Q.Assertion (A): The $\alpha$-hydrogen atom in carbonyl compounds is less acidic.
Reason (R): The anion formed after the loss of the $\alpha$-hydrogen atom in carbonyl compounds is resonance stabilised.

The key here is to recall the specific reactions and properties of acetic acid. Each option tests a different fact: a reduction reaction, acid strength comparison, a test for acidity, and a test for reducing sugars. We need to spot the one that doesn't match reality.

1. Option (A): Acetic acid on reaction with HI and red P at 473 K gives iodoethane

   This describes the reduction of a carboxylic acid to an alkane. Red phosphorus and hydroiodic acid (HI) at high temperature are a strong reducing agent. They first convert the –COOH group to –CH₃, but the mechanism involves replacing the –OH with iodine, then reducing the iodine to hydrogen. For acetic acid (CH₃COOH), the product is ethane (CH₃CH₃), not iodoethane (CH₃CH₂I). Iodoethane would require stopping at the alkyl iodide stage, but under these conditions, the reduction goes all the way to the alkane. So this statement is false.

2. Option (B): Acetic acid is a weaker acid than formic acid

   This is true. Formic acid (HCOOH) has a pKa of about 3.75, while acetic acid (CH₃COOH) has a pKa of about 4.76. The methyl group in acetic acid is electron-donating (inductive effect), which destabilizes the conjugate base (acetate ion) by increasing electron density, making it a weaker acid. Formic acid’s hydrogen has no such donating effect, so it is stronger.

3. Option (C): Acetic acid gives effervescence with aqueous NaHCO₃ solution

   This is true. Acetic acid is a carboxylic acid, and it reacts with sodium bicarbonate to produce carbon dioxide gas, which causes effervescence:

   $$$$

[!TLDR]

The stated acidity order (Cl>F>Br>CH$_3$) is wrong because para-fluorobenzoic acid is actually a weaker acid than para-bromobenzoic acid, while the Reason (F has the largest $-I$ effect) is a correct statement — so option (C).

Concept

In CBSE/NCERT aromatic chemistry, a para substituent affects benzoic-acid acidity through both its inductive ($-I$) and resonance ($\pm M$) effects. Electron-withdrawing groups strengthen the acid (stabilise the carboxylate); electron-donating groups (like $-CH_3$) weaken it. Halogens are $-I$ (acid-strengthening) but also weak $+M$ donors, and at the para position the $+M$ donation is felt directly.

Solution

The four acids are para-substituted benzoic acids with substituents Br [A], Cl [B], CH$_3$ [C], F [D]. Their measured strengths (pK$_a$ in water) are approximately:

Lower pK$_a$ = stronger acid, giving the real order

The Assertion claims Cl $>$ F $>$ Br $>$ CH$_3$, i.e. it places F above Br. This is incorrect: although fluorine has the strongest $-I$ pull, at the para position its lone pairs donate electron density into the ring by resonance ($+M$), destabilising the carboxylate and making p-F-benzoic acid a weaker acid than p-Br-benzoic acid. So the Assertion order is wrong. …

Concept & Intuition

Acidity in carboxylic acids depends on the stability of the conjugate base (the carboxylate anion). Electron-withdrawing groups (EWGs) like –NO₂ pull electron density away from the carboxylate, dispersing its negative charge and making the acid stronger. Electron-donating groups (EDGs) like –OCH₃ do the opposite, destabilizing the anion and weakening the acid. The more EWGs and the closer they are to the –COOH group, the greater the effect.

Step-by-step reasoning

1. Identify the substituent effects

   • Benzoic acid (B) has no substituent — it’s the reference.
   • 4-Methoxybenzoic acid (C) has –OCH₃ at the para position. Methoxy is an electron-donating group (resonance donor), so it decreases acidity relative to benzoic acid.
   • 4-Nitrobenzoic acid (D) has –NO₂ at the para position. Nitro is a strong electron-withdrawing group (both inductive and resonance), so it increases acidity.
   • 3,4-Dinitrobenzoic acid (A) has two –NO₂ groups at the meta and para positions. Both withdraw electrons, and their effects are additive.

2. Compare acid strengths qualitatively

   • (C) is the weakest because of electron donation.
   • (B) is stronger than (C) but weaker than any nitro-substituted acid.
   • (D) is stronger than (B) because one –NO₂ stabilizes the carboxylate.
   • (A) has two –NO₂ groups, so it should be stronger than (D). The meta nitro also withdraws electrons inductively, and the para nitro does so via both induction and resonance. Together, they create a highly stabilized conjugate base.

3. Quantitative check (pKa values)

   • Benzoic acid: pKa ≈ 4.20
   • 4-Methoxybenzoic acid: pKa ≈ 4.47 (less acidic)
   • 4-Nitrobenzoic acid: pKa ≈ 3.41 (more acidic) …

Step 1 — The concept: what a repeat unit tells you.

For an addition polymer of a vinyl monomer $\mathrm{CH_2{=}CH{-}X}$, the double bond opens and the chain grows as

So the group $X$ hanging off the CH carbon in the drawn repeat unit is the substituent of the monomer. Structures (a), (b) and (c) are all of this $-(\mathrm{CH_2-CHX})_n-$ type, so they must come from the three vinyl monomers; the amide unit (d) must come from the remaining monomer.

Step 2 — Match 4. Propene → (a).

$\mathrm{CH_2{=}CH{-}CH_3}$ has $X = \mathrm{CH_3}$, so it gives $-(\mathrm{CH_2-CH(CH_3)})_n-$ = polypropene = structure (a) (the $\mathrm{CH_3}$-bearing unit). ⇒ 4-a

Step 3 — Match 2. Vinyl chloride → (c).

$\mathrm{CH_2{=}CHCl}$ has $X = \mathrm{Cl}$, giving $-(\mathrm{CH_2-CHCl})_n-$ = PVC = structure (c) (the Cl-bearing unit). ⇒ 2-c

Step 4 — Match 3. Styrene → (b).

$\mathrm{CH_2{=}CH{-}C_6H_5}$ has $X = \mathrm{C_6H_5}$, giving $-(\mathrm{CH_2-CH(C_6H_5)})_n-$ = polystyrene = structure (b) (the phenyl-bearing unit). ⇒ 3-b

Step 5 — Match 1. Caprolactam → (d). …

The acidic strength of an alcohol is governed by how well the resulting alkoxide $\mathrm{RCH_2O^-}$ is stabilised.

• $\mathrm{CH_3-}$ is weakly electron donating ($+I$), destabilising the alkoxide $\Rightarrow$ ethanol is the weakest acid.
• $\mathrm{CCl_3-}$ is strongly electron withdrawing ($-I$), stabilising the negative charge. …

The key is to understand what "acidity" means in each of the three classical theories. Arrhenius and Brønsted-Lowry both define acids in terms of protons ($\mathrm{H^+}$). Lewis, on the other hand, defines an acid as an electron-pair acceptor — a much broader definition that includes molecules like $\mathrm{BF_3}$.

$\mathrm{BF_3}$ has only six electrons in its valence shell (boron has three, each fluorine contributes one in a single bond). This makes it electron-deficient and highly eager to accept a lone pair from a base (like $\mathrm{NH_3}$ or $\mathrm{F^-}$). It has no proton to donate, so it cannot fit the Arrhenius or Brønsted-Lowry definitions.

Let’s check each option systematically.

1. Arrhenius concept — An acid must produce $\mathrm{H^+}$ in water. $\mathrm{BF_3}$ does not have a hydrogen atom to release, so it is not an Arrhenius acid. This option is out.

2. Brønsted-Lowry concept — An acid must donate a proton ($\mathrm{H^+}$) to a base. Again, $\mathrm{BF_3}$ has no proton to give. It is not a Brønsted-Lowry acid. This option is also out.

3. Lewis concept — An acid is any species that can accept an electron pair. $\mathrm{BF_3}$ has an incomplete octet on boron, so it readily accepts a lone pair from a Lewis base (e.g., $\mathrm{BF_3 + NH_3 \rightarrow F_3B-NH_3}$). This fits perfectly. $\mathrm{BF_3}$ is a classic example of a Lewis acid.

4. Brønsted-Lowry as well as Lewis — Since $\mathrm{BF_3}$ fails the Brønsted-Lowry test, this combined option is incorrect. …