Which of the following are secondary bromides? (Two or more than two options may be correct.) (i) mathrm(CH₃)₂

NCERT Exemplar · Q40

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

(i)

\mathrm{(CH_3)_2CHBr}

(ii)

\mathrm{(CH_3)_3C\,CH_2Br}

(iii)

\mathrm{CH_3CH(Br)CH_2CH_3}

(iv)

\mathrm{(CH_3)_2CBrCH_2CH_3}

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Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (acidified potassium dichromate): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones. (Not to be confused with Jones reagent, which is specifically CrO₃ dissolved in dilute aqueous H₂SO₄, often used in acetone — a related but distinct oxidant with the same general 1°→acid / 2°→ketone outcome.) …

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

Alcohol oxidation is a fundamental reaction in organic chemistry, and understanding why it proceeds as it does is crucial for Indian board exams (Class 12, JEE, NEET). Let's break it down step-by-step.

1. The Core Idea: Loss of Hydrogen

Oxidation in organic chemistry means loss of hydrogen (or gain of oxygen). For alcohols, this happens at the carbon bearing the –OH group.

Primary alcohol ( $R-CH_2OH$ ): Has two hydrogens on the carbon attached to –OH.
Secondary alcohol ( $R_2CHOH$ ): Has one hydrogen on that carbon.
Tertiary alcohol ( $R_3COH$ ): Has zero hydrogens on that carbon.

Key insight: The number of hydrogens on the carbon with –OH determines if and how far oxidation can go.

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

When a primary alcohol ( $R-CH_2OH$ ) is oxidized, the first product is an aldehyde ( $R-CHO$ ).

Why? The oxidizing agent (like $K_2Cr_2O_7$ / $H_2SO_4$ or $PCC$ ) removes two hydrogens:

One from the –OH group
One from the carbon atom

The carbon–oxygen bond becomes a double bond ( $C=O$ ), forming the aldehyde.

R-CH_2OH \xrightarrow{[O]} R-CHO + H_2O

But why stop here? The aldehyde still has one hydrogen on the carbonyl carbon. If a strong oxidant is present, it can remove that hydrogen too.

Step 2: Carboxylic acid formation

With excess strong oxidant (e.g., $K_2Cr_2O_7$ / $H_2SO_4$ , heat), the aldehyde is further oxidized to a carboxylic acid ( $R-COOH$ ).

R-CHO \xrightarrow{[O]} R-COOH

Why does this happen? The aldehyde's carbonyl carbon is electrophilic (partially positive). Water (from the reaction medium) adds to it, forming a gem-diol intermediate. The oxidant then removes two more hydrogens, giving the acid.

Exam tip: To stop at the aldehyde, use a mild oxidant like PCC (pyridinium chlorochromate) in anhydrous conditions — no water means no gem-diol formation.

3. Why Secondary Alcohols Give Ketones (and Stop)

A secondary alcohol ( $R_2CHOH$ ) has only one hydrogen on the carbon with –OH. Oxidation removes:

One hydrogen from –OH
One hydrogen from the carbon

This forms a ketone ( $R_2C=O$ ).

R_2CHOH \xrightarrow{[O]} R_2C=O + H_2O

Why does it stop here? The ketone has no hydrogen on the carbonyl carbon. Without that hydrogen, further oxidation (under normal conditions) is impossible — you'd need to break a $C-C$ bond, which requires much harsher conditions.

Key result: Secondary alcohols cannot be oxidized further than ketones under standard conditions.

4. Why Tertiary Alcohols Do NOT Oxidize

A tertiary alcohol ( $R_3COH$ ) has zero hydrogens on the carbon bearing –OH.

What happens if you try? The oxidant cannot remove any hydrogen from that carbon. The only possible reaction would be breaking a $C-C$ bond, which doesn't happen under normal oxidation conditions.

Result: Tertiary alcohols are resistant to oxidation under mild to moderate conditions. They require strong heating with powerful oxidants (like $K_2Cr_2O_7$ / $H_2SO_4$ , heat) to break carbon–carbon bonds — this is destructive oxidation, not useful for synthesis.

5. The "Why" in One Table

Alcohol Type	Hydrogens on C–OH	Product	Why?
Primary ( $1^\circ$ )	2	Aldehyde → Carboxylic acid	Two hydrogens available; aldehyde still has one more

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A secondary bromide has the bromine atom attached to a carbon that is itself bonded to exactly two other carbons (and one hydrogen). Checking each structure: (i) is secondary, (ii) is primary, (iii) is secondary, (iv) is tertiary. So the correct options are (i) and (iii).

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.

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

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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” …

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

Concept understanding — Alcohol Oxidation

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Alcohol Oxidation: Why the Reactions Work the Way They Do

1. The Core Idea: Loss of Hydrogen

2. Why Primary Alcohols Give Aldehydes (Then Carboxylic Acids)

Step 1: Aldehyde formation

Step 2: Carboxylic acid formation

3. Why Secondary Alcohols Give Ketones (and Stop)

4. Why Tertiary Alcohols Do NOT Oxidize

5. The "Why" in One Table

Method: Classification of Alkyl Halides by Carbon Substitution

Steps

Applying to each option

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

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

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

Common Mistakes in Identifying Secondary Bromides

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

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

✗ Mistake 3: Misreading the Bond Line or Condensed Formula

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

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

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