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Problems · Example 9.7

Q.Write IUPAC names of the following compounds:

(i) (CH3)2CH–CH=CH–CH2–CH=CH–CH(CH3)–CH2–CH3
(ii) CH2=CH–CH=CH–CH=CH–CH=CH2 (drawn in the textbook as a zig-zag skeletal chain of eight carbons with four alternating double bonds)
(iii) CH2=C(CH2CH2CH3)2
(iv) CH3–CH(CH3)–CH=C(CH2CH3)–CH2–CH(CH3)–CH2–CH2–CH2–CH3
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✓ Free question

  1. 2,8-Dimethyl-3,6-decadiene;
  2. 1,3,5,7-Octatetraene;
  3. 2-n-Propylpent-1-ene;
  4. 4-Ethyl-2,6-dimethyl-dec-4-ene. In each case, the parent is the longest chain containing the double bond(s), numbered to give them the lowest locants.

For alkene nomenclature (§9.3.2) three rules do all the work: (1) the parent chain is the longest continuous chain that contains the double bond(s); (2) numbering starts from the end nearer a double bond (lowest locants for the unsaturation); (3) substituents are then cited alphabetically with their locants.

(i) (CH3)2CH−CH=CH−CH2−CH=CH−CH(CH3)−CH2−CH3(CH_3)_2CH-CH=CH-CH_2-CH=CH-CH(CH_3)-CH_2-CH_3

Tracing the longest chain through both double bonds — starting inside the isopropyl group and running to the terminal CHX3\ce{CH3} of the ethyl end — gives ten carbons. Numbering from the isopropyl end places the double bonds at C3–C4 and C6–C7 (locants 3, 6) and leaves one methyl on C2 and one on C8.

Name: 2,8-Dimethyl-3,6-decadiene.

(ii) CH2=CH−CH=CH−CH=CH−CH=CH2CH_2=CH-CH=CH-CH=CH-CH=CH_2

The textbook draws this one as a plain zig-zag skeletal chain: eight carbons with four alternating (conjugated) double bonds. Numbering from either end gives the double bonds locants 1, 3, 5, 7.

Name: 1,3,5,7-Octatetraene.

(iii) CH2=C(CH2CH2CH3)2CH_2=C(CH_2CH_2CH_3)_2

The doubly bonded C2 carries two identical n-propyl groups. The parent must contain the C=C, so take the CHX2=C\ce{CH2=C} unit plus one propyl group: a five-carbon pent-1-ene chain, with the second propyl group left as a substituent on C2.

Name: 2-n-Propylpent-1-ene.

(iv) CH3−CH(CH3)−CH=C(CH2CH3)−CH2−CH(CH3)−CH2−CH2−CH2−CH3CH_3-CH(CH_3)-CH=C(CH_2CH_3)-CH_2-CH(CH_3)-CH_2-CH_2-CH_2-CH_3

The longest chain through the double bond has ten carbons: from the methyl end, C2 carries a methyl, the double bond spans C3–C4, C4 also carries an ethyl group, and C6 carries the second methyl. Numbering from this end gives the double bond locant 4 (from the other end it would be 6).

Name: 4-Ethyl-2,6-dimethyl-dec-4-ene.

Watch out

In (iv) it is tempting to run the chain into the ethyl branch at C4 — but that path gives a shorter chain through the double bond. Always compare every path that contains the C=C and keep the longest.

✓Final answer

  1. 2,8-Dimethyl-3,6-decadiene;
  2. 1,3,5,7-Octatetraene;
  3. 2-n-Propylpent-1-ene;
  4. 4-Ethyl-2,6-dimethyl-dec-4-ene.

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