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MHT-CET Chemistry · Alkenes

Alkenes: IUPAC Names, Isomers, Hybridisation and Stability

An alkene is named from the longest chain that contains its C=C, numbered to give the double bond the lowest locant; its stability grows with the number of alkyl groups on the double-bonded carbons.

All Alkenes notesMHT-CET Chemistry strategy

Why this matters

13 PYQs, none HARD. Five are IUPAC names of a drawn alkene, four are isomers, sp³ counts and the alkadiene, and four put substituted alkenes in stability order. Three cards.

Concept 1 of 3: Naming a Substituted Alkene

Find the longest chain that passes through BOTH carbons of the double bond — it may not be the longest chain overall. Number it from the end that gives the C=C the lower locant, then list the substituents alphabetically with their numbers.

Definition

  • Parent chain: the longest chain containing the C=C; suffix -ene with the locant of its first carbon.
  • Numbering: lowest locant to the double bond first, then to the substituents.
  • Order: substituents alphabetical — bromo before methyl, chloro before methyl.
  • A name ending in 'but-3-ene' or 'pent-3-ene' is a numbering error — the double bond should have got the lower number.

Name pattern

(locant-substituent)alpha + parent-n-ene,n as low as possible\text{(locant-substituent)}_{\text{alpha}}\ +\ \text{parent}\text{-}n\text{-ene},\quad n\ \text{as low as possible}

Worked example

Name CH₃–CH=CH–C(CH₃)(Br)–CH₃.
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q61Moderate

Example 1 · Alkenes · Nomenclature, Hybridization and Stability of Alkenes

What is the IUPAC name of following compound?

Taking the longest chain that skips the double bond

Options like '4-bromo-4-ethyl-3-methylbut-3-ene' come from a short chain through the C=C with an ethyl branch left over. The parent must contain the C=C AND be as long as possible — hex-3-ene here.
Drill 4 more on naming a substituted alkene

Concept 2 of 3: Isomers, Hybridisation and Alkadienes

Draw the carbon skeletons first, then place the double bond in every distinct position — that counts the structural isomers. Each double-bonded carbon is sp², every other carbon is sp³. Cis–trans isomers need two DIFFERENT groups on each end of the C=C.

Definition

  • C₅H₁₀ alkenes: 5 structural isomers — pent-1-ene, pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene.
  • Hybridisation: C=C carbons sp²; the rest sp³. 2-Methylbut-2-ene has 3 sp³ carbons.
  • Cis–trans: impossible if either C=C carbon carries two identical groups — but-1-ene (=CH₂) cannot; but-2-ene can.
  • Alkadiene: two C=C — isoprene (2-methylbuta-1,3-diene).

Worked example

How many sp³ carbons in one molecule of 2-methylbut-2-ene?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q64Easy

Example 2 · Alkenes · Nomenclature, Hybridization and Stability of Alkenes

What is the number of moles of sp3sp^{3} hybrid carbon atoms in one mole of 2-Methylbut-2-ene?
Drill 3 more on isomers, hybridisation and alkadienes

Concept 3 of 3: Stability of Substituted Alkenes

Every alkyl group attached to a double-bonded carbon donates electron density into the π bond (hyperconjugation), so more alkyl groups mean a more stable alkene. Count the R groups on the two sp² carbons and rank.

Definition

  • More alkyl groups on the C=C → more stable: R₂C=CR₂ > R₂C=CHR > R₂C=CH₂ ≈ RCH=CHR > RCH=CH₂ > CH₂=CH₂.
  • Reason: hyperconjugation (and the +I effect) of the alkyl groups.
  • This is also why elimination gives the more substituted alkene (Saytzeff).

Stability order

R2C=CR2>R2C=CHR>R2C=CH2>RCH=CH2>CH2=CH2\mathrm{R_2C{=}CR_2 > R_2C{=}CHR > R_2C{=}CH_2 > RCH{=}CH_2 > CH_2{=}CH_2}

Worked example

Order by stability: (I) (CH₃)₂C=C(CH₃)₂, (II) (CH₃)₂C=CH₂, (III) (CH₃)₂C=CHCH₃.
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q59Moderate

Example 3 · Alkenes · Nomenclature, Hybridization and Stability of Alkenes

Identify the correct stability order of following alkenes. (I) (CH3)2C=C(CH3)2(CH_3)_2C=C(CH_3)_2 (II) (CH3)2C=CH2(CH_3)_2C=CH_2 (III) (CH3)2C=CHCH3(CH_3)_2C=CHCH_3
Drill 3 more on stability of substituted alkenes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Naming a Substituted Alkene

    Name pattern

    (locant-substituent)alpha + parent-n-ene,n as low as possible\text{(locant-substituent)}_{\text{alpha}}\ +\ \text{parent}\text{-}n\text{-ene},\quad n\ \text{as low as possible}
  • Stability of Substituted Alkenes

    Stability order

    R2C=CR2>R2C=CHR>R2C=CH2>RCH=CH2>CH2=CH2\mathrm{R_2C{=}CR_2 > R_2C{=}CHR > R_2C{=}CH_2 > RCH{=}CH_2 > CH_2{=}CH_2}

Watch out for (1)

Test yourself on Alkenes

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Alkenes chapter testDrill all 13 questions