Match each bromoalkene structure in Column I with its IUPAC name in Column II. Column I (the four structures exactly as the Exemplar prints them): (i) $\mathrm{CH_3{-}CH{=}CH{-}CH(Br){-}CH_3}$ ![Printed structure (i): 4-bromopent-2-ene — pent-2-ene chain with Br on C4](/figures/haloalkanes-and-haloarenes/4-bromopent-2-ene-as-printed-exemplar-q83.svg) (ii) $\mathrm{CH_3{-}CH{=}C(CH_3){-}CH(Br){-}CH_3}$ ![Printed structure (ii): 4-bromo-3-methylpent-2-ene — pent-2-ene chain with a 3-methyl branch and Br on C4](/figures/haloalkanes-and-haloarenes/4-bromo-3-methylpent-2-ene-as-printed-exemplar-q83.svg) (iii) $\mathrm{CH_3{-}CH{=}CH{-}CH(Br){-}CH_3}$ ![Printed structure (iii): identical to structure (i) — 4-bromopent-2-ene, exactly as the book prints it](/figures/haloalkanes-and-haloarenes/4-bromopent-2-ene-as-printed-exemplar-q83.svg) (iv) $\mathrm{BrCH_2{-}C(CH_3){=}CH{-}CH_2{-}CH_3}$ ![Printed structure (iv): 1-bromo-2-methylpent-2-ene — Br on the terminal C1, methyl on C2, double bond C2=C3](/figures/haloalkanes-and-haloarenes/1-bromo-2-methylpent-2-ene-as-printed-exemplar-q83.svg) Column II (names): (a) 4-Bromopent-2-ene (b) 4-Bromo-3-methylpent-2-ene (c) 1-Bromo-2-methylbut-2-ene (d) 1-Bromo-2-methylpent-2-ene > [!NOTE] > This question carries a printing error in the NCERT Exemplar itself: the book's > printed structure (iii) is an exact duplicate of structure (i), and no printed > structure corresponds to name (c) at all — yet the book's own answer key reads > (i)→(a), (ii)→(c), (iii)→(b), (iv)→(d), which does not match its own drawings. > We show the page exactly as printed and match what is actually drawn.

> [!TLDR] > Naming each printed structure honestly gives (i) → (a), (ii) → (b), (iii) → (a) again (the book prints (iii) identical to (i)), and (iv) → (d). Name (c) matches nothing that is printed — the Exemplar's own key for this question contradicts its own drawings, a known printing error. **How each printed structure is named:** 1. **Structure (i): $CH_3–CH{=}CH–CH(Br)–CH_3$.** Five carbons, double bond between C2 and C3 (numbering from the end that gives the alkene the lower locant), Br on C4 → **4-bromopent-2-ene = (a)**. 2. **Structure (ii): $CH_3–CH{=}C(CH_3)–CH(Br)–CH_3$.** The same pent-2-ene skeleton with a methyl branch on C3 and Br on C4 → **4-bromo-3-methylpent-2-ene = (b)**. 3. **Structure (iii): $CH_3–CH{=}CH–CH(Br)–CH_3$.** The book prints this identical to structure (i) — so as drawn it is again **4-bromopent-2-ene = (a)**. (The key's claim that (iii) → (b) would require a 3-methyl branch the printed drawing simply does not have.) 4. **Structure (iv): $BrCH_2–C(CH_3){=}CH–CH_2–CH_3$.** Number from the Br end: Br on C1, methyl on C2, double bond C2=C3, five-carbon chain → **1-bromo-2-methylpent-2-ene = (d)**. …

Match each bromoalkene structure in Column I with its IUPAC name in Column II. Column I (the four structures e

NCERT Exemplar · Q83

Q. Match each bromoalkene structure in Column I with its IUPAC name in Column II. Column I (the four structures exactly as the Exemplar prints them):

(i)

\mathrm{CH_3{-}CH{=}CH{-}CH(Br){-}CH_3}

Printed structure (i): 4-bromopent-2-ene — pent-2-ene chain with Br on C4 — Figure

(ii)

\mathrm{CH_3{-}CH{=}C(CH_3){-}CH(Br){-}CH_3}

Printed structure (ii): 4-bromo-3-methylpent-2-ene — pent-2-ene chain with a 3-methyl branch and Br on C4 — Figure

(iii)

\mathrm{CH_3{-}CH{=}CH{-}CH(Br){-}CH_3}

Printed structure (iii): identical to structure (i) — 4-bromopent-2-ene, exactly as the book prints it — Figure

(iv)

\mathrm{BrCH_2{-}C(CH_3){=}CH{-}CH_2{-}CH_3}

Printed structure (iv): 1-bromo-2-methylpent-2-ene — Br on the terminal C1, methyl on C2, double bond C2=C3 — Figure

Column II (names):

(a) 4-Bromopent-2-ene

(b) 4-Bromo-3-methylpent-2-ene

(d) 1-Bromo-2-methylpent-2-ene

[!NOTE]
This question carries a printing error in the NCERT Exemplar itself: the book's
printed structure

(iii) is an exact duplicate of structure (i), and no printed

structure corresponds to name

(i)→(a), (ii)→(c), (iii)→(b), (iv)→(d), which does not match its own drawings.
We show the page exactly as printed and match what is actually drawn.

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Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

IUPAC nomenclature is a systematic way to name a compound so that its name alone tells you its exact structure, with no ambiguity. Every organic name follows the same underlying recipe, whatever the functional group.

The Recipe

Identify the principal characteristic group. If the molecule has a functional group senior enough to be named as a suffix (carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine, and so on down the seniority order), that group decides the suffix and must be included in the parent chain. A halogen is never senior enough to be a suffix — it is always named as a prefix ("halo-"), whatever else is present.
Choose the parent chain. The parent is the longest continuous carbon chain that contains the principal characteristic group (if there is one). Among chains of the same length, the one with the most substituents wins.
Number the chain. Number from whichever end gives the LOWEST LOCANT to the principal characteristic group first. If there is no principal group (e.g. a simple haloalkane, or an alkene with a halogen substituent), lowest locant goes to the site of unsaturation (double/triple bond) first, then to substituents as a set.

Watch out

When two numbering directions give the SAME locant for the principal group/unsaturation (a genuine tie), the tie-break is the lowest locant SET for the substituents as a group — compare the two sets at their first point of difference. Only if the sets are themselves tied does the alphabetically-first substituent get the lower number.
Name and cite the substituents as prefixes, in alphabetical order (ignoring multiplying prefixes like di-/tri- but not ignoring structural prefixes like iso-/cyclo-), each with its own locant.
Assemble the name: locants + substituent prefixes (alphabetical) + parent chain name + suffix (if any).

Worked Example

$CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3$ : the longest chain is 5 carbons (pentane), no principal characteristic group (just a halogen substituent), so number for the lowest locant set. From the left: Cl at C2, methyl at C3 → set {2,3}. From the right: methyl at C3, Cl at C4 → set {3,4}. {2,3} is lower, so numbering from the left wins: 2-chloro-3-methylpentane. …

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

IUPAC nomenclature is not a single formula, but a system of rules designed to give every organic compound a unique, unambiguous name. The "why" behind these rules lies in clarity, consistency, and communication — ensuring that a chemist in Tokyo and one in Toronto draw the same structure from the same name.

1. The Core Principle: The Longest Carbon Chain

Rule: Identify the longest continuous chain of carbon atoms. This becomes the parent chain (e.g., pentane, hexane).

Why?

The longest chain represents the backbone of the molecule.
It gives the most stable, fundamental name — shorter chains would be branches, not the main structure.
Example: In a molecule with 5 carbons in a row and a 2-carbon branch, calling it "pentane" (not "ethane") tells you the core skeleton is 5 carbons long.

Key idea: The parent chain is the maximum continuous path — not necessarily the one that looks "straight" on paper.

2. Numbering: Lowest Locants (The "First Point of Difference" Rule)

Rule: Number the parent chain so that substituents get the smallest possible numbers. When there's a tie, compare the first point of difference.

Why?

This ensures reproducibility — two chemists will always number the same way.
It avoids ambiguity: "2-methylpentane" is unambiguous; "3-methylpentane" would be a different compound.
The first point of difference rule: If you have substituents at positions 2,4 and 3,5, choose 2,4 because 2 < 3 (the first number is smaller).

Example:

For a methyl group on carbon 2 vs. carbon 4 of a 5-carbon chain:
- 2-methylpentane (correct)
- 4-methylpentane (wrong — higher number)

3. Alphabetical Order of Substituents

Rule: List substituents in alphabetical order (ignoring prefixes like di-, tri-, sec-, tert- but not iso-).

Why?

Alphabetical order is a universal sorting convention — no need to remember priority based on size or complexity.
It makes names searchable and predictable.
Example: "3-ethyl-2-methylpentane" (e before m) — not "2-methyl-3-ethylpentane".

Exception: Prefixes like iso- and neo- are considered part of the name (e.g., isopropyl comes before methyl because "i" < "m").

4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes

Rule: Number the chain so that the double or triple bond gets the lowest possible number, even if it means giving a substituent a higher number.

Why?

The functional group (alkene/alkyne) is more important than alkyl substituents.
The bond position defines the compound's reactivity and geometry.
Example: In pent-2-ene (not pent-3-ene), the double bond is between carbons 2 and 3 — the lower number (2) is used.

Priority order:

Principal functional group (e.g., -OH, -COOH, C=C)
Multiple bonds
Substituents (alkyl, halo, etc.)

5. The "Suffix" and "Prefix" System …

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Naming each printed structure honestly gives (i) → (a), (ii) → (b), (iii) → (a) again (the book prints (iii) identical to (i)), and (iv) → (d). Name (c) matches nothing that is printed — the Exemplar's own key for this question contradicts its own drawings, a known printing error.

How each printed structure is named:

Structure (i): $CH_3–CH{=}CH–CH(Br)–CH_3$ . Five carbons, double bond between C2 and C3 (numbering from the end that gives the alkene the lower locant), Br on C4 → 4-bromopent-2-ene = (a).
Structure (ii): $CH_3–CH{=}C(CH_3)–CH(Br)–CH_3$ . The same pent-2-ene skeleton with a methyl branch on C3 and Br on C4 → 4-bromo-3-methylpent-2-ene = (b).
Structure (iii): $CH_3–CH{=}CH–CH(Br)–CH_3$ . The book prints this identical to structure (i) — so as drawn it is again 4-bromopent-2-ene = (a). (The key's claim that (iii) → (b) would require a 3-methyl branch the printed drawing simply does not have.)
Structure (iv): $BrCH_2–C(CH_3){=}CH–CH_2–CH_3$ . Number from the Br end: Br on C1, methyl on C2, double bond C2=C3, five-carbon chain → 1-bromo-2-methylpent-2-ene = (d). …

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Q.Match each bromoalkene structure in Column I with its IUPAC name in Column II. Column I (the four structures exactly as the Exemplar prints them):

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Q.Match each bromoalkene structure in Column I with its IUPAC name in Column II.
Column I (the four structures exactly as the Exemplar prints them):

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