as shown below,
The molecular formula of the given compound is identified as
The molecular formula for the given structure is identified.
Want to see more full solutions like this?
Chapter 13 Solutions
Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card
- Five isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compoundarrow_forward7. The three sets of replicate results below were accumulated for the analysis of the same sample. Pool these data to obtain the most efficient estimate of the mean analyte content and the standard deviation. Lead content/ppm: Set 1 Set 2 Set 3 1. 9.76 9.87 9.85 2. 9.42 9.64 9.91 3. 9.53 9.71 9.42 9.81 9.49arrow_forwardDraw the Zaitsev product famed when 2,3-dimethylpentan-3-of undergoes an El dehydration. CH₂ E1 OH H₁PO₁ Select Draw Templates More QQQ +H₂Oarrow_forward
- Complete the clean-pushing mechanism for the given ether synthesia from propanol in concentrated sulfurica140°C by adding any mining aloms, bands, charges, nonbonding electron pairs, and curved arrows. Draw hydrogen bonded to cayan, when applicable. ore 11,0 HPC Step 1: Draw curved arrows Step 2: Complete the intend carved Q2Q 56 QQQ Step 3: Complete the intermediate and add curved Step 4: Modify the structures to draw the QQQ QQQarrow_forward6. In an experiment the following replicate set of volume measurements (cm3) was recorded: (25.35, 25.80, 25.28, 25.50, 25.45, 25.43) A. Calculate the mean of the raw data. B. Using the rejection quotient (Q-test) reject any questionable results. C. Recalculate the mean and compare it with the value obtained in 2(a).arrow_forwardA student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T G OH де OH This transformation can't be done in one step.arrow_forward
- Macmillan Leaming Draw the major organic product of the reaction. 1. CH3CH2MgBr 2. H+ - G Select Draw Templates More H о QQarrow_forwardDraw the condensed structure of 3-hydroxy-2-butanone. Click anywhere to draw the first atom of your structure.arrow_forwardGive the expected major product of reaction of 2,2-dimethylcyclopropane with each of the following reagents. 2. Reaction with dilute H₂SO, in methanol. Select Draw Templates More CHC Erase QQQ c. Reaction with dilute aqueous HBr. Select Drew Templates More Era c QQQ b. Reaction with NaOCH, in methanol. Select Draw Templates More d. Reaction with concentrated HBr. Select Draw Templates More En a QQQ e. Reaction with CH, Mg1, then H*, H₂O 1. Reaction with CH,Li, then H', H₂Oarrow_forward
- Write the systematic name of each organic molecule: structure O OH OH name X ☐arrow_forwardMacmillan Learning One of the molecules shown can be made using the Williamson ether synthesis. Identify the ether and draw the starting materials. А со C Strategy: Review the reagents, mechanism and steps of the Williamson ether synthesis. Determine which of the molecules can be made using the steps. Then analyze the two possible disconnection strategies and deduce the starting materials. Identify the superior route. Step 6: Put it all together. Complete the two-step synthesis by selecting the reagents and starting materials. C 1. 2. Answer Bank NaH NaOH NaOCH, снен, сен, он Сиси, Сне (СН), СОН (Сн, Свarrow_forwardWrite the systematic name of each organic molecule: structure CH3 O CH3-CH-CH-C-CH3 OH HV. CH3-C-CH-CH2-CH3 OH CH3 O HO—CH, CH–CH—C CH3 OH 오-오 name X G ☐arrow_forward
- Organic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
- Chemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub CoChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning




