Concept explainers
Videos
(a)
Interpretation:
The member of the given pair which is expected to have high boiling point has to be identified.
Concept Introduction:
The derivative of alkane with halogen instead of one or more hydrogen atoms is known as halogenated alkane. If the same is present in cycloalkane, then it is known as halogenated cycloalkane. They are product of reaction between alkane/cycloalkane with halogens.
Physical properties of halogenated
Boiling point of halogenated hydrocarbon is usually higher than the corresponding hydrocarbon. This is because, there is a polarity difference between carbon and halogen atom. This result in increased dipole‑dipole interations.
General trend considering the boiling point and melting point of halogenated hydrocarbon are,
- Melting and boiling points increase with the increase in size of alkyl groups that is present. Due to the increasing intermolecular forces, the melting and boiling point increases.
- As the size of halogen atom increases, the melting point and boiling point also increases.
Halogenated hydrocarbons do not possess hydrogen bonding capability. Therefore, solubility of halogenated hydrocarbon is limited.
(b)
Interpretation:
The member of the given pair which is expected to have high boiling point has to be identified.
Concept Introduction:
The derivative of alkane with halogen instead of one or more hydrogen atoms is known as halogenated alkane. If the same is present in cycloalkane, then it is known as halogenated cycloalkane. They are product of reaction between alkane/cycloalkane with halogens.
Physical properties of halogenated alkanes:
Boiling point of halogenated hydrocarbon is usually higher than the corresponding hydrocarbon. This is because, there is a polarity difference between carbon and halogen atom. This result in increased dipole‑dipole interations.
General trend considering the boiling point and melting point of halogenated hydrocarbon are,
- Melting and boiling points increase with the increase in size of alkyl groups that is present. Due to the increasing intermolecular forces, the melting and boiling point increases.
- As the size of halogen atom increases, the melting point and boiling point also increases.
Halogenated hydrocarbons do not possess hydrogen bonding capability. Therefore, solubility of halogenated hydrocarbon is limited.
(c)
Interpretation:
The member of the given pair which is expected to have high boiling point has to be identified.
Concept Introduction:
The derivative of alkane with halogen instead of one or more hydrogen atoms is known as halogenated alkane. If the same is present in cycloalkane, then it is known as halogenated cycloalkane. They are product of reaction between alkane/cycloalkane with halogens.
Physical properties of halogenated alkanes:
Boiling point of halogenated hydrocarbon is usually higher than the corresponding hydrocarbon. This is because, there is a polarity difference between carbon and halogen atom. This result in increased dipole‑dipole interations.
General trend considering the boiling point and melting point of halogenated hydrocarbon are,
- Melting and boiling points increase with the increase in size of alkyl groups that is present. Due to the increasing intermolecular forces, the melting and boiling point increases.
- As the size of halogen atom increases, the melting point and boiling point also increases.
Halogenated hydrocarbons do not possess hydrogen bonding capability. Therefore, solubility of halogenated hydrocarbon is limited.
(d)
Interpretation:
The member of the given pair which is expected to have high boiling point has to be identified.
Concept Introduction:
The derivative of alkane with halogen instead of one or more hydrogen atoms is known as halogenated alkane. If the same is present in cycloalkane, then it is known as halogenated cycloalkane. They are product of reaction between alkane/cycloalkane with halogens.
Physical properties of halogenated alkanes:
Boiling point of halogenated hydrocarbon is usually higher than the corresponding hydrocarbon. This is because, there is a polarity difference between carbon and halogen atom. This result in increased dipole‑dipole interations.
General trend considering the boiling point and melting point of halogenated hydrocarbon are,
- Melting and boiling points increase with the increase in size of alkyl groups that is present. Due to the increasing intermolecular forces, the melting and boiling point increases.
- As the size of halogen atom increases, the melting point and boiling point also increases.
Halogenated hydrocarbons do not possess hydrogen bonding capability. Therefore, solubility of halogenated hydrocarbon is limited.
Trending nowThis is a popular solution!
Chapter 12 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- Na :S f. F NO2arrow_forwardQ1: For each molecule, assign each stereocenter as R or S. Circle the meso compounds. Label each compound as chiral or achiral. + CI OH woཡི།༠w Br H مه D CI ပ။ Br H, Br Br H₂N OMe R IN Ill N S H CI Br CI CI D OH H 1/111arrow_forwardDraw the two products of the reaction. H₂C. CH₂ H :0: CH3 CH₂ +1arrow_forward
- 93 = Volume 93 = 5.32× 10 3 -23 ст a √ 1073 5.32× 10 3 cm³arrow_forwardASP.....arrow_forwardQuestion 7 (10 points) Identify the carboxylic acid present in each of the following items and draw their structures: Food Vinegar Oranges Yogurt Sour Milk Pickles Acid Structure Paragraph ✓ BI UAE 0118 + v Task: 1. Identify the carboxylic acid 2. Provide Name 3. Draw structure 4. Take a picture of your table and insert Add a File Record Audio Record Video 11.arrow_forward
- Check the box under each structure in the table that is an enantiomer of the molecule shown below. If none of them are, check the none of the above box under the table. Molecule 1 Molecule 2 IZ IN Molecule 4 Molecule 5 ZI none of the above ☐ Molecule 3 Х IN www Molecule 6 NH Garrow_forwardHighlight each chiral center in the following molecule. If there are none, then check the box under the drawing area. There are no chiral centers. Cl Cl Highlightarrow_forwardA student proposes the following two-step synthesis of an ether from an alcohol A: 1. strong base A 2. R Is the student's proposed synthesis likely to work? If you said the proposed synthesis would work, enter the chemical formula or common abbreviation for an appropriate strong base to use in Step 1: If you said the synthesis would work, draw the structure of an alcohol A, and the structure of the additional reagent R needed in Step 2, in the drawing area below. If there's more than one reasonable choice for a good reaction yield, you can draw any of them. ☐ Click and drag to start drawing a structure. Yes No ロ→ロ 0|0 G Х D : ☐ பarrow_forward
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning