Alkenes and Alkynes (Senior 5 Chemistry): free lessons | Ganzaa

Alkenes and Alkynes

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Senior 5 · Chemistry

Alkenes and Alkynes

5 lessons

Lesson 1 of 5

Alkenes: Bonding, Names and E/Z Isomers

22 min

Seeing depends on a double bond

At the back of your eye, each light-sensitive rod cell holds a molecule called retinal. In the dark, retinal has one C=C bond in the cis arrangement, which puts a kink in its long chain. When a photon of light is absorbed, it supplies enough energy to break the π bond for an instant; the chain swings round and the bond re-forms in the trans arrangement. That change of shape is the first step of the nerve signal that lets you see. Ordinary C=C bonds stay locked in one arrangement at room temperature, and this lesson shows why that gives alkenes their stereoisomers.

Definition

Alkene: an unsaturated hydrocarbon containing at least one C=C double bond. Alkenes with one C=C have the general formula $\mathrm{C_nH_{2n}}$ . Example: propene, $\mathrm{CH_3CH{=}CH_2}$ .

Diene: a hydrocarbon with two C=C bonds. When the double bonds alternate with a single bond (C=C–C=C), the diene is conjugated. Example: buta-1,3-diene, $\mathrm{CH_2{=}CH{-}CH{=}CH_2}$ , used to make synthetic rubber.

Electrophile: a species that accepts a pair of electrons to form a new covalent bond; it is attracted to regions of high electron density. Examples: $\mathrm{H^+}$ , $\mathrm{Br^{\delta+}}$ in a polarised $\mathrm{Br_2}$ molecule.

The double bond

Each carbon of a C=C bond is $s p^{2}$ hybridised (Unit 1): the two carbons and the four atoms joined to them lie in one plane, with angles of about 120°. The π bond, a cloud of electron density above and below that plane, has two consequences:

Restricted rotation, which gives geometric (E/Z) isomers.
Reactivity: the π electrons are less tightly held than σ electrons and stick out from the molecule, so they attract electrophiles. This is why alkenes add reagents across the double bond (Lesson 3), while alkanes do not.

Naming alkenes

Choose the longest chain that contains the C=C bond, and change "-ane" to "-ene".
Number from the end nearer the double bond; the locant is the lower-numbered carbon of the C=C: $\mathrm{CH_3CH{=}CHCH_2CH_3}$ is pent-2-ene, not pent-3-ene.
Two double bonds: "-diene", adding "a" to the stem: buta-1,3-diene, penta-1,4-diene.
In a ring, C1 and C2 are the double-bond carbons, numbered to give substituents the lowest locants: 3-methylcyclohexene.
If the alkene has stereoisomers, put (E) or (Z) in front: (E)-pent-2-ene.

Figure 1.2: Skeletal formulae of some alkenes. In (Z)-pent-2-ene the two parts of the chain are on the same side of the C=C.

Definition

Cahn–Ingold–Prelog (CIP) priority rules, used for E/Z: on each carbon of the C=C, (1) the atom directly attached with the higher atomic number has priority; (2) if those atoms are the same, compare the sets of atoms attached to them, highest first, at the first point of difference. Example: –CH₂Cl (Cl, H, H) beats –CH₂CH₃ (C, H, H), which beats –CH₃ (H, H, H).

E/Z when the first atoms tie

Figure 1.1: Assigning E and Z when both groups on a carbon start with a carbon atom. Compare the next atoms out: –CH₂CH₃ has (C, H, H), –CH₃ has (H, H, H), so ethyl wins.

Physical properties

Alkenes are non-polar or only slightly polar, with London forces between molecules, so their physical properties are close to those of the alkanes with the same number of carbons: $\mathrm{C_2}$ to $\mathrm{C_4}$ are gases, they are insoluble in water and they float on it. Each alkene boils a few degrees lower than its alkane, because it has two fewer electrons.

Compound	Boiling point / °C	Compound	Boiling point / °C
ethane	−89	ethene	−104
propane	−42	propene	−48
butane	−1	but-1-ene	−6
(E)-1,2-dichloroethene	48	(Z)-1,2-dichloroethene	60

The last row shows a stereoisomer effect. In the Z isomer both polar C–Cl bonds point the same way, so the molecule has a permanent dipole and extra dipole–dipole attractions; in the E isomer the two dipoles cancel. Alkenes burn with a smokier flame than alkanes, because they have a higher ratio of carbon to hydrogen.

Worked example

Problem. Name: (a) $\mathrm{CH_2{=}C(CH_3)CH_2CH_3}$ ; (b) $\mathrm{CH_3CH{=}CHCH(CH_3)_2}$ ; (c) $\mathrm{CH_2{=}CHCH{=}CHCH_3}$ .

Solution. (a) Four-carbon chain containing C=C, numbered from the left: C=C at C1, methyl at C2: 2-methylbut-1-ene. (b) Five carbons; numbering from the left puts C=C at C2 (from the right it would be C3); the methyl is then on C4: 4-methylpent-2-ene, which has E and Z forms. (c) Five carbons, double bonds starting at C1 and C3: penta-1,3-diene.

Worked example

Problem. Which of these show E/Z isomerism? (a) hex-3-ene; (b) 2-methylpent-2-ene, $\mathrm{(CH_3)_2C{=}CHCH_2CH_3}$ ; (c) 1-bromo-1-chloroethene, $\mathrm{CH_2{=}CBrCl}$ ; (d) 1-bromo-2-chloroethene. For (d), name the isomer in which Br and Cl are on the same side.

Solution. (a) Yes: each C=C carbon carries H and $\mathrm{C_2H_5}$ . (b) No: C2 carries two $\mathrm{CH_3}$ groups. (c) No: C1 carries two H atoms. (d) Yes: each carbon carries a halogen and an H. The halogen is the higher priority on each carbon, so Br and Cl on the same side is (Z)-1-bromo-2-chloroethene.

Warning

Do not number from the end nearer a branch. In alkenes the double bond outranks alkyl branches for low numbers: $\mathrm{CH_2{=}CHCH_2CH(CH_3)_2}$ is 4-methylpent-1-ene, not 2-methylpent-4-ene.

Tip

For E/Z, work on ONE carbon at a time: pick its higher-priority group, then do the same on the other carbon. Same side = Z ("zame zide"); opposite = E.

CrashCourse: E/Z Alkenes, Electrophilic Addition, and Carbocations, Crash Course Organic Chemistry #14 (14 min). The first part shows how priority rules assign E and Z; the rest previews the addition reactions of Lesson 3.

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Practice

Name: (a) $\mathrm{CH_3CH_2CH{=}CH_2}$ ; (b) $\mathrm{(CH_3)_2C{=}C(CH_3)_2}$ ; (c) $\mathrm{CH_2{=}C(CH_3)CH{=}CH_2}$ .
Write the structural formula of (a) 3,3-dimethylbut-1-ene; (b) cyclopentene; (c) (E)-but-2-ene, as a displayed formula.
How many σ and π bonds are there in buta-1,3-diene?
Explain why but-1-ene has no E/Z isomers but but-2-ene has two.
Assign E or Z: in 1-chloro-2-methylbut-1-ene, $\mathrm{ClCH{=}C(CH_3)CH_2CH_3}$ , the Cl atom and the ethyl group are on opposite sides.
(Z)-1,2-dichloroethene boils at 60 °C and the E isomer at 48 °C. Explain the difference.

Show answersHide answers

(a) but-1-ene; (b) 2,3-dimethylbut-2-ene; (c) 2-methylbuta-1,3-diene (isoprene).
(a) $\mathrm{CH_2{=}CHC(CH_3)_3}$ ; (b) a five-membered ring with one C=C, $\mathrm{C_5H_8}$ ; (c) the two $\mathrm{CH_3}$ groups on opposite sides of the C=C, each carbon also carrying one H.
9 σ (6 C–H and 3 C–C) and 2 π.
In but-1-ene, C1 carries two H atoms, so swapping them makes no new compound. In but-2-ene each C=C carbon carries H and $\mathrm{CH_3}$ , so the two $\mathrm{CH_3}$ groups can be on the same side (Z) or opposite sides (E).
On C1: Cl > H. On C2: ethyl (C, H, H) > methyl (H, H, H). Cl and ethyl are the higher-priority groups and are on opposite sides: E.
In the Z isomer the two C–Cl dipoles point the same way, giving a polar molecule with dipole–dipole forces as well as London forces; in the E isomer the dipoles cancel.

Sources and credits

Written by Ganzaa.
Video: E/Z Alkenes, Electrophilic Addition, & Carbocations: Crash Course Organic Chemistry #14, CrashCourse, embedded from YouTube.
Boiling points are typical data-book values.
The text, diagrams, table, worked examples and Practice items were written and drawn by Ganzaa.

1Seeing depends on a double bond

2The double bond

3Naming alkenes

4E/Z when the first atoms tie

5Physical properties

Seeing depends on a double bond

The double bond

Naming alkenes

E/Z when the first atoms tie

Physical properties