Alkenes and Alkynes: Quick Revision

A brief note on Alkenes and Alkynes. Read it before a test. Try each question at the end before you open its answer.

Shape and isomers

• Alkene: $\text{C}_n\text{H}_{2n}$, $\text{sp}^2$, flat. Alkyne: $\text{C}_n\text{H}_{2n-2}$, $\text{sp}$, linear.
• Cis-trans (E and Z) isomers need two different groups on each carbon of the double bond.
• Trans is more stable and melts higher. Cis is slightly polar and boils higher.
• More alkyl groups on the double bond: a more stable alkene, and less heat on hydrogenation.

Making them

• Alkynes with Lindlar's catalyst: cis alkene. Sodium in liquid ammonia: trans alkene.
• Alkyl halide with alcoholic KOH: elimination. Saytzeff: the more substituted alkene is the main product.
• Alcohol with concentrated $\text{H}_{2}\text{SO}_{4}$ at about 443 K: alkene. A lower temperature gives an ether.
• Vicinal dihalide with zinc: alkene. Calcium carbide with water: ethyne.

Addition to alkenes

• $\text{Br}_{2}$ in an inert solvent loses its colour: the test for a double bond.
• Order of reactivity: $\text{HI} > \text{HBr} > \text{HCl}$. Markovnikov: the more stable carbocation forms. A cation can shift to a more stable one.
• A peroxide with $\text{HBr}$ only: a radical chain, anti-Markovnikov. Propene gives 1-bromopropane.
• Dilute acid and water: an alcohol, by Markovnikov's rule.

Oxidation and alkynes

• Cold, dilute, alkaline $\text{KMnO}_{4}$: a diol. Hot, acidified: the bond is cut to acids and ketones.
• Ozonolysis with zinc and water: $\text{CH}_{2}\text{=}$ gives methanal, $\text{RCH}\text{=}$ an aldehyde, $\text{R}_{2}\text{C}\text{=}$ a ketone.
• A terminal alkyne: sodium, sodium amide, ammoniacal silver nitrate (white precipitate) and copper(I) chloride (red precipitate).
• Hydration of an alkyne: a ketone, or ethanal from ethyne. Three ethyne molecules make benzene.

A worked example

An alkene of formula $\text{C}_{6}\text{H}_{12}$ gives ethanal and butanal on ozonolysis with zinc and water. Find the alkene.

1. Remove the oxygen. Ethanal is $\text{CH}_{3}\text{CHO}$ and butanal is $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CHO}$. Take the oxygen atoms away from the two $\text{CHO}$ carbon atoms.
2. Join them. Put a double bond between those two carbon atoms: $\text{CH}_{3}\text{CH}\text{=}\text{CHCH}_{2}\text{CH}_{2}\text{CH}_{3}$, hex-2-ene.
3. Check. It has $2 + 4 = 6$ carbon atoms and the formula $\text{C}_{6}\text{H}_{12}$. Each carbon of its double bond has two different groups, so it exists as cis and trans.

Practice with new numbers

A program to try

The program finds the rings and pi bonds of four formulas, and the moles of hydrogen that an open-chain compound of each would take up.

Mistakes to avoid

• The peroxide effect with HCl or HI. It works only with HBr.
• Aqueous KOH for elimination. It gives substitution. The solvent must be alcohol.
• Hot permanganate for a diol. It breaks the bond.
• Cis-trans in but-1-ene. One carbon holds two hydrogen atoms.
• A ketone from every alkyne hydration. Ethyne gives ethanal.

Check yourself

1. Which alkene is the main product from 2-bromobutane with alcoholic KOH?

2. What do you see when ethene is bubbled into bromine in tetrachloromethane?

3. What does ozonolysis of 2-methylpropene with zinc and water give?

4. Which reagent gives a white precipitate with propyne but not with but-2-yne?

5. What does the hydration of ethyne give?

Sources

• OpenStax Chemistry 2e, chapter 20, section 20.1 "Hydrocarbons".
• NCERT Chemistry, Class 11, Part II, the chapter "Hydrocarbons".