Q.Write structures of the following compounds:

Concept: Structural Isomerism — drawing correct carbon skeletons and placing substituents per IUPAC naming.

Reasoning steps:

1. 2-Chloro-3-methylpentane: Pentane backbone (5 carbons). Chlorine on C2, methyl on C3.

   Structure: $\text{CH}_3\text{CHClCH}(\text{CH}_3)\text{CH}_2\text{CH}_3$

2. 1-Chloro-4-ethylcyclohexane: Cyclohexane ring. Chlorine at C1, ethyl at C4 (trans/cis not specified — draw one, e.g., trans).

   Structure: ethyl group on C4 opposite to Cl on C1.

3. 4-tert-Butyl-3-iodoheptane: Heptane chain. Iodine on C3, tert-butyl ($-\text{C}(\text{CH}_3)_3$) on C4.

   Structure: $\text{CH}_3\text{CH}_2\text{CHI}\,\text{CH}[\text{C}(\text{CH}_3)_3]\,\text{CH}_2\text{CH}_2\text{CH}_3$

4. 1,4-Dibromobut-2-ene: But-2-ene (4 carbons, double bond between C2 and C3). Bromines on C1 and C4.

   Structure: $\text{BrCH}_2\text{CH}=\text{CHCH}_2\text{Br}$ (geometric isomers possible — draw trans as common).

5. 1-Bromo-4-sec-butyl-2-methylbenzene: Benzene ring. Bromine at C1, sec-butyl ($-\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$) at C4, methyl at C2.

The two ring compounds — (ii) and (v) — are drawn below (the same verified figure used in the detailed answer): …

Why this approach works

IUPAC nomenclature is a systematic code. Every part of the name tells you exactly where to put atoms and bonds.

• The parent name (pentane, cyclohexane, heptane, butene, benzene) gives the longest carbon chain or the ring.
• Prefixes (chloro, methyl, ethyl, etc.) are substituents attached at the numbered positions.
• Suffixes (-ane, -ene, -ol, etc.) tell you about unsaturation or functional groups.
• Stereodescriptors (like tert., sec.) describe the branching of the substituent itself.

If you misread the parent chain or the locant numbers, the structure will be wrong. So we go step by step.

1. 2‑Chloro‑3‑methylpentane

Parent chain: pentane → 5‑carbon straight chain.

Substituents:

• Chloro at carbon 2
• Methyl at carbon 3

Number the pentane chain from the end that gives the lowest locants to the substituents. Here, both 2 and 3 are already the smallest possible.

   Cl
    |
CH3-CH-CH-CH2-CH3
      |
     CH3

Condensed: $\mathrm{CH_3CHClCH(CH_3)CH_2CH_3}$

2. 1‑Chloro‑4‑ethylcyclohexane

Parent: cyclohexane (6‑carbon ring).

Substituents:

• Chloro at position 1
• Ethyl at position 4

Number the ring so that the substituents get the lowest possible numbers. Here, 1 and 4 are the smallest set (ring structure shown in the diagram above).

The ethyl group is a straight chain of two carbons attached to the ring.

3. 4‑tert.‑Butyl‑3‑iodoheptane

Parent: heptane → 7‑carbon straight chain.

Substituents:

• Iodo at carbon 3
• tert.‑Butyl at carbon 4

The tert.‑butyl group is $\mathrm{-C(CH_3)_3}$ — a central carbon bonded to three methyl groups.

Number the heptane chain from the end that gives the lower locant to the first substituent (iodo at 3, tert.‑butyl at 4).

        I
        |
CH3-CH2-CH-CH-CH2-CH2-CH3
          |
          C(CH3)3

Condensed: $\mathrm{CH_3CH_2CHI\,CH[C(CH_3)_3]\,CH_2CH_2CH_3}$

4. 1,4‑Dibromobut‑2‑ene

Parent: but‑2‑ene → 4‑carbon chain with a double bond between carbons 2 and 3.

Substituents:

• Bromo at carbon 1
• Bromo at carbon 4

Number the chain so that the double bond gets the lowest possible number (it is already at 2). The bromines go at the ends.

Method: IUPAC Nomenclature to Structure Drawing

Method Name: Backbone-First Construction Method

This method works by first building the parent chain (backbone), then adding substituents at the correct positions.

Steps for Each Compound

(i) 2-Chloro-3-methylpentane

Step 1: Identify the parent chain — pentane (5-carbon straight chain).

Step 2: Number the carbons from the end nearest the first substituent.

Step 3: Add substituents at the correct positions:

• Chloro at C-2
• Methyl at C-3

Structure:

    Cl   CH₃
     |    |
CH₃—C—C—CH₂—CH₃
     |    |
     H    H

Or in condensed form:

$\text{CH}_3\text{—CH(Cl)—CH(CH}_3\text{)—CH}_2\text{—CH}_3$

(ii) 1-Chloro-4-ethylcyclohexane

Step 1: Parent chain is cyclohexane (6-carbon ring).

Step 2: Number the ring carbons — start at the carbon bearing the chloro group (C-1).

Step 3: Add substituents:

• Chloro at C-1
• Ethyl at C-4

Structure:

(iii) 4-tert. Butyl-3-iodoheptane

Step 1: Parent chain — heptane (7-carbon straight chain).

Step 2: Number from the end nearest the first substituent.

Step 3: Add substituents:

• Iodo at C-3
• tert-Butyl at C-4 (tert-butyl = $\text{—C(CH}_3\text{)}_3$)

Structure:

         I   H
         |   |
CH₃—CH₂—C—C—CH₂—CH₂—CH₃
         |   |
         H   C(CH₃)₃

Condensed:

$\text{CH}_3\text{—CH}_2\text{—CH(I)—CH[C(CH}_3\text{)}_3\text{]—CH}_2\text{—CH}_2\text{—CH}_3$

(iv) 1,4-Dibromobut-2-ene

Step 1: Parent chain — but-2-ene (4-carbon chain with a double bond between C-2 and C-3).

Step 2: Number from the end nearest the double bond. …

Here are the most common mistakes students make when drawing structures from IUPAC names, along with clear strategies to avoid them.

1. Misidentifying the Parent Chain (For Acyclic Compounds)

• The Mistake: For (i) 2-Chloro-3-methylpentane, students often draw a hexane chain because they count the carbon in the -methyl substituent as part of the main chain. They might draw CH3-CHCl-CH(CH3)-CH2-CH3 but then think the parent is hexane.
• Why it happens: Confusing a substituent (methyl) with a main chain carbon. The name ends in -pentane, so the longest continuous carbon chain has exactly 5 carbons.
• How to Avoid:
  • Rule: The parent chain is the longest continuous carbon chain. The suffix (e.g., -pentane, -heptane) tells you exactly how many carbons are in this chain.
  • Action: First, draw a straight chain of the parent length (5 carbons for pentane). Then, attach the substituents (chloro at C-2, methyl at C-3) to this pre-drawn chain.

2. Incorrect Numbering of the Parent Chain

• The Mistake: For (iii) 4-tert. Butyl-3-iodoheptane, students might number the heptane chain from the wrong end, placing the -iodo group at C-4 and the tert-butyl group at C-3.
• Why it happens: Forgetting the lowest locant rule. When numbering, you give the first point of difference the lowest number. Here, 3-iodo is lower than 4-iodo, so numbering must start from the end closest to the iodine.
• How to Avoid:
  • Rule: Number the parent chain so that the substituents get the lowest possible set of locants.
  • Action: Before drawing, write the parent chain (7 carbons). Try numbering from left to right, then right to left. Choose the numbering that gives the smallest number to the first substituent encountered. In this case, iodine at position 3 is correct.

3. Misinterpreting Complex Substituents (e.g., tert. butyl, sec. butyl)

• The Mistake: For (iii) 4-tert. Butyl-3-iodoheptane, students draw a straight chain of 4 carbons for the tert-butyl group. For (v) 1-Bromo-4-sec. butyl-2-methylbenzene, they draw a straight chain for sec. butyl.
• Why it happens: Not knowing the structure of common alkyl groups.
  • tert-butyl = -C(CH3)3 (a central carbon bonded to three methyl groups).
  • sec-butyl = -CH(CH3)CH2CH3 (a central carbon bonded to one methyl and one ethyl group).
• How to Avoid:
  • Memorize the common alkyl groups: methyl, ethyl, propyl, isopropyl, butyl, sec-butyl, tert-butyl, isobutyl.
  • Draw the substituent separately: Before attaching it to the main chain, draw the exact carbon skeleton of the substituent. For tert-butyl, draw a carbon with three methyl groups attached.

4. Forgetting to Show the Correct Position of Double Bonds (For Alkenes)

• The Mistake: For (iv) 1,4-Dibromobut-2-ene, students draw the double bond between C-1 and C-2, or they forget to show the E/Z isomerism (geometric isomerism) if required.
• Why it happens: The locant -2-ene means the double bond starts at carbon 2. Also, but-2-ene can exist as cis (Z) or trans (E) isomers.
• How to Avoid:
  • Rule: The locant for the double bond is the lower-numbered carbon of the double bond.
  • Action: Draw the 4-carbon chain. Place the double bond between C-2 and C-3. Then, place the bromine atoms on C-1 and C-4. If the question doesn't specify E or Z, you can draw either one, but be aware that both are possible.

5. Incorrectly Placing Substituents on a Ring (For Cyclic Compounds)

• The Mistake: For (ii) 1-Chloro-4-ethylcyclohexane, students place the ethyl group on C-1 and the chloro group on C-4, or they forget that the ring itself is the parent chain.
• Why it happens: Confusing the numbering of the ring. The substituents must be placed at the positions indicated by the locants.
• How to Avoid:
  • Rule: For monosubstituted rings, the substituent is at position 1. For disubstituted rings, number the ring so that the substituents get the lowest possible locants. …