JEE Chemistry Organic Chemistry Some Basic Principles and Techniques Notes Set 14

Monochlorination of Alkanes

In the presence of light, chlorine replaces one hydrogen of an alkane or cycloalkane by chlorine (free-radical substitution). When the molecule has more than one type of hydrogen the product is a mixture of position isomers. The number of structural monochloro isomers equals the number of chemically different (non-equivalent) types of hydrogen.
Examples: propane 2, butane 2, isobutane 2, n-pentane 3, 2-methylbutane 4, neopentane 1, methylcyclohexane 5, cyclohexane 1. Among the hexanes: hexane 3, 2-methylpentane 5, 3-methylpentane 4, 2,2-dimethylbutane 3 and 2,3-dimethylbutane 2. 2,2,3,3-Tetramethylbutane (C8H18) gives only one.
In aromatic hydrocarbons only side-chain hydrogens react under these conditions; ring hydrogens do not. Toluene gives benzyl chloride; ethylbenzene gives two side-chain products.

Catalytic Hydrogenation

With H2 and Ni, Pt or Pd, every C=C and C≡C bond is hydrogenated; the reaction cannot be stopped at an intermediate stage, and the product is always a saturated compound. Moles of H2 taken up per mole of compound equals the number of pi bonds (a triple bond counts 2).
The aromatic ring is stable at room temperature and is reduced only at higher temperature and pressure (benzene takes 3 mol of H2). Positional isomers of an alkene or alkyne give the same alkane (but-1-yne, but-2-yne and buta-1,3-diene all give butane).
The carbon skeleton does not change during hydrogenation or monochlorination, so the two reactions can be combined to deduce the structure of an unknown hydrocarbon.

Ozonolysis

Ozone cleaves C=C bonds (and, at higher temperature, aromatic double bonds) oxidatively and shows where the double bond was. Reductive work-up (O3, then Zn with H2O or acetic acid, or dimethyl sulphide) gives aldehydes and ketones; oxidative work-up (O3, then H2O2) gives carboxylic acids and ketones.
A =CH2 end gives methanal, a =CHR end gives an aldehyde, a =CR2 end gives a ketone. Alkynes are cleaved to carboxylic acids (but-2-yne gives two molecules of ethanoic acid).
To find the original alkene, remove the oxygen atoms from the carbonyl compounds and join the carbonyl carbons by double bonds. Examples: but-2-ene gives 2 CH3CHO; 2-methylpropene gives HCHO and (CH3)2CO; cyclohexene gives hexanedial; hexa-2,4-diene gives 2 CH3CHO and OHC–CHO.

Detection of Elements (Lassaigne Test)

The organic compound is fused with sodium metal; the elements are converted into ionic sodium salts (NaCN, Na2S, NaX) and the extract is tested.
Nitrogen: boil the extract with FeSO4, then add FeCl3 and acidify; green or Prussian blue colour due to ferric ferrocyanide Fe4[Fe(CN)6]3. Carbon must be present; ammonium nitrate (no carbon) fails.
Sulphur: sodium nitroprusside gives a purple colour (Na4[Fe(CN)5NOS]); lead acetate gives a black precipitate. Nitrogen and sulphur together form NaSCN and give a blood-red colour with FeCl3.
Halogens: acidify with HNO3 and add AgNO3. White precipitate soluble in NH4OH is Cl; dull yellow, partly soluble, is Br; yellow, insoluble, is I. Phosphorus: oxidise to phosphate and add magnesia mixture or ammonium molybdate; Mg2P2O7 (white) is formed on ignition.

Tests for Functional Groups

Alkenes and alkynes: decolourise bromine water and cold alkaline dilute KMnO4 (Baeyer reagent); both react with ozone. Terminal alkynes give a white precipitate with ammoniacal AgNO3 and a red precipitate with ammoniacal Cu2Cl2.
Alcohols: Na metal gives H2. Lucas reagent (conc. HCl + anhydrous ZnCl2) distinguishes classes: tertiary gives turbidity at once, secondary in about 5 minutes, primary after about 30 minutes or on heating. Allyl and benzyl alcohols react instantly because their carbocations are stabilised by delocalisation.
Phenols and enols: coloured complex with neutral FeCl3. Aldehydes and ketones: yellow-orange precipitate with 2,4-DNP. Aldehydes: Tollens reagent (silver mirror or black precipitate), Fehling solution (red Cu2O, only aliphatic aldehydes), Schiff reagent (pink colour restored). Methyl ketones, ethanal and CH3CH(OH)R: iodoform test with I2/NaOH (yellow precipitate of CHI3).
Carboxylic acids: turn blue litmus red and give CO2 effervescence with NaHCO3. Esters: pink phenolphthalein colour in NaOH fades on heating. Amides: heated with conc. NaOH they evolve NH3.
Amines: primary amines give the carbylamine reaction (CHCl3 + KOH, foul smell of isocyanide). Aliphatic primary amines liberate N2 with NaNO2/HCl; aromatic primary amines give a diazonium salt that couples with beta-naphthol to form an orange-red dye. Secondary amines give a nitrosoamine (Liebermann test). Nitro compounds: reduction to the hydroxylamine, which reduces Tollens reagent (Mulliken test). Carbohydrates: Molisch test (violet ring). Amino acids: ninhydrin (blue-violet colour).

Separation of Binary Mixtures

Chemical separation uses the solubility of one component in a particular solvent. Water dissolves low-molecular-mass compounds that can hydrogen bond strongly (sugars, polyols, diacids, amino acids) while most aromatic compounds are insoluble.
Aqueous NaHCO3 dissolves carboxylic acids (including aromatic acids) by salt formation; aqueous NaOH dissolves phenols (and acids); dilute HCl dissolves amines. Neutral compounds (hydrocarbons, haloarenes, nitro compounds, esters) dissolve in none of them. For a mixture of a carboxylic acid and a phenol, NaHCO3 takes up only the acid because phenols are weaker acids than carbonic acid. After the extraction the dissolved component is recovered by neutralisation.

Worked Examples

Worked Example 1. A hydrocarbon X, C5H8, takes up 2 mol of H2 over Ni to give Y, and Y gives three monochloro products. Identify the possibilities for X.
Answer:
DU of C5H8 = 6 − 4 = 2, so X has two pi bonds (or one ring and one pi bond).
Y is C5H12; three monochloro products means Y is pentane (2-methylbutane gives 4, neopentane 1).
X is pent-1-yne, pent-2-yne, penta-1,2-, 1,3-, 1,4- or 2,3-diene (any unbranched five-carbon chain with two pi bonds).

Worked Example 2. An alkene on reductive ozonolysis gives only one product, propanone. Identify the alkene.
Answer:
Each half of the double bond gives (CH3)2C=O.
Join the two carbonyl carbons by a double bond: (CH3)2C=C(CH3)2.
2,3-Dimethylbut-2-ene.

Worked Example 3. Two compounds, A and B, both of formula C3H6O, are distinguished: A gives a silver mirror with Tollens reagent; B does not but gives an iodoform precipitate. Identify them.
Answer:
A is an aldehyde: propanal.
B is a ketone with a CH3CO group: propanone.
A = CH3CH2CHO, B = CH3COCH3.