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🧪 NEET Chemistry❓ Question

IUPAC naming of branched alkenes, which chain do I pick when the double bond isn't in the longest chain?

Rahul DubeyDropper Sample

12 Sept 2026 · 1 min read

Third attempt, so I'm rebuilding organic from zero with a pencil and a very old notebook. Nomenclature is the part I thought I knew and it turns out I don't.

Two structures from a practice sheet:

A: CH2=CH−CH(CH3)−CH2−CH2−CH3CH_2=CH-CH(CH_3)-CH_2-CH_2-CH_3 B: CH3−CH2−CH2−C(=CH2)−CH2−CH3CH_3-CH_2-CH_2-C(=CH_2)-CH_2-CH_3

For A I wrote 3-methylhex-1-ene. For B I wrote 2-propylbut-1-ene because I looked at the double-bonded carbon, saw a propyl on one side and an ethyl on the other, and used the shorter arm plus the =CH2 as the main chain, treating the longer arm as the branch. The key says 2-ethylpent-1-ene. Why isn't the longest chain "hexane"? Counting along the straight path from the propyl end to the ethyl end I get six carbons, but that path doesn't pass through =CH2.

So what is the actual rule? Does the main chain have to include the double bond even if a longer chain exists? And what about numbering when there's both a substituent and a double bond, which wins?

Also, for the classification part, I keep forgetting which general formula goes with which family. A quick way to check?

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7 answers & replies

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  • [+]Vivek Mishra12thSample· 1mo ago

    I think for A you should number from the side where the methyl is closest, so it will be 4-methylhex-5-ene? Substituent gets the lowest number, then the double bond. This is what I learned for alkanes.

    • [+]Tanvi Deshmukh12thSample· 1mo ago

      Vivek, for alkenes the double bond wins. The order is:

      1. Main chain = the longest chain that contains the double bond (that's the part Rahul is missing).
      2. Number it so the double bond gets the lowest locant, before thinking about substituents.
      3. Name substituents alphabetically with locants.

      So A is hex-1-ene with a methyl on C-3: 3-methylhex-1-ene. And "hex-5-ene" can never be right since you'd just number from the other end. For B, the longest chain through the C=C is =CH2, C, then the propyl arm: five carbons. That gives pent-1-ene, with the ethyl at C-2: 2-ethylpent-1-ene. The double bond must be in the main chain.

      • [+]Vivek Mishra12thSample· 1mo ago

        ohh. So for alkenes the double bond decides the numbering first. Got it, thanks didi.

      • [+]Rahul DubeyDropperSample· 1mo agoAuthor

        Got it, the double bond is the boss of the main chain, then of the numbering. Funny how I read this three times and never saw it. Thanks all.

  • [+]Harsh Gupta12thSample· 1mo ago

    Hexane is sad because it was the longest chain and didn't get picked 😔 jokes apart, tanvi is right for NCERT / NEET. Newer IUPAC recommendations changed this for some cases (they allow the longest chain to be the parent and the double bond becomes a "methylidene" substituent), but NEET uses the older NCERT rule: chain must include the double bond. Write 2-ethylpent-1-ene in the exam, don't be a hero.

  • [+]Ishita GoyalDropperSample· 1mo ago

    Quick formula check for families (n = number of carbons):

    • Alkane CnH2n+2C_nH_{2n+2}
    • Alkene or cycloalkane CnH2nC_nH_{2n}
    • Alkyne CnH2n−2C_nH_{2n-2}

    If a molecule has a formula with 2n, it has one double bond or one ring. For B, C7H14C_7H_{14}: 7 carbons, 14 hydrogens = 2n2n, matches an alkene. And one more NCERT-type tip: when a molecule has both a double and a triple bond (enyne), the chain must contain both, numbering gives the lowest locants to the multiple bonds together, and if there's a choice the double bond gets the lower number. CH2=CH−C≡CHCH_2=CH-C\equiv CH is but-1-en-3-yne.

  • [+]Neel Chatterjee12thSample· 1mo ago

    Tiebreakers for alkanes, in case the sheet moves on to them:

    • If two chains are the same length, pick the one with more substituents.
    • Alphabetical order ignores di, tri, tetra: "ethyl" comes before "dimethyl" because e is before m.
    • Lowest locant set at the first point of difference.

    Example: CH3−C(CH3)2−CH(C2H5)−CH2−CH3CH_3-C(CH_3)_2-CH(C_2H_5)-CH_2-CH_3 is 3-ethyl-2,2-dimethylpentane. Ethyl first even though it is numbered 3 and the methyls are 2,2.

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