Organic Chemistry

Concept: Aniline ( ) can be prepared from benzene by introducing a nitro group followed by reduction. This is a standard two-step aromatic substitution and reduction sequence.
Step 1: Nitration of Benzene Benzene is treated with a mixture of concentrated nitric acid and concentrated sulfuric acid (nitrating mixture) to form nitrobenzene.
Step 2: Reduction of Nitrobenzene Nitrobenzene is reduced to aniline using reducing agents such as:

• Sn/HCl or Fe/HCl (followed by NaOH)
• Catalytic hydrogenation (H /Pd)

Overall Conversion: Conclusion: Thus, benzene can be converted to aniline in two steps: nitration followed by reduction of the nitro group.

Concept: A zwitterion is a molecule that contains both a positive and a negative charge on different atoms but is overall electrically neutral. Such species are commonly observed in amino acids due to the presence of both acidic and basic functional groups. Definition: A zwitterion is a dipolar ion that has equal positive and negative charges within the same molecule, resulting in a net zero charge. Example: Glycine Glycine is the simplest amino acid with the structure: In aqueous solution, glycine undergoes an internal acid-base reaction:

• The amino group ( ) accepts a proton to become .
• The carboxyl group ( ) loses a proton to become .

Thus, the zwitterionic form of glycine is:
Explanation:

• Contains both positive ( ) and negative ( ) charges.
• Overall charge remains zero.
• This form predominates at the isoelectric point (neutral pH).

Significance:

• Explains high melting points of amino acids
• Responsible for their amphoteric nature
• Affects solubility and conductivity in solution

Conclusion: Thus, a zwitterion is a dipolar ion with both positive and negative charges in the same molecule, as seen in glycine where represents its zwitterionic form.

Concept: The iodoform test is used to identify compounds containing the methyl ketone group ( ) or alcohols that can be oxidized to it. Ethanol is the only primary alcohol that gives a positive iodoform test because it can be oxidized to acetaldehyde. Procedure: Add iodine solution and sodium hydroxide (NaOH) to the given alcohol and warm gently. Observation:

• Ethanol: Produces a yellow precipitate of iodoform ( ) with a characteristic antiseptic smell.
• Methanol: Does not produce any yellow precipitate (negative test).

Reason:

• Ethanol is oxidized to acetaldehyde ( ), which contains the required methyl carbonyl group and gives the iodoform reaction.
• Methanol is oxidized to formaldehyde ( ), which does not contain the methyl group required for the test.

Conclusion: Thus, ethanol can be distinguished from methanol because ethanol gives a positive iodoform test (yellow precipitate), while methanol does not.