Comprehension:
Nucleophilic Addition Reactions
Organic compounds containing carbon-oxygen double bond (>C=O ) are called carbonyl group, which is one of the most important functional group in organic chemistry. In aldehydes, the carbonyl group is bonded to a carbon and hydrogen while in ketones, it is bonded to two carbon atoms.The carbonyl compounds in which carbon of carbonyl group is bonded to carbon or hydrogen and oxygen of hydroxyl moiety (-OH) are known as carboxylic acids , while in compounds where carbon is attached to carbon or hydrogen and nitrogen of -NH2 moiety or to halogens are called amides and acyl halides respectively. Aldehydes, ketones, and carboxylic acids are widespread in plants and animal kingdom. They add fragrance and flavour to nature, for example, vanillin ( from vanilla bean), salicylaldehyde ( from meadow sweet ) and cinnamaldehyde ( cinnamon) have pleasant fragrances.
The carbonyl carbon atom is sp2-hybridised and forms three sigma ( σ ) bonds. The fourth valence electron of carbon remains in its p-orbital and forms a π-bond with oxygen by overlap with p-orbital of an oxygen. In addition , the oxygen atom also has two non bonding electron pairs. Thus, the carbonyl carbon and the three atoms attached to it lie in the same plane and the π-electron cloud is above and below this plane. The bond angles are approximately 120° and trigonal coplanar structure.
The carbon-oxygen double bond is polarised due to higher electronegativity of oxygen relative to carbon. Hence, the carbonyl carbon is an electrophilic (Lewis acid), and carbonyl oxygen , a nucleophilic (Lewis base) centre. Carbonyl compounds have substantial dipole moments and are more polar than ethers. Aldehyde and ketone can be prepared by oxidation of alcohols, dehydrogenation of alcohols, and ozonolysis of alkenes etc.
Which of the following represents Hell-Volhard-Zelinsky reaction?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → structure 3
In the Hell–Volhard–Zelinsky (HVZ) reaction, a carboxylic acid that has an α-hydrogen is treated with chlorine or bromine in the presence of a small amount of red phosphorus, followed by water. The halogen replaces an α-hydrogen, the one on the carbon next to –COOH:
\( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}/\text{red P}} \ce{R-CH(X)-COOH} \)
Phosphorus and bromine first form \( \ce{PBr3} \), which converts the acid into an acyl bromide; the acyl bromide is brominated at the α-carbon, and water then hydrolyses –COBr back to –COOH.
For 3-methylbutanoic acid, \( \ce{(CH3)2CH-CH2-COOH} \), the α-carbon is the \( \ce{CH2} \) next to –COOH:
\( \displaystyle \ce{(CH3)2CH-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{Br2}/\text{red P}} \ce{(CH3)2CH-CH(Br)-COOH} \)
The product is 2-bromo-3-methylbutanoic acid, which is shown in structure 3.
Why the other options are incorrect
Option (1) structure 1 — Incorrect. The bromine is at the correct carbon, but the product is still the acyl bromide (–CO–Br); step (ii), water, hydrolyses it to the acid.
Option (2) structure 2 — Incorrect. The bromine is on a terminal methyl carbon, far from –COOH; HVZ halogenation takes place only at the α-carbon.
Option (4) structure 4 — Incorrect. The bromine is on the β-carbon (the branched carbon), not on the α-carbon.
NCERT (2026-27 edition):
Substitution Reactions in the Hydrocarbon Part
