ChemistryClass 11CBSE

How Chemists Work Out an Unknown Alkene by Cutting It in Half

Name alkenes and alkynes and their cis-trans isomers, prepare them from alkynes, alkyl halides, dihalides, alcohols and calcium carbide, predict addition products with Markovnikov's rule and the peroxide effect, and use ozonolysis and the acidity of terminal alkynes.

What makes alkenes and alkynes so much more reactive than alkanes?

Alkanes hold their electrons tightly in sigma bonds and react only under harsh conditions. Alkenes and alkynes have pi bonds — electron clouds exposed above and below the carbon chain — that attract electrophiles and let new atoms add on.

That reactivity makes them the starting point for plastics, alcohols and countless other compounds.

This part covers structure and isomerism, preparation, electrophilic addition, and oxidation, ozonolysis and the acidity of terminal alkynes.

How do you name alkenes and alkynes, identify cis-trans isomers and describe their double and triple bonds?

**Alkenes, C

_{n}

_{2 n}

, contain a C=C bond of one sigma and one pi bond between sp

^{2}

carbons; alkynes, C

_{n}

_{2 n - 2}

, contain a C

\equiv

C bond of one sigma and two pi bonds between sp carbons; the chain is numbered to give the multiple bond the lowest locant, and alkenes whose double-bonded carbons each carry two different groups show cis-trans isomerism.

Bonds.** C=C is

134

pm long with

120°

angles; C

\equiv

C is

120

pm long and linear.

Worked example 1 — naming.

- CH

_{3}

CH=CHCH

_{3}

— but-2-ene
- CH

_{2}

=C(CH

_{3}

)CH

_{3}

— 2-methylprop-1-ene
- CH

\equiv

CCH

_{2}

_{3}

— but-1-yne; CH

_{3}

\equiv

CCH

_{3}

— but-2-yne

**Worked example 2 — isomers of C

_{4}

_{8}

with a double bond. But-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene — four in all. But-1-ene has no cis-trans form because one carbon carries two hydrogens.

Cis versus trans. Cis-but-2-ene is polar and boils slightly higher; trans-but-2-ene is less polar and packs better, with a higher melting point.

An everyday example. The "trans fats" found in some fried and processed foods get their name from the trans arrangement around the double bonds in their chains.

The substance. The pi bond prevents rotation about C=C**, which is exactly why cis and trans forms can exist as separate compounds.

How are alkenes and alkynes prepared, including ethyne from calcium carbide?

**Alkenes are made by partial hydrogenation of alkynes, by removing HX from alkyl halides with alcoholic KOH, by removing X

_{2}

from vicinal dihalides with zinc, or by dehydrating alcohols with concentrated sulphuric acid; alkynes are made from calcium carbide and water, or by removing two HX from vicinal dihalides.

Alkenes:

- From alkynes**: H

_{2}

with Lindlar's catalyst gives a cis alkene; sodium in liquid ammonia gives a trans alkene
- Dehydrohalogenation: CH

_{3}

_{2}

Br + alcoholic KOH

\to

_{2}

=CH

_{2}

; 2-bromobutane gives mainly but-2-ene, the more substituted alkene (Saytzeff rule)
- Dehalogenation: CH

_{2}

BrCH

_{2}

Br + Zn

\to

_{2}

=CH

_{2}

+ ZnBr

_{2}

- Dehydration: C

_{2}

_{5}

OH with concentrated H

_{2}

_{4}

at about

443

\to

_{2}

=CH

_{2}

+ H

_{2}

O

Alkynes:

- From calcium carbide: CaC

_{2}

+ 2H

_{2}

\to

Ca(OH)

_{2}

+ C

_{2}

_{2}

- From vicinal dihalides: two successive eliminations with alcoholic KOH and then sodamide

Worked example — ethyne yield.

6.4

g of CaC

_{2}

(

64

g/mol) is

0.1

mol, giving

0.1

mol of ethyne:

0.1 \times 26 = 2.6 g 0.1 \times 22.4 = 2.24 L at STP

An everyday example. Old carbide lamps dripped water onto calcium carbide, and the ethyne released burned with a bright flame.

The substance. Elimination favours the more substituted alkene, because it is more stable.

How do you predict electrophilic addition products with Markovnikov's rule and the peroxide effect?

**Electrophiles add across C=C and C

\equiv

C bonds; when HX adds to an unsymmetrical alkene, the hydrogen goes to the carbon that already has more hydrogens (Markovnikov's rule) because this gives the more stable carbocation, but HBr in the presence of peroxide adds the opposite way through a free-radical mechanism.

Additions to alkenes:

- Hydrogen**: CH

_{2}

=CH

_{2}

+ H

_{2}

\to

_{3}

_{3}

over Ni
- Halogens: CH

_{2}

=CH

_{2}

+ Br

_{2}

\to

_{2}

BrCH

_{2}

Br — reddish-brown bromine water turns colourless, a test for unsaturation
- Water: propene with dilute acid gives propan-2-ol

Worked example 1 — Markovnikov addition. CH

_{3}

CH=CH

_{2}

+ HBr:

- H

^{+}

adds to the end carbon, forming the secondary carbocation CH

_{3}

^{+}

_{3}

, more stable than the primary CH

_{3}

_{2}

_{2}^{+}

- Br

^{-}

attacks it, giving 2-bromopropane

Worked example 2 — peroxide effect. With HBr and a peroxide, a Br

^{∙}

radical adds first to the end carbon to form the more stable secondary radical, and the product is 1-bromopropane.

Alkynes add twice: ethyne with water over HgSO

_{4}

and H

_{2}

_{4}

gives an unstable enol that becomes ethanal; propyne gives propanone.

An everyday example. Shaking a gas with bromine water in the school laboratory — if the colour vanishes, the gas contains double or triple bonds.

The substance. The peroxide effect works only with HBr — not with HCl or HI.

How do oxidation and ozonolysis work, and what makes terminal alkynes acidic?

Cold dilute alkaline permanganate adds two OH groups to an alkene, hot acidic permanganate splits it, and ozonolysis cuts the C=C bond into two carbonyl compounds whose structures reveal the parent alkene; terminal alkynes are weakly acidic because their sp carbon holds the C–H electrons tightly.

Oxidation.

- Baeyer's reagent (cold dilute alkaline KMnO

_{4}

): CH

_{2}

=CH

_{2}

\to

HOCH

_{2}

_{2}

OH, and the purple colour disappears
- **Acidic KMnO

_{4}

: but-2-ene splits into two molecules of ethanoic acid

Ozonolysis.** O

_{3}

followed by Zn and water replaces the C=C with two C=O groups.

Worked example 1 — forward. Propene gives ethanal and methanal; 2-methylbut-2-ene gives propanone and ethanal.

Worked example 2 — finding the alkene. Ozonolysis gives propanone and methanal. Remove the oxygens and join the two carbonyl carbons with a double bond: (CH

_{3}

)

_{2}

C=CH

_{2}

, 2-methylpropene. Each mole of alkene uses

1

mol, or

48

g, of ozone.

Acidity of terminal alkynes. The sp carbon has

50%

s-character, so:

- HC

\equiv

CH + Na

\to

\equiv

^{-}

^{+}

\frac{1}{2}

_{2}

- Terminal alkynes give a white precipitate with ammoniacal silver nitrate; but-2-yne, with no terminal hydrogen, does not

Polymerisation. Ethene polymerises to polythene; three ethyne molecules passed through a red-hot iron tube form benzene.

An everyday example. Polythene carry bags are made from ethene molecules joined into long chains.

The substance. Only hydrogens on a triple-bonded carbon are acidic enough to react with sodium — alkenes and alkanes do not.

Exam tip

What earns full marks on alkenes and alkynes?

For every addition to an unsymmetrical alkene, draw the carbocation or radical first and pick the more stable one before writing the product.

- Isomerism: cis-trans needs two different groups on each double-bonded carbon
- Preparation: Lindlar gives cis, Na in liquid NH

_{3}

gives trans; Saytzeff favours more substituted alkenes
- Markovnikov: H to the carbon with more H; peroxide reverses this for HBr only
- Ozonolysis: join the two carbonyl carbons to find the alkene
- Terminal alkynes: acidic; white precipitate with ammoniacal AgNO

_{3}

The trap. Applying the peroxide effect to HCl. Only HBr adds anti-Markovnikov in the presence of peroxides.

Did you know

How does a gas help mangoes ripen?

Ripening fruits such as mangoes and bananas give off tiny amounts of ethene, the simplest alkene. Ethene acts as a plant hormone, switching on the changes that soften the fruit, sweeten it and alter its colour.

That is why a raw mango kept in a closed paper bag with a ripe banana ripens faster — the banana's ethene builds up around it.

The same small molecule,

CH_{2}

CH_{2}

, is also the building block of polythene — one alkene doing two very different jobs.

Exam relevance

How are alkenes and alkynes tested in JEE Main and NEET?

Alkenes and alkynes are among the most frequently used topics in Hydrocarbons for both JEE Main and NEET, and JEE Advanced builds multi-step reaction sequences on them.

What gets asked. Markovnikov and anti-Markovnikov products, working out an alkene from its ozonolysis products, cis-trans isomer counts, Lindlar versus sodium-ammonia reduction, Saytzeff products of elimination, and distinguishing terminal from internal alkynes. These reactions return in haloalkanes, alcohols and aldehydes in Class 12.

Question types. Reaction-product multiple-choice questions, conversion problems and match-the-column lists.

The trap that costs marks. Applying the peroxide effect to HCl or HI.

Key takeaways

What must you be able to do from this part?

- Structure and isomers: C

_{4}

_{8}

gives but-1-ene, cis- and trans-but-2-ene and 2-methylpropene; the pi bond blocks rotation
- Preparation: Lindlar gives cis, Na in NH

_{3}

gives trans;

6.4

g CaC

_{2}

gives

2.6

g of ethyne
- Addition: propene + HBr gives 2-bromopropane; with peroxide, 1-bromopropane; ethyne + water gives ethanal
- Oxidation and alkynes: propanone + methanal on ozonolysis means 2-methylpropene; terminal alkynes react with sodium and ammoniacal AgNO

_{3}

An alkene C

_{5}

_{10}

gives propanone and ethanal on ozonolysis. Identify it, and predict its product with HBr with and without peroxide.