Q.The formation of cyanohydrin from acetone is an example of a) nucleophilic substitution b) electrophilic substitution c) electrophilic addition d) Nucleophilic addition

The cyanide ion is a nucleophile that adds directly across acetone's polarised carbonyl double bond (no group is displaced), giving the addition product cyanohydrin — a textbook nucleophilic addition, typical of the carbonyl group's reactivity.

Acetone's carbonyl carbon is electrophilic (partial positive charge, due to oxygen's higher electronegativity polarising the $C=O$ bond). The cyanide ion, $CN^-$ (generated from HCN with a trace of base/KCN), is a good nucleophile that attacks this electrophilic carbon; the pi electrons of the $C=O$ bond shift onto oxygen, and after protonation of the resulting alkoxide, the tetrahedral addition product — 2-hydroxy-2-methylpropanenitrile (acetone cyanohydrin) — is formed: $CH_3COCH_3 + HCN \rightarrow (CH_3)_2C(OH)CN$ This is a nucleophilic addition (not substitution, since no leaving group departs from the carbonyl carbon; and not electrophilic, since the attacking species is a nucleophile, not an electrophile) — the general and defining reaction pattern of aldehydes and ketones toward nucleophiles.

Acetone condenses with 2,4-DNP to give acetone 2,4-dinitrophenylhydrazone, a classic carbonyl-identification derivative.

The carbonyl group of acetone reacts with the primary amino group of 2,4-dinitrophenylhydrazine (2,4-DNP) via nucleophilic addition–elimination (condensation), losing a molecule of water, to form a hydrazone:

$CH_3COCH_3 + H_2N-NH-C_6H_3(NO_2)_2 \rightarrow (CH_3)_2C=N-NH-C_6H_3(NO_2)_2 + H_2O$

The product, acetone 2,4-dinitrophenylhydrazone, is an orange/yellow crystalline solid with a sharp melting point, and this reaction (Brady's test) is used as a qualitative/quantitative test for aldehydes and ketones.

Ketones react with hydroxylamine to form oximes, a classic carbonyl-to-C=N condensation.

Identifying the reaction: Conversion of the carbonyl group C=O of acetone into C=N-OH is oxime formation, a nucleophilic addition-elimination reaction typical of aldehydes/ketones with primary-amine-like nucleophiles.

Reagent: Hydroxylamine, $NH_2OH$ (commonly used as its hydrochloride salt, $NH_2OH\cdot HCl$, buffered with a mild base such as sodium acetate to free the nucleophilic amine).

Equation:

$CH_3COCH_3 + NH_2OH \longrightarrow (CH_3)_2C=NOH + H_2O$

(Acetoxime — the nitrogen lone pair attacks the carbonyl carbon, forming a tetrahedral intermediate that then loses water to give the C=N double bond.)

Formaldehyde condenses with excess ammonia in a well-known reaction to form hexamethylenetetramine (urotropine).

Reasoning: Six molecules of formaldehyde react with four molecules of ammonia in a cyclic condensation to give hexamethylenetetramine (also called hexamine or urotropine), a cage-like heterocyclic compound used as a urinary antiseptic and as a precursor in the manufacture of certain resins/explosives (RDX):

$6HCHO + 4NH_3 \longrightarrow (CH_2)_6N_4\;(\text{hexamethylenetetramine}) + 6H_2O$

Bakelite, by contrast, is the phenol-formaldehyde condensation polymer (not formed from ammonia), so it is not the correct answer here.

The carbonyl carbon of acetone is electrophilic, so the nucleophilic $CN^-$ (from HCN) adds to it, and after protonation of the resulting alkoxide the $C=O$ double bond becomes a single bond bearing both $-OH$ and $-CN$ — an addition, driven by a nucleophile, not a substitution.

Acetone, $(CH_3)_2C=O$, has a polarised carbonyl carbon ($\delta^+$) due to the electronegative oxygen. The cyanide ion $CN^-$ acts as a nucleophile, attacking this carbon and adding across the double bond; protonation of the resulting alkoxide by HCN/$H^+$ gives the cyanohydrin, $(CH_3)_2C(OH)CN$. Because a nucleophile adds to (rather than substitutes at) an unsaturated carbon, with no leaving group departing, this is nucleophilic addition, not substitution; nor is it electrophilic, since the attacking species (cyanide) is electron-rich.