Organic Chemistry

Dehydrogenation: Dehydrogenation refers to the removal of hydrogen atoms from a molecule, resulting in the formation of a double bond. This process is often used to convert saturated hydrocarbons into unsaturated hydrocarbons. The most common example is the dehydrogenation of alkanes to alkenes. For instance, the dehydrogenation of ethane (C2H6) produces ethene (C2H4):

C2H6 → C2H4 + H2

Dehydrogenation reactions are typically carried out using catalysts, such as platinum or palladium, at high temperatures. These catalysts facilitate the breaking of the C-H bonds and promote the formation of double bonds. Dehydrohalogenation: Dehydrohalogenation involves the elimination of a halogen atom (usually chlorine, bromine, or iodine) and a hydrogen atom from an organic molecule, leading to the formation of a double bond. This process is commonly used to synthesize alkenes or alkynes from alkyl halides. The reaction is often conducted using a strong base, such as potassium hydroxide (KOH) or sodium ethoxide (NaOCH2CH3), which abstracts a proton from the molecule, resulting in the removal of a halogen atom.

For example, the dehydrohalogenation of 2-chloroethane (C2H5Cl) with potassium hydroxide can yield ethene (C2H4): C2H5Cl + KOH → C2H4 + KCl + H2O

It's important to note that dehydrohalogenation can also occur in the presence of a strong base without the simultaneous removal of a proton, resulting in the formation of an alkene.

Both dehydrogenation and dehydrohalogenation reactions are significant in organic chemistry as they provide synthetic pathways to produce unsaturated compounds, which are important in the production of polymers, pharmaceuticals, and various other chemical compounds.

The compound that undergoes electrophilic substitution most readily among the given options is determined by the electron-donating or electron-withdrawing nature of the substituents on the aromatic ring. Electrophilic substitution reactions occur more readily in aromatic compounds that have electron-donating groups, which activate the ring by increasing electron density. Let's evaluate the options:

• Nitrobenzene: Contains a nitro group (-NO₂) that is electron-withdrawing, deactivating the benzene ring towards electrophilic substitution.
• Toluene: Contains a methyl group (-CH₃), which is electron-donating. This increases the electron density on the benzene ring, activating it for electrophilic substitution reactions.
• Benzene: A standard aromatic compound with no substituents. It is less reactive than toluene due to the lack of electron-donating groups.
• Benzoic acid: Contains a carboxyl group (-COOH), which is electron-withdrawing, thus deactivating the benzene ring.

Among these, Toluene undergoes electrophilic substitution most readily, due to the electron-donating methyl group that enhances the reactivity of the aromatic ring towards electrophiles.

To determine which compound has the highest boiling point,
we need to consider the intermolecular forces that affect boiling points,
such as hydrogen bonding, molecular weight, and molecular structure.
Step 1: Identify intermolecular forces
- All the given compounds (methanol, ethanol, propanol, and water)
are capable of forming hydrogen bonds because they have an -OH group,
which allows hydrogen bonding between molecules.
- Boiling point increases with stronger intermolecular forces
and higher molecular weight, but hydrogen bonding strength depends
on the molecule's structure.
Step 2: Compare molecular weights and hydrogen bonding
- Methanol (CH OH): Molecular weight 32 g/mol, 1 carbon.
- Ethanol (C H OH): Molecular weight 46 g/mol, 2 carbons.
- Propanol (C H OH): Molecular weight 60 g/mol, 3 carbons.
- Water (H O): Molecular weight 18 g/mol, no carbon,
but two hydrogen atoms available for hydrogen bonding per molecule.
Step 3: Analyze hydrogen bonding
- Water has a unique structure with two hydrogen atoms
and two lone pairs on oxygen, allowing it to form an extensive network of hydrogen bonds.
This results in a higher boiling point despite its lower molecular weight.
- Alcohols (methanol, ethanol, propanol) also form hydrogen bonds,
but the effect is less extensive due to the presence of the alkyl group,
which reduces the hydrogen bonding efficiency per molecule as the carbon chain length increases.
Step 4: Compare boiling points
- Methanol: Boiling point 64.7°C
- Ethanol: Boiling point 78.4°C
- Propanol: Boiling point 97.4°C
- Water: Boiling point 100°C
The boiling point increases with molecular weight among alcohols
due to stronger van der Waals forces, but water’s extensive hydrogen bonding network
gives it the highest boiling point.
Step 5: Conclusion
Despite having a lower molecular weight, water has the highest boiling point (100°C)
due to its strong and extensive hydrogen bonding network,
making (c) Water (H O) the correct answer.

To solve the problem, we need to identify the reagent that can convert an aldehyde to a primary alcohol.

1. Understanding the Reaction:
- Aldehydes can be reduced to primary alcohols by reducing agents.
- Oxidizing agents typically convert aldehydes to carboxylic acids.
- Some reagents oxidize aldehydes without reducing them.

2. Analysis of Each Reagent:
- PCC (Pyridinium chlorochromate): An oxidizing agent that oxidizes primary alcohols to aldehydes but does not reduce aldehydes.
- (Sodium borohydride): A mild reducing agent that reduces aldehydes to primary alcohols.
- (Potassium permanganate): A strong oxidizing agent, converts aldehydes to carboxylic acids.
- (Nitric acid): An oxidizing agent, not a reducing agent.

3. Conclusion:
Only is capable of reducing aldehydes to primary alcohols.

Final Answer:
The correct reagent is (Sodium borohydride).