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(a)
Interpretation:
The structure of
Concept introduction:
Answer to Problem 21.1P
The structure of
Explanation of Solution
In
Figure 1
At the fifth carbon, there is a cyano group
Figure 2
The above skeleton is completed by adding hydrogens as per the tetravalency of carbon as shown below.
Figure 3
The structure of
(b)
Interpretation:
The structure of isopropyl valerate is to be stated.
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of isopropyl valerate is shown below.
Explanation of Solution
The structure of isopropyl valerate consists of eight carbons, sixteen hydrogen atoms and two oxygen atoms. Isopropyl valerate is a common name of isopropyl pentanoate. It is a five carbon carboxylic acid. The structure of pentanoic acid is shown below.
Figure 4
The hydrogen of carboxylic group is replaced by an isopropyl group to form an ester. The structure of isopropyl valerate is shown below.
Figure 5
The structure of isopropyl valerate is shown in Figure 5.
(c)
Interpretation:
The structure of ethyl methyl malonate is to be stated.
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of ethyl methyl malonate is shown below.
Explanation of Solution
The structure of ethyl methyl malonate consists of a eight carbons, fourteen hydrogen and four oxygen. Ethyl methyl malonate consists of propane dioic acid as parent chain.
The structure of propane dioic acid is shown below.
Figure 6
In the above structure, at first position hydrogen is substituted by a methyl group and at the last carboxyl group an ethyl group replaces the hydrogen of the carboxyl group. The structure of ethyl methyl malonate is shown below.
Figure 7
The structure of ethyl methyl malonate is shown in Figure 6.
(d)
Interpretation:
The structure of cyclohexyl acetate is to be stated.
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of cyclohexyl acetate is shown below.
Explanation of Solution
Cyclohexyl acetate consists of a cyclohexane ring and an acetate group. The acetate group is given by the formula
Figure 8
The structure of cyclohexyl acetate is shown in Figure 8
(e)
Interpretation:
The structure of
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of
Explanation of Solution
The structure of
Figure 9
The structure of
(f)
Interpretation:
The structure of
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of
Explanation of Solution
Lactones are cyclic esters. The structure of
Figure 10
The structure of
(g)
Interpretation:
The structure of glutarimide is to be stated.
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of glutarimide is shown below.
Explanation of Solution
The structure of glutarimide consists of a piperidine ring. There are two carbonyl group present adjacent to the nitrogen group. The structure of a piperidine ring is shown below.
Figure 11
Therefore, the structure of glutarimide is shown below.
Figure 12
The structure of glutarimide is shown in Figure 12.
(h)
Interpretation:
The structure of
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of
Explanation of Solution
There is carboxyl group present in
Figure 13
The structure of
(i)
Interpretation:
The structure of
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Answer to Problem 21.1P
The structure of
Explanation of Solution
In
Figure 14
In the above structure there is a ethoxy carbonyl group at the third position. Therefore, the structure of
Figure 15
The structure of
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Chapter 21 Solutions
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- Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100arrow_forwardPart IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.arrow_forwardFor each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.arrow_forward
- A molecule shows peaks at 1379, 1327, 1249, 739 cm-1. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule. In the diagram imagine exciting an electron, what are its various options for getting back to the ground state? What process would promote radiation less decay? What do you expect for the lifetime of an electron in the T1 state? Why is phosphorescence emission weak in most substances? What could you do to a sample to enhance the likelihood that phosphorescence would occur over radiationless decay?arrow_forwardRank the indicated C—C bonds in increasing order of bond length. Explain as why to the difference.arrow_forwardUse IUPAC rules to name the following alkanearrow_forward
- Please correct answer and don't use hand ratingarrow_forwardPlease correct answer and don't use hand ratingarrow_forwardThe SN 1 mechanism starts with the rate-determining step which is the dissociation of the alkyl halide into a carbocation and a halide ion. The next step is the rapid reaction of the carbocation intermediate with the nucleophile; this step completes the nucleophilic substitution stage. The step that follows the nucleophilic substitution is a fast acid-base reaction. The nucleophile now acts as a base to remove the proton from the oxonium ion from the previous step, to give the observed product. Draw a curved arrow mechanism for the reaction, adding steps as necessary. Be sure to include all nonzero formal charges. Cl: Add/Remove step G Click and drag to start drawing a structure.arrow_forward
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