Q.Which of the following are secondary bromides? (Two or more than two options may be correct.)

Concept: Alkyl Halide Classification — A secondary bromide has the bromine on a carbon that is bonded to exactly two other carbons (and one hydrogen).

Step 1: Identify the carbon bearing the Br in each compound.

Step 2: Count how many carbon atoms are directly attached to that carbon.

Step 3: If the count is 2, it is secondary; if 1 → primary; if 3 → tertiary; if 0 → methyl.

• (i) $(CH_3)_2CHBr$ — Br is on a carbon with two methyl groups → secondary.
• (ii) $(CH_3)_3C\,CH_2Br$ — Br is on a $CH_2$ group attached to a tertiary carbon → primary. …

The classification of alkyl halides (here, bromides) as primary, secondary, or tertiary depends entirely on the carbon that carries the halogen — not on the overall size or branching of the molecule. That carbon’s “degree” is simply the number of other carbon atoms directly attached to it.

• If that carbon is bonded to one other carbon → primary (1°)
• If bonded to two other carbons → secondary (2°)
• If bonded to three other carbons → tertiary (3°)

This is a pure structural definition, and it’s the first thing to check in any such problem. Let’s apply it to each option.

1. Option (i): $\mathrm{(CH_3)_2CHBr}$

   The carbon bearing Br is the middle one: it has two methyl groups ($\mathrm{CH_3}$) attached, plus one H. That’s two carbon neighbours. So it is a secondary bromide.

2. Option (ii): $\mathrm{(CH_3)_3C\,CH_2Br}$

   Here the Br is on a $\mathrm{CH_2}$ group. That carbon is attached to one other carbon (the bulky $\mathrm{C(CH_3)_3}$ group) and two hydrogens. Only one carbon neighbour → primary bromide.

3. Option (iii): $\mathrm{CH_3CH(Br)CH_2CH_3}$

   The Br is on the second carbon of a straight chain. That carbon is bonded to a $\mathrm{CH_3}$ on one side and a $\mathrm{CH_2CH_3}$ on the other — two carbon neighbours. So it is a secondary bromide.

4. Option (iv): $\mathrm{(CH_3)_2CBrCH_2CH_3}$ …

Method: Classification of Alkyl Halides by Carbon Substitution

This method identifies whether a bromide is primary (1°), secondary (2°), or tertiary (3°) based on the carbon atom bonded to the bromine.

Steps

1. Locate the carbon directly attached to bromine — this is the α-carbon.
2. Count how many carbon atoms are directly bonded to this α-carbon (ignore hydrogen atoms).
3. Classify:
   • 1 carbon attached → primary (1°) bromide
   • 2 carbons attached → secondary (2°) bromide
   • 3 carbons attached → tertiary (3°) bromide

Applying to each option

(i) $\mathrm{(CH_3)_2CHBr}$

• α-carbon: the CH (the one with Br)
• Attached carbons: two methyl groups ($\mathrm{CH_3}$)
• Count = 2 → secondary (2°) bromide

(ii) $\mathrm{(CH_3)_3C\,CH_2Br}$

• α-carbon: the $\mathrm{CH_2}$ (the one with Br)
• Attached carbons: only the $\mathrm{C(CH_3)_3}$ group
• Count = 1 → primary (1°) bromide

(iii) $\mathrm{CH_3CH(Br)CH_2CH_3}$

• α-carbon: the CH (the one with Br)
• Attached carbons: $\mathrm{CH_3}$ on left, $\mathrm{CH_2CH_3}$ on right …

Common Mistakes in Identifying Secondary Bromides

✗ Mistake 1: Confusing the Carbon Bearing –Br with the “Main” Carbon Chain

Students often look at the longest chain or the most substituted carbon in the molecule, instead of focusing only on the carbon directly attached to bromine.

Example: In option (ii) $\mathrm{(CH_3)_3C\,CH_2Br}$, the carbon with Br is $\mathrm{CH_2}$ — that’s a primary carbon (attached to only one other carbon). Yet many call it secondary because the molecule has a bulky tertiary butyl group.

✓ How to avoid:

• Circle the carbon that holds the Br atom.
• Count how many other carbon atoms are directly bonded to that circled carbon.
  • 1 bond → primary (1°)
  • 2 bonds → secondary (2°)
  • 3 bonds → tertiary (3°)

✗ Mistake 2: Forgetting that “Secondary” Refers to the Carbon, Not the Molecule

A “secondary bromide” means the carbon with Br is secondary, not that the molecule looks “branchy” or has a secondary alkyl group elsewhere.

Example: Option (iv) $\mathrm{(CH_3)_2CBrCH_2CH_3}$ — the carbon with Br is attached to three other carbons (two methyls + one ethyl). That’s tertiary, not secondary.

✓ How to avoid:

• Always ask: “How many carbons is the Br-carbon directly touching?”
• Ignore the rest of the molecule’s shape.

✗ Mistake 3: Misreading the Bond Line or Condensed Formula

In condensed formulas like $\mathrm{CH_3CH(Br)CH_2CH_3}$, students sometimes miss that the Br is in parentheses — meaning it’s a branch on the middle carbon.

Example: Option (iii) $\mathrm{CH_3CH(Br)CH_2CH_3}$ — the carbon with Br is the second carbon, bonded to:

• $\mathrm{CH_3}$ (left)
• $\mathrm{CH_2CH_3}$ (right)

That’s two carbon neighbours → secondary.

✓ How to avoid:

• Rewrite the condensed formula as a skeletal structure if needed.
• For $\mathrm{RCH(Br)R'}$, the carbon is always secondary (unless R or R' is H).

✗ Mistake 4: Selecting Only One Option When the Question Says “Two or More” …