Concept explainers
Videos
Draw the structure corresponding to each name.
a. 3-methyihexanoic acid
b. 3-hydroxy-4-methylheptanoic acid
c. p-nitrobenzoic acid
d. sodium hexanoate
e. m-ethylbenzoic acid
f. propyl decanoate
(a)
Interpretation:
The structure of 3-methylhexanoic acid should be drawn.
Concept Introduction:
The structure of organic compound is drawn in order to represent the arrangement of atoms in which they are present in a molecule.
Answer to Problem 44P
Explanation of Solution
An organic compound in which carboxy functional group that is -COOH is bonded to the carbon atom is said to be a carboxylic acid. The general formula for carboxylic acid is RCOOH or RCO2H.
The IUPAC names are given to the carboxylic acid according to the following steps:
1. The parent (longest) alkane chain is identified.
2. The ending of the parent chain from alkane (-e) is changed to -oic acid for a carboxylic acid group.
3. The numbering is of the chain is done in such a way that carbonyl carbon gets the smaller number.
4. Name should be written in alphabetical order and other substituents are shown by the number.
For number of carbons atoms chain, the prefix is given as:
Carbon-1 meth
Carbon-2 eth
Carbon-3 prop
Carbon-4 but
Carbon-5 pent
Carbon-6 hex
Carbon-7 hept
Carbon-8 oct
Carbon-9 non
Carbon-10 dec
The given name is 3-methylhexanoic acid where the parent chain is hexane that is 6 carbon atom chain having a methyl substituent at carbon-3. So, the structure of 3-methylhexanoic acid is:
(b)
Interpretation:
The structure of 3-hydroxy-4-methylheptanoic acid should be drawn.
Concept Introduction:
The structure of organic compound is drawn in order to represent the arrangement of atoms in which they are present in a molecule.
Answer to Problem 44P
Explanation of Solution
An organic compound in which carboxy functional group that is -COOH is bonded to the carbon atom is said to be a carboxylic acid. The general formula for carboxylic acid is RCOOH or RCO2H.
The IUPAC names are given to the carboxylic acid according to the following steps:
1. The parent (longest) alkane chain is identified.
2. The ending of the parent chain from alkane (-e) is changed to -oic acid for a carboxylic acid group.
3. The numbering is of the chain is done in such a way that carbonyl carbon gets the smaller number.
4. Name should be written in alphabetical order and other substituents are shown by the number.
For number of carbons atoms chain, the prefix is given as:
Carbon-1 meth
Carbon-2 eth
Carbon-3 prop
Carbon-4 but
Carbon-5 pent
Carbon-6 hex
Carbon-7 hept
Carbon-8 oct
Carbon-9 non
Carbon-10 dec
The given name is 3-hydroxy-4-methylheptanoic acid where the parent chain is heptane that is 7 carbon atom chain having a hydroxy substituent at carbon-3 and a methyl substituent at carbon-4. So, the structure of 3-hydroxy-4-methylheptanoic acid is:
(c)
Interpretation:
The structure of p-nitrobenzoic acid should be drawn.
Concept Introduction:
The structure of organic compound is drawn in order to represent the arrangement of atoms in which they are present in a molecule.
Answer to Problem 44P
Explanation of Solution
An organic compound in which carboxy functional group that is -COOH is bonded to the carbon atom is said to be a carboxylic acid. The general formula for carboxylic acid is RCOOH or RCO2H.
The IUPAC names are given to the carboxylic acid according to the following steps:
1. The parent (longest) alkane chain is identified.
2. The ending of the parent chain from alkane (-e) is changed to -oic acid for a carboxylic acid group.
3. The numbering is of the chain is done in such a way that carbonyl carbon gets the smaller number.
4. Name should be written in alphabetical order and other substituents are shown by the number.
For number of carbons atoms chain, the prefix is given as:
Carbon-1 meth
Carbon-2 eth
Carbon-3 prop
Carbon-4 but
Carbon-5 pent
Carbon-6 hex
Carbon-7 hept
Carbon-8 oct
Carbon-9 non
Carbon-10 dec
The given name is p-nitrobenzoic acid where the parent chain is benzene having a nitro substituent at carbon-4. So, the structure of p-nitrobenzoic acid is:
(d)
Interpretation:
The structure of sodium hexanoate should be drawn.
Concept Introduction:
The structure of organic compound is drawn in order to represent the arrangement of atoms in which they are present in a molecule.
Answer to Problem 44P
Explanation of Solution
An organic compound in which carboxy functional group that is -COOH is bonded to the carbon atom is said to be a carboxylic acid. The general formula for carboxylic acid is RCOOH or RCO2H. When H of hydroxyl group present in carboxylic acid is replaced by an atom then it results in the formation of respective salt.
The IUPAC names are given to the carboxylic acid salt according to the following steps:
- The parent (longest) carbon chain is identified.
- The name of metal is written first from which the salt is made up of.
- The ending of the for a carboxylic acid group is changed to -oate for naming salt of carboxylic acid.
- The numbering is of the chain is done in such a way that carbonyl carbon gets the smaller number.
- Name should be written in alphabetical order and other substituents are shown by the number.
For number of carbons atoms chain, the prefix is given as:
Carbon-1 meth
Carbon-2 eth
Carbon-3 prop
Carbon-4 but
Carbon-5 pent
Carbon-6 hex
Carbon-7 hept
Carbon-8 oct
Carbon-9 non
Carbon-10 dec
The given name is sodium hexanoate where the parent chain is hexane having 6 carbon atoms and metal is sodium. So, the structure of sodium hexanoate is:
(e)
Interpretation:
The structure of m-ethylbenzoic acid should be drawn.
Concept Introduction:
The structure of organic compound is drawn in order to represent the arrangement of atoms in which they are present in a molecule.
Answer to Problem 44P
Explanation of Solution
An organic compound in which carboxy functional group that is -COOH is bonded to the carbon atom is said to be a carboxylic acid. The general formula for carboxylic acid is RCOOH or RCO2H.
The IUPAC names are given to the carboxylic acid according to the following steps:
1. The parent (longest) alkane chain is identified.
2. The ending of the parent chain from alkane (-e) is changed to -oic acid for a carboxylic acid group.
3. The numbering is of the chain is done in such a way that carbonyl carbon gets the smaller number.
4. Name should be written in alphabetical order and other substituents are shown by the number.
For number of carbons atoms chain, the prefix is given as:
Carbon-1 meth
Carbon-2 eth
Carbon-3 prop
Carbon-4 but
Carbon-5 pent
Carbon-6 hex
Carbon-7 hept
Carbon-8 oct
Carbon-9 non
Carbon-10 dec
The given name is m-ethylbenzoic acid where the parent chain is benzene having an ethyl substituent at carbon-3. So, the structure of m-ethylbenzoic acid is:
(f)
Interpretation:
The structure of propyl decanoate should be drawn.
Concept Introduction:
The structure of organic compound is drawn in order to represent the arrangement of atoms in which they are present in a molecule.
Answer to Problem 44P
Explanation of Solution
An organic compound in which carboxy functional group that is -COOH is bonded to the carbon atom is said to be a carboxylic acid. The general formula for carboxylic acid is RCOOH or RCO2H. When -H of the carboxylic acid is replaced by an alkyl or aryl group (-R') then it results in the formation of an ester having general formula RCOOR'.
The reaction which results in the formation of at least one ester along with water on heating acids with alcohols is said to be esterification.
So, in order to give the IUPAC name to the esters, the following steps are followed:
- The alkyl substituent from the alcohol is named first.
- The name of the parent chain from carboxylic acid part is replaced as carboxylate.
In order to write the common name of the esters, the common of acids are written from which the ester has been formed.
The given name is propyl decanoate where propyl name is derived from propanol and decanoate is derived from the name decanoic acid. So, the structure of propyl decanoate is:
Want to see more full solutions like this?
Chapter 17 Solutions
GENERAL,ORGANIC, & BIOLOGICAL CHEM-ACCES
- 1. Write the dissociation reaction then calculate the pH for the following STRONG substances. a. 2.5x103 M HBr b.5.6x10 M NaOHarrow_forward74. A contour map for an atomic orbital of hydrogen is shown below for the xy and xz planes. Identify the type (s, p, d, f, g . . .) of orbital. axis x axis z axis Cooo xy planearrow_forwardA buffer is prepared by adding 0.50 mol of acetic acid (HC2H3O2) and 0.75 mol of sodium acetate (NaC2H3O2) to enough water to form 2.00L solution. (pKa for acetic acid is 4.74) Calculate the pH of the buffer.arrow_forward
- Modify the given carbon skeleton to draw the major product of the following reaction. If a racemic mixture of enantiomers is expected, draw both enantiomers. Note: you can select a structure and use Copy and Paste to save drawing time. HBr کی CH3 کی Edit Drawingarrow_forwardSort the following into the classification for a reaction that is NOT at equilibrium versus a reaction system that has reached equilibrium. Drag the appropriate items to their respective bins. View Available Hint(s) The forward and reverse reactions proceed at the same rate. Chemical equilibrium is a dynamic state. The ratio of products to reactants is not stable. Reset Help The state of chemical equilibrium will remain the same unless reactants or products escape or are introduced into the system. This will disturb the equilibrium. The concentration of products is increasing, and the concentration of reactants is decreasing. The ratio of products to reactants does not change. The rate at which products form from reactants is equal to the rate at which reactants form from products. The concentrations of reactants and products are stable and cease to change. The reaction has reached equilibrium. The rate of the forward reaction is greater than the rate of the reverse reaction. The…arrow_forwardPlace the following characteristics into the box for the correct ion. Note that some of the characteristics will not be placed in either bin. Use your periodic table for assistance. Link to Periodic Table Drag the characteristics to their respective bins. ▸ View Available Hint(s) This anion could form a neutral compound by forming an ionic bond with one Ca²+. Reset Help This ion forms ionic bonds with nonmetals. This ion has a 1- charge. This is a polyatomic ion. The neutral atom from which this ion is formed is a metal. The atom from which this ion is formed gains an electron to become an ion. The atom from which this ion is formed loses an electron to become an ion. This ion has a total of 18 electrons. This ion has a total of 36 electrons. This ion has covalent bonds and a net 2- charge. This ion has a 1+ charge. Potassium ion Bromide ion Sulfate ionarrow_forward
- U Consider the following graph containing line plots for the moles of Product 1 versus time (minutes) and the moles of Product 2 versus time in minutes. Choose all of the key terms/phrases that describe the plots on this graph. Check all that apply. ▸ View Available Hint(s) Slope is zero. More of Product 1 is obtained in 12 minutes. Slope has units of moles per minute. plot of minutes versus moles positive relationship between moles and minutes negative relationship between moles and minutes Slope has units of minutes per moles. More of Product 2 is obtained in 12 minutes. can be described using equation y = mx + b plot of moles versus minutes y-intercept is at (12,10). y-intercept is at the origin. Product Amount (moles) Product 1 B (12,10) Product 2 E 1 Time (minutes) A (12,5)arrow_forwardSolve for x, where M is molar and s is seconds. x = (9.0 × 10³ M−². s¯¹) (0.26 M)³ Enter the answer. Include units. Use the exponent key above the answer box to indicate any exponent on your units. ▸ View Available Hint(s) ΜΑ 0 ? Units Valuearrow_forwardLearning Goal: This question reviews the format for writing an element's written symbol. Recall that written symbols have a particular format. Written symbols use a form like this: 35 Cl 17 In this form the mass number, 35, is a stacked superscript. The atomic number, 17, is a stacked subscript. "CI" is the chemical symbol for the element chlorine. A general way to show this form is: It is also correct to write symbols by leaving off the atomic number, as in the following form: atomic number mass number Symbol 35 Cl or mass number Symbol This is because if you write the element symbol, such as Cl, you know the atomic number is 17 from that symbol. Remember that the atomic number, or number of protons in the nucleus, is what defines the element. Thus, if 17 protons are in the nucleus, the element can only be chlorine. Sometimes you will only see 35 C1, where the atomic number is not written. Watch this video to review the format for written symbols. In the following table each column…arrow_forward
- need help please and thanks dont understand only need help with C-F Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal…arrow_forwardneed help please and thanks dont understand only need help with C-F Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal…arrow_forwardPlease correct answer and don't used hand raitingarrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles 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 Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY