Q.Write IUPAC names of the following:

Concept: Structural Isomerism — these are all bromoalkenes; the key is to number the chain so that the double bond gets the lowest locant, then the bromine (or, if both directions tie on the double bond, whichever direction gives the substituents the lower locants).

Step 1: Identify the longest carbon chain that includes the double bond — check carefully whether a methyl group on the double-bond carbon is really part of the chain or just a branch.

Step 2: Number from the end nearer the double bond (if there is a tie, number from the end that gives the lower locant to the bromine).

Step 3: Assign locants to the double bond and the bromine substituent.

Step 4: Write the name as: (locant)-bromo-(locant)-methyl-(locant)-ene, as applicable.

1. $CH_3CH=CHCH(Br)CH_3$ Chain: 5 carbons, double bond between C2–C3, Br on C4. Name: 4-bromopent-2-ene
2. $CH_2=C(CH_3)CH(Br)CH_3$ Chain: 4 carbons, double bond between C1–C2, methyl on C2, Br on C3. Name: 3-bromo-2-methylbut-1-ene
3. $CH_3CH=C(CH_3)CH(Br)CH_3$ Chain: 5 carbons, double bond between C2–C3, methyl on C3, Br on C4. Name: 4-bromo-3-methylpent-2-ene
4. $CH_3CH=C(CH_3)CH_2Br$ …

(i) $CH_3CH=CHCH(Br)CH_3$

Numbering from the left gives the double bond locant 2; from the right it would be locant 3. Left wins outright (2 < 3, no tie). Br sits on C4: 4-bromopent-2-ene.

(ii) $CH_2=C(CH_3)CH(Br)CH_3$

The double bond is terminal, so it is always locant 1 from that end — no other numbering is possible. Methyl on C2, bromo on C3, alphabetical order (bromo before methyl): 3-bromo-2-methylbut-1-ene.

(iii) $CH_3CH=C(CH_3)CH(Br)CH_3$

Numbering from the left gives the double bond locant 2; from the right it would be locant 3. Left wins outright. Methyl on C3, bromo on C4: 4-bromo-3-methylpent-2-ene.

(iv) $CH_3CH=C(CH_3)CH_2Br$

Both numbering directions give the double bond the SAME locant (2) — a genuine tie, since this is only a 4-carbon chain. When the suffix locant ties, the win goes to whichever direction gives the LOWER locant SET to the substituents. Numbering from the $CH_2Br$ end: Br@C1, methyl@C2 — set $\{1,2\}$. Numbering from the $CH_3$ end (as in the naive "closer to the double bond" reading): methyl@C3, Br@C4 — set $\{3,4\}$. $\{1,2\}$ is lower, so Br takes C1: 1-bromo-2-methylbut-2-ene. …

Method: IUPAC Naming of Alkenes with Substituents

Method Name: Longest Chain + Lowest Locant + Priority Functional Group method.

Steps:

1. Identify the principal functional group — here, the alkene ($C=C$) is the priority group (suffix -ene). Bromine is a substituent (prefix bromo-).

2. Select the longest carbon chain that includes the double bond. Number the chain from the end nearest the double bond so that the $C=C$ gets the lowest possible locant. If both directions give the double bond the SAME locant (a genuine tie on a short chain), the direction is then decided by whichever gives the lower locant set to the substituents.

3. Number the substituents (bromo, methyl, etc.) according to their positions on the numbered chain.

4. Write the name as:

   • Locant(s) for substituents (alphabetical order) + substituent names
   • Locant of double bond + -ene
   • Parent alkane name (replace -ane with -ene)

Solutions:

(i) $CH_3CH=CHCH(Br)CH_3$

• Longest chain with $C=C$: 5 carbons → pentene
• Number from left (nearest double bond): $C=C$ at position 2
• Bromine at carbon 4
• Name: 4-bromopent-2-ene

(ii) $CH_2=C(CH_3)CH(Br)CH_3$

• Longest chain: 4 carbons → butene
• Double bond at position 1 (number from left)
• Methyl at carbon 2, bromine at carbon 3
• Name: 3-bromo-2-methylbut-1-ene

(iii) $CH_3CH=C(CH_3)CH(Br)CH_3$

• Longest chain: 5 carbons → pentene
• Double bond at position 2
• Methyl at carbon 3, bromine at carbon 4
• Name: 4-bromo-3-methylpent-2-ene

(iv) $CH_3CH=C(CH_3)CH_2Br$

• Longest chain: 4 carbons → butene (the methyl is a branch, not part of the main chain — this molecule is shorter than (iii), which has an extra $CH(Br)CH_3$ at the end)
• Both numbering directions give the double bond the SAME locant (2) — a genuine tie on this 4-carbon chain. Numbering from the $CH_2Br$ end gives the substituents the lower locant set (bromo@1, methyl@2) versus numbering from the other end (methyl@3, bromo@4), so the $CH_2Br$ end is C1.
• Bromine at carbon 1, methyl at carbon 2, double bond at position 2
• Name: 1-bromo-2-methylbut-2-ene

(v) $CH_3CH=CHCH_2Br$

• Longest chain: 4 carbons → butene …

This is a classic trap in Structural Isomerism and IUPAC Nomenclature. Students often lose marks here not because they don’t know the rules, but because they rush or misread the structure.

Below are the most common mistakes and how to avoid each, with the correct IUPAC names for all six compounds.

Mistake #1: Choosing the wrong parent chain (longest carbon chain)

Why it happens:

Students see a double bond or a bromine and instinctively pick a chain that includes them, but they miscount carbons when a methyl branch is present — usually by folding the branch into the main chain and counting one carbon too many.

Example from the list:

For (iv) $CH_3CH=C(CH_3)CH_2Br$ and (vi) $CH_2=C(CH_3)CH_2Br$, the methyl group sitting on the double-bond carbon is a branch, not a chain carbon — it is easy to mistakenly fold it into the main chain and count one carbon too many.

• (iv)'s longest chain is only 4 carbons (a butene), not 5 — the chain dead-ends at the $CH_2Br$ carbon, and the methyl is a substituent.
• (vi)'s longest chain is only 3 carbons (a propene), not 4, for the same reason.

How to avoid:

• Always number the carbons in the condensed formula before naming, and check whether a group drawn off to the side is really a continuation of the chain or a dead-end branch.
• The parent chain must be the longest continuous carbon chain that includes the double bond (for alkenes) — a one-carbon branch off the double-bond carbon almost never extends the chain.

Correct parent chain for (iv): but-2-ene (4 carbons). Correct parent chain for (vi): prop-1-ene (3 carbons).

Mistake #2: Wrong numbering direction (lowest locants for double bond)

Why it happens:

Students number from the end closest to the first substituent (bromine), forgetting that the double bond gets priority in numbering.

Example from the list:

For (i) $CH_3CH=CHCH(Br)CH_3$

• If you number from the Br side (right to left): double bond at C-3, Br at C-2.
• If you number from the other end (left to right): double bond at C-2, Br at C-4.
• The double bond locant is lower from the left (2 < 3), so the left-to-right numbering wins outright, even though it gives bromine the higher-looking number (4).

How to avoid:

• Rule: For alkenes, the double bond gets the lowest possible locant — even if it means a substituent gets a higher number.
• Always number from the end that gives the double bond the smaller number, and only fall back on the substituent locants as a tiebreaker if both directions give the double bond the SAME number.

Correct for (i):

4-bromopent-2-ene (double bond at C-2, bromine at C-4).

Mistake #3: Forgetting to specify the position of the double bond in the name

Why it happens:

Students write “pentene” without a number, or put the number in the wrong place.

How to avoid:

• The locant for the double bond is placed just before the “-ene” suffix (e.g., pent-2-ene).
• For IUPAC, the number is written as a hyphenated prefix: pent-2-ene or 2-pentene (both accepted, but pent-2-ene is more modern).

Mistake #4: Mixing up similar-looking structures

Why it happens: …