Q.Give plausible explanation for each of the following:

1. Concept: Acidity depends on the stability of the conjugate base after losing a proton. In alcohols, the conjugate base is an alkoxide ion ($RO^-$), where the negative charge is localised on a highly electronegative oxygen. In amines, the conjugate base is an amide ion ($R_2N^-$), where nitrogen is less electronegative than oxygen and holds the negative charge less effectively. The alkoxide ion is therefore more stable than the amide ion, making alcohols more willing to lose a proton — i.e., more acidic. Amines are weaker acids.
2. Concept: Boiling point depends on the strength of intermolecular hydrogen bonding. Primary amines ($RNH_2$) have two N–H bonds, allowing them to form extensive intermolecular hydrogen bonds. Tertiary amines ($R_3N$) have no N–H bond, so they cannot hydrogen-bond with each other — only weak dipole-dipole and van der Waals forces operate. Stronger hydrogen bonding in primary amines requires more energy to overcome, giving them higher boiling points.
3. Concept: Base strength depends on the availability of the lone pair on nitrogen. …

(i) Why are amines less acidic than alcohols of comparable molecular masses?

Concept first: Acidity is about how easily a compound gives up a proton ($H^+$). When an acid loses $H^+$, it forms a conjugate base. The more stable that conjugate base, the stronger the acid. So to compare acidity of amines ($R-NH_2$) and alcohols ($R-OH$), we compare the stability of their conjugate bases: the amide ion ($R-NH^-$) and the alkoxide ion ($R-O^-$).

1. Electronegativity of the atom holding the negative charge.

   Oxygen is more electronegative than nitrogen (3.44 vs 3.04 on Pauling scale). In the alkoxide ion, the negative charge is held by a more electronegative atom, which stabilises the charge better. The amide ion has the negative charge on a less electronegative nitrogen, so it is less stable.

2. Resonance stabilisation (or lack thereof).

   In an alkoxide ion, the negative charge on oxygen can be delocalised into the alkyl group only through inductive effects — no resonance. In an amide ion, there is also no significant resonance stabilisation. So the key difference remains the electronegativity factor.

3. Inductive effect of the alkyl group.

   Alkyl groups are electron-donating (+I effect). They push electron density toward the negatively charged atom, which destabilises the conjugate base. This effect is similar for both amines and alcohols of comparable molecular mass, so it doesn't tip the scale.

Conclusion: The alkoxide ion is more stable than the amide ion because oxygen holds negative charge better. Hence alcohols are stronger acids than amines of comparable molecular mass.

(ii) Why do primary amines have higher boiling point than tertiary amines?

Concept first: Boiling point depends on the strength of intermolecular forces. For amines, the key forces are hydrogen bonding (between N–H of one molecule and lone pair of another) and van der Waals forces (which increase with molecular size and surface area).

1. Hydrogen bonding capability.

   A primary amine ($R-NH_2$) has two N–H bonds, so it can form two hydrogen bonds per molecule (as donor). A tertiary amine ($R_3N$) has no N–H bond — it can only act as an acceptor (via the lone pair on nitrogen), not as a donor. So primary amines form extensive intermolecular hydrogen bonding networks; tertiary amines cannot.

2. Effect on boiling point.

   Breaking hydrogen bonds requires extra energy. Primary amines thus have significantly higher boiling points than tertiary amines of comparable molecular mass. For example, $n$-propylamine (b.p. 49°C) vs. trimethylamine (b.p. 3°C) — both have molecular mass ~59 g/mol.

3. What about secondary amines?

   Secondary amines ($R_2NH$) have one N–H bond, so they fall in between: higher boiling than tertiary, but lower than primary (because they form fewer hydrogen bonds per molecule).

Conclusion: Primary amines have higher boiling points than tertiary amines because they can form intermolecular hydrogen bonds (as donors), while tertiary amines cannot.

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Concept first: Basicity is about the availability of the lone pair on nitrogen to accept a proton ($H^+$). The more available (electron-rich) the lone pair, the stronger the base. Anything that delocalises or withdraws electron density from the nitrogen lone pair reduces basicity.

1. Resonance in aromatic amines. In aniline ($C_6H_5NH_2$), the lone pair on nitrogen is in conjugation with the $\pi$-electrons of the benzene ring. This delocalises the lone pair into the ring, making it less available for protonation. The resonance structures show the lone pair being shared with the ortho and para positions of the ring. …

Here is a clear, concept-first solution for each part of your question.

Method: Inductive Effect & Solvation Analysis

This method explains basicity and acidity by analyzing how electron density is distributed around the reactive site (N or O) and how the resulting ion is stabilized.

(i) Why are amines less acidic than alcohols of comparable molecular masses?

Step 1: Identify the acidic proton.

• In an alcohol ($R-OH$), the acidic proton is on the oxygen.
• In an amine ($R-NH_2$), the acidic proton is on the nitrogen.

Step 2: Compare the stability of the conjugate base.

• When an alcohol loses $H^+$, it forms an alkoxide ion ($R-O^-$).
• When an amine loses $H^+$, it forms an amide ion ($R-NH^-$).

Step 3: Apply the concept of electronegativity.

• Oxygen is more electronegative than Nitrogen.
• A more electronegative atom can better stabilize a negative charge by pulling the electron density towards itself.
• Therefore, the alkoxide ion ($R-O^-$) is more stable than the amide ion ($R-NH^-$).

Step 4: Conclude the acidity order.

• A stronger acid has a more stable conjugate base.
• Since $R-O^-$ is more stable, $R-OH$ (alcohol) is a stronger acid.
• Since $R-NH^-$ is less stable, $R-NH_2$ (amine) is a weaker acid.

Final Answer: Amines are less acidic than alcohols because oxygen is more electronegative than nitrogen, allowing the alkoxide conjugate base to stabilize the negative charge better than the amide ion.

(ii) Why do primary amines have higher boiling points than tertiary amines?

Step 1: Identify the key intermolecular force.

• Boiling point depends on the strength of intermolecular forces. For amines, the most important force is hydrogen bonding.

Step 2: Count the available hydrogen atoms for bonding.

• A primary amine ($RNH_2$) has two N-H bonds. This allows it to form two hydrogen bonds with neighboring molecules (as a donor).
• A tertiary amine ($R_3N$) has no N-H bonds. It can only accept hydrogen bonds (via the lone pair on N) but cannot donate them.

Step 3: Compare the extent of hydrogen bonding.

• Primary amines can form an extensive network of hydrogen bonds (both donating and accepting).
• Tertiary amines can only form weak dipole-dipole interactions and act as acceptors, resulting in significantly weaker overall intermolecular attraction.

Step 4: Conclude the boiling point order.

• Stronger intermolecular forces require more energy to overcome, leading to a higher boiling point.

Final Answer: Primary amines have higher boiling points than tertiary amines because primary amines have N-H bonds that allow them to form strong intermolecular hydrogen bonds, while tertiary amines lack N-H bonds and can only form weaker dipole-dipole interactions.

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Step 1: Identify the basic site.

• Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton ($H^+$). …

Here is a breakdown of the common mistakes students make on this exact question, along with the correct conceptual reasoning required for Indian exams (JEE, NEET, Class 12).

(i) Why are amines less acidic than alcohols of comparable molecular masses?

The Concept (The "Why")

Acidity is the ability to donate a proton ($H^+$). To compare acidity, we compare the stability of the conjugate base formed after losing $H^+$.

• Alcohol loses $H^+$ to give Alkoxide ion ($R-O^-$).
• Amine loses $H^+$ to give Amide ion ($R-NH^-$).

Common Mistake #1: Comparing the wrong atom.

• The Error: Students compare the electronegativity of N vs O in the original molecule (amine vs alcohol).
• The Fix: You must compare the stability of the conjugate base (the anion). Oxygen is more electronegative than Nitrogen. Therefore, the negative charge is better stabilized on Oxygen ($RO^-$) than on Nitrogen ($RNH^-$). A more stable conjugate base means a stronger acid.
• Result: $RO^-$ is more stable than $RNH^-$. Therefore, Alcohols are stronger acids than amines.

Common Mistake #2: Confusing Basicity with Acidity.

• The Error: Students think "Amines are basic, so they must be more acidic." This is backwards.
• The Fix: Basicity is the ability to accept a proton. Acidity is the ability to donate a proton. They are opposite properties. Amines are good bases (they want to grab $H^+$), so they are poor acids (they don't want to give $H^+$ away).

How to avoid: Always draw the conjugate base. The atom holding the negative charge determines stability. O > N in electronegativity, so $RO^-$ > $RNH^-$ in stability.

(ii) Why do primary amines have higher boiling point than tertiary amines?

The Concept (The "Why")

Boiling point depends on the strength of intermolecular forces. The key force here is Hydrogen bonding.

Common Mistake #1: Forgetting the number of H-atoms on Nitrogen.

• The Error: Students say "Primary amines have more H-bonding because they have more H atoms." This is partially correct but incomplete.
• The Fix: For a molecule to form a strong H-bond, it needs a H atom attached to N (donor) and a lone pair on N (acceptor).
  • Primary ($1^\circ$): Has two H atoms on N. Can form extensive H-bonds (donor and acceptor).
  • Tertiary ($3^\circ$): Has zero H atoms on N. It can only accept H-bonds (using its lone pair), but cannot donate them. This makes its intermolecular attraction much weaker.
• Result: $1^\circ$ amines have strong intermolecular H-bonding, requiring more energy to boil. $3^\circ$ amines have only weak dipole-dipole and van der Waals forces.

Common Mistake #2: Ignoring molecular mass/surface area.

• The Error: Students think tertiary amines (which are more branched) have higher boiling points because they have more carbon atoms.
• The Fix: For comparable molecular masses, branching decreases surface area and decreases van der Waals forces. So, even if $3^\circ$ amines had H-bonding (which they don't), their shape would lower the boiling point. The dominant factor here is the absence of H-bond donation in $3^\circ$ amines.

How to avoid: Count the number of $N-H$ bonds. More $N-H$ bonds = stronger H-bonding = higher boiling point. $1^\circ > 2^\circ > 3^\circ$.

(iii) Why are aliphatic amines stronger bases than aromatic amines?

The Concept (The "Why")

Basicity depends on the availability of the lone pair on Nitrogen to accept a proton ($H^+$).

Common Mistake #1: Only thinking about Inductive Effect (+I) of alkyl groups.

• The Error: Students correctly say alkyl groups are electron-donating (+I), which increases electron density on N, making aliphatic amines stronger bases. But they stop there.
• The Fix: This is only half the story. The real killer is the Resonance Effect in aromatic amines (e.g., Aniline).

Common Mistake #2: Forgetting the Resonance in Aniline.

• The Error: Students don't draw the resonance structures of aniline. …