
(a)
Interpretation:
The IUPAC name for the alcohol has to be assigned.
Concept Introduction:
IUPAC rules for naming alcohols that contain single hydroxyl group:
- • Longest carbon chain has to be identified that contains hydroxyl group also. The chain name is obtained by replacing the letter “-e” in
alkane with “-ol”. If the compound contains a unsaturated bond, then the respective name has to be changed with regard to alkane. - • The numbering has to be given so that the hydroxyl group gets the least numbering.
- • Name and location of any other substituent present in the chain has to be identified.
- • If in a ring the hydroxyl group is present, then that carbon is numbered 1 and the numbering then proceeds counterclockwise or clockwise in a way that substituents present if any gets the least numbering.
- • Hydroxyl group as a substituent in a molecule is named as hydroxy group rather than hydroxyl group.
- • If the compound contains bulky groups on same side of the double bond, then it is a cis isomer and if the bulkyl groups are present on opposite side of the double bond, then it is a trans isomer.
- • In case of cycloalkane compounds, if the substitutions are present on same side of the ring of carbon atoms, it is a cis isomer. If the substitutions are present above and below the ring, then it is a trans isomer.
IUPAC rules for naming alcohols that contain more than one hydroxyl group:
- • The same rules said above is followed but the prefix di-, tri-, tetra etc is added corresponding to the number of hydroxyl groups that is present.
(a)

Answer to Problem 3.28EP
The IUPAC name of the given alcohol is trans-2-penten-1-ol.
Explanation of Solution
Given structure of alcohol is,
The longest carbon chain in the above structure with the hydroxyl group is found to be a five carbon chain. Hence, the parent alkane is pentane. As there is a double bond present in the carbon chain, the name is converted as pentene. The hydroxyl group is found to be present on first carbon atom. Therefore, the name of the given alcohol can be given as 2-penten-1-ol.
Looking for the stereochemistry, the given structure has bulky group on opposite side of the double bond and hence this is a trans isomer. This has to be included in the IUPAC name. Therefore, IUPAC name of the given compound is trans-2-penten-1-ol.
IUPAC name for the given alcohol is trans-2-penten-1-ol.
(b)
Interpretation:
The IUPAC name for the alcohol has to be assigned.
Concept Introduction:
IUPAC rules for naming alcohols that contain single hydroxyl group:
- • Longest carbon chain has to be identified that contains hydroxyl group also. The chain name is obtained by replacing the letter “-e” in alkane with “-ol”. If the compound contains a unsaturated bond, then the respective name has to be changed with regard to alkane.
- • The numbering has to be given so that the hydroxyl group gets the least numbering.
- • Name and location of any other substituent present in the chain has to be identified.
- • If in a ring the hydroxyl group is present, then that carbon is numbered 1 and the numbering then proceeds counterclockwise or clockwise in a way that substituents present if any gets the least numbering.
- • Hydroxyl group as a substituent in a molecule is named as hydroxy group rather than hydroxyl group.
- • If the compound contains bulky groups on same side of the double bond, then it is a cis isomer and if the bulkyl groups are present on opposite side of the double bond, then it is a trans isomer.
- • In case of cycloalkane compounds, if the substitutions are present on same side of the ring of carbon atoms, it is a cis isomer. If the substitutions are present above and below the ring, then it is a trans isomer.
IUPAC rules for naming alcohols that contain more than one hydroxyl group:
- • The same rules said above is followed but the prefix di-, tri-, tetra etc is added corresponding to the number of hydroxyl groups that is present.
(b)

Answer to Problem 3.28EP
The IUPAC name of the given alcohol is cis-2-penten-1-ol.
Explanation of Solution
Given structure of alcohol is,
The longest carbon chain in the above structure with the hydroxyl group is found to be a five carbon chain. Hence, the parent alkane is pentane. As there is a double bond present in the carbon chain, the name is converted as pentene. The hydroxyl group is found to be present on first carbon atom and the double bond between second and third carbon atom. Therefore, the name of the given alcohol can be given as 2-penten-1-ol.
Looking for the stereochemistry, the given structure has bulky group on same side of the double bond and hence this is a cis isomer. This has to be included in the IUPAC name. Therefore, IUPAC name of the given compound is cis-2-penten-1-ol.
IUPAC name for the given alcohol is cis-2-penten-1-ol.
(c)
Interpretation:
The IUPAC name for the alcohol has to be assigned.
Concept Introduction:
IUPAC rules for naming alcohols that contain single hydroxyl group:
- • Longest carbon chain has to be identified that contains hydroxyl group also. The chain name is obtained by replacing the letter “-e” in alkane with “-ol”. If the compound contains a unsaturated bond, then the respective name has to be changed with regard to alkane.
- • The numbering has to be given so that the hydroxyl group gets the least numbering.
- • Name and location of any other substituent present in the chain has to be identified.
- • If in a ring the hydroxyl group is present, then that carbon is numbered 1 and the numbering then proceeds counterclockwise or clockwise in a way that substituents present if any gets the least numbering.
- • Hydroxyl group as a substituent in a molecule is named as hydroxy group rather than hydroxyl group.
- • If the compound contains bulky groups on same side of the double bond, then it is a cis isomer and if the bulkyl groups are present on opposite side of the double bond, then it is a trans isomer.
- • In case of cycloalkane compounds, if the substitutions are present on same side of the ring of carbon atoms, it is a cis isomer. If the substitutions are present above and below the ring, then it is a trans isomer.
IUPAC rules for naming alcohols that contain more than one hydroxyl group:
- • The same rules said above is followed but the prefix di-, tri-, tetra etc is added corresponding to the number of hydroxyl groups that is present.
(c)

Answer to Problem 3.28EP
The IUPAC name of the given alcohol is trans-2-methylcyclohexanol.
Explanation of Solution
Given structure of alcohol is,
The longest carbon chain in the above structure with the hydroxyl group is found to be a six carbon cyclic chain. Hence, the parent cycloalkane is cyclohexane. As a hydroxyl group is present in the ring, the name can be given as cyclohexanol. Looking for the substituents present in the ring, it is found that a methyl group is present in the second position. Therefore, the name of the given alcohol can be given as 2-methylcyclohexanol.
Looking for the stereochemistry, the given structure has substituents on above and below the ring of carbon atoms and hence this is a trans isomer. This has to be included in the IUPAC name. Therefore, IUPAC name of the given compound is trans-2-methylcyclohexanol.
IUPAC name for the given alcohol is trans-2-methylcyclohexanol.
(d)
Interpretation:
The IUPAC name for the alcohol has to be assigned.
Concept Introduction:
IUPAC rules for naming alcohols that contain single hydroxyl group:
- • Longest carbon chain has to be identified that contains hydroxyl group also. The chain name is obtained by replacing the letter “-e” in alkane with “-ol”. If the compound contains a unsaturated bond, then the respective name has to be changed with regard to alkane.
- • The numbering has to be given so that the hydroxyl group gets the least numbering.
- • Name and location of any other substituent present in the chain has to be identified.
- • If in a ring the hydroxyl group is present, then that carbon is numbered 1 and the numbering then proceeds counterclockwise or clockwise in a way that substituents present if any gets the least numbering.
- • Hydroxyl group as a substituent in a molecule is named as hydroxy group rather than hydroxyl group.
- • If the compound contains bulky groups on same side of the double bond, then it is a cis isomer and if the bulkyl groups are present on opposite side of the double bond, then it is a trans isomer.
- • In case of cycloalkane compounds, if the substitutions are present on same side of the ring of carbon atoms, it is a cis isomer. If the substitutions are present above and below the ring, then it is a trans isomer.
IUPAC rules for naming alcohols that contain more than one hydroxyl group:
- • The same rules said above is followed but the prefix di-, tri-, tetra etc is added corresponding to the number of hydroxyl groups that is present.
(d)

Answer to Problem 3.28EP
The IUPAC name of the given alcohol is cis-3-bromocyclohexanol.
Explanation of Solution
Given structure of alcohol is,
The longest carbon chain in the above structure with the hydroxyl group is found to be a six carbon cyclic chain. Hence, the parent cycloalkane is cyclohexane. As a hydroxyl group is present in the ring, the name can be given as cyclohexanol. Looking for the substituents present in the ring, it is found that a bromo is present in the third position. Therefore, the name of the given alcohol can be given as 3-bromocyclohexanol.
Looking for the stereochemistry, the given structure has substituents on same side the ring of carbon atoms and hence this is a cis isomer. This has to be included in the IUPAC name. Therefore, IUPAC name of the given compound is cis-3-bromocyclohexanol.
IUPAC name for the given alcohol is cis-3-bromocyclohexanol.
Want to see more full solutions like this?
Chapter 3 Solutions
Organic And Biological Chemistry
- Homework 13 (Ch17) Question 4 of 4 (1 point) | Question Attempt: 2 of 2 ✓ 1 ✓ 2 = 3 4 Time Remaining: 4:25:54 Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction free energy of the following chemical reaction: 2CH3OH (g)+302 (g) → 2CO2 (g) + 4H₂O (g) Round your answer to zero decimal places. ☐ kJ x10 ☐ Subm Check 2020 Hill LLC. All Rights Reserved. Terms of Use | Privacy Cearrow_forwardIdentifying the major species in weak acid or weak base equilibria Your answer is incorrect. • Row 2: Your answer is incorrect. • Row 3: Your answer is incorrect. • Row 6: Your answer is incorrect. 0/5 The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HF is a weak acid. acids: HF 0.1 mol of NaOH is added to 1.0 L of a 0.7M HF solution. bases: 0.13 mol of HCl is added to 1.0 L of a solution that is 1.0M in both HF and KF. Exponent other: F acids: HF bases: F other: K 1 0,0,... ? 000 18 Ararrow_forwardUsing reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NOCI (g) 2NO (g) + Cl2 (g) AGº =41. kJ Now suppose a reaction vessel is filled with 4.50 atm of nitrosyl chloride (NOCI) and 6.38 atm of chlorine (C12) at 212. °C. Answer the following questions about this system: ? rise Under these conditions, will the pressure of NOCI tend to rise or fall? x10 fall Is it possible to reverse this tendency by adding NO? In other words, if you said the pressure of NOCI will tend to rise, can that be changed to a tendency to fall by adding NO? Similarly, if you said the pressure of NOCI will tend to fall, can that be changed to a tendency to rise by adding NO? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO needed to reverse it. Round your answer to 2 significant digits. 0.035 atm ✓ G 00. 18 Ararrow_forward
- Highlight each glycosidic bond in the molecule below. Then answer the questions in the table under the drawing area. HO- HO- -0 OH OH HO NG HO- HO- OH OH OH OH NG OHarrow_forward€ + Suppose the molecule in the drawing area below were reacted with H₂ over a platinum catalyst. Edit the molecule to show what would happen to it. That is, turn it into the product of the reaction. Also, write the name of the product molecule under the drawing area. Name: ☐ H C=0 X H- OH HO- H HO- -H CH₂OH ×arrow_forwardDraw the Haworth projection of the disaccharide made by joining D-glucose and D-mannose with a ẞ(1-4) glycosidic bond. If the disaccharide has more than one anomer, you can draw any of them. Click and drag to start drawing a structure. Xarrow_forward
- Epoxides can be opened in aqueous acid or aqueous base to produce diols (molecules with two OH groups). In this question, you'll explore the mechanism of epoxide opening in aqueous acid. 2nd attempt Be sure to show all four bonds at stereocenters using hash and wedge lines. 0 0 Draw curved arrows to show how the epoxide reacts with hydronium ion. 100 +1: 1st attempt Feedback Be sure to show all four bonds at stereocenters using hash and wedge lines. See Periodic Table See Hint H A 5 F F Hr See Periodic Table See Hintarrow_forward03 Question (1 point) For the reaction below, draw both of the major organic products. Be sure to consider stereochemistry. > 1. CH₂CH₂MgBr 2. H₂O 3rd attempt Draw all four bonds at chiral centers. Draw all stereoisomers formed. Draw the structures here. e 130 AN H See Periodic Table See Hint P C Brarrow_forwardYou may wish to address the following issues in your response if they are pertinent to the reaction(s) you propose to employ:1) Chemoselectivity (why this functional group and not another?) 2) Regioselectivity (why here and not there?) 3) Stereoselectivity (why this stereoisomer?) 4) Changes in oxidation state. Please make it in detail and draw it out too in what step what happens. Thank you for helping me!arrow_forward
- 1) Chemoselectivity (why this functional group and not another?) 2) Regioselectivity (why here and not there?) 3) Stereoselectivity (why this stereoisomer?) 4) Changes in oxidation state. Everything in detail and draw out and write it.arrow_forwardCalculating the pH at equivalence of a titration 3/5 Izabella A chemist titrates 120.0 mL of a 0.7191M dimethylamine ((CH3)2NH) solution with 0.5501 M HBr solution at 25 °C. Calculate the pH at equivalence. The pk of dimethylamine is 3.27. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added. pH = ☐ ✓ 18 Ar Boarrow_forwardAlcohols can be synthesized using an acid-catalyzed hydration of an alkene. An alkene is combined with aqueous acid (e.. sulfuric acid in water). The reaction mechanism typically involves a carbocation intermediate. > 3rd attempt 3343 10 8 Draw arrows to show the reaction between the alkene and hydronium ion. that 2nd attempt Feedback 1st attempt تعمال Ju See Periodic Table See Hint F D Ju See Periodic Table See Hintarrow_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 for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning



