Q.What will be the conjugate bases for the following Brönsted acids: HF, H 2SO4 and HCO3 – ?

Concept & Intuition

This is a classic Lewis acid–base reaction. BF₃ has an empty $p$ orbital on boron (it’s $sp^2$ hybridized, trigonal planar) and needs a pair of electrons. NH₃ has a lone pair on nitrogen (it’s $sp^3$ hybridized, trigonal pyramidal) and can donate it. When they combine 1:1, the lone pair forms a coordinate covalent bond from N to B. That bond fills boron’s octet, forcing boron to rehybridize from $sp^2$ to $sp^3$ to accommodate four bonds. Nitrogen ends up with four bonds and no lone pair, so it too is $sp^3$ and tetrahedral.

Step-by-step reasoning

1. Identify the reactants’ initial hybridization and geometry

   • BF₃: Boron has 3 valence electrons, forms 3 σ bonds with fluorine. No lone pairs. It’s electron-deficient (only 6 electrons around B). To minimize repulsion, it uses $sp^2$ hybridization → trigonal planar (bond angles 120°).
   • NH₃: Nitrogen has 5 valence electrons, forms 3 σ bonds with hydrogen, and has 1 lone pair. That’s 4 electron domains → $sp^3$ hybridization. The lone pair repels more strongly than bonds, so the geometry is trigonal pyramidal (bond angles ~107°), but the electron-domain geometry is tetrahedral.

2. What happens in the 1:1 reaction?

   The lone pair on NH₃ attacks the empty $p$ orbital on B in BF₃, forming a coordinate (dative) bond N → B. Boron now has 4 bonds (three B–F and one B–N) → 8 electrons around B. Nitrogen’s lone pair is now used in the N–B bond, so after the reaction N has four σ bonds (three N–H and one N–B) and zero lone pairs — that is, 4 bonding domains and no lone pair.

3. Determine hybridization after adduct formation

   • Boron: 4 bonding domains → $sp^3$ hybridization. Geometry: tetrahedral (all bonds equivalent in space, angles ~109.5°).
   • Nitrogen: 4 bonding domains, no lone pairs → also $sp^3$ hybridization. Geometry: tetrahedral (not pyramidal, because there’s no lone pair to distort the shape). The adduct BF₃–NH₃ is drawn with a B–N bond, and both centers are tetrahedral.

4. Match with the options

   • (A) says B: $sp^3$, tetrahedral ; N: $sp^3$, tetrahedral → matches.
   • (B) says B: $sp^2$, trigonal planar → wrong, B gains a fourth bond.
   • (C) says N: $sp^3$, pyramidal → wrong, N has no lone pair in the adduct.
   • (D) says N: $sp^2$, trigonal planar → wrong, N has four bonds.

The concept first

Step one is the standard way to climb from an acid to an amide. Ammonia first simply neutralises the acid to give the ammonium salt; the salt is then thermally dehydrated:

Step two is the Hofmann bromamide reaction, and the idea worth holding on to is that it is a degradation — the product has one carbon fewer than the starting amide, because the carbonyl carbon is expelled. Why? The nitrogen is brominated and deprotonated; loss of $\mathrm{Br^-}$ leaves an electron-deficient nitrogen, and the alkyl group sitting on the neighbouring carbon migrates onto that nitrogen. The carbon it left behind is now an isocyanate carbon, and hydroxide hydrolyses it away as carbonate. Net effect: $\mathrm{R\!-\!CO\!-\!NH_2 \rightarrow R\!-\!NH_2}$.

This is a hugely useful synthetic tool: it is the standard way to make a primary amine with one carbon less, and it always gives a primary amine, never a mixture.

Step-by-step

1. Form X.

2. Brominate the nitrogen. $\mathrm{CH_3CONH_2 + Br_2 + NaOH \rightarrow CH_3CONHBr}$ ($N$-bromoacetamide).
3. Deprotonate and lose bromide. $\mathrm{CH_3CO\bar{N}Br \rightarrow CH_3CO\ddot{N}{:}\ (nitrene\text{-}like) + Br^-}$.
4. 1,2-Migration. The methyl group migrates from carbon to nitrogen, giving methyl isocyanate:

5. Hydrolysis. $\mathrm{CH_3NCO + 2NaOH \rightarrow CH_3NH_2 + Na_2CO_3}$.
6. Overall balanced equation.

So $\mathbf{Y = CH_3NH_2}$, methylamine — a primary amine with one carbon less than acetamide.

7. Reject the others: $\mathrm{CH_3CH_2Br}$ and $\mathrm{CH_3COONa}$ are not amines at all (and no reagent here makes a C–Br bond on carbon); $\mathrm{(CH_3)_2NH}$ is secondary — Hofmann degradation never gives a secondary amine.

ANSWER: C

Concept & Intuition

Kjeldahl’s method converts organic nitrogen into ammonia (NH₃). The ammonia is trapped in a known volume of standard acid (here H₂SO₄). Some of the acid reacts with NH₃; the leftover acid is then titrated with standard base (NaOH). The difference between the initial moles of acid and the moles of base needed gives the moles of NH₃, and hence the moles of nitrogen. From that, we compute the percentage of nitrogen in the sample. Then we compare that percentage with the theoretical nitrogen percentage of each candidate compound.

1. Find moles of H₂SO₄ initially taken

   Volume = 60 mL = 0.060 L, concentration = 0.1 M.

   Moles of H₂SO₄ = $0.060 \times 0.1 = 0.0060$ mol.

2. Find moles of NaOH used for back-titration

   Volume = 20 mL = 0.020 L, concentration = 0.1 M.

   Moles of NaOH = $0.020 \times 0.1 = 0.0020$ mol.

3. Relate NaOH to unreacted H₂SO₄

   The reaction is:

So 1 mol H₂SO₄ reacts with 2 mol NaOH.

Moles of unreacted H₂SO₄ = $\frac{0.0020}{2} = 0.0010$ mol.

4. Find moles of H₂SO₄ that reacted with NH₃

   Reacted H₂SO₄ = initial – unreacted = $0.0060 - 0.0010 = 0.0050$ mol.

5. Find moles of NH₃ (and hence moles of N)

   The reaction of NH₃ with H₂SO₄ is:

So 1 mol H₂SO₄ reacts with 2 mol NH₃.

Moles of NH₃ = $2 \times 0.0050 = 0.010$ mol.

Therefore moles of nitrogen = 0.010 mol.

6. Mass of nitrogen in the sample

   Mass = moles × atomic mass = $0.010 \times 14 = 0.14$ g.

7. Percentage of nitrogen in the compound

   Sample mass = 0.933 g.

8. Compute theoretical %N for each option

   • (A) C₆H₅CH₂NH₂ (benzylamine): Molar mass = 107 g/mol, N = 14 → %N = $14/107 \approx 13.1\%$
   • (B) C₆H₅NH₂ (aniline): Molar mass = 93 g/mol, N = 14 → %N = $14/93 \approx 15.05\%$
   • (C) CH₃CH₂NH₂ (ethylamine): Molar mass = 45 g/mol, N = 14 → %N = $14/45 \approx 31.1\%$
   • (D) Acetamide (CH₃CONH₂): Molar mass = 59 g/mol, N = 14 → %N = $14/59 \approx 23.7\%$

   Only aniline gives ~15.0% nitrogen, matching the experimental value.

The key to this question is recalling the structure of asparagine. Amino acids all share a common backbone: a central carbon (the α-carbon) bonded to an amino group ($-\mathrm{NH}_2$), a carboxylic acid group ($-\mathrm{COOH}$), a hydrogen atom, and a variable side chain (the "R group"). For asparagine, the side chain is $-\mathrm{CH}_2-\mathrm{CONH}_2$, which contains a carboxamide group — that is, a carbonyl ($\mathrm{C=O}$) attached to an amine ($-\mathrm{NH}_2$). So the three functional groups present are: the amino group, the carboxylic acid group, and the carboxamide group.

Let’s examine each option carefully.

1. Option (A): $-\mathrm{NH}_2,\ -\mathrm{COOH},\ =\mathrm{NH}$

   The third group here is an imine ($=\mathrm{NH}$), which is a double bond between carbon and nitrogen. Asparagine does not have an imine; its side chain has a carbonyl ($\mathrm{C=O}$) and a separate amine ($-\mathrm{NH}_2$), not a C=N bond. So this is incorrect.

2. Option (B): $-\mathrm{NH}_2,\ -\mathrm{COOH},\ \underset{|}{\mathrm{C}}-\mathrm{NH}_2$

   The third symbol $\underset{|}{\mathrm{C}}-\mathrm{NH}_2$ represents a carbon attached to an amine — but in context, the "C" here is part of a carbonyl group (the double-bonded oxygen is implied in the notation). This is the standard way to depict a carboxamide group ($-\mathrm{CONH}_2$). So this matches asparagine exactly: an amino group, a carboxylic acid, and a carboxamide side chain. This is correct.

3. Option (C): $-\mathrm{NH}-\underset{|}{\mathrm{C}},\ -\mathrm{COOH},\ -\underset{|}{\mathrm{C}}-\mathrm{Cl}$

   This contains a chloro group ($-\mathrm{C}-\mathrm{Cl}$), which is not present in asparagine. Also, the first group is a secondary amine ($-\mathrm{NH}-$), whereas asparagine has a primary amine ($-\mathrm{NH}_2$). So this is wrong.

4. Option (D): $-\mathrm{NH}_2,\ -\mathrm{COOH},\ -\mathrm{OH}$

   The third group is a hydroxyl ($-\mathrm{OH}$), which is found in amino acids like serine or threonine, but not in asparagine. So this is incorrect.

The concept first

A base (Lewis/Bronsted) is something that donates a lone pair to a proton. So the question "which N is the strongest base?" is really "whose lone pair is most willing to leave home?"

All three nitrogens here have exactly one lone pair; what differs is the hybrid orbital it occupies:

Why s-character matters. An $s$ orbital is spherical and penetrates close to the nucleus; a $p$ orbital has a node at the nucleus. So the more $s$-character a hybrid orbital has, the closer its electrons sit to the positively charged nucleus, the lower their energy, and the more tightly bound — i.e. the less available for donation. This is the same idea that makes an $sp$ carbon (acetylene) more acidic and its conjugate base more stable; here we are seeing the basicity face of the same coin.

Step-by-step

Step 1 — Assign hybridisation at nitrogen.

• $\mathrm{RNH_2}$: N makes three $\sigma$ bonds (to R, H, H) and holds one lone pair $\Rightarrow$ four electron domains $\Rightarrow$ $sp^{3}$.
• $\mathrm{RN{=}CHR^{1}}$: N makes one $\sigma$ bond to R, one $\sigma$ (+ one $\pi$) to C, and holds one lone pair $\Rightarrow$ three $\sigma$/lone-pair domains $\Rightarrow$ $sp^{2}$.
• $\mathrm{RC\equiv N}$: N makes one $\sigma$ (+ two $\pi$) to C and holds one lone pair $\Rightarrow$ two domains $\Rightarrow$ $sp$.

Step 2 — Rank the lone-pair availability.

So availability, and therefore basicity, decreases in that same order.

Step 3 — Translate to the three compounds.

A sanity check with real $pK_{a}$ values of the conjugate acids (higher $pK_{aH}$ = stronger base): $\mathrm{CH_3NH_3^{+}}\approx 10.6$; a protonated imine $\approx 7$; a protonated nitrile $\approx -10$. The nitrile is a spectacularly poor base — exactly as the $sp$ argument predicts.

Step 4 — Read the options.

Decreasing basicity = III > I > II $\Rightarrow$ option (B).

Common mistake

Assuming the triple bond makes the nitrile "electron rich" and therefore basic. The $\pi$ electrons are not the ones being donated — the lone pair in the $sp$ orbital is, and being $50\%$ s in character it is held very tightly, making nitriles the weakest bases of the three.

ANSWER: B

Concept & Intuition

Bond angles in molecules with a central atom are largely governed by VSEPR theory (Valence Shell Electron Pair Repulsion). Electron pairs — whether bonding or lone — repel each other and arrange themselves as far apart as possible. Crucially, lone pairs occupy more space than bonding pairs because they are held closer to the nucleus and are not “shared” with another atom. This means lone pairs exert stronger repulsion, squeezing the bonding pairs closer together and reducing the bond angle. So, the more lone pairs on the central atom, the smaller the bond angle.

Let’s apply this to the three species.

1. Identify the central atom and its electron groups

   All three have nitrogen as the central atom. Count the total number of electron groups (bonding pairs + lone pairs) around N:

   • $\mathrm{NH}_4^+$: N forms 4 single bonds (to H) and has no lone pairs. Total = 4 electron groups → tetrahedral geometry. Bond angle = $109.5^\circ$.
   • $\mathrm{NH}_3$: N forms 3 single bonds and has 1 lone pair. Total = 4 electron groups → trigonal pyramidal geometry. The lone pair repels the bonding pairs, reducing the angle from $109.5^\circ$ to about $107^\circ$.
   • $\mathrm{NH}_2^-$: N forms 2 single bonds and has 2 lone pairs. Total = 4 electron groups → bent geometry. Two lone pairs push the bonding pairs even closer, giving an angle of about $104.5^\circ$.

2. Compare the bond angles

   From the above:

   • $\mathrm{NH}_4^+$: $109.5^\circ$ (no lone pairs)
   • $\mathrm{NH}_3$: $107^\circ$ (one lone pair)
   • $\mathrm{NH}_2^-$: $104.5^\circ$ (two lone pairs)

   So the order is: $\mathrm{NH}_4^+ > \mathrm{NH}_3 > \mathrm{NH}_2^-$.

3. Match with the options

   Option (A) states exactly this order.

Concept and Intuition

Van der Waals forces are the weak, temporary attractions between molecules. They include:

• London dispersion forces (present in all molecules, stronger in larger, more polarizable molecules)
• Dipole-dipole interactions (present in polar molecules)
• Hydrogen bonding (a special, strong dipole-dipole interaction when H is bonded to N, O, or F)

To find which gas has the highest van der Waals force, we must compare both molecular size and polarity. A common pitfall is to only consider molecular weight, but polarity—especially hydrogen bonding—can dominate.

Step-by-Step Reasoning

1. Identify molecular properties

   • CO₂: Linear, nonpolar (dipoles cancel). Only London forces. Molar mass = 44 g/mol.
   • NH₃: Trigonal pyramidal, polar. Has N–H bonds → hydrogen bonding possible. Molar mass = 17 g/mol.
   • CH₄: Tetrahedral, nonpolar. Only London forces. Molar mass = 16 g/mol.
   • N₂O: Linear, but asymmetric (N≡N–O) → slightly polar. Molar mass = 44 g/mol. No hydrogen bonding.

2. Rank the strength of intermolecular forces

   • Hydrogen bonding (NH₃) is typically much stronger than ordinary dipole-dipole or London forces.
   • Even though CO₂ and N₂O are heavier (more electrons → stronger London forces), their total van der Waals forces are still weaker than the hydrogen bonds in NH₃.
   • CH₄ is the smallest and nonpolar → weakest.

3. Confirm with boiling points (a practical measure of van der Waals strength)

   • NH₃ boils at –33.3 °C
   • N₂O boils at –88.5 °C
   • CO₂ sublimes at –78.5 °C
   • CH₄ boils at –161.5 °C The much higher boiling point of NH₃ confirms its strongest intermolecular attraction.

The van der Waals equation corrects the ideal gas law for two real-gas effects: finite molecular size (constant ‘b’) and intermolecular attractions (constant ‘a’). The constant ‘a’ is directly related to how strongly molecules pull on each other — the stronger the attractive forces, the larger the value of ‘a’. So to find which gas has the maximum ‘a’, we need to compare the intermolecular forces in $\mathrm{H_2}$, $\mathrm{He}$, $\mathrm{CO_2}$, and $\mathrm{NH_3}$.

1. Identify the type of intermolecular forces in each gas.

   • $\mathrm{H_2}$ and $\mathrm{He}$ are nonpolar and small. Their only attractions are weak London dispersion forces.
   • $\mathrm{CO_2}$ is also nonpolar (linear molecule, no net dipole), so it too relies on dispersion forces — but because it has more electrons than $\mathrm{H_2}$ or $\mathrm{He}$, its dispersion forces are stronger.
   • $\mathrm{NH_3}$ is polar and, crucially, has N–H bonds. This allows hydrogen bonding — a much stronger intermolecular force than ordinary dipole-dipole or dispersion.

2. Rank the strength of attractions.

   Hydrogen bonding > dipole-dipole > dispersion. Among these four, only $\mathrm{NH_3}$ can hydrogen-bond. Even though $\mathrm{CO_2}$ has larger dispersion forces than $\mathrm{H_2}$ or $\mathrm{He}$, hydrogen bonding in $\mathrm{NH_3}$ is far stronger. So $\mathrm{NH_3}$ has the strongest intermolecular attraction.

3. Connect to the van der Waals constant ‘a’.

   A larger ‘a’ means a greater correction for attraction — i.e., the gas deviates more from ideality because molecules stick together more. Since $\mathrm{NH_3}$ has the strongest attractions, it has the largest ‘a’.

Concept and Intuition

Boric acid is often misunderstood because its formula $H_3BO_3$ looks like a triprotic acid (like $H_3PO_4$). But the key is to look at its structure and bonding. In the solid state, $H_3BO_3$ molecules are planar and link together via hydrogen bonds to form sheets — a classic layered structure. However, it does not donate three protons; it acts as a Lewis acid (accepting $OH^-$) rather than a Brønsted acid that loses three $H^+$ ions. Let’s examine each statement.

Step-by-step reasoning

1. Statement (I): $H_3BO_3$ is a liquid

   Boric acid is a white crystalline solid at room temperature (melting point ~171 °C). It is not a liquid.

   → False.

2. Statement (II): $H_3BO_3$ forms a layer structure

   In the solid state, each boron atom is $sp^2$ hybridized, giving a planar $BO_3$ unit. These units are linked by O–H···O hydrogen bonds between the hydroxyl groups of adjacent molecules, creating infinite two-dimensional sheets (layers) that stack on top of each other via weak van der Waals forces.

   → True.

3. Statement (III): In $H_3BO_3$, each $BO_3$ unit is joined by hydrogen bonds

   Yes — the hydrogen atom of one $B–OH$ group forms a hydrogen bond with an oxygen atom of a neighboring $BO_3$ unit. This is the glue that holds the layers together.

   → True.

4. Statement (IV): $H_3BO_3$ is tribasic

   This is the classic trap. Boric acid does not release three $H^+$ ions in water. Instead, it acts as a monobasic Lewis acid:

It accepts a hydroxide ion (from water), leaving a free proton. Only one proton is effectively donated per molecule. It is monobasic, not tribasic.

→ False.

Conclusion

Only statements (II) and (III) are correct.

Concept and Intuition:

This problem involves two key chemical reactions: the formation of a metal nitride and its subsequent hydrolysis.

1. Formation of Nitrides: Many active metals, especially those in Groups 1, 2, and 13, react directly with nitrogen gas at high temperatures to form ionic nitrides. In these compounds, the metal typically exhibits its characteristic positive oxidation state, and nitrogen is in the -3 oxidation state as the nitride ion ($N^{3-}$). Aluminium, being a Group 13 metal, forms aluminium nitride (AlN).
2. Hydrolysis of Nitrides: Ionic nitrides are strong bases and react vigorously with water. The nitride ion ($N^{3-}$), being highly basic, readily accepts protons from water molecules to form ammonia gas ($NH_3$). Simultaneously, the metal cation combines with hydroxide ions from water to form the corresponding metal hydroxide. This reaction is a classic test for the presence of ionic nitrides.

Step-by-step solution:

1. First Reaction: Formation of Compound (A) The first step in the sequence is the reaction of aluminium (Al) with nitrogen gas (N$_2$) upon heating ($\Delta$). Aluminium is a reactive metal, and it combines directly with nitrogen at high temperatures to form aluminium nitride.

Therefore, compound (A) is aluminium nitride (AlN).

2. Second Reaction: Hydrolysis of Compound (A) to form (B) and (C)

Compound (A), aluminium nitride (AlN), then reacts with water (H$_2$O). As discussed in the concept, metal nitrides undergo hydrolysis. The nitride ion ($N^{3-}$) reacts with water to produce ammonia gas ($NH_3$), and the aluminium ion ($Al^{3+}$) reacts to form aluminium hydroxide ($Al(OH)_3$).

The problem states that (B) is a precipitate (ppt) and (C) is a gas (g).
*   Aluminium hydroxide ($Al(OH)_3$) is an insoluble white precipitate. This matches the description for compound (B).
*   Ammonia ($NH_3$) is a gas at standard conditions. This matches the description for compound (C).

> [!WARNING]
> It is crucial to remember that the hydrolysis of metal nitrides produces ammonia ($NH_3$), not oxides of nitrogen (like NO or NO$_2$) or elemental nitrogen ($N_2$). The nitrogen in the nitride is in the -3 oxidation state, which is converted to the -3 oxidation state in ammonia.

3. Identifying Compound (C)

From the hydrolysis reaction, we have identified that the gas produced, compound (C), is ammonia ($NH_3$).

Comparing this with the given options:
(A) NO$_2$
(B) NH$_3$
(C) NO
(D) N$_2$

Our identified compound (C) is $NH_3$, which corresponds to option (B).

The concept: where does the missing carbon go?

The balanced equation is the fingerprint of the reaction:

The mechanism explains every product:

1. Deprotonation of the amide N–H by $\mathrm{OH^-}$.
2. N-bromination: the amide anion attacks $\mathrm{Br_2}$ giving $\mathrm{R{-}CO{-}NHBr}$ (releasing one $\mathrm{Br^-} \rightarrow \mathrm{NaBr}$).
3. Second deprotonation gives the bromamide anion $\mathrm{R{-}CO{-}\overset{-}{N}Br}$.
4. Concerted migration: the R group migrates from the carbonyl carbon to nitrogen as $\mathrm{Br^-}$ leaves (the second $\mathrm{NaBr}$), producing an isocyanate $\mathrm{R{-}N{=}C{=}O}$. This is the step that shortens the chain: R is now bonded to N, and the old carbonyl carbon is left holding only N and O.
5. Alkaline hydrolysis of the isocyanate gives a carbamate, which loses $\mathrm{CO_3^{2-}}$ to leave the free amine:

So the carbonyl carbon ends up as sodium carbonate — that is precisely why the amine has one carbon fewer, and why the by-product must be $\mathrm{Na_2CO_3}$ and never $\mathrm{NaHCO_3}$ or free $\mathrm{Br_2}$ (all four NaOH are consumed, and both bromines end up as $\mathrm{NaBr}$).

Step-by-step

1. Count the carbons of the amide. $\mathrm{CH_3CH_2CH_2CH_2CONH_2}$: four carbons in the chain + the carbonyl carbon = pentanamide, $\mathrm{C_5}$, so $\mathrm{R} = \mathrm{CH_3CH_2CH_2CH_2{-}}$ (butyl).
2. Apply the rule ($\mathrm{C_n}$ amide $\rightarrow \mathrm{C_{n-1}}$ amine):

This immediately kills options (A) and (C), which show a five-carbon amine (pentan-1-amine) — the commonest slip in this question.

3. Fix the by-products by balancing. With 4 NaOH: 2 Na go into $\mathrm{Na_2CO_3}$, 2 Na into $\mathrm{2NaBr}$, and the remaining hydrogens/oxygens give $\mathrm{2H_2O}$:

Option (D) wrongly leaves $\mathrm{Br_2}$ unreacted and demands $2\mathrm{Na_2CO_3}$ (four Na from carbonate alone — impossible with only 4 NaOH plus the NaBr requirement).

4. Only option (B) has both the right amine and the correctly balanced by-products.

ANSWER: B

Concept and Intuition

Carboxylic acids, like acetic acid ($\mathrm{CH_3COOH}$), are acidic due to the presence of the carboxyl group. Ammonia ($\mathrm{NH_3}$) is a weak base and also a nucleophile due to the lone pair on the nitrogen atom.

When an acid reacts with a base, an acid-base neutralization reaction occurs, forming a salt. In this case, acetic acid donates a proton to ammonia, forming ammonium acetate. This is the initial product.

However, in organic chemistry, when carboxylic acids react with ammonia, the ultimate goal is often the formation of an amide. Amides are derivatives of carboxylic acids where the hydroxyl group is replaced by an amino group. The ammonium salt formed in the first step is an intermediate that, upon heating, can undergo a dehydration reaction (loss of a water molecule) to form the corresponding amide. This is a common method for synthesizing primary amides.

Step-by-step Solution

1. Initial Acid-Base Reaction: Acetic acid reacts with ammonia in an acid-base neutralization reaction. The acidic proton from the carboxyl group of acetic acid is transferred to the ammonia molecule, forming an ammonium ion ($\mathrm{NH_4^+}$) and an acetate ion ($\mathrm{CH_3COO^-}$). These ions combine to form ammonium acetate.

This product, ammonium acetate, corresponds to option (D).

2. Dehydration upon Heating:

When ammonium acetate is heated, it undergoes a dehydration reaction. A molecule of water is eliminated from the ammonium acetate salt. Specifically, a hydrogen atom from the ammonium ion and a hydroxyl group from the acetate ion combine to form water. The remaining fragments then link to form an amide.

This product, acetamide, is a primary amide.

> [!IMPORTANT]
> In organic chemistry, when a carboxylic acid reacts with ammonia, and no specific conditions (like "at room temperature") are mentioned, the formation of the amide via heating the intermediate ammonium salt is generally implied as the final product.

3. Identifying the Correct Option:

Comparing the final product, acetamide ($\mathrm{CH_3CONH_2}$), with the given options:

(A) $\mathrm{CH_3CONH_2}$ (Acetamide)

(B) $\mathrm{CH_3CONHCH_3}$ (N-methylacetamide, formed from methylamine)

(C) $\mathrm{CH_3CN}$ (Acetonitrile, formed by dehydration of acetamide with strong dehydrating agents)

(D) $\mathrm{CH_3COONH_4}$ (Ammonium acetate, the intermediate salt)

The reaction ultimately yields acetamide.