JEE Chemistry Organic Chemistry Some Basic Principles and Techniques Notes Set 22

Catalytic Hydrogenation: Counting Unsaturation

Alkenes, alkynes and other unsaturated aliphatic compounds add hydrogen over Ni, Pt or Pd at room temperature; the carbon skeleton does not change and the reaction cannot normally be stopped part way (an alkyne goes straight to the alkane). One mole of compound uses as many moles of H\(_2\) as it has \( \pi \) bonds (a C≡C counts as two). Benzene rings resist hydrogenation at room temperature and need high temperature and pressure.
This gives two uses: the moles of H\(_2\) absorbed show the number of double or triple bonds, and different unsaturated compounds with the same skeleton give the same alkane. To count the alkenes that give a given alkane, place a double bond at every C-C bond where both carbons carry hydrogen (for example, 3 alkenes give 2-methylbutane and 6 give 3-methylhexane).

Monohalogenation: Counting Isomers

An alkane or cycloalkane reacts with Cl\(_2\) or Br\(_2\) in light by a free radical substitution of one C-H by C-X. The number of structural monochloro isomers equals the number of different kinds of hydrogen (non-equivalent hydrogens) in the molecule. In aromatic hydrocarbons only the side-chain hydrogens react; the ring hydrogens are inert under these conditions.
Examples: pentane 3, 2-methylbutane 4, 2,3-dimethylbutane 2, neopentane and 2,2,3,3-tetramethylbutane 1, ethylcyclohexane 6, methylcyclopentane 4.

Ozonolysis

Ozone cleaves C=C (and C≡C) completely. After reductive work-up (Zn/H\(_2\)O or Me\(_2\)S) each alkene carbon becomes a carbonyl: =CH\(_2\) gives HCHO, =CHR gives an aldehyde, =CR\(_2\) gives a ketone. After oxidative work-up (H\(_2\)O\(_2\)) aldehydes are oxidised to carboxylic acids (HCHO to CO\(_2\)) while ketones are unchanged. An alkyne gives two carboxylic acids (or a 1,2-dicarbonyl under mild conditions). A ring double bond gives a single dicarbonyl chain. To find the original alkene, join the carbonyl carbons of the fragments by a double bond.

Tests for Unsaturation and Acidic Hydrogen

One mole of a compound liberates 1/2 mole of H\(_2\) per acidic hydrogen. Internal alkynes (R-C≡C-R) do not react with sodium or the silver and copper reagents.

Alcohols, Phenols, Nitro Compounds and Halides

1. Ceric ammonium nitrate: yellow to red with alcohols. Lucas reagent (anhydrous ZnCl\(_2\) + conc. HCl): tertiary alcohols give turbidity immediately, secondary in about 5 minutes, primary only on heating. Victor Meyer test (red P/I\(_2\), AgNO\(_2\), HNO\(_2\), NaOH): primary gives red, secondary blue, tertiary colourless.
2. Phenols give violet, blue or green colours with neutral FeCl\(_3\). Phenols are more acidic than alcohols but weaker than carbonic acid, so they do not liberate CO\(_2\) from NaHCO\(_3\).
3. Nitro compounds: reduction with Zn and NH\(_4\)Cl gives a hydroxylamine, which reduces Tollens' reagent (Mulliken-Barker test).
4. Alkyl halides: warm with aqueous NaOH, acidify with HNO\(_3\) and add AgNO\(_3\): white (Cl), pale yellow (Br) or yellow (I) precipitate. The halogen on a benzene ring does not respond.

Aldehydes and Ketones

1. 2,4-DNP gives an orange-yellow precipitate with both aldehydes and ketones (carbonyl test).
2. Tollens' reagent \( [\ce{Ag(NH3)2}]^+ \) gives a silver mirror and Fehling's solution a red precipitate of Cu\(_2\)O with aldehydes (aliphatic) but not with ketones; benzaldehyde does not respond to Fehling's. Schiff's reagent is restored to magenta by aldehydes.
3. Iodoform test (I\(_2\)/NaOH): a yellow precipitate of CHI\(_3\) from methyl ketones (CH\(_3\)CO-), ethanal and alcohols with the CH\(_3\)CH(OH)- group. Methanol, methanal and propanal do not respond.
4. NaHSO\(_3\) addition and the benzoin-type reactions are further characteristic of aldehydes.

Acids, Esters, Amides and Amines

1. Carboxylic acids (and strongly acidic phenols such as picric acid) liberate CO\(_2\) from NaHCO\(_3\); simple phenols do not. Esters have a fruity smell and are neutral; an amide warmed with NaOH evolves NH\(_3\).
2. Carbylamine test: primary amines (aliphatic and aromatic) with CHCl\(_3\) and alcoholic KOH give a foul-smelling isocyanide RNC. Secondary and tertiary amines do not.
3. Hinsberg test with benzenesulphonyl chloride: primary amines give a sulphonamide that dissolves in alkali; secondary amines give an insoluble sulphonamide; tertiary amines do not react. Nitrous acid: aliphatic primary amines give N\(_2\); aromatic primary amines give a diazonium salt that couples with 2-naphthol to a red-orange azo dye; secondary amines give a yellow oily N-nitrosoamine.

Qualitative Analysis of Elements

Lassaigne's test: fuse the compound with sodium metal, plunge into water and filter; C, N, S, X form NaCN, Na\(_2\)S, NaX (and NaSCN if both N and S are present).
1. Nitrogen: boil with FeSO\(_4\), acidify with dilute H\(_2\)SO\(_4\), add FeCl\(_3\): Prussian blue Fe\(_4\)[Fe(CN)\(_6\)]\(_3\). If sulphur is also present, a blood-red [Fe(SCN)]\(^{2+}\) forms. A compound with no carbon (like hydrazine) gives no test.
2. Sulphur: acidify with acetic acid, add lead acetate: black PbS. Alternatively sodium nitroprusside gives violet. Acetic acid is used because H\(_2\)SO\(_4\) would give white PbSO\(_4\).
3. Halogens: boil with dilute HNO\(_3\) (to remove CN\(^-\) and S\(^{2-}\)), then add AgNO\(_3\): AgCl white, AgBr pale yellow, AgI yellow.
4. Beilstein test: a green flame with a copper wire from volatile copper halides (not for fluorides).

Quantitative Analysis

1. Carbon and hydrogen (Liebig): burn the compound in oxygen with CuO; H\(_2\)O is absorbed in anhydrous CaCl\(_2\) (or Mg(ClO\(_4\))\(_2\)) and CO\(_2\) in KOH solution. \( \%C=\frac{12}{44}\times\frac{m_{CO_2}}{m}\times100 \) and \( \%H=\frac{2}{18}\times\frac{m_{H_2O}}{m}\times100 \).
2. Nitrogen by Dumas: the compound is heated with CuO in CO\(_2\); N\(_2\) is collected over KOH. Use the pressure of dry nitrogen (total minus aqueous tension) in PV = nRT.
3. Nitrogen by Kjeldahl: digest with conc. H\(_2\)SO\(_4\) (K\(_2\)SO\(_4\) and a Hg or CuSO\(_4\) catalyst) to (NH\(_4\))\(_2\)SO\(_4\), add NaOH to release NH\(_3\), absorb in a measured excess of standard acid and back-titrate. \( \%N=\frac{1.4\times M\times V_{acid\ used}\times n}{m} \) with the acid used in mL and n the basicity. It does not work for nitro, azo and diazo compounds or for N in a ring (pyridine).
4. Halogens (Carius): heat with fuming HNO\(_3\) and AgNO\(_3\) in a sealed tube; \( \%X=\frac{\text{atomic mass of X}}{\text{molar mass of AgX}}\times\frac{m_{AgX}}{m}\times100 \). Sulphur is weighed as BaSO\(_4\) and phosphorus as Mg\(_2\)P\(_2\)O\(_7\). Oxygen is found by difference.

Worked Example 1. 0.5 g of a compound gives ammonia in Kjeldahl's method that neutralises 10 mL of 1 M H\(_2\)SO\(_4\). Find the percentage of nitrogen.
Answer:
10 mL of 1 M H\(_2\)SO\(_4\) = 10 mmol, which neutralises 20 mmol NH\(_3\).
N = 20 mmol x 14 = 280 mg. Percentage = 280/500 x 100 = 56%.

Worked Example 2. How many structural monochloro products does 2-methylbutane give?
Answer:
The H atoms are of four kinds: the two CH\(_3\) groups on C2 (equivalent), the CH on C2, the CH\(_2\), and the terminal CH\(_3\).
Hence 4 isomers.