1. Would Ch3-OH or CH3-CH2-CH2-CH2-CH2-OH have a higher boiling point? Fully explain ypur answer.
2. Explain why a tertiary alcohol will not undergo an oxidation reaction. Give an example.
3. Compare and contyrast similarities/differences of condensation and substitution reactions. Provide examples(s) to illustrate their differences.
4. A student is given a contianer with a pure organic compound. OUtline a procedure that the student could perform to explore the type of a functional group on the prganic compound.
5. HOW WOULD YOU DRAW 2-METHYL-2-AMINOBUTANE and
2-AMINOPENTANE ....TO DRAW THE STRUCTURAL DIAGRAM
1. I would expect n-pentanol to have a higher boiling point because it has a much higher molar mass as well as some hydrogen bonding (although not nearly as much as with CH3OH).
2. Here is a site where you can read about the oxidiation of primary, secondary, and tertiary alcohols.
http://www.chemguide.co.uk/organicprops/alcohols/oxidation.html
I think you can find the others in your text or your notes.
Determine All the products of the acid -catalyzed deghydration of 2-pentanol. Show the mechanism for one product.
1. To determine which compound would have a higher boiling point between Ch3-OH (methanol) and CH3-CH2-CH2-CH2-CH2-OH (pentanol), we need to consider the molecular structure and intermolecular forces.
Methanol (CH3-OH) is a smaller molecule than pentanol (CH3-CH2-CH2-CH2-CH2-OH). Smaller molecules tend to have weaker intermolecular forces. In this case, methanol can form hydrogen bonds due to the presence of an -OH group, which creates stronger intermolecular forces compared to the London dispersion forces within pentanol.
Pentanol has more carbon atoms in its chain, giving it a larger molecular size. Larger molecules have stronger London dispersion forces, which can contribute to a higher boiling point.
Overall, the pentanol molecule with its longer carbon chain and stronger London dispersion forces will have a higher boiling point compared to methanol.
2. A tertiary alcohol is characterized by having the -OH group attached to a carbon atom that is bonded to three other carbon atoms. Due to this structural arrangement, tertiary alcohols are highly resistant to oxidation reactions.
In oxidation reactions, the -OH group in alcohols is typically converted into a carbonyl group (C=O), resulting in the formation of an aldehyde or a ketone. However, in a tertiary alcohol, there are no hydrogen atoms bonded to the carbon atom adjacent to the -OH group. As a result, there are no hydrogen atoms available for oxidation.
For example, 2-methyl-2-propanol (also known as tertiary butyl alcohol) cannot be oxidized because its tertiary carbon center lacks adjacent hydrogen atoms required for the oxidation process.
3. Let's compare and contrast condensation and substitution reactions:
Similarities:
- Both condensation and substitution reactions involve the breaking and forming of chemical bonds.
- Both reactions typically occur between organic compounds.
- Both reactions can result in the formation of new molecules.
Differences:
Condensation Reaction:
- In a condensation reaction, two molecules combine to form a larger molecule while simultaneously releasing a small molecule, such as water or an alcohol.
- It involves the elimination of a small molecule, leading to the formation of a new covalent bond between the remaining atoms.
- An example is the esterification reaction between an alcohol and a carboxylic acid to produce an ester, with the elimination of water.
Substitution Reaction:
- In a substitution reaction, one functional group in a molecule is replaced by another functional group.
- It involves the replacement of an atom or a group of atoms by another atom or group of atoms.
- An example is the halogenation of an alkane, where a hydrogen atom is replaced by a halogen atom, such as chlorine, resulting in the formation of an alkyl halide.
4. To explore the type of functional group present in a pure organic compound, a student could perform the following procedure:
1. Determine the compound's physical properties: Measure the compound's melting point, boiling point, density, solubility, or other relevant physical properties. Compare these values with known compounds to identify potential functional groups.
2. Conduct a functional group test: Perform specific tests for different functional groups. For example, the student could conduct a test for alcohols using a chromic acid test, or a test for carbonyl groups using a 2,4-DNP test.
3. Analyze spectroscopic data: Utilize techniques like infrared spectroscopy (IR), nuclear magnetic resonance (NMR), or mass spectrometry (MS) to gather data on the compound's molecular structure. Compare the obtained spectroscopic data with known functional groups to deduce the presence of specific functional groups.
5. To draw the structural diagrams for 2-methyl-2-aminobutane and 2-aminopentane, follow these steps:
2-Methyl-2-aminobutane:
- Start with a butane chain, consisting of four carbon atoms in a straight line.
- Place a methyl group (CH3) on the second carbon atom.
- Attach an amino group (-NH2) to the fourth carbon atom.
- The final structure should have a butane chain with a methyl group on the second carbon and an amino group on the fourth carbon.
2-Aminopentane:
- Begin with a pentane chain, consisting of five carbon atoms in a straight line.
- Attach an amino group (-NH2) to the second carbon atom.
- The final structure should resemble a pentane chain with an amino group on the second carbon.