Chapter 12: Aldehydes, Ketones & Carboxylic Acids
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What is a carbonyl group? How do aldehydes differ structurally from ketones?
Carbonyl Group: A functional group consisting of a carbon atom double-bonded to an oxygen atom ($>C=O$).
- Aldehydes: The carbonyl carbon is bonded to at least one hydrogen atom (and one alkyl/aryl group). Their general formula is $R-CHO$. The carbonyl group is always at the end of the carbon chain.
- Ketones: The carbonyl carbon is bonded to two alkyl or aryl groups. Their general formula is $R-CO-R'$. The carbonyl group is always within the carbon chain.
2. Explain the nature of the carbonyl group. Why is it polar?
In the carbonyl group ($>C=O$), the carbon is $sp^2$ hybridized, forming a planar triangular structure.
Polarity: Oxygen is more electronegative than carbon. Therefore, the $\pi$-electron cloud of the $C=O$ double bond is pulled towards oxygen. This gives oxygen a partial negative charge ($\delta^-$) and the carbon a partial positive charge ($\delta^+$).
Because of this, the carbonyl carbon acts as an electrophile (Lewis acid) and the oxygen acts as a nucleophile (Lewis base).
3. Describe the Rosenmund Reduction. What is the role of $BaSO_4$?
Rosenmund Reduction: It is the preparation of an aldehyde by the catalytic hydrogenation of an acyl chloride (acid chloride) over a Palladium catalyst supported on Barium sulfate.
$$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$$
Role of $BaSO_4$: It acts as a catalytic poison (poisoned with sulfur or quinoline) to partially deactivate the Palladium catalyst. This prevents the further reduction of the newly formed aldehyde into a primary alcohol.
4. Describe the Stephen Reaction.
Stephen Reaction: Nitriles (Cyanides) are reduced to corresponding imine hydrochlorides using Stannous chloride ($SnCl_2$) in the presence of hydrochloric acid ($HCl$). The imine is then subjected to acid hydrolysis to yield an aldehyde.
1. $R-C \equiv N + 2[H] \xrightarrow{SnCl_2/HCl} R-CH=NH \cdot HCl \text{ (Imine hydrochloride)}$
2. $R-CH=NH \cdot HCl + H_2O \xrightarrow{\Delta} R-CHO \text{ (Aldehyde)} + NH_4Cl$
5. Explain the Etard Reaction for the preparation of benzaldehyde.
Etard Reaction: Toluene is selectively oxidized to benzaldehyde using Chromyl chloride ($CrO_2Cl_2$) in a non-polar solvent like $CS_2$.
Chromyl chloride oxidizes the methyl group to form a brown chromium complex, which prevents further oxidation to benzoic acid. Upon acid hydrolysis, this complex decomposes to yield benzaldehyde.
$C_6H_5CH_3 \xrightarrow{CrO_2Cl_2 \text{ in } CS_2} \text{Chromium Complex} \xrightarrow{H_3O^+} C_6H_5CHO \text{ (Benzaldehyde)}$
6. Write a short note on the Gattermann-Koch Formylation reaction.
Gattermann-Koch Reaction: When benzene or its derivatives are treated with a mixture of carbon monoxide ($CO$) and hydrogen chloride ($HCl$) gas in the presence of an anhydrous Lewis acid catalyst (like $AlCl_3$) and a trace of cuprous chloride ($CuCl$), benzaldehyde is formed.
$$C_6H_6 + CO + HCl \xrightarrow{\text{Anhyd. } AlCl_3 / CuCl} C_6H_5CHO + HCl$$
7. How are ketones prepared from acyl chlorides using dialkyl cadmium?
Grignard reagents react with Cadmium chloride to form Dialkyl cadmium, which then reacts with acyl chlorides to produce ketones.
1. $2R-MgX + CdCl_2 \rightarrow R_2Cd \text{ (Dialkyl cadmium)} + 2Mg(X)Cl$
2. $2R'-COCl + R_2Cd \rightarrow 2R'-CO-R \text{ (Ketone)} + CdCl_2$
Dialkyl cadmium is less reactive than Grignard reagents, so it stops at the ketone stage without further nucleophilic addition.
8. How are ketones prepared from nitriles?
Nitriles react with Grignard reagents in dry ether to form an imine magnesium complex intermediate, which upon acid hydrolysis yields a ketone.
$R-C \equiv N + R'-MgX \xrightarrow{\text{ether}} R-C(R')=N-MgX$
$R-C(R')=N-MgX + H_2O \xrightarrow{H^+} R-CO-R' \text{ (Ketone)} + NH_3 + Mg(OH)X$
9. Write the reaction for the preparation of acetophenone by Friedel-Crafts acylation.
Benzene reacts with acetyl chloride in the presence of anhydrous $AlCl_3$ (Lewis acid catalyst) to form acetophenone.
$$C_6H_6 + CH_3COCl \xrightarrow{\text{Anhyd. } AlCl_3} C_6H_5COCH_3 \text{ (Acetophenone)} + HCl$$
10. How will you prepare Acetaldehyde from Ethyne (Acetylene)?
By the Hydration of Alkynes. Ethyne reacts with water in the presence of 40% $H_2SO_4$ and 1% Mercuric sulfate ($HgSO_4$) at 333 K to form an unstable enol intermediate (vinyl alcohol), which tautomerizes to acetaldehyde.
$HC \equiv CH + H_2O \xrightarrow{H_2SO_4 / HgSO_4} [CH_2=CH-OH] \xrightarrow{\text{tautomerization}} CH_3CHO \text{ (Acetaldehyde)}$
(Note: All other alkynes give ketones).
11. Compare the boiling points of aldehydes/ketones with alkanes and alcohols of comparable molecular mass.
Order: Alcohols > Aldehydes/Ketones > Alkanes.
- Aldehydes and ketones have higher boiling points than alkanes because they are polar molecules and have dipole-dipole intermolecular interactions.
- They have lower boiling points than alcohols because, unlike alcohols, they lack an $O-H$ bond and therefore cannot form strong intermolecular hydrogen bonds among themselves.
12. Discuss the solubility of aldehydes and ketones in water.
Lower members (like methanal, ethanal, propanone) are soluble in water in all proportions. This is because the polar oxygen atom of the carbonyl group can form hydrogen bonds with water molecules.
However, solubility decreases rapidly as the length of the alkyl chain increases, because the bulky non-polar, hydrophobic hydrocarbon part hinders hydrogen bonding.
13. Describe the mechanism of Nucleophilic Addition to a carbonyl group.
Step 1: The nucleophile ($Nu^-$) attacks the electrophilic carbonyl carbon (which is $sp^2$ hybridized and planar) from a direction perpendicular to the plane. The $\pi$-electrons of the $C=O$ bond shift to the electronegative oxygen atom, forming a tetrahedral alkoxide intermediate ($sp^3$ hybridized).
Step 2: The alkoxide intermediate rapidly captures a proton ($H^+$) from the reaction medium to form the electrically neutral addition product.
14. Aldehydes are generally more reactive than ketones towards nucleophilic addition. Give two reasons.
- Steric Reason: Ketones have two bulky alkyl groups around the carbonyl carbon, which physically block the approach of the incoming nucleophile. Aldehydes have only one alkyl group and a small hydrogen atom, offering less steric hindrance.
- Electronic Reason (+I effect): Alkyl groups are electron-donating. In ketones, two alkyl groups push electron density towards the carbonyl carbon, reducing its positive charge ($\delta^+$) and making it less electrophilic compared to aldehydes.
15. Write the reaction for the addition of Hydrogen Cyanide (HCN) to Acetaldehyde. Name the product.
The reaction occurs in the presence of a base catalyst to generate the $CN^-$ nucleophile.
$$CH_3CHO + HCN \xrightarrow{OH^-} CH_3-CH(OH)-CN$$
The product is Acetaldehyde cyanohydrin.
16. How do aldehydes and ketones react with Sodium bisulfite ($NaHSO_3$)? What is the utility of this reaction?
Aldehydes and aliphatic methyl ketones react with saturated aqueous $NaHSO_3$ to form a crystalline, water-soluble bisulfite addition compound.
$>C=O + NaHSO_3 \rightleftharpoons >C(OH)-SO_3Na$
Utility: This reaction is reversible. Heating the crystalline adduct with dilute mineral acid or aqueous alkali regenerates the original aldehyde/ketone. Therefore, it is used for the separation and purification of aldehydes and ketones from non-carbonyl compounds.
17. What are Acetals and Hemiacetals? How are they formed?
Aldehydes react with one equivalent of monohydric alcohol in the presence of dry $HCl$ gas to form alkoxyalcohols known as hemiacetals (unstable).
$R-CHO + R'OH \xrightarrow{HCl_{(g)}} R-CH(OH)(OR') \text{ [Hemiacetal]}$
Hemiacetals react with a second equivalent of alcohol to form gem-dialkoxy compounds known as acetals.
$R-CH(OH)(OR') + R'OH \xrightarrow{HCl_{(g)}} R-CH(OR')_2 \text{ [Acetal]} + H_2O$
18. Why is dry HCl gas used in the formation of acetals?
Dry $HCl$ gas acts as a catalyst by protonating the carbonyl oxygen, which increases the electrophilicity of the carbonyl carbon, facilitating the nucleophilic attack of the alcohol.
Furthermore, since the reaction is reversible and water is a by-product, avoiding an aqueous acid ensures the equilibrium shifts forward to form the acetal rather than hydrolyzing it back to the aldehyde.
19. How do Ketones react with ethylene glycol?
Ketones generally do not react with monohydric alcohols easily due to steric hindrance. However, they react with dihydric alcohols like ethylene glycol in the presence of dry HCl or p-toluenesulfonic acid to form cyclic ethylene glycol ketals.
This reaction is widely used to "protect" the ketone group during organic synthesis.
20. Write the reaction of a carbonyl compound with Hydroxylamine ($NH_2OH$) and Hydrazine ($NH_2NH_2$). Name the products.
These are nucleophilic addition-elimination reactions (loss of water).
- With Hydroxylamine: $
- With Hydrazine: $
21. What is the 2,4-DNP test?
Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (2,4-DNP or Brady's reagent) to form yellow, orange, or red precipitates of 2,4-dinitrophenylhydrazones.
This is a characteristic test used to detect the presence of a carbonyl group in an organic compound.
22. Describe the Clemmensen Reduction.
Clemmensen Reduction: The carbonyl group ($>C=O$) of aldehydes and ketones is directly reduced to a methylene group ($>CH_2$) to form an alkane when treated with Zinc amalgam ($Zn-Hg$) and concentrated Hydrochloric acid ($HCl$).
$$>C=O + 4[H] \xrightarrow{Zn-Hg / \text{conc. } HCl} >CH_2 + H_2O$$
23. Describe the Wolff-Kishner Reduction.
Wolff-Kishner Reduction: The carbonyl group is reduced to a methylene group to form an alkane. The carbonyl compound is first treated with hydrazine ($NH_2NH_2$) to form a hydrazone. The hydrazone is then heated with a strong base like KOH or potassium tert-butoxide in a high-boiling solvent like ethylene glycol.
$>C=O \xrightarrow{NH_2NH_2, \ -H_2O} >C=N-NH_2 \xrightarrow{KOH/ \text{ethylene glycol}, \ \Delta} >CH_2 + N_2 \uparrow$
24. What is Tollens' Reagent? Describe Tollens' test (Silver Mirror Test).
Tollens' Reagent: It is a freshly prepared ammoniacal silver nitrate solution, containing the complex ion $[Ag(NH_3)_2]^+$.
Test: When warmed with an aldehyde, Tollens' reagent acts as a mild oxidizing agent. It oxidizes the aldehyde to a carboxylate ion, while the silver ion ($Ag^+$) is reduced to metallic silver, which deposits on the inner wall of the test tube as a bright silver mirror.
$R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \xrightarrow{\Delta} R-COO^- + 2Ag \downarrow \text{(mirror)} + 4NH_3 + 2H_2O$
Ketones do not give this test.
25. Describe Fehling's Test. Which compounds give a positive test?
Fehling's solution is a mixture of Fehling A (aqueous $CuSO_4$) and Fehling B (alkaline sodium potassium tartrate / Rochelle salt). The active oxidizing agent is $Cu^{2+}$.
Test: When heated with an aliphatic aldehyde, it oxidizes the aldehyde to a carboxylate ion and gets reduced to Copper(I) oxide ($Cu_2O$), which forms a red-brown precipitate.
$R-CHO + 2Cu^{2+} + 5OH^- \xrightarrow{\Delta} R-COO^- + Cu_2O \downarrow \text{(red-brown)} + 3H_2O$
Ketones and Aromatic aldehydes (like benzaldehyde) do not give this test.
26. What is the Haloform (Iodoform) test? Which functional groups give a positive Iodoform test?
When compounds containing a specific structural unit are heated with Iodine and NaOH (or NaOI), they form a yellow precipitate of Iodoform ($CHI_3$).
Conditions for positive test: The compound must have a methyl ketone group ($CH_3-CO-$) or be an alcohol that can be oxidized to a methyl ketone ($CH_3-CH(OH)-$ group).
Examples: Acetaldehyde, Acetone, Ethanol, Propan-2-ol.
27. Why are $\alpha$-hydrogens of aldehydes and ketones acidic in nature?
The $\alpha$-hydrogens (hydrogens attached to the carbon adjacent to the carbonyl group) are acidic due to two reasons:
- The strong electron-withdrawing inductive effect (-I effect) of the carbonyl ($>C=O$) group weakens the $C-H$ bond.
- When the base removes the $\alpha$-proton, the resulting enolate anion is highly stabilized by resonance, as the negative charge is delocalized onto the electronegative oxygen atom.
28. Describe the Aldol Condensation reaction. Write the equation for Acetaldehyde.
Aldol Condensation: Aldehydes and ketones having at least one $\alpha$-hydrogen undergo a reaction in the presence of dilute alkali (like dil. NaOH or $Ba(OH)_2$) to form $\beta$-hydroxy aldehydes (aldols) or $\beta$-hydroxy ketones (ketols).
Upon heating, the aldol easily loses a water molecule ($\alpha,\beta$-elimination) to form an $\alpha,\beta$-unsaturated carbonyl compound.
For Acetaldehyde:
$2CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO \text{ (Acetaldol)}$
$CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, \ -H_2O} CH_3-CH=CH-CHO \text{ (But-2-enal)}$
29. Briefly explain the mechanism of Aldol Condensation.
Step 1: The base ($OH^-$) removes an acidic $\alpha$-hydrogen from one aldehyde molecule to form a resonance-stabilized enolate ion.
Step 2: The enolate ion acts as a nucleophile and attacks the electrophilic carbonyl carbon of a second un-ionized aldehyde molecule to form an alkoxide intermediate.
Step 3: The alkoxide intermediate extracts a proton from water to yield the final aldol product and regenerates the $OH^-$ catalyst.
30. What is Cross-Aldol Condensation? Give an example.
When aldol condensation is carried out between two different aldehydes and/or ketones, it is called cross-aldol condensation. If both molecules contain $\alpha$-hydrogens, a complex mixture of four products is formed.
To make it synthetically useful, one compound should have no $\alpha$-hydrogen (e.g., Benzaldehyde or Formaldehyde).
Example: Benzaldehyde + Acetophenone $\xrightarrow{\text{dil. } NaOH, \ \Delta}$ Benzalacetophenone (Chalcone) + $H_2O$.
31. Describe the Cannizzaro Reaction. Write the equation for Formaldehyde.
Cannizzaro Reaction: Aldehydes which do not have an $\alpha$-hydrogen atom (like Formaldehyde, Benzaldehyde) undergo self-oxidation and reduction (disproportionation) when heated with concentrated alkali (like 50% NaOH).
One molecule is oxidized to a carboxylic acid salt, while the other is reduced to a primary alcohol.
For Formaldehyde:
$$2HCHO \xrightarrow{\text{conc. } NaOH, \ \Delta} CH_3OH \text{ (Methanol)} + HCOONa \text{ (Sodium formate)}$$
32. What is Cross-Cannizzaro reaction? Give an example.
When the Cannizzaro reaction is carried out between two different aldehydes (both lacking $\alpha$-hydrogens), it is called a Cross-Cannizzaro reaction.
In a reaction between formaldehyde ($HCHO$) and another aldehyde (like benzaldehyde), formaldehyde is always oxidized to formate ion (since it is more reactive to nucleophilic attack), while the other aldehyde is reduced to the corresponding alcohol.
$C_6H_5CHO + HCHO \xrightarrow{\text{conc. } NaOH} C_6H_5CH_2OH \text{ (Benzyl alcohol)} + HCOONa$
33. Write the reaction for electrophilic substitution in Benzaldehyde (e.g., Nitration). Is $-CHO$ activating or deactivating?
The aldehyde group ($-CHO$) is an electron-withdrawing group via resonance (-R effect). Therefore, it is ring deactivating and meta-directing.
Nitration: When benzaldehyde is treated with a nitrating mixture, the electrophile ($NO_2^+$) attacks the meta position.
$C_6H_5CHO + HNO_3 \xrightarrow{\text{conc. } H_2SO_4, \ 273-283 K} \text{m-Nitrobenzaldehyde} + H_2O$
34. How are Carboxylic acids prepared from primary alcohols and aldehydes?
Primary alcohols and aldehydes are easily oxidized to carboxylic acids using strong oxidizing agents such as acidic/alkaline Potassium permanganate ($KMnO_4$) or acidic Potassium dichromate ($K_2Cr_2O_7 / H_2SO_4$) or Jones reagent ($CrO_3 / H_2SO_4$).
$R-CH_2OH \xrightarrow{[O]} R-COOH$
$R-CHO \xrightarrow{[O]} R-COOH$
35. Describe the preparation of carboxylic acids by the hydrolysis of Nitriles (Cyanides) and Amides.
Nitriles are hydrolyzed to amides and then to carboxylic acids by boiling with mineral acids ($H^+$) or aqueous alkalis ($OH^-$).
$R-C \equiv N \xrightarrow{H_2O / H^+} R-CONH_2 \text{ (Amide)}$
$R-CONH_2 \xrightarrow{H_2O / H^+, \ \Delta} R-COOH + NH_4^+$
The reaction can be stopped at the amide stage using mild conditions (like $H_2O_2/OH^-$).
36. How will you prepare a carboxylic acid using a Grignard reagent?
Grignard reagents react with dry ice (solid Carbon dioxide, $O=C=O$) in dry ether to form the magnesium salt of a carboxylic acid, which upon acid hydrolysis gives the corresponding carboxylic acid.
$R-MgX + O=C=O \xrightarrow{\text{dry ether}} R-COO-MgX$
$R-COO-MgX + H_2O \xrightarrow{H^+} R-COOH + Mg(OH)X$
37. How is Benzoic acid prepared from Toluene?
Aromatic carboxylic acids can be prepared by vigorous oxidation of alkylbenzenes with alkaline $KMnO_4$ or acidic $K_2Cr_2O_7$. Regardless of the length of the side chain (as long as it has at least one benzylic hydrogen), the entire alkyl group is oxidized to a $-COOH$ group.
$C_6H_5CH_3 \xrightarrow{1. \ KMnO_4 / KOH, \ \Delta} C_6H_5COO^-K^+$
$C_6H_5COO^-K^+ \xrightarrow{2. \ H_3O^+} C_6H_5COOH \text{ (Benzoic acid)}$
38. Why do carboxylic acids have higher boiling points than aldehydes, ketones, and even alcohols of comparable molecular mass?
Carboxylic acids have both a carbonyl oxygen ($>C=O$) and a hydroxyl group ($-OH$). They form very strong intermolecular hydrogen bonds, more extensive than in alcohols.
In fact, in the vapor phase or in aprotic solvents, most carboxylic acids exist as cyclic dimers held together by two hydrogen bonds per pair of molecules. Breaking these strong dimers requires significantly higher thermal energy, leading to high boiling points.
39. Discuss the solubility of carboxylic acids in water.
Simple aliphatic carboxylic acids up to four carbon atoms (methanoic to butanoic acid) are miscible in water in all proportions because they can form extensive hydrogen bonds with water molecules.
As the number of carbon atoms increases, the size of the hydrophobic (water-repelling) alkyl chain increases, and the solubility rapidly decreases. Higher acids are virtually insoluble in water.
40. Why are carboxylic acids stronger acids than phenols?
A carboxylic acid releases $H^+$ to form a carboxylate ion ($R-COO^-$). The carboxylate ion is highly stabilized by resonance where the negative charge is delocalized equally over two highly electronegative oxygen atoms.
Phenol releases $H^+$ to form a phenoxide ion. Although it is resonance stabilized, the negative charge spends most of its time on the less electronegative carbon atoms of the ring (ortho/para). Since dispersing a negative charge on two electronegative oxygens is more stabilizing than on carbons, the carboxylate ion is much more stable, making carboxylic acids stronger acids than phenols.
41. What is the effect of an Electron Withdrawing Group (EWG) on the acidity of a carboxylic acid?
An EWG (like $-Cl, -F, -NO_2, -CN$) exerts a -I (inductive) effect. It pulls electron density away from the $O-H$ bond, facilitating the release of the $H^+$ ion. Furthermore, it stabilizes the resulting carboxylate anion by dispersing its negative charge.
Therefore, EWG increases the acidic strength of a carboxylic acid. ($pK_a$ value decreases).
42. What is the effect of an Electron Donating Group (EDG) on the acidity of a carboxylic acid?
An EDG (like alkyl groups: $-CH_3, -C_2H_5$) exerts a +I effect. It pushes electron density towards the carboxyl group, strengthening the $O-H$ bond and making it harder to release $H^+$. It also destabilizes the resulting carboxylate anion by intensifying its negative charge.
Therefore, EDG decreases the acidic strength. Thus, formic acid ($HCOOH$) is a stronger acid than acetic acid ($CH_3COOH$).
43. Arrange the following in increasing order of acidic strength: Acetic acid, Fluoroacetic acid, Chloroacetic acid, Bromoacetic acid. Give reasons.
Order: Acetic acid < Bromoacetic acid < Chloroacetic acid < Fluoroacetic acid.
Reason: Halogens are EWGs and increase acidity via the -I effect. The strength of the -I effect depends on the electronegativity of the halogen: $F > Cl > Br$. Therefore, Fluorine stabilizes the carboxylate ion the most, making fluoroacetic acid the strongest among them.
44. Describe the esterification reaction. Write its mechanism.
Carboxylic acids react with alcohols in the presence of concentrated $H_2SO_4$ to form esters. It is a reversible reaction.
$RCOOH + R'OH \xrightleftharpoons{H^+} RCOOR' + H_2O$
Mechanism (Nucleophilic Acyl Substitution):
- Protonation of the carbonyl oxygen of the acid.
- Nucleophilic attack by the alcohol oxygen on the carbonyl carbon to form a tetrahedral intermediate.
- Proton transfer within the intermediate.
- Elimination of a water molecule.
- Deprotonation to yield the ester.
45. How do carboxylic acids react with $PCl_3$, $PCl_5$, and $SOCl_2$?
The hydroxyl group ($-OH$) of the carboxylic acid is replaced by a chlorine atom to form acid chlorides (acyl chlorides).
- $3RCOOH + PCl_3 \rightarrow 3RCOCl + H_3PO_3$
- $RCOOH + PCl_5 \rightarrow RCOCl + POCl_3 + HCl$
- $RCOOH + SOCl_2 \xrightarrow{\Delta} RCOCl + SO_{2(g)} + HCl_{(g)}$ (Thionyl chloride is the preferred reagent as byproducts are gases).
46. Write the reaction of acetic acid with ammonia. How is an amide formed?
Carboxylic acids react with ammonia ($NH_3$) to first form an ammonium salt. Upon heating at high temperatures, this salt loses a molecule of water to form an amide.
$CH_3COOH + NH_3 \rightleftharpoons CH_3COO^-NH_4^+ \text{ (Ammonium acetate)}$
$CH_3COO^-NH_4^+ \xrightarrow{\Delta} CH_3CONH_2 \text{ (Acetamide)} + H_2O$
47. Name the reagent used to reduce a carboxylic acid directly to a primary alcohol.
Carboxylic acids are notoriously difficult to reduce. They require strong reducing agents.
The reagent used is Lithium aluminum hydride ($LiAlH_4$) followed by acid hydrolysis, or Diborane ($B_2H_6$).
Note: $NaBH_4$ (Sodium borohydride) is a milder reducing agent and cannot reduce carboxylic acids.
48. What is Decarboxylation? Write the reaction for sodium acetate.
Decarboxylation: The process of removal of a molecule of carbon dioxide from a carboxylic acid (or its salt). It is achieved by heating the sodium salt of the acid with soda lime (a mixture of $NaOH$ and $CaO$ in 3:1 ratio).
It yields an alkane containing one carbon atom less than the parent acid.
$$CH_3COONa + NaOH \xrightarrow{CaO, \ \Delta} CH_4 \text{ (Methane)} + Na_2CO_3$$
49. Explain the Hell-Volhard-Zelinsky (HVZ) reaction.
HVZ Reaction: Carboxylic acids having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus to give $\alpha$-halo carboxylic acids.
$$R-CH_2-COOH \xrightarrow{1. \ X_2/\text{Red } P \quad 2. \ H_2O} R-CH(X)-COOH$$
Where $X = Cl$ or $Br$. (This reaction relies on the enolizability of the acid).
50. Write the reaction for electrophilic substitution in Benzoic acid (e.g., Nitration). Is $-COOH$ activating or deactivating?
The carboxyl group ($-COOH$) is strongly electron-withdrawing (-R effect). Thus, it is ring deactivating and meta-directing. Also, benzoic acid does not undergo Friedel-Crafts reaction because the carboxyl group bonds strongly with the Lewis acid catalyst ($AlCl_3$).
Nitration:
$$C_6H_5COOH + HNO_3 \xrightarrow{\text{conc. } H_2SO_4, \ \Delta} \text{m-Nitrobenzoic acid} + H_2O$$
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