Predict connectivity, regiochemistry and stereochemistry as three separate decisions.
JEE Main · JEE Advanced · NEET UG55 min study guide12 indexed sections
BY THE END, YOU CAN
Explain polar and radical addition pathways without confusing their intermediates.
Choose alkene hydration and oxidation methods by orientation and stereochemical outcome.
Distinguish complete hydrogenation from selective cis/trans alkyne reduction.
Track enol–carbonyl tautomerization after alkyne hydration.
Use terminal-alkyne acidity and acetylide alkylation with realistic substrate restrictions.
BEFORE YOU START
σ and π bonding, hybridization and functional groups.
Carbocation/radical stability and electron-flow notation.
E/Z and basic stereochemical vocabulary.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
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No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A π bond offers a reactive electron pair, but the whole mechanism decides the product
An alkene has one σ and one π bond between its unsaturated carbons; an alkyne has one σ and two perpendicular π bonds. Addition commonly replaces a π bond with new σ bonds to incoming groups. The original carbon–carbon σ framework usually remains unless the reaction is explicitly an oxidative cleavage.
Regiochemistry asks which atom or group attaches to which carbon of an unsymmetrical multiple bond. Stereochemistry asks from which faces groups add and how the resulting stereoisomers relate. Chemoselectivity asks which functional group reacts when several are present. A correct answer to one question does not automatically answer the others.
Markovnikov and anti-Markovnikov describe orientation in suitable unsymmetrical additions, not syn versus anti facial addition. Hydroboration–oxidation is commonly anti-Markovnikov in orientation and syn in its net H/OH addition. Bromination is anti in facial addition even when a symmetrical alkene makes orientation irrelevant.
Before using a reagent mnemonic, identify whether the pathway involves a free carbocation, a bridged ion, a concerted addition or a radical chain. That choice predicts whether rearrangement is plausible and whether stereospecificity is expected. A mechanism is more transferable than a product list detached from its conditions.
02 / UNDERSTAND & APPLY
2. Polar HX addition follows carbocation stability and can rearrange
In the common polar addition of HCl, HBr or HI to an alkene, the π electron pair forms a bond to H while the H–X bond pair moves to X. The resulting carbocation is then attacked by halide. For a simple unsymmetrical alkene, the favored protonation usually forms the more stable carbocation, giving the familiar Markovnikov orientation.
The rule is best understood through the intermediate rather than simply “hydrogen goes to the carbon with more hydrogens.” Resonance-stabilized allylic or benzylic cations, neighboring groups and rearrangements can dominate over a crude primary/secondary/tertiary count. First draw the actual candidate cations.
A hydride or alkyl shift can occur if it produces a more stable carbocation before nucleophile capture. The carbon skeleton may then change. Concerted hydroboration and bridged-ion pathways do not normally permit the same free-carbocation rearrangement, which is one reason alternative reagents can give different products.
A planar carbocation can be attacked from either face. If a new stereocenter forms from an achiral substrate under achiral conditions, a racemic outcome is often expected, but existing stereocenters or ion-pair effects can modify the ratio. Polar HX addition is not generally assigned one universal syn or anti outcome.
ORIGINAL WORKED EXAMPLE
Polar HBr orientation
Predict the principal simple polar-addition product of HBr with pent-1-ene, excluding radical initiation and unusual rearrangements.
Protonation at terminal C1 places the carbocation at C2, a secondary center.
The alternative gives a primary C1 carbocation and is less favorable.
Bromide capture at C2 gives CH₃CH(Br)CH₂CH₂CH₃.
C2 is stereogenic, so the ideal achiral planar-cation model gives a racemic product.
Answer 2-Bromopentane, racemic in the stated model.
03 / UNDERSTAND & APPLY
3. Radical HBr addition reverses the orientation through a different intermediate
With suitable radical initiation, HBr can add through a chain mechanism. A bromine radical adds to the alkene so that the new carbon radical is relatively stable. That carbon radical then abstracts H from HBr, producing the alkyl bromide and regenerating Br• for propagation.
For a simple terminal alkene, Br• commonly bonds to the terminal carbon and leaves the radical on the more substituted internal carbon. Hydrogen is delivered to that internal radical in the next step. Thus bromine ends up at the less substituted carbon, the anti-Markovnikov product.
The orientation is not caused by a less stable carbocation becoming preferred. No carbocation is required in this radical chain. Single-headed fishhook arrows, rather than ordinary two-electron curved arrows, are appropriate for individual electron movements in a detailed radical mechanism.
The textbook peroxide effect is characteristic of HBr among the usual hydrogen halides. HCl and HI do not support the same favorable pair of chain-propagation steps under analogous conditions. Do not generalize that any peroxide makes every HX addition anti-Markovnikov.
ORIGINAL WORKED EXAMPLE
Change only the reaction pathway
Compare products of pent-1-ene with HBr under ordinary polar conditions and under suitable radical-chain conditions.
The polar route favors a secondary carbocation at C2 and gives 2-bromopentane.
The radical route places Br at terminal C1 while leaving the more stable carbon radical at C2.
Hydrogen abstraction then gives 1-bromopentane.
The starting carbon skeleton is unchanged; only the orientation of H and Br differs.
Addition of Br₂ or Cl₂ to an alkene is commonly described through a cyclic halonium ion rather than a freely rotating open carbocation. The alkene attacks the electrophilic halogen as the X–X bond breaks, and a halogen lone pair helps bridge the two carbon atoms. Nucleophilic opening occurs preferentially from the opposite face.
In a nonnucleophilic medium, halide opens the bridge to give a vicinal dihalide with anti addition. On a cycloalkene, the added halogens therefore appear trans in the usual product. For an acyclic substrate, determining whether the product is meso or racemic requires drawing the initial E/Z geometry and the stereospecific addition.
In a water-rich medium, water can compete with halide to open the halonium ion. Deprotonation then gives a halohydrin with OH and X on adjacent carbons. For a typical unsymmetrical alkene, water attacks the more substituted carbon of the unsymmetrical bridge, so OH goes to that carbon and X to the less substituted one.
The halohydrin remains an anti-addition product in the simple bridged-ion model. The absence of an ordinary free carbocation also explains why the usual hydride/alkyl rearrangements are not expected. Changing the nucleophilic solvent can therefore change the functional groups installed without changing the initial electrophilic activation.
ORIGINAL WORKED EXAMPLE
Predict a bromohydrin
For pent-1-ene in a brominating water-rich medium that gives the standard bromohydrin, predict connectivity.
A bromonium bridge forms across C1 and C2.
Water preferentially opens at the more substituted C2.
After deprotonation, C2 bears OH and terminal C1 bears Br.
The product is BrCH₂CH(OH)CH₂CH₂CH₃; the mechanistic H/OH versus Br orientation must not be confused with polar HBr addition.
Answer 1-Bromopentan-2-ol, with anti addition in the bridged-ion model.
05 / UNDERSTAND & APPLY
5. Three hydration routes solve different selectivity problems
Acid-catalyzed hydration adds H and OH across an alkene through protonation, carbocation formation, water attack and deprotonation. It commonly gives Markovnikov orientation and can rearrange if a more stable carbocation is accessible. The equilibrium also depends on water activity and competing dehydration.
Oxymercuration followed by demercuration gives Markovnikov alcohol connectivity while avoiding the usual freely rearranging carbocation pathway. The initial bridged intermediate controls water attack, but the complete two-stage sequence should not be labeled universally stereospecific anti addition because the reduction stage can complicate stereochemistry.
Hydroboration forms C–B and C–H bonds in a concerted addition. Boron commonly attaches to the less substituted carbon, and B and H add syn. Subsequent oxidation replaces the C–B bond by C–O with retention at that carbon, giving net syn H/OH addition and anti-Markovnikov alcohol orientation in the standard simple-alkene case.
These routes often retain the same carbon count but produce different alcohol positions or different stereochemical outcomes. A route choice should state both reagent stages where needed. Writing only “oxidation” after an alkene does not distinguish hydration, dihydroxylation, epoxidation or cleavage.
ORIGINAL WORKED EXAMPLE
Make two positional alcohol isomers
Compare the standard hydration products of 2-methylpropene by acid-catalyzed hydration and hydroboration–oxidation.
Acid protonation favors the tertiary carbocation at the substituted carbon; water capture gives (CH₃)₃COH.
Hydroboration places boron at terminal CH₂ and hydrogen at the substituted carbon.
Oxidation replaces C–B by C–OH, giving (CH₃)₂CHCH₂OH.
The products are 2-methylpropan-2-ol and 2-methylpropan-1-ol, both with four carbons.
6. Syn reduction, syn dihydroxylation and epoxide opening are distinct
Catalytic hydrogenation adds H₂ across an alkene, usually with syn delivery from a metal surface in the elementary stereochemical model. One mole of an isolated C=C bond consumes one mole of H₂ on complete reduction. Other reducible groups can also react depending on catalyst and conditions, so hydrogen uptake is not automatically a unique double-bond count in an unknown complex molecule.
Suitable osmium-based oxidation followed by workup or mild dilute permanganate conditions can give syn vicinal diols. The two OH groups are installed on the same face in the standard concerted/intermediate model. Oxidant severity matters: stronger permanganate conditions can cleave the multiple bond instead of stopping at a diol.
A peroxyacid can epoxidize an alkene by a concerted oxygen-transfer step, preserving the alkene’s relative substituent geometry in the epoxide. Subsequent acid- or base-catalyzed ring opening requires backside attack at the carbon being opened and can lead to anti-related oxygen substitution. The regioselectivity of unsymmetrical epoxide opening depends on the conditions.
For a simple cycloalkene, syn dihydroxylation gives a cis-1,2-diol, whereas epoxidation followed by appropriate hydrolytic opening gives a trans-1,2-diol. The ring makes facial relationships easy to see. In flexible acyclic chains, draw configurations rather than treating cis/trans as a universal label for saturated products.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Schematic ring projection, not a chair conformation. The figure compares relative facial stereochemistry; it does not assign absolute configurations or a particular enantiomer ratio.
ORIGINAL WORKED EXAMPLE
Choose a diol stereochemical route
Starting from cyclopentene, which conceptual route gives cis-1,2-cyclopentanediol and which gives the trans diol?
Syn dihydroxylation places both OH groups on the same face of the original double bond, giving the cis diol.
Epoxidation first forms a three-membered oxygen bridge.
Backside hydrolytic opening places the two oxygen-derived groups on opposite faces, giving the trans diol in the standard model.
The choice is stereochemical, not a change in carbon count.
Answer Syn dihydroxylation: cis diol. Epoxidation then hydrolysis: trans diol.
07 / UNDERSTAND & APPLY
7. Cleavage changes the carbon skeleton into separate carbonyl fragments
Ozonolysis cleaves a C=C bond and converts each alkene carbon into a carbonyl carbon. Under a reductive workup, a carbon bearing hydrogen becomes an aldehyde and a carbon bearing two carbon substituents becomes a ketone. A terminal CH₂ carbon gives formaldehyde in that reductive product accounting.
An oxidative workup can convert initially formed aldehydes onward to carboxylic acids. Vigorous permanganate cleavage likewise gives products determined by the substitution at each alkene carbon, with terminal carbon potentially oxidized to carbon dioxide under strong conditions. Always state the oxidant and workup before deciding the final oxidation level.
Reverse structure problems can reconnect the carbonyl carbons of cleavage products to reconstruct a possible alkene. Remove each carbonyl oxygen and join those carbons by a double bond, preserving their substituents. If the cleavage products could originate from multiple arrangements or the starting material contains several double bonds, additional information may be needed.
The atom count is an important check: the sum of carbon atoms in all organic and inorganic products must account for the original substrate. A cyclic alkene can open into one molecule with two carbonyl termini rather than two separate molecules. Do not automatically split every cleavage into two disconnected products.
ORIGINAL WORKED EXAMPLE
Reconstruct a possible alkene
A simple acyclic alkene gives only propanone and propanal under reductive ozonolysis. Reconstruct its carbon skeleton.
Propanone’s carbonyl carbon carries two methyl groups.
Propanal’s carbonyl carbon carries hydrogen and ethyl.
Join those two carbonyl carbons with C=C and remove the oxygen labels.
The structure is (CH₃)₂C=CHCH₂CH₃, 2-methylpent-2-ene, with six total carbons.
Answer 2-Methylpent-2-ene.
08 / UNDERSTAND & APPLY
8. Alkyne hydration first gives an enol, then a carbonyl
An alkyne can undergo addition once to give an alkene derivative and again to give a saturated derivative. Addition of one equivalent of HX to a simple terminal alkyne commonly places X at the internal carbon; excess HX can give a geminal dihalide with both halogens on that carbon. Product mixtures and regioselectivity require more care for unsymmetrical internal alkynes.
Halogen addition can give a dihaloalkene after one addition and a tetrahalide after two. The exact stereochemical distribution depends on substrate and conditions, so a general reaction map should not promise one exclusive isomer where the reagent system does not justify it.
Mercury-assisted acidic hydration of a terminal alkyne commonly gives a Markovnikov enol that tautomerizes to a methyl ketone. Acetylene is the special simplest case giving ethanal. For an unsymmetrical internal alkyne, more than one carbonyl regioisomer can result.
Hydroboration–oxidation using a suitable bulky borane can give the opposite hydration orientation for a terminal alkyne, leading through an enol to an aldehyde. The borane choice helps avoid undesired multiple addition. The enol is usually an intermediate in the final carbonyl-product problem, not the final isolated answer.
Keto–enol tautomerization moves a proton and shifts the π bond through acid- or base-catalyzed steps. Tautomers are constitutional isomers in equilibrium, not resonance contributors: atomic connectivity, especially the H attachment, changes. Resonance arrows are therefore not the proper symbol between an enol and ketone.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Conceptual product map. The enol–carbonyl equilibrium commonly favors the displayed carbonyl under the usual conditions; each route requires the stated reagent class and workup.
ORIGINAL WORKED EXAMPLE
Two hydration outcomes from one terminal alkyne
Predict final carbonyl products of pent-1-yne under standard mercury-assisted acidic hydration and under suitable bulky-borane hydroboration–oxidation.
The terminal alkyne is HC≡CCH₂CH₂CH₃.
Markovnikov hydration gives an enol with OH at C2, then tautomerization gives CH₃COCH₂CH₂CH₃.
Opposite-orientation hydroboration–oxidation gives the terminal enol that tautomerizes to CH₃CH₂CH₂CH₂CHO.
Both products retain five carbons; they differ as ketone versus aldehyde.
Answer Pent-2-one and pentanal, respectively.
09 / UNDERSTAND & APPLY
9. The reduction method selects alkane, cis alkene or trans alkene
Ordinary catalytic hydrogenation with sufficient H₂ and an active catalyst can reduce an alkyne fully to an alkane, consuming two equivalents of hydrogen per triple bond. An alkene intermediate need not accumulate because it can continue reacting on the catalyst.
A suitably poisoned catalyst such as a Lindlar-type system permits partial hydrogenation to a cis alkene from an appropriate internal alkyne, corresponding to syn hydrogen addition. The method and controlled reaction extent matter; ordinary unmodified active catalyst conditions are not the same selective stopping point.
Dissolving-metal reduction in the standard alkali-metal/liquid-ammonia model gives a trans alkene from an appropriate internal alkyne through stepwise electron and proton transfers. It is not surface-catalyzed syn hydrogenation. Terminal alkene products do not have cis/trans or E/Z labels if one carbon bears two identical hydrogens.
For a symmetrical internal alkyne such as hex-3-yne, these three outcomes are especially clear: full hydrogenation gives hexane, selective syn reduction gives (Z)-hex-3-ene, and the dissolving-metal model gives (E)-hex-3-ene. Stereochemical names must still be assigned from actual substituent priorities.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Reaction-class comparison for a simple internal alkyne, with Et explicitly defined. Selective stopping conditions matter; the drawings show final alkene geometry rather than mechanistic trajectories.
ORIGINAL WORKED EXAMPLE
Select the alkene geometry
How can hex-3-yne be converted conceptually into (Z)-hex-3-ene rather than (E)-hex-3-ene?
The target has the two ethyl groups on the same side of the double bond.
Partial syn hydrogenation with a suitable poisoned catalyst gives the cis, here Z, alkene.
The standard dissolving-metal reduction instead gives the trans, here E, alkene.
Full catalytic hydrogenation would overshoot to hexane.
Answer Use the selective syn partial-hydrogenation route, not full hydrogenation or the trans-selective dissolving-metal route.
10 / UNDERSTAND & APPLY
10. A terminal alkyne can become a carbon nucleophile
A terminal alkyne R–C≡C–H has an acidic hydrogen on an sp carbon. Its conjugate base places the lone pair in an orbital with relatively high s character, stabilizing it more than a corresponding simple sp² or sp³ carbanion. Typical terminal-alkyne pKa values are around 25, so ordinary hydroxide is not a quantitatively effective deprotonating reagent in the usual equilibrium comparison.
A sufficiently strong base such as amide ion can form an acetylide. Compare acid pKa values in a consistent solvent framework rather than assuming every negatively charged reagent is strong enough. The acetylide is both a strong base and a carbon nucleophile.
Acetylide alkylation with a methyl or suitable primary alkyl halide can form a new C–C bond by SN2 substitution. Secondary and tertiary electrophiles commonly give competing or dominant elimination because the acetylide is strongly basic and backside substitution is hindered. An unactivated aryl or vinyl halide is not an ordinary SN2 partner.
The carbon count is additive: acetylide R–C≡C⁻ plus R′–X gives R–C≡C–R′. This can convert a terminal alkyne into an internal one while preserving the triple bond. The route is different from electrophilic addition across the triple bond, which lowers the unsaturation.
ORIGINAL WORKED EXAMPLE
Build a longer alkyne by substitution
Predict the product of the acetylide derived from prop-1-yne reacting with bromoethane by the intended SN2 pathway.
Prop-1-yne is CH₃C≡CH; deprotonation gives CH₃C≡C⁻.
The acetylide carbon attacks the primary carbon of CH₃CH₂Br and bromide leaves.
The product is CH₃C≡CCH₂CH₃.
It has five carbons and is pent-2-yne; the triple bond is retained.
Answer Pent-2-yne.
11 / UNDERSTAND & APPLY
11. Preparation and multistep sequences need elimination and stoichiometric checks
Alkenes can be prepared by suitable alcohol dehydration or alkyl-halide dehydrohalogenation. Regioselectivity, E/Z preference and rearrangement depend on whether the mechanism is E1, E2 or another pathway. These elimination mechanisms receive a dedicated lesson; the present reaction map should not treat every loss of HX as identical.
Alkynes can be formed by two eliminations from suitable vicinal or geminal dihalides with sufficiently strong base. The second elimination from a vinyl-halide intermediate is more demanding than an ordinary alkyl-halide elimination. For a terminal alkyne, extra strong base may deprotonate the product, requiring a later protonating workup to recover the neutral alkyne.
For hydrogen-uptake problems, a simple isolated alkene requires one H₂ and an alkyne two H₂ for complete saturation. Conjugated systems, aromatic rings and other reducible groups require the stated catalyst and reaction extent before translating uptake into structure. A benzene ring is not simply three ordinary isolated alkenes.
In every sequence, redraw the structure after each step, keep carbon count visible, identify any new stereocenters, and decide whether the next reagent reacts with the new group or an untouched group. Do not apply the original substrate’s reaction rule after the functional group has already changed.
Markovnikov/anti-Markovnikov is not syn/anti.
Peroxide-mediated orientation reversal is not a universal HX rule.
A free carbocation can rearrange; a concerted or bridged pathway has different constraints.
Oxymercuration–demercuration does not promise a universal overall anti stereochemistry.
Specify oxidative versus reductive cleavage workup.
After alkyne hydration, check the carbonyl tautomer rather than stopping at the enol.
Acetylide alkylation requires a suitable SN2 electrophile.
12 / UNDERSTAND & APPLY
12. Eight solved reaction checks
Each problem assumes the standard textbook transformation specified, without additional competing functional groups.
ORIGINAL PRACTICE · 1
How many moles H₂ are required for complete reduction of 0.200 mol of an acyclic compound containing one isolated alkene and one isolated alkyne?
Reveal answer & reasoning
0.600 mol H₂.
Each molecule needs one equivalent for C=C and two for C≡C, for three total. Multiply 0.200×3.
ORIGINAL PRACTICE · 2
Does radical HBr addition to propene favor 1-bromopropane or 2-bromopropane?
Reveal answer & reasoning
1-Bromopropane.
Br• adds to the terminal carbon to leave a secondary carbon radical; H abstraction completes the anti-Markovnikov product.
ORIGINAL PRACTICE · 3
What relative bromine positions result from standard anti bromination of cyclohexene?
Reveal answer & reasoning
Trans-1,2-dibromocyclohexane.
Bromonium opening occurs from the opposite face, putting the added bromines on opposite ring faces.
ORIGINAL PRACTICE · 4
What is the standard hydroboration–oxidation product of but-1-ene?
Reveal answer & reasoning
Butan-1-ol.
Boron initially bonds to the less substituted terminal carbon; oxidation replaces C–B by C–OH with retention.
ORIGINAL PRACTICE · 5
What final product follows standard mercury-assisted acidic hydration of but-1-yne?
Reveal answer & reasoning
Butan-2-one.
Markovnikov enol formation is followed by tautomerization to CH₃COCH₂CH₃.
ORIGINAL PRACTICE · 6
Does but-1-ene have an E/Z label?
Reveal answer & reasoning
No.
Its terminal double-bond carbon has two identical hydrogen substituents, so the requirement of two different groups on each alkene carbon is not met.
ORIGINAL PRACTICE · 7
Why is tert-butyl bromide a poor choice for acetylide SN2 alkylation?
The acetylide does not make a tertiary carbon an unhindered SN2 center merely because it is a strong nucleophile.
ORIGINAL PRACTICE · 8
Are CH₂=CHOH and CH₃CHO resonance forms?
Reveal answer & reasoning
No; they are enol and aldehyde tautomers.
A hydrogen changes attachment and the bond connectivity changes, whereas resonance contributors keep atom positions and connectivity fixed while redistributing electrons.
KEEP THE BIG IDEAS
Quick revision
Choose the mechanism before the product mnemonic.
Track connectivity, orientation and stereochemistry separately.
Hydration route determines alcohol position and rearrangement risk.
Cleavage workup determines the final oxidation level.
Alkyne reduction can stop at different stereochemical stages only with the appropriate method.
Terminal-alkyne nucleophilicity is created by deprotonation and used through compatible substitution chemistry.
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