Easy Practice — Set 1

Quick Recap — Alkanes, Alkenes, Alkynes & Arenes

• Formulae: alkanes $C_nH_{2n+2}$ ($sp^3$, saturated, substitution); alkenes $C_nH_{2n}$ ($sp^2$, addition); alkynes $C_nH_{2n-2}$ ($sp$, addition, terminal ones acidic).
• Arenes: benzene is aromatic (planar, 6 $\pi$ electrons, Hückel) and undergoes electrophilic substitution.
• Markovnikov's rule: in HX addition, H goes to the carbon bearing more H; the peroxide effect (anti-Markovnikov) applies to HBr only.
• Tests for unsaturation: decolourising bromine water and Baeyer's reagent (cold dilute $KMnO_4$).

Beyond-NCERT JEE Essentials

High-yield named reactions, and the regio-/stereo-selectivity JEE loves to test. Structures and reagents are in plain text; a wrong product is worse than none, so every outcome below is checked.

A. Alkanes — building the chain and breaking C-H bonds

• Wurtz reaction: 2 R-X + 2 Na (dry ether) -> R-R + 2 NaX. Couples two alkyl halides into a symmetrical alkane with an even number of carbons. [JEE Tip] Two different halides give a statistical mixture of THREE alkanes (R-R, R'-R', R-R'), so Wurtz is useless for odd-carbon or unsymmetrical alkanes; tertiary halides mostly eliminate.
• Kolbe electrolysis: electrolysis of aqueous R-COONa. At the anode two R groups couple: 2 R-COO(-) -> R-R + 2 CO2, while H2 + NaOH appear at the cathode. Sodium acetate -> ethane. [JEE Tip] Like Wurtz, it JOINS two R groups (carbon count doubles) but starts from a carboxylate salt.
• Decarboxylation (soda-lime): R-COONa + NaOH -(CaO, heat)-> R-H + Na2CO3. Sodium acetate -> methane. [JEE Tip] This REMOVES one carbon (as carbonate) - the mirror image of Kolbe. Remember the contrast: acetate -> methane (soda-lime) but -> ethane (Kolbe).
• Controlled oxidation: CH4 + O2 (Cu, 523 K, 100 atm) -> CH3OH; CH4 + O2 (Mo2O3) -> HCHO; an alkane bearing a tertiary C-H + KMnO4 -> the tertiary alcohol.
• Free-radical halogenation: initiation/propagation/termination chain; ease of H-abstraction follows radical stability $3^\circ > 2^\circ > 1^\circ$. [JEE Tip] Chlorination is fast but UNSELECTIVE (per-H reactivity roughly $1 : 3.8 : 5$ for $1^\circ : 2^\circ : 3^\circ$), so it gives every isomer; bromination is slow but HIGHLY SELECTIVE (roughly $1 : 82 : 1600$) and cleanly attacks the most substituted position. Fluorination is explosive and iodination is reversible (needs an oxidant to pull off HI) - only Cl2 and Br2 are synthetically useful.

B. Alkenes — regio- and stereochemistry of addition

• Markovnikov (HX): H to the carbon already carrying more H's; X to the more substituted carbon (via the more stable carbocation). CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 (2-bromopropane). [JEE Tip] Because a free carbocation forms, hydride/alkyl SHIFTS can occur: (CH3)2CH-CH=CH2 + HCl gives mainly (CH3)2CCl-CH2-CH3 after a 2-degree -> 3-degree hydride shift.
• Peroxide (Kharasch) effect: HBr + R-O-O-R adds by a radical chain -> ANTI-Markovnikov. CH3-CH=CH2 + HBr/peroxide -> CH3-CH2-CH2Br (1-bromopropane). [JEE Tip] This works for HBr ONLY. HCl (bond too strong) and HI (bond too weak) cannot sustain the radical chain, so both stay Markovnikov whether or not peroxide is present.
• Hydroboration-oxidation (B2H6, then H2O2/OH-): net anti-Markovnikov addition of water, SYN, with NO rearrangement (concerted, no carbocation); boron takes the less hindered carbon. CH3-CH=CH2 -> CH3-CH2-CH2OH (propan-1-ol).
• Oxymercuration-demercuration (Hg(OAc)2/H2O, then NaBH4): Markovnikov addition of water, again NO rearrangement (goes through a bridged mercurinium ion, not a free cation). CH3-CH=CH2 -> CH3-CHOH-CH3 (propan-2-ol). [JEE Tip] The clean pair to memorise: oxymercuration = Markovnikov alcohol, hydroboration = anti-Markovnikov alcohol, BOTH rearrangement-free; only direct acid hydration (H2O/H+) can rearrange.
• Ozonolysis (O3, then work-up): cleaves C=C into two carbonyl fragments. Reductive (Zn/H2O or Me2S): =CH- becomes an aldehyde, =CR2 becomes a ketone. Oxidative (H2O2): the aldehyde ends are oxidised further to -COOH; ketones survive unchanged. Example, (CH3)2C=CH-CH3: reductive -> acetone + acetaldehyde; oxidative -> acetone + acetic acid. [JEE Tip] A double-bond carbon that carries an H becomes an aldehyde (reductive) or acid (oxidative); a fully substituted one becomes a ketone either way - use this to reason backwards to the alkene.
• Baeyer's reagent (cold dilute alkaline KMnO4): SYN dihydroxylation -> a cis vicinal diol; decolourising the purple colour is the test for unsaturation. CH2=CH2 -> HOCH2-CH2OH (ethane-1,2-diol). [JEE Tip] COLD dilute KMnO4 = cis-diol; HOT/concentrated KMnO4 CLEAVES the C=C instead (=CH- -> COOH, =CR2 -> ketone, terminal =CH2 -> CO2). Same reagent, opposite outcome - read the conditions.
• Epoxidation (a peroxyacid, e.g. mCPBA): delivers a single O across the C=C -> an epoxide (syn, stereospecific: a cis-alkene gives a cis-epoxide). [JEE Tip] Epoxide + H3O+ opens to the TRANS (anti) diol, whereas Baeyer's/OsO4 gives the CIS diol - so the route chosen fixes the diol's stereochemistry.

C. Alkynes

• Terminal-alkyne acidity: in $R-C\equiv C-H$ the sp C-H (50% s-character) is weakly acidic. With Na -> a sodium acetylide + 1/2 H2; with ammoniacal AgNO3 -> a silver acetylide (white/grey precipitate); with ammoniacal Cu2Cl2 -> a copper acetylide (red precipitate). [JEE Tip] These precipitates form ONLY with terminal alkynes, so they are the classic way to tell a terminal from an internal alkyne (an internal alkyne has no acidic H).
• Hydration (dilute H2SO4 / HgSO4, Markovnikov): water adds Markovnikov -> an enol -> tautomerises to a carbonyl. Ethyne -> acetaldehyde (CH3CHO); every OTHER terminal alkyne -> a methyl ketone, e.g. propyne $CH_3-C\equiv CH$ -> CH3-CO-CH3 (acetone). [JEE Tip] Ethyne is the ONLY alkyne that gives an aldehyde - flag it as the special case.
• Controlled reduction: H2 over Lindlar's catalyst (Pd-CaCO3, poisoned) -> the CIS (Z) alkene (syn addition of H2); Na (or Li) in liquid NH3 -> the TRANS (E) alkene (anti addition). [JEE Tip] "Lindlar = cis, sodium/ammonia = trans." Ordinary Pd/Pt/Ni with excess H2 reduces all the way to the alkane.

D. Aromatic — electrophilic substitution (EAS) and directing effects

• The five electrophiles: nitration (conc. HNO3/H2SO4, electrophile NO2+), sulfonation (oleum/SO3, reversible), halogenation (Cl2 or Br2 with FeX3/AlCl3), Friedel-Crafts alkylation (R-X/AlCl3) and acylation (RCOCl/AlCl3). [JEE Tip] Prefer ACYLATION over alkylation: alkylation suffers carbocation rearrangement and polysubstitution, whereas acylation neither rearranges nor over-substitutes (the acyl product is deactivated, so it stops at one).
• Directing effects - two rules to memorise:

1. o/p-directors, activating: -OH, -OR, -NH2, -NHR, -NR2, -NHCOR, -R (alkyl), -C6H5 (they release electrons by a lone pair or +I).
2. m-directors, deactivating: -NO2, -CN, -CHO, -COR, -COOH, -COOR, -SO3H, -N+R3, -CF3 (strong -M/-I withdrawal).

• [JEE Tip] The single exception: the HALOGENS (-F, -Cl, -Br, -I) are o/p-directing but DEACTIVATING (inductive withdrawal slows the ring, yet lone-pair donation still steers ortho/para). Bottom line: "every o/p-director activates EXCEPT the halogens; every m-director deactivates."
• [JEE Tip] A strong activator can over-react: phenol + Br2 (in water) -> 2,4,6-tribromophenol directly, no catalyst needed. Bulky electrophiles or substituents favour para over ortho on steric grounds.

Worked Examples — Beyond-NCERT Essentials

Example 1 - Hydroboration vs acid hydration (regiochemistry). But-1-ene, CH3-CH2-CH=CH2, is treated (i) with B2H6 then H2O2/OH-, and (ii) with dilute H2SO4/H2O. Give each major product. Reasoning: Hydroboration adds water anti-Markovnikov and syn with no carbocation, so -OH lands on the terminal (less substituted) carbon -> butan-1-ol, CH3CH2CH2CH2OH. Acid hydration is Markovnikov (through the more stable 2-degree carbocation), so -OH goes to C2 -> butan-2-ol, CH3CH2CHOHCH3. Answer: (i) butan-1-ol; (ii) butan-2-ol. (Oxymercuration would also give butan-2-ol, but with no risk of rearrangement.)

Example 2 - The rearrangement trap. 3-methylbut-1-ene, CH2=CH-CH(CH3)2, is treated with HCl. What is the major product, and how does hydroboration differ? Reasoning: Markovnikov protonation puts H on C1 and a 2-degree carbocation on C2. A hydride shift from the adjacent tertiary carbon gives a more stable 3-degree carbocation, which Cl- traps -> 2-chloro-2-methylbutane, CH3CH2-CCl(CH3)-CH3 (the un-rearranged 2-chloro-3-methylbutane is only minor). Hydroboration-oxidation has no carbocation and so cannot rearrange: it would give 3-methylbutan-1-ol. Answer: Major = 2-chloro-2-methylbutane (via rearrangement); hydroboration gives 3-methylbutan-1-ol cleanly.

Example 3 - Ozonolysis, forward and backward. (a) Give the reductive and oxidative ozonolysis products of 2-methylbut-2-ene, (CH3)2C=CH-CH3. (b) An unknown C5H10 alkene gives only acetone and acetaldehyde on reductive ozonolysis - identify it. Reasoning: Cleave the C=C. The (CH3)2C= end has no H, so it becomes a ketone (acetone) in both work-ups. The =CH-CH3 end has an H, so it becomes acetaldehyde under reductive work-up but is oxidised to acetic acid under H2O2. For (b), rejoin the carbonyl carbons of acetone, (CH3)2C=O, and acetaldehyde, CH3CHO, by a double bond -> (CH3)2C=CH-CH3, the same 2-methylbut-2-ene. Answer: (a) reductive -> acetone + acetaldehyde; oxidative -> acetone + acetic acid. (b) 2-methylbut-2-ene.

Example 4 - Choosing cis vs trans. From but-2-yne, $CH_3-C\equiv C-CH_3$, which reagent gives cis-but-2-ene and which gives trans-but-2-ene? Reasoning: H2 over Lindlar's catalyst (poisoned Pd) delivers both hydrogens to the same face (syn addition) -> cis-but-2-ene. Dissolving-metal reduction, Na in liquid NH3, proceeds through a trans-oriented vinyl radical-anion -> trans-but-2-ene. Answer: Lindlar/H2 -> cis-but-2-ene; Na/liquid NH3 -> trans-but-2-ene.

Example 5 - Directing effects: predict the major product. Where does the incoming group go when (a) toluene and (b) nitrobenzene are mononitrated? Reasoning: In toluene the -CH3 is a weak activator and an o/p-director, so nitration is faster than for benzene and gives a mixture of ortho- and para-nitrotoluene (ortho is in fact the major single isomer). In nitrobenzene the -NO2 is a strong deactivator and a m-director, so nitration is slower than for benzene and gives mainly m-dinitrobenzene. Answer: (a) ortho- and para-nitrotoluene (o/p mixture); (b) meta-dinitrobenzene.

Example 6 - Terminal vs internal alkyne (a distinguishing test). How would you chemically tell but-1-yne, $CH_3CH_2-C\equiv CH$, from but-2-yne, $CH_3-C\equiv C-CH_3$? Reasoning: But-1-yne is terminal, so its acidic C-H reacts with ammoniacal AgNO3 to throw down a white/grey silver-acetylide precipitate (and a red precipitate with ammoniacal Cu2Cl2). But-2-yne is internal with no acidic H, so it gives no precipitate. Answer: Add ammoniacal AgNO3 (or Cu2Cl2): but-1-yne gives a precipitate, but-2-yne does not.

Example 7 - Markovnikov hydration of an alkyne. What forms when propyne, $CH_3-C\equiv CH$, is treated with dilute H2SO4/HgSO4, and how does ethyne differ? Reasoning: Water adds Markovnikov (-OH to the more substituted, internal carbon), giving the enol CH3-C(OH)=CH2, which tautomerises to the ketone propan-2-one (acetone), CH3-CO-CH3. Ethyne is symmetric and one carbon shorter, so its enol tautomerises to acetaldehyde - the only alkyne that yields an aldehyde. Answer: Propyne -> acetone (propan-2-one); ethyne -> acetaldehyde.

Example 8 - Halogenation selectivity. 2-methylbutane, (CH3)2CH-CH2-CH3, is treated separately with Br2/hv and with Cl2/hv. Which gives one clean major product? Reasoning: Bromination is highly selective for the weakest (3-degree) C-H, so it substitutes the single tertiary hydrogen -> mainly 2-bromo-2-methylbutane, (CH3)2CBr-CH2-CH3. Chlorination is fast but unselective, abstracting 1-degree, 2-degree and 3-degree hydrogens at comparable per-H rates, so it gives a mixture of all four monochloro isomers. Answer: Bromination -> chiefly 2-bromo-2-methylbutane (clean); chlorination -> a mixture of monochlorides.