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Mechanism of Some Important Reactions of Alkenes

ChemistryHydrocarbonsFor JEE aspirants

The mechanisms of alkene reactions all start the same way: the electron-rich bond attacks an electrophile. What happens next decides the stereochemistry. Halogen addition and halohydrin formation go through a bridged halonium ion and are therefore anti and stereospecific. Hydroxylation with cold KMnO4 or OsO4 goes through a cyclic ester and is syn. Epoxidation followed by hydrolysis delivers the anti diol, and ozonolysis cuts the double bond cleanly into two carbonyl fragments.

Key Reactions - Quick Reference
  1. Halogenation (anti): alkene bromonium ion vicinal dibromide
  2. Halohydrin (anti): alkene OH on the more substituted carbon
  3. Syn hydroxylation: alkene cis-diol (Baeyer's test)
  4. Anti hydroxylation: alkene epoxide trans-diol
  5. Reductive ozonolysis: alkene aldehydes and ketones
  6. Oxidative cleavage: , ,

1. Mechanism of Halogen Addition

Bromine has no permanent dipole, so how does an alkene attack it? As the electrons approach, they push the Br-Br bonding electrons towards the far bromine. The near bromine becomes , the far one , and the weakened bond breaks heterolytically.

Formation of the bromonium ion from an alkene and bromine Bromine has no permanent dipole, but as the pi electrons of the alkene approach they push the bromine bromine bonding electrons towards the far atom. The near bromine becomes delta positive and the far one delta negative, the bond breaks heterolytically, bromide ion leaves, and a three membered bromonium ion is formed in which bromine carries the positive charge. Step 1: the pi bond makes its own electrophile H H H H alkene Br Br δ+ δ− polarised as the alkene comes close Br H H H H + bromonium ion three-membered ring + Br− Br2 has no permanent dipole of its own. The approaching π cloud pushes the Br–Br bonding electrons onto the far atom, so the near Br turns δ+ and the far one leaves as Br−. The alkene therefore creates the electrophile it needs.
Figure 1: Step 1. The electrons polarise Br, expel Br and close a three-membered bromonium ion.

Step 2: the halide opens the ring

Bromide ion opens the bromonium ion by backside attack The bridging bromine covers one face of the two carbon unit completely, so bromide ion can only attack from the opposite face. The ring opens and the two bromine atoms finish on opposite faces of the former double bond, which makes the overall addition anti. Step 2: bromide can only reach the far face Br H H H H + top face blocked bromonium ion Br- attacks from below, 180° from the bridge anti opening Br Br H H vicinal dibromide one Br in front, one behind The bridging Br covers one face, so the only trajectory left is from the other side. That single geometric fact is the whole reason halogen addition is anti.
Figure 2: Step 2. Br attacks from the back, so the two bromines end up anti.
Why not a normal carbocation? In the halonium ion every atom, including bromine, has a filled octet. But the three-membered ring is strained and the positive charge sits on an electronegative atom, so the ion is still very electrophilic and is opened quickly by any nucleophile present.
Bridged bromonium ion compared with an open carbocation In the bromonium ion every atom including bromine has a complete octet and the positive charge sits on bromine rather than carbon. An open carbocation would leave carbon with only six electrons and, being planar, could be attacked from either face, which would give a mixture of syn and anti products. Experiment shows only anti product, so the bridged ion must be the real intermediate. Why a bridged ion and not an open carbocation? What really forms What does NOT form Br H H + CH2 CH2 Br + every atom keeps a full octet positive charge sits on Br, not C one face is permanently blocked explains anti and stereospecificity carbon would have only six electrons a planar cation can be hit from either face that would give syn and anti mixed experiment says only anti is seen The ring is still strained and the charge sits on an electronegative atom, so the bromonium ion is highly electrophilic and is opened by any nucleophile around. Stable enough to control stereochemistry, reactive enough to be short lived.
Figure 3: The bridged ion keeps every octet full and blocks one face. An open cation would be planar and give both syn and anti, which is not what is observed.

Stereochemical proof: anti addition

Bromination of cyclopentene gives only the trans dibromide Cyclopentene reacts with bromine to give only trans one two dibromocyclopentane and never the cis isomer. Bromide ion can open the bromonium ion at either carbon with equal probability, so the two mirror image trans products form in equal amounts as a racemic pair. This result is direct evidence for anti addition through a bridged intermediate. Stereochemical proof: cyclopentene plus bromine cyclopentene Br2 CCl4 Br Br trans-1,2-dibromocyclopentane the only product + Br Br its mirror image equal amounts No cis product is ever isolated. That is the experimental fact a mechanism has to explain. Bromide can attack either carbon of the bromonium ion with equal probability, so the two mirror-image trans products form in equal amounts: a racemic pair. Anti addition explains it; a free carbocation cannot.
Figure 4: Cyclopentene plus Br gives only the trans-1,2-dibromide, as a racemic pair. The cis isomer is never formed.
Stereospecific bromination of cis and trans but two ene Bromination of cis but two ene gives a racemic pair of two three dibromobutane, while bromination of trans but two ene gives the meso compound. Both reactions are anti additions, but because the starting geometries differ the products differ too. A reaction in which different stereoisomers of the reactant give different stereoisomers of the product is called stereospecific. Stereospecific: different alkene, different product CH3 H CH3 H cis-but-2-ene Br2 Br Br CH3 CH3 Br Br CH3 CH3 + racemic pair (optically inactive mixture) CH3 H H CH3 trans-but-2-ene Br2 Br Br CH3 CH3 meso compound (internal mirror plane)
Figure 5: Different starting stereoisomers give different product stereoisomers, which is the signature of a stereospecific reaction. Note that both rows are anti additions; only the starting geometry changed.
Remark: A stereospecific reaction is one in which different stereoisomers of the reactant give different stereoisomers of the product. Bromination of the but-2-enes is the textbook demonstration, and it is the strongest evidence that a bridged intermediate exists.

2. Mechanism of Halohydrin Formation

Run the same halogenation in water and the solvent, present in vast excess, becomes the nucleophile instead of the halide ion.

Mechanism of halohydrin formation showing regiochemistry In an unsymmetrical halonium ion the bond to the more substituted carbon is stretched further, so that carbon carries more of the positive charge. Water, present in huge excess as the solvent, attacks that carbon from the back face. The hydroxyl group therefore lands on the more substituted carbon and the halogen on the other, and the addition is anti. Halohydrin: water beats halide simply by numbers X CH3 H H H + δ+ unsymmetrical halonium ion more C–X stretching here, so more + charge here H2O attacks the carbon that carries more charge H2O – H+ OH CH3 X halohydrin OH on the more substituted C Nucleophile water, because it is the solvent and swamps the halide ion Stereochemistry anti — water enters the face opposite X Regiochemistry OH to the more substituted carbon, X to the less substituted one Worked case propene + Cl2 in water gives 1-chloropropan-2-ol
Figure 6: Water opens the halonium ion at the more substituted carbon, so OH lands there and X on the other. No free carbocation means no rearrangement.
  • The addition is anti, since the ring is still opened from the back side.
  • There is no free carbocation, so rearrangement never happens.
  • The OH goes to the more substituted carbon because that carbon carries more of the developing positive charge.
  • Example: propene plus Cl2 in water gives 1-chloropropan-2-ol, the propylene chlorohydrin.

3. Syn Hydroxylation (Baeyer's Reagent)

Cold dilute alkaline KMnO4 converts an alkene into a vicinal diol, and the purple colour fading to a brown MnO2 precipitate is Baeyer's test for unsaturation.

Syn hydroxylation of an alkene by cold alkaline permanganate Cold dilute alkaline potassium permanganate adds across the double bond through a five membered cyclic manganate ester in which the metal holds both oxygen atoms on the same face of the alkene. Hydrolysis of that ester releases the vicinal diol with both hydroxyl groups on the same face, so hydroxylation is a syn addition. The purple colour fading to brown manganese dioxide is Baeyer's test for unsaturation. Baeyer's reagent: both OH groups from one face R H R H KMnO4 cold, dilute, OH- O O Mn O O R R H H cyclic manganate ester Mn holds both oxygens on the SAME face of the C=C H2O hydrolysis OH OH cis-diol (glycol) syn addition Both C–O bonds are made in the same cyclic step, so both oxygens must arrive on one face. OsO4 then NaHSO3 does exactly the same job and in better yield. Baeyer's test: the purple KMnO4 fades and brown MnO2 settles out, which is the classic bench test for unsaturation.
Figure 7: Baeyer's reagent delivers both OH groups to the same face through a cyclic ester, so hydroxylation is syn.
Stereochemical outcome of syn hydroxylation of cis and trans alkenes Syn hydroxylation of cis but two ene gives the meso diol, while syn hydroxylation of trans but two ene gives the racemic diol. This is exactly the opposite of what anti bromination gives from the same two alkenes, which shows that the stereochemistry of the reagent, not of the alkene alone, controls the answer. Syn addition inverts the bromination pattern CH3 H CH3 H cis-but-2-ene cold KMnO4 OH OH CH3 CH3 meso diol (internal mirror plane) CH3 H H CH3 trans-but-2-ene cold KMnO4 OH OH CH3 CH3 OH OH CH3 CH3 + racemic diol (equal enantiomers) Same alkene, opposite answer: Br2 turns cis into racemic, KMnO4 turns cis into meso.
Figure 8: cis-alkene gives the meso diol; trans-alkene gives the racemic diol. Compare with Figure 5 and note that the answers swap.
Cross-check the two additions: anti addition of Br2 turns cis-but-2-ene into the racemate; syn addition of KMnO4 turns the same cis-but-2-ene into the meso diol. The stereochemistry of the reagent, not of the alkene alone, controls the answer.

4. Anti Hydroxylation via Epoxidation

A peroxyacid, a carboxylic acid carrying an extra oxygen in a -O-O- linkage, hands that oxygen to the double bond in a single concerted step.

Epoxidation of an alkene by a peroxyacid A peroxyacid carries an extra oxygen in a peroxide linkage and hands that oxygen to the double bond in a single concerted step through a butterfly shaped transition state. Because everything happens at once, the oxygen is delivered to one face only and the addition is syn. The carboxylic acid is released as the by product. Epoxidation: one concerted 'butterfly' step R H R H O H O R'C O peroxyacid R'CO3H: the O carrying H is the one delivered one step O R R H H epoxide (oxirane) O added to one face: syn + R'COOH Common peroxyacids you must recognise in a question stem: PFA HCO3H PAA CH3CO3H PBA C6H5CO3H MCPBA m-ClC6H4CO3H TFPAA CF3CO3H Selectivity: the more substituted (more nucleophilic) double bond is epoxidised first.
Figure 9: One step, one face. The epoxide keeps the alkene geometry, which matters for the next step.
Acid catalysed hydrolysis of an epoxide to the anti diol Acid protonates the epoxide oxygen and turns it into a good leaving group. Water then attacks one of the ring carbons from the face opposite the oxygen bridge, so that carbon is inverted. The two hydroxyl groups therefore finish on opposite faces and the overall sequence of epoxidation followed by hydrolysis is an anti hydroxylation. Acid opens the epoxide from the back: anti diol O epoxide H3O+ O H + protonated: now a much better leaving group H2O anti attack then –H+ OH OH anti diol Two syn-style steps in a row do NOT give a syn product. Step 1 peroxyacid delivers O to one face: syn Step 2 water attacks the opposite face: inversion at that carbon Net result the two OH groups end up anti, on opposite faces
Figure 10: Peroxyacids give the epoxide in one syn step; acid hydrolysis then opens it from the back, so the overall result is the anti diol.
AbbreviationFull nameFormula
PFAPerformic acidHCO3H
PAAPeracetic acidCH3CO3H
PBAPerbenzoic acidC6H5CO3H
MCPBAmeta-Chloroperbenzoic acidClC6H4CO3H
TFPAATrifluoroperacetic acidCF3CO3H
Selectivity: When a molecule has two double bonds, the more substituted one is more nucleophilic and is epoxidised first. A ring double bond usually beats an open-chain one for the same reason.

5. Mechanism of Ozonolysis

Ozone cleaves the double bond under far milder conditions than permanganate, and both aldehydes and ketones survive the work-up.

Mechanism of ozonolysis of an alkene Ozone adds across the double bond to give an unstable molozonide in which the carbon carbon bond is still present. That ring rearranges to the ozonide, in which the carbon carbon bond has been broken and replaced by oxygen bridges. Reductive work up with zinc and water then releases two carbonyl compounds, and the zinc destroys the hydrogen peroxide that would otherwise oxidise any aldehyde to a carboxylic acid. Ozonolysis: two ring rearrangements, then a cut alkene + O3 molozonide ozonide carbonyls R H R H O3 O O O unstable C–C still intact rearr. O O O C–C now broken explosive if isolated Zn H2O RCHO + RCHO The work-up decides what you isolate: Zn / H2O (reductive) aldehydes and ketones survive untouched H2O only (oxidative) the H2O2 formed oxidises aldehydes to acids (CH3)2S or H2/Pd same job as Zn, common in modern questions Zinc is there purely to destroy the hydrogen peroxide before it can oxidise the aldehyde.
Figure 11: Molozonide, then ozonide, then work-up. Zn/HO protects the aldehyde; water alone would push it on to the acid.

Reading a structure back from ozonolysis products

  1. Write both carbonyl fragments obtained.
  2. Remove the oxygen from each carbonyl carbon.
  3. Join those two carbons with a double bond.
  4. Check the molecular formula of the alkene you have rebuilt against the data given.
How to deduce an alkene structure from its ozonolysis products Ozonolysis cuts the double bond and puts an oxygen on each of the two carbons that were joined by it. To work backwards, write both carbonyl fragments, remove the oxygen from each carbonyl carbon, join those two carbons with a double bond, and check the molecular formula. If only one fragment is obtained the alkene was symmetrical about the double bond. Reading the alkene back from ozonolysis products Forward: cut the C=C CH3 CH3 CH3 cut here O3 Zn/H2O CH3COCH3 + CH3CHO ketone aldehyde Backward: rebuild the alkene 1 Write both carbonyl fragments you were given. 2 Delete the oxygen from each carbonyl carbon. 3 Join those two carbons with a double bond. 4 Check the molecular formula against the data. Two different fragments means an unsymmetrical alkene; one fragment means a symmetrical one.
Figure 12: Ozonolysis questions are almost always run in reverse. Delete the two oxygens, glue the carbons back together with a double bond.

6. Oxidative Cleavage with Hot KMnO4

Products of oxidative cleavage of an alkene with hot permanganate Hot concentrated permanganate breaks the double bond completely. A terminal carbon carrying two hydrogens is oxidised all the way to carbon dioxide. A carbon carrying one hydrogen becomes a carboxylic acid. A carbon carrying no hydrogen has nothing left to lose and stops at the ketone. Hot KMnO4 cleaves: read the product off the H count =CH2 2 H on that carbon hot KMnO4 CO2 burns off as CO2 =CHR 1 H on that carbon hot KMnO4 RCOOH one H left, so it becomes the acid =CR2 0 H on that carbon hot KMnO4 R2C=O no H to lose, so it stops here Count the hydrogens on each alkene carbon first. Everything else follows from that.
Figure 13: Hot KMnO cleaves the C=C. Read the product off how many hydrogens each alkene carbon was carrying.
Cold dilute against hot concentrated permanganate Cold dilute alkaline permanganate hydroxylates the double bond and gives a vicinal diol with the carbon skeleton intact. Hot concentrated acidic permanganate cleaves the double bond completely and gives carboxylic acids, ketones or carbon dioxide. Only the conditions differ, so the conditions must be read carefully. Same reagent, two completely different answers Cold, dilute, alkaline Hot, concentrated, acidic R H R H R H R H KMnO4 KMnO4 OH OH vicinal diol, C=C kept syn hydroxylation RCOOH + RCOOH molecule cut in two oxidative cleavage Temperature and concentration alone decide whether the C=C survives. Read the conditions in the question before you write a single product.
Figure 14: Temperature alone decides whether KMnO hydroxylates or cleaves.

7. The Whole Oxidation Picture

Summary of the four oxidation outcomes of an alkene Cold dilute alkaline permanganate gives the cis diol by syn hydroxylation. A peroxyacid followed by acid hydrolysis gives the trans diol by anti hydroxylation. Ozonolysis with reductive work up cuts the double bond to aldehydes and ketones. Hot concentrated permanganate cuts it to carboxylic acids, ketones or carbon dioxide. The first two keep the carbon skeleton whole and the last two break it. One double bond, four oxidation outcomes R H R H cold dil. KMnO4 / OH- cis-diol (syn) C=C kept RCO3H then H3O+ trans-diol (anti) C=C kept O3 then Zn/H2O aldehydes + ketones C=C cut hot conc. KMnO4 acids, ketones, CO2 C=C cut Top row keeps the carbon skeleton whole. Bottom row cuts it in half. Left column is syn or reductive; right column is anti or fully oxidative.
Figure 15: One double bond, four outcomes. Match the reagent to the product you need, and check whether the skeleton survives.
Summary table of syn and anti stereochemistry in alkene reactions Reactions that go through a bridged intermediate such as a halonium ion or a protonated epoxide are anti, because the nucleophile must attack the opposite face. Reactions that go through a cyclic ester or a metal surface or a four centre transition state are syn, because both new bonds are made on the same face. Addition of a hydrogen halide in the dark goes through an open planar carbocation and gives a mixture. The one table that settles every stereochemistry question Reaction Intermediate Faces Result X2 addition bridged halonium opposite ANTI X2 / H2O bridged halonium opposite ANTI epoxide + H3O+ protonated epoxide opposite ANTI cold KMnO4 cyclic manganate ester same SYN OsO4 cyclic osmate ester same SYN H2 / Pt adsorbed on metal same SYN BH3 then H2O2 four-centre TS same SYN HX in the dark open carbocation either MIXED Two words decide everything: bridged means anti, cyclic ester means syn.
Figure 16: Bridged intermediate means anti; cyclic ester or surface means syn; open carbocation means a mixture.

Solved Examples

Solved Example 1
Explain why bromine adds to cis-but-2-ene to give a racemic mixture but to trans-but-2-ene to give a meso compound.
Solution

Both reactions go through a bromonium ion, and the bromide must attack from the face opposite the bridging bromine, so the addition is anti in both cases.

Starting from the cis alkene, anti delivery puts the two methyl groups in a relationship that generates a pair of enantiomers, the racemate.

Starting from the trans alkene, the same anti delivery generates a molecule with an internal mirror plane, the meso compound, which is optically inactive.

Since different stereoisomers give different products, the reaction is stereospecific.

Solved Example 2
An alkene C8H16 on ozonolysis followed by oxidative work-up gives a single carboxylic acid, butanoic acid. Identify the alkene.
Solution

Degree of unsaturation , which is the one C=C and no ring.

Only one acid is obtained, so the two fragments must be identical and the double bond must sit exactly at the centre of the chain.

Butanoic acid is CH3CH2CH2COOH. Delete the oxygen from each carboxyl carbon and join the two carbons with a double bond:

CH3CH2CH2-CH=CH-CH2CH2CH3

Answer: oct-4-ene (cis or trans; ozonolysis cannot tell them apart). The formula checks out as C8H16.

Solved Example 3
A hydrocarbon C16H26 on ozonolysis followed by hydrolysis gives only CH3(CH2)4CO2H and succinic acid. What is the hydrocarbon?
Solution

Degree of unsaturation .

The products are acids, so oxidative work-up was used and each cleaved carbon carried a hydrogen or was part of a triple bond. Succinic acid, HOOC-CH2CH2-COOH, is a two-ended fragment, so it must have sat between two unsaturated linkages.

Assembling: CH3(CH2)4-CC-CH2-CH2-CC-(CH2)4CH3.

Two triple bonds account for all four degrees of unsaturation, and the formula checks out as C16H26.

Solved Example 4
Give the product when cyclohexene is treated with (a) cold dilute alkaline KMnO4 and (b) perbenzoic acid followed by dilute H2SO4.
Solution

(a) Permanganate goes through the cyclic manganate ester, so both OH groups arrive on the same face: cis-cyclohexane-1,2-diol. On a ring, cis and meso are the same molecule here, so it is optically inactive.

(b) The peroxyacid gives cyclohexene oxide, and acid then opens that epoxide by backside attack: trans-cyclohexane-1,2-diol, obtained as a racemic pair.

Same substrate, same element added, opposite stereochemistry. The intermediate decides, not the reagent's formula.

Solved Example 5
An alkene on ozonolysis with reductive work-up gives only propanone, (CH3)2CO. Identify the alkene and state what hot KMnO4 would have given instead.
Solution

One fragment only, so the alkene is symmetrical. Removing the oxygen from propanone leaves (CH3)2C, and joining two such units gives (CH3)2C=C(CH3)2, 2,3-dimethylbut-2-ene.

With hot KMnO4 the answer is the same, because each alkene carbon carries no hydrogen. A carbon with no hydrogen has nothing further to lose, so oxidation stops at the ketone and you again get two molecules of propanone.

Solved Example 6
Propene is treated with Br2 in water containing dissolved NaCl. Three organic products are formed. Explain.
Solution

The bromonium ion forms first, and then whichever nucleophile is present can open it.

Water gives the bromohydrin 1-bromopropan-2-ol, chloride gives 1-bromo-2-chloropropane, and the bromide released in the first step gives 1,2-dibromopropane.

In every case the nucleophile attacks the more substituted carbon and does so from the back face, so all three products are anti additions. This experiment is itself strong evidence for a bridged intermediate: a free carbocation would not give such clean regiochemistry.

Solved Example 7
Distinguish, using a single reagent, between but-1-ene and butane.
Solution

Add cold dilute alkaline KMnO4 (Baeyer's reagent) to each.

But-1-ene decolourises the purple solution and a brown precipitate of MnO2 appears, because the alkene is hydroxylated to butane-1,2-diol.

Butane has no bond and gives no reaction, so the purple colour persists.

Bromine water would work equally well, but remember that neither test is specific to alkenes: alkynes, phenols and aldehydes also respond.

Common Mistakes to Avoid

Watch out
  • Drawing an open carbocation for halogen addition. The intermediate is bridged, which is exactly why the addition is anti.
  • Assuming epoxidation plus hydrolysis is syn because epoxidation itself is syn. The ring opening inverts one carbon, so the overall result is the anti diol.
  • Swapping the meso and racemic answers. Anti addition to cis-but-2-ene gives the racemate; syn addition to cis-but-2-ene gives the meso compound.
  • Forgetting the work-up in ozonolysis. Zn/H2O keeps aldehydes as aldehydes; water alone lets the H2O2 oxidise them to acids.
  • Writing a carboxylic acid from a =CR2 carbon under hot KMnO4. With no hydrogen on that carbon the oxidation stops at the ketone.
  • Forgetting that a terminal =CH2 is lost as CO2 under hot KMnO4, so that carbon disappears from the organic product.
  • Calling a reaction stereospecific when it is only stereoselective. Stereospecific needs different reactant stereoisomers to give different product stereoisomers.
  • Using hot concentrated KMnO4 when the question asked for a diol. Read the temperature and concentration before writing anything.

Frequently Asked Questions

Why is halogen addition to an alkene anti and not syn?

The intermediate is a bridged halonium ion in which the halogen sits over one face of the two carbon unit. The nucleophile can only reach the other face, so the two new groups end up on opposite sides. There is no open planar carbocation that could be attacked from either face.

What does stereospecific mean, and how is it different from stereoselective?

A stereospecific reaction is one in which different stereoisomers of the reactant give different stereoisomers of the product, as in bromination of cis and trans but-2-ene. A stereoselective reaction gives one stereoisomer in excess from a single starting material, which is a weaker statement.

Why does bromination of cis-but-2-ene give the racemate but trans-but-2-ene give the meso compound?

Both are anti additions, so the two bromines must arrive on opposite faces. Applying that to the cis geometry produces a pair of enantiomers in equal amounts, while applying it to the trans geometry produces a molecule with an internal mirror plane, which is the meso compound.

Why is hydroxylation with cold KMnO4 syn?

Permanganate forms a five membered cyclic manganate ester in which the metal holds both oxygen atoms on the same face of the double bond. Both carbon oxygen bonds are made in that one cyclic step, so both hydroxyl groups must end up on the same face.

Epoxidation is syn, so why is the overall hydroxylation anti?

Epoxidation delivers the oxygen to one face, but the second step is an acid catalysed ring opening in which water attacks the opposite face of one ring carbon. That inversion puts the two hydroxyl groups on opposite faces, so the two step sequence gives the trans diol.

What is the role of zinc in ozonolysis?

Hydrolysis of the ozonide also produces hydrogen peroxide, which would oxidise any aldehyde on to a carboxylic acid. Zinc destroys that hydrogen peroxide as it forms, so the aldehydes survive. Dimethyl sulfide or hydrogen over palladium does the same job.

How do I work out an alkene structure from its ozonolysis products?

Write both carbonyl fragments, delete the oxygen from each carbonyl carbon, and join those two carbons with a double bond. Then check the molecular formula against the data. If only one fragment is obtained the alkene was symmetrical about the double bond.

What is the difference between cold dilute and hot concentrated KMnO4?

Cold dilute alkaline permanganate hydroxylates the double bond and gives a vicinal diol with the carbon skeleton intact. Hot concentrated acidic permanganate cleaves the double bond completely and gives carboxylic acids, ketones or carbon dioxide depending on how many hydrogens each alkene carbon carried.

Why does water and not bromide open the halonium ion in halohydrin formation?

Water is the solvent and is present in enormous excess compared with the bromide ion that was just released. On simple collision frequency water wins, so the product is the halohydrin rather than the vicinal dihalide.

Previous year questions on Mechanism of Some Important Reactions of Alkenes

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