Q.Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen (four bond-line structures (a)–(d) are given).

Expanding Bond-Line Formulas

Concept: Bond-Line (Skeletal) Notation

In bond-line structures, carbon atoms sit at each vertex and line-end, hydrogens on carbon are implicit (added to satisfy tetravalency), and only heteroatoms and their hydrogens are shown explicitly.

Expansion Rules

1. Locate all carbons: each junction, bend, and terminus represents one carbon atom.
2. Count bonds: each carbon forms four bonds total.
3. Add hydrogens: $\text{H atoms on C} = 4 - (\text{bonds shown})$.
4. Preserve heteroatoms: oxygen, nitrogen, halogens remain as drawn with their explicit hydrogens.

Since the four structures (a)–(d) are not provided in your question, I'll demonstrate the method with a typical example:

Example: The bond-line \/\/O represents:

• Three carbons (two vertices + one end). …

Bond-line structures are a shorthand chemists use to avoid cluttering diagrams with the most common atoms. The convention is simple but powerful: every vertex (corner or end) represents a carbon atom, and hydrogen atoms attached to carbon are implied—never drawn. All other atoms (O, N, Cl, etc.) must be shown explicitly, along with their hydrogens.

To expand a bond-line formula, you reverse this process. The key principle is that carbon always forms exactly four bonds. Count the bonds already shown in the structure (C–C, C=C, C–O, etc.), and the remaining valences are filled by hydrogen atoms.

Since you've provided four structures labeled (a)–(d) but haven't included the actual images, I'll demonstrate the expansion process with common examples that cover the typical cases you'll encounter. You can apply this method to your specific structures.

The Expansion Process

1. Identify every carbon atom.

   Each line terminus and each vertex (bend or junction) in the bond-line formula represents one carbon. A straight chain with three bends has four carbons; a benzene ring has six.

2. Count the explicit bonds on each carbon.

   Single lines are single bonds, double lines are double bonds, triple lines are triple bonds. Include bonds to other carbons and to any heteroatoms (O, N, etc.) shown.

3. Add hydrogen atoms to complete the octet.

   Carbon needs four bonds. If a carbon already has two bonds showing, attach two hydrogens. If it has three bonds, attach one hydrogen. If it has four bonds (like the central carbon in a branched chain), attach zero hydrogens.

4. Write out the full structural formula.

   Draw each carbon as "C" and each hydrogen as "H," with lines representing all bonds.

Worked Examples

Example (a): A simple alkane chain

Bond-line:

A zigzag with four vertices (looks like: ⌐⌐⌐)

Expansion:

• Carbons: Four vertices = four carbons, $\ce{C1–C2–C3–C4}$.
• Bonds on each carbon:
  • $\ce{C1}$ (left end): one bond to $\ce{C2}$ → needs 3 H → $\ce{CH3}$
  • $\ce{C2}$: two bonds (to $\ce{C1}$ and $\ce{C3}$) → needs 2 H → $\ce{CH2}$
  • $\ce{C3}$: two bonds → $\ce{CH2}$
  • $\ce{C4}$ (right end): one bond → $\ce{CH3}$

Full structure:

This is butane, $\ce{C4H10}$.

Example (b): A branched alkane

Bond-line:

A T-shape: vertical line meeting a horizontal line at the centre

Expansion:

• Carbons: Three line segments meet at one vertex, plus two ends = four carbons total.
  • One carbon at the junction (centre)
  • Three carbons at the ends
• Bonds:
  • Each terminal carbon: one bond → $\ce{CH3}$
  • Central carbon: three bonds (to three other carbons) → needs 1 H → $\ce{CH}$

Full structure:

This is 2-methylpropane (isobutane), $\ce{C4H10}$.

Example (c): An alkene with a double bond

Bond-line:

A chain with a double line (=) between two vertices

Expansion:

Suppose we have three carbons with a double bond between the first two: $\ce{C=C–C}$

• Bonds:
  • $\ce{C1}$: one double bond to $\ce{C2}$ (counts as 2), so 2 bonds used → needs 2 H → $\ce{CH2}$
  • $\ce{C2}$: one double bond to $\ce{C1}$ (2 bonds) + one single bond to $\ce{C3}$ (1 bond) = 3 bonds → needs 1 H → $\ce{CH}$
  • $\ce{C3}$: one bond to $\ce{C2}$ → needs 3 H → $\ce{CH3}$

Full structure:

This is propene, $\ce{C3H6}$.

Example (d): A cyclic structure (cyclohexane)

Bond-line:

A hexagon

Expansion:

• Carbons: Six vertices = six carbons arranged in a ring. …