a)(Z)-2-Ethyl-2-buten-1-ol
Interpretation:
To draw the structure of (Z)-2-ethyl-2-buten-1-ol.
Concept introduction:
The longest continuous carbon chain which contains the hydroxyl group and any other
E-Z notation is used to represent the arrangement of different groups around the double bond. Applying the sequence rules the groups attached to the double bond are ranked as either higher or lower. If the groups of similar ranking on each carbon are arranged on the same side of the double bond then Z configuration is assigned. If the groups of similar rankings on each carbon are arranged on the opposite sides of the double bond then E configuration is assigned.
To draw:
The structure of (Z)-2-ethyl-2-buten-1-ol.
b)3-Cyclohexene-1-ol
Interpretation:
To draw the structure of 3-cyclohexene-1-ol.
Concept introduction:
In naming cyclic alcohols the parent name is derived from the cycloalkene ring by replacing –e of the cycloalkene with –ol. The ring is numbered from the carbon with –OH in such a way that lowest number possible is given to the carbons in the double bond.
To draw:
The structure of 3-cyclohexene-1-ol.
c) trans-3-Chlorocycloheptanol
Interpretation:
To draw the structure of trans-3-chlorocycloheptanol.
Concept introduction:
In naming cyclic alcohols the parent name is derived from the cycloalkene ring by replacing –e of the cycloalkene with –ol. The ring is numbered from the carbon with –OH in such a way that lowest number possible is given to the carbons in the double bond.
To draw:
The structure of trans-3-chlorocycloheptanol.
d) 1, 4-Pentanediol
Interpretation:
To draw the structure of 1, 4-pentanediol.
Concept introduction:
Alcohols are named as derivatives of the parent
To draw:
The structure of 1, 4-pentanediol.
e) 2, 6-Dimethylphenol
Interpretation:
To draw the structure of 2, 6-dimethylphenol.
Concept introduction:
Hydroxybenzenes are named as derivatives of phenol. If the ring contains other substituents also, the ring is numbered from the carbon with –OH in such a way that lowest number possible is given to the carbon with the substituent. While writing the name the substituents are arranged in the alphabetical order.
To draw:
The structure of 2, 6-dimethylphenol.
f) o-(2-hydroxyethyl) phenol.
Interpretation:
To draw the structure of o-(2-hydroxyethyl) phenol.
Concept introduction:
Hydroxybenzenes are named as derivatives of phenol. If the ring contains other substituent also the ring is numbered from the carbon with –OH in such a way that lowest number possible is given to the carbon with the substituent. While writing the name the substituents are arranged in the alphabetical order.
To draw:
The structure of o-(2-hydroxyethyl) phenol.
Trending nowThis is a popular solution!
Chapter 17 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
- Show work. Don't give Ai generated solutionarrow_forwardNonearrow_forwardTransmitance 3. Which one of the following compounds corresponds to this IR spectrum? Point out the absorption band(s) that helped you decide. OH H3C OH H₂C CH3 H3C CH3 H3C INFRARED SPECTRUM 0.8- 0.6 0.4- 0.2 3000 2000 1000 Wavenumber (cm-1) 4. Consider this compound: H3C On the structure above, label the different types of H's as A, B, C, etc. In table form, list the labeled signals, and for each one state the number of hydrogens, their shifts, and the splitting you would observe for these hydrogens in the ¹H NMR spectrum. Label # of hydrogens splitting Shift (2)arrow_forward
- Nonearrow_forwardDraw the Lewis structure of C2H4Oarrow_forwarda) 5. Circle all acidic (and anticoplanar to the Leaving group) protons in the following molecules, Solve these elimination reactions, and identify the major and minor products where appropriate: 20 points + NaOCH3 Br (2 productarrow_forward
- Nonearrow_forwardDr. Mendel asked his BIOL 260 class what their height was and what their parent's heights were. He plotted that data in the graph below to determine if height was a heritable trait. A. Is height a heritable trait? If yes, what is the heritability value? (2 pts) B. If the phenotypic variation is 30, what is the variation due to additive alleles? (2 pts) Offspring Height (Inches) 75 67.5 60 52.5 y = 0.9264x + 4.8519 55 60 65 MidParent Height (Inches) 70 75 12pt v V Paragraph B IUA > AT2 v Varrow_forwardExperiment: Each team will be provided with 5g of a mixture of acetanilide and salicylic acid. You will divide it into three 1.5 g portions in separate 125 mL Erlenmeyer flasks savıng some for melting point analysis. Dissolve the mixture in each flask in ~60mL of DI water by heating to boiling on a hotplate. Take the flasks off the hotplate once you have a clear solution and let them stand on the bench top for 5 mins and then allow them to cool as described below. Sample A-Let the first sample cool slowly to room temperature by letting it stand on your lab bench, with occasional stirring to promote crystallization. Sample B-Cool the second sample 1n a tap-water bath to 10-15 °C Sample C-Cool the third sample in an ice-bath to 0-2 °C Results: weight after recrystalization and melting point temp. A=0.624g,102-115° B=0.765g, 80-105° C=1.135g, 77-108 What is the percent yield of A,B, and C.arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY