Q.Write bond-line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Concept: Chain Isomerism — compounds with the same molecular formula but different carbon skeletons.

Step 1: Isopropyl alcohol

The common name is propan-2-ol. The OH group is on the middle carbon of a three-carbon chain.

Bond-line: a zigzag of three carbons; a line from the middle carbon to an OH group.

    OH
    |
   C-C-C

Step 2: 2,3-Dimethylbutanal

The parent chain is butanal (4-carbon aldehyde). Methyl groups are on carbons 2 and 3.

Bond-line: a four-carbon chain with the aldehyde at the end; a branch (CH₃) on the second and third carbons.

   CH₃  CH₃
    |    |
   C-C-C-C=O

Step 3: Heptan-4-one

A seven-carbon chain with a ketone at carbon 4.

Bond-line: a zigzag of seven carbons; a double bond to oxygen on the fourth carbon.

   C-C-C-C-C-C-C
        ||
        O

Let’s start with the core idea. Chain isomerism is all about the carbon backbone — the same atoms rearranged into different branching patterns. When you’re asked to write bond-line formulas, you’re essentially drawing the skeleton of the molecule, where each vertex (or bend) represents a carbon atom, and hydrogens are implied. The key is to first identify the parent chain and then place the functional groups correctly.

We’ll tackle each compound one by one.

1. Isopropyl alcohol

   The common name “isopropyl alcohol” tells us it’s a three-carbon alcohol with the –OH group on the middle carbon. The systematic name is propan-2-ol.

   • The parent chain is propane (3 carbons).
   • The –OH group is attached to carbon number 2.
   • In bond-line notation, we draw a zigzag of three carbons: a central carbon with a line to an –OH group.
   • The bond-line formula looks like a V shape (two carbons in a line, then a branch for the –OH).
   Tip

   Remember: In bond-line drawings, each end and bend is a carbon. For isopropyl alcohol, the central carbon is the one with the –OH, so the structure is a simple “V” with an O attached to the vertex.

   Bond-line formula:

     OH
     |
   C-C-C
   

   (In proper bond-line, it’s a zigzag: a line from left to right with a bend at the middle, and an –OH group sticking out from the bend.)

2. 2,3-Dimethylbutanal

   This is an aldehyde (suffix “-al”) with a 4-carbon parent chain (butanal) and two methyl groups at positions 2 and 3.

   • The parent chain is butane: C1–C2–C3–C4.
   • The aldehyde group (–CHO) is at C1 (always at the end for an aldehyde).
   • At C2, there’s a methyl group (–CH₃). At C3, another methyl group.
   • So the carbon skeleton has a total of 6 carbons: 4 in the main chain + 2 methyl branches.
   • In bond-line, draw a 4-carbon zigzag. At the second carbon (from the left), draw a line upward (or downward) for the methyl branch. At the third carbon, draw another line for the second methyl branch. The leftmost carbon is the aldehyde carbon, which we show as a double bond to oxygen (C=O) and a single bond to H (often written as –CHO).
   Watch out

   A common mistake is to forget that the aldehyde carbon is part of the main chain. Count carefully: the –CHO group is carbon #1, not a branch. So the main chain has 4 carbons, not 3.

   Bond-line formula:

     CH₃   CH₃
      |     |
   C-C-C-C=O
     (with H on the aldehyde carbon)
   

   In standard bond-line, it’s a zigzag of 4 carbons, with a methyl branch at the second and third carbons, and a double bond to O at the end.

3. Heptan-4-one

   This is a ketone (suffix “-one”) with a 7-carbon parent chain (heptane) and the carbonyl group (C=O) at carbon number 4.

   • The parent chain is heptane: C1–C2–C3–C4–C5–C6–C7.
   • The carbonyl group is at C4, so the structure is symmetric: three carbons on each side of the C=O.
   • In bond-line, draw a zigzag of 7 carbons. At the fourth carbon (the middle one), draw a double bond to an oxygen atom.
   • No branches — it’s a straight chain with a ketone group in the middle.
   Note

   For ketones, the carbonyl carbon is part of the main chain, not a branch. The numbering starts from the end nearest the carbonyl, but here it’s symmetric, so position 4 is unambiguous.

   Bond-line formula:

   C-C-C-C-C-C-C
         ||
         O
   

   In proper bond-line, it’s a straight zigzag of 7 carbons, with a double bond to O at the fourth carbon.