WBJEE Organic Chemistry Syllabus

Define homolytic and heterolytic bond fission and the species each produces.

Homolytic fission: each atom takes one electron, giving free radicals ($\ce{A-B -> A^. + B^.}$); favoured by non-polar conditions/heat/light. Heterolytic fission: one atom takes both electrons, giving ions ($\ce{A-B -> A+ + B-}$); favoured by polar conditions.

State the principle of crystallisation as a purification method.

Crystallisation purifies solids by dissolving the impure compound in a hot solvent in which it is sparingly soluble when cold; on cooling the pure compound crystallises out while soluble impurities remain in the mother liquor (and insoluble ones are filtered off hot).

On what principle does simple distillation work, and when is it used?

Simple distillation separates a liquid from non-volatile impurities, or two liquids whose boiling points differ appreciably (roughly more than $25\text{--}30^{\circ}\text{C}$) and that boil below $150^{\circ}\text{C}$. The more volatile component vaporises first and is condensed.

What is fractional distillation and where is it applied?

Fractional distillation uses a fractionating column to separate liquids whose boiling points are close. Repeated vaporisation–condensation along the column enriches the vapour in the more volatile component. Used in petroleum refining and separating air/aqueous mixtures.

Explain steam distillation and the condition a compound must satisfy.

Steam distillation separates steam-volatile substances that are immiscible with and do not react with water. The mixture boils when the sum of the vapour pressures equals atmospheric pressure ($p_{\text{total}} = p_{\text{water}} + p_{\text{compound}}$), so it distils below $100^{\circ}\text{C}$. Example: purification of aniline.

What is the principle of differential extraction (solvent extraction)?

An organic compound is separated from an aqueous solution by shaking with an immiscible organic solvent in which it is more soluble; the compound distributes preferentially into the organic layer (governed by the distribution/partition coefficient), which is then separated and evaporated.

State the principle behind chromatography and name its two broad types.

Chromatography separates components by their differential distribution between a stationary phase and a moving mobile phase. Two types: adsorption chromatography (e.g. column, thin-layer) based on differential adsorption, and partition chromatography (e.g. paper) based on differential partition.

In thin-layer chromatography, define the retardation factor $R_f$.

$$R_f = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}}$$ It is characteristic of a compound under given conditions and is always less than 1.

Describe the Lassaigne's (sodium fusion) test for detecting N, S and halogens.

The organic compound is fused with sodium metal, converting N, S, X into ionic $\ce{NaCN}$, $\ce{Na2S}$ and $\ce{NaX}$. The aqueous extract gives: Prussian blue with $\ce{FeSO4}/\ce{Fe^{3+}}$ (nitrogen), black/violet with lead acetate / sodium nitroprusside (sulphur), and characteristic AgX precipitate with $\ce{AgNO3}$ (halogens).

How is nitrogen estimated by Duma's method?

In Duma's method the compound is heated with $\ce{CuO}$ in a $\ce{CO2}$ atmosphere; nitrogen is liberated as $\ce{N2}$, collected over $\ce{KOH}$ solution and its volume measured. $$\%\,\ce{N} = \frac{28}{22400}\times\frac{V_{STP}}{W}\times 100$$ where $W$ is the mass of compound.

State the formula for percentage of nitrogen in Kjeldahl's method.

$$\%\,\ce{N} = \frac{1.4 \times M \times V}{W}$$ where $M$ is molarity of acid, $V$ is volume of acid neutralised by the liberated $\ce{NH3}$ (in mL), and $W$ is the mass of compound in grams. Not applicable to N in rings or $\ce{-NO2}$/azo groups.

In Liebig's method for carbon and hydrogen, what is C converted to, H converted to, and how is each absorbed?

The compound is burnt in excess $\ce{O2}$ over $\ce{CuO}$: carbon $\to \ce{CO2}$ (absorbed in $\ce{KOH}$) and hydrogen $\to \ce{H2O}$ (absorbed in anhydrous $\ce{CaCl2}$). Masses give: $\%\ce{C}=\frac{12}{44}\times\frac{m_{\ce{CO2}}}{W}\times100$, $\%\ce{H}=\frac{2}{18}\times\frac{m_{\ce{H2O}}}{W}\times100$.

Give the general molecular formulas for alkanes, alkenes and alkynes.

Alkanes: $C_nH_{2n+2}$ (saturated). Alkenes: $C_nH_{2n}$ (one double bond). Alkynes: $C_nH_{2n-2}$ (one triple bond).

Write the Wurtz reaction for preparation of alkanes.

Alkyl halides react with sodium in dry ether to give a symmetrical alkane: $$\ce{2R-X + 2Na ->[\text{dry ether}] R-R + 2NaX}$$ It is used to make alkanes with an even number of carbons / double the alkyl chain.

What is the Kolbe electrolysis reaction?

Electrolysis of an aqueous solution of a sodium or potassium salt of a carboxylic acid gives an alkane at the anode: $$\ce{2RCOONa + 2H2O ->[\text{electrolysis}] R-R + 2CO2 + 2NaOH + H2}$$ ($\ce{H2}$ liberated at the cathode).

State Markovnikov's rule and Anti-Markovnikov (peroxide/Kharasch) effect.

Markovnikov's rule: in addition of $\ce{HX}$ to an unsymmetrical alkene, the negative part ($\ce{X}$) adds to the carbon bearing fewer hydrogens (forming the more stable carbocation). Anti-Markovnikov: with $\ce{HBr}$ in the presence of peroxides, addition is reversed (free-radical mechanism); occurs only with $\ce{HBr}$.

What is ozonolysis and what does it reveal?

Ozonolysis is the cleavage of a $\ce{C=C}$ double bond by ozone followed by reductive workup ($\ce{Zn}/\ce{H2O}$) to give carbonyl compounds (aldehydes/ketones). It is used to locate the position of the double bond in an alkene.

How are alkynes prepared from vicinal and geminal dihalides?

By dehydrohalogenation with alcoholic $\ce{KOH}$ followed by sodamide ($\ce{NaNH2}$): a vicinal/geminal dihalide loses two molecules of $\ce{HX}$ to give an alkyne. Terminal alkynes ($\ce{R-C#CH}$) have acidic hydrogens removable by $\ce{NaNH2}$ or sodium.

State Hückel's rule for aromaticity.

A planar, cyclic, fully conjugated system is aromatic if it contains $(4n+2)$ delocalised $\pi$ electrons, where $n = 0, 1, 2, \dots$. Benzene with $6$ $\pi$ electrons ($n=1$) is the classic example.

Name the four common electrophilic substitution reactions of benzene and their electrophiles.

Nitration ($\ce{NO2+}$), halogenation ($\ce{X+}$, e.g. $\ce{Cl+}$ from $\ce{Cl2}/\ce{FeCl3}$), sulphonation ($\ce{SO3}$/$\ce{^+SO3H}$), and Friedel–Crafts alkylation/acylation ($\ce{R+}$ / $\ce{RCO+}$ using anhydrous $\ce{AlCl3}$).

Classify ortho/para- and meta-directing groups in electrophilic aromatic substitution.

Ortho/para directors (mostly activating, $+M$/$+I$): $\ce{-OH}$, $\ce{-NH2}$, $\ce{-OR}$, $\ce{-CH3}$, and halogens (deactivating but o/p-directing). Meta directors (deactivating, $-M$/$-I$): $\ce{-NO2}$, $\ce{-COOH}$, $\ce{-CHO}$, $\ce{-SO3H}$, $\ce{-CN}$, $\ce{-^+NR3}$.

Compare $S_N1$ and $S_N2$ mechanisms (molecularity, kinetics, stereochemistry, substrate preference).

$S_N1$: two steps via a carbocation, first-order rate $= k[\text{substrate}]$, racemisation, favoured by $3^{\circ}$ halides and polar protic solvents. $S_N2$: one concerted step, second-order rate $= k[\text{substrate}][\text{Nu}]$, inversion of configuration (Walden inversion), favoured by $1^{\circ}$ halides and strong nucleophiles.

Give the order of reactivity of alkyl halides in $S_N1$ and in $S_N2$ reactions.

$S_N1$ reactivity: $3^{\circ} > 2^{\circ} > 1^{\circ} > \ce{CH3X}$ (stability of carbocation). $S_N2$ reactivity: $\ce{CH3X} > 1^{\circ} > 2^{\circ} > 3^{\circ}$ (decreasing steric hindrance).

State Saytzeff's rule for elimination reactions.

In a dehydrohalogenation (E1/E2), the major product is the more substituted (more stable) alkene, formed by preferential removal of the $\beta$-hydrogen from the carbon with the fewer hydrogens.

Compare E1 and E2 elimination mechanisms.

E1: two steps via carbocation, first-order rate $= k[\text{substrate}]$, favoured by $3^{\circ}$ substrates and weak bases. E2: single concerted step, second-order rate $= k[\text{substrate}][\text{base}]$, requires anti-periplanar $\beta$-H, favoured by strong bases like alcoholic $\ce{KOH}$.

Why are haloarenes much less reactive than haloalkanes towards nucleophilic substitution?

In haloarenes the $\ce{C-X}$ bond has partial double-bond character due to resonance with the ring, and the carbon is $sp^{2}$ (more electronegative, holding the bond pair tighter). These shorten and strengthen the $\ce{C-X}$ bond, and the ring repels incoming nucleophiles, so substitution is difficult without strong activation.

How are alcohols prepared by hydroboration–oxidation, and what is the regiochemistry?

Alkene reacts with diborane ($\ce{B2H6}$) then alkaline $\ce{H2O2}$: $$\ce{3RCH=CH2 + (BH3)2 -> (RCH2CH2)3B ->[H2O2/OH-] 3RCH2CH2OH}$$ It gives anti-Markovnikov (less-substituted) alcohols by syn addition of $\ce{H}$ and $\ce{OH}$.

Compare the acidity of alcohols, water and phenol, and explain phenol's higher acidity.

Acidity order: phenol > water > alcohols ($\text{p}K_a$ roughly $10$, $15.7$, $16\text{--}18$). Phenol is more acidic because the phenoxide ion is resonance-stabilised by delocalisation of the negative charge into the ring, whereas alkoxide ions are destabilised by the $+I$ effect of alkyl groups.

Distinguish the products of oxidation of primary, secondary and tertiary alcohols.

Primary alcohols oxidise to aldehydes then carboxylic acids; secondary alcohols oxidise to ketones (same carbon number); tertiary alcohols resist oxidation under normal conditions but, with strong oxidants, undergo C–C cleavage to give a mixture of smaller acids/ketones.

What is the Williamson ether synthesis and why are primary halides preferred?

An alkoxide reacts with an alkyl halide by $S_N2$ to give an ether: $$\ce{R-O^- Na+ + R'-X -> R-O-R' + NaX}$$ Primary (unhindered) halides are preferred because secondary/tertiary halides favour elimination (alkene formation) over substitution.