(a)
Interpretation:
The IUPAC name for the given molecule has to be assigned including the prefix cis- or trans-.
Concept Introduction:
- The suffix –ane has to be replaced with the suffix –ene. This is used to indicate the presence of double bond.
- The longest continuous chain of carbon atoms has to be chosen that contains both carbon atoms of the double bond.
- The parent carbon chain has to be numbered in a way so that the numbering begins at the end near to the double bond. In case if the double bond is equidistant from both ends, then numbering has to be done from the end that is closer to substituents.
- The position of the double bond has to be given a single number which is lower‑numbered carbon atom that is present in the double bond.
- Suffixes like –diene, -triene, -tetrene, and so on are used when the compound contains more than one double bond.
- In case of cycloalkenes which do not have any substitution, the numbering is not needed to locate the double bond because the bond is assumed to be between the carbons 1 and 2.
- In case if substituents are present in cycloalkene, then the double‑bonded carbon
atoms are numbered 1 and 2 in a direction where the substituent gets the lower number. - If the cycloalkenes contain more than one double bond, then one double bond is assigned the numbers 1 and 2 followed by the other double bond so that the lowest number possible is given.
Structural formula where a line represent carbon‑carbon bond and the carbon atom is considered to be present in each point and the end of lines is known as Line-angle structural formula. To indicate a double bond between carbon atom, double line is used.
If stereoisomers are possible for the alkene, the stereoinformation is shown in the IUPAC name by adding prefix cis- or trans- with respect to the groups present on the same side of double bond or opposite side of double bond.
(a)
Answer to Problem 13.50EP
The IUPAC name for the given compound is chloroethene.
Explanation of Solution
Given compound is,
The longest continuous carbon chain present in the given compound is two carbon atoms. Hence, the parent
As the given compound contains a double bond, the suffix –ane is replaced by –ene. Therefore, the name obtained is ethene.
Numbering has to be given in a way that the carbon atoms present in the double bond gets the least numbering followed by least numbering to the substituents. Therefore, the name of the given compound is chloroethene. This is because both carbon atoms are identical.
In order to include the stereo information, the groups attached to the double‑bonded carbon atoms are looked into.
The carbon atom bears identical groups that are attached to the double bond. Therefore, stereoisomerism is not possible. Hence, the IUPAC name can be given as chloroethene.
IUPAC name for the given molecule is assigned.
(b)
Interpretation:
The IUPAC name for the given molecule has to be assigned including the prefix cis- or trans-.
Concept Introduction:
IUPAC nomenclature for alkene: There are about eight rules to be followed in giving IUPAC name for alkene.
- The suffix –ane has to be replaced with the suffix –ene. This is used to indicate the presence of double bond.
- The longest continuous chain of carbon atoms has to be chosen that contains both carbon atoms of the double bond.
- The parent carbon chain has to be numbered in a way so that the numbering begins at the end near to the double bond. In case if the double bond is equidistant from both ends, then numbering has to be done from the end that is closer to substituents.
- The position of the double bond has to be given a single number which is lower‑numbered carbon atom that is present in the double bond.
- Suffixes like –diene, -triene, -tetrene, and so on are used when the compound contains more than one double bond.
- In case of cycloalkenes which do not have any substitution, the numbering is not needed to locate the double bond because the bond is assumed to be between the carbons 1 and 2.
- In case if substituents are present in cycloalkene, then the double‑bonded carbon atoms are numbered 1 and 2 in a direction where the substituent gets the lower number.
- If the cycloalkenes contain more than one double bond, then one double bond is assigned the numbers 1 and 2 followed by the other double bond so that the lowest number possible is given.
Structural formula where a line represent carbon‑carbon bond and the carbon atom is considered to be present in each point and the end of lines is known as Line-angle structural formula. To indicate a double bond between carbon atom, double line is used.
If stereoisomers are possible for the alkene, the stereo information is shown in the IUPAC name by adding prefix cis- or trans- with respect to the groups present on the same side of double bond or opposite side of double bond.
(b)
Answer to Problem 13.50EP
The IUPAC name for the given compound is cis-1,2-dibromoethene.
Explanation of Solution
Given compound is,
The longest continuous carbon chain present in the given compound is two carbon atoms. Hence, the parent alkane is ethane.
As the given compound contains a double bond, the suffix –ane is replaced by –ene. Therefore, the name obtained is ethene.
Numbering has to be given in a way that the carbon atoms present in the double bond gets the least numbering. This is followed by the substituents present in the given molecule. It is found that an bromine atom is present on the second carbon atom and bromine atom is present on the first carbon atom. As the substituents are same, prefix di- is added. Therefore, the name of the given compound is 1,2-dibromoethene.
In order to include the stereo information, the groups attached to the double‑bonded carbon atoms are looked into.
The bulky groups are present on same side of the double bond. Hence, the configuration of the given molecule is cis-. Therefore, the IUPAC name can be given as cis-1,2-dibromoethene.
IUPAC name for the given molecule is assigned.
(c)
Interpretation:
The IUPAC name for the given molecule has to be assigned including the prefix cis- or trans-.
Concept Introduction:
IUPAC nomenclature for alkene: There are about eight rules to be followed in giving IUPAC name for alkene.
- The suffix –ane has to be replaced with the suffix –ene. This is used to indicate the presence of double bond.
- The longest continuous chain of carbon atoms has to be chosen that contains both carbon atoms of the double bond.
- The parent carbon chain has to be numbered in a way so that the numbering begins at the end near to the double bond. In case if the double bond is equidistant from both ends, then numbering has to be done from the end that is closer to substituents.
- The position of the double bond has to be given a single number which is lower‑numbered carbon atom that is present in the double bond.
- Suffixes like –diene, -triene, -tetrene, and so on are used when the compound contains more than one double bond.
- In case of cycloalkenes which do not have any substitution, the numbering is not needed to locate the double bond because the bond is assumed to be between the carbons 1 and 2.
- In case if substituents are present in cycloalkene, then the double‑bonded carbon atoms are numbered 1 and 2 in a direction where the substituent gets the lower number.
- If the cycloalkenes contain more than one double bond, then one double bond is assigned the numbers 1 and 2 followed by the other double bond so that the lowest number possible is given.
Structural formula where a line represent carbon‑carbon bond and the carbon atom is considered to be present in each point and the end of lines is known as Line-angle structural formula. To indicate a double bond between carbon atom, double line is used.
If stereoisomers are possible for the alkene, the stereoinformation is shown in the IUPAC name by adding prefix cis- or trans- with respect to the groups present on the same side of double bond or opposite side of double bond.
(c)
Answer to Problem 13.50EP
The IUPAC name for the given compound is trans-2-butene.
Explanation of Solution
Given compound is,
The longest continuous carbon chain present in the given compound is four carbon atoms. Hence, the parent alkane is butane.
As the given compound contains a double bond, the suffix –ane is replaced by –ene. Therefore, the name obtained is butene.
Numbering has to be given so that the double bond gets the least numbering. This gives the name as 2-butene.
Regarding the stereo informaton, the groups attached to the double‑bonded carbon atom has to be considered.
In this case, the methyl groups are present on opposite side of the double bond. Hence, trans- has to be included before the IUPAC name. The IUPAC name can be given as, trans-2-butene.
IUPAC name for the given molecule is assigned.
(d)
Interpretation:
The IUPAC name for the given molecule has to be assigned including the prefix cis- or trans-.
Concept Introduction:
IUPAC nomenclature for alkene: There are about eight rules to be followed in giving IUPAC name for alkene.
- The suffix –ane has to be replaced with the suffix –ene. This is used to indicate the presence of double bond.
- The longest continuous chain of carbon atoms has to be chosen that contains both carbon atoms of the double bond.
- The parent carbon chain has to be numbered in a way so that the numbering begins at the end near to the double bond. In case if the double bond is equidistant from both ends, then numbering has to be done from the end that is closer to substituents.
- The position of the double bond has to be given a single number which is lower‑numbered carbon atom that is present in the double bond.
- Suffixes like –diene, -triene, -tetrene, and so on are used when the compound contains more than one double bond.
- In case of cycloalkenes which do not have any substitution, the numbering is not needed to locate the double bond because the bond is assumed to be between the carbons 1 and 2.
- In case if substituents are present in cycloalkene, then the double‑bonded carbon atoms are numbered 1 and 2 in a direction where the substituent gets the lower number.
- If the cycloalkenes contain more than one double bond, then one double bond is assigned the numbers 1 and 2 followed by the other double bond so that the lowest number possible is given.
Structural formula where a line represent carbon‑carbon bond and the carbon atom is considered to be present in each point and the end of lines is known as Line-angle structural formula. To indicate a double bond between carbon atom, double line is used.
If stereoisomers are possible for the alkene, the stereoinformation is shown in the IUPAC name by adding prefix cis- or trans- with respect to the groups present on the same side of double bond or opposite side of double bond.
(d)
Answer to Problem 13.50EP
The IUPAC name for the given compound is 1,1-dibromoethene.
Explanation of Solution
Given compound is,
The longest continuous carbon chain present in the given compound is two carbon atoms. Hence, the parent alkane is ethane.
As the given compound contains a double bond, the suffix –ane is replaced by –ene. Therefore, the name obtained is ethene.
Numbering has to be given in a way that the carbon atoms present in the double bond gets the least numbering. Therefore, the name of the given compound is ethene. The substituent present on the longest carbon chain are bromine atoms that is present on the first carbon atom. Therefore, the name of the given molecule can be given as,
In order to include the stereo information, the groups attached to the double‑bonded carbon atoms are looked into.
The bulky groups are present on same carbon atom of the double bond. Hence, isomerism is not possible. Therefore, the IUPAC name of the given molecule is 1,1-dibromoethene.
IUPAC name for the given molecule is assigned.
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Chapter 13 Solutions
Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
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- For each reaction below, decide if the first stable organic product that forms in solution will create a new C - C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 tu ? ? OH Will the first product that forms in this reaction create a new CC bond? Yes No Will the first product that forms in this reaction create a new CC bond? Yes No C $ ©arrow_forwardAs the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C-C bond as its major product: 1. MgCl ? 2. H₂O* If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new CC bond. G marrow_forwardIncluding activity coefficients, find [Hg22+] in saturated Hg2Br2 in 0.00100 M NH4 Ksp Hg2Br2 = 5.6×10-23.arrow_forward
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- 3-Oxo-butanenitrile and (E)-2-butenal are mixed with sodium ethoxide in ethanol. Draw and name the structures of the products.arrow_forwardWhat is the reason of the following(use equations if possible) a.) In MO preperation through diazotization: Addition of sodium nitrite in acidfied solution in order to form diazonium salt b.) in MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at low pH c.) In MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at pH 4.5 d.) Avoiding not cooling down the reaction mixture when preparing the diazonium salt e.) Cbvcarrow_forwardA 0.552-g sample of an unknown acid was dissolved in water to a total volume of 20.0 mL. This sample was titrated with 0.1103 M KOH. The equivalence point occurred at 29.42 mL base added. The pH of the solution at 10.0 mL base added was 3.72. Determine the molar mass of the acid. Determine the Ka of the acid.arrow_forward
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