Write the structures of the isomers of alcohols with molecular formula $C_4H_{10}O$. Which one of these exhibits optical activity?

Write the structures of the isomers of alcohols with molecular formula C₄H₁₀O. Which one of these exhibits opt

NCERT Exemplar · Q35

C_4H_{10}O

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Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

IUPAC nomenclature is a systematic way to name a compound so that its name alone tells you its exact structure, with no ambiguity. Every organic name follows the same underlying recipe, whatever the functional group.

The Recipe

Identify the principal characteristic group. If the molecule has a functional group senior enough to be named as a suffix (carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine, and so on down the seniority order), that group decides the suffix and must be included in the parent chain. A halogen is never senior enough to be a suffix — it is always named as a prefix ("halo-"), whatever else is present.
Choose the parent chain. The parent is the longest continuous carbon chain that contains the principal characteristic group (if there is one). Among chains of the same length, the one with the most substituents wins.
Number the chain. Number from whichever end gives the LOWEST LOCANT to the principal characteristic group first. If there is no principal group (e.g. a simple haloalkane, or an alkene with a halogen substituent), lowest locant goes to the site of unsaturation (double/triple bond) first, then to substituents as a set.

Watch out

When two numbering directions give the SAME locant for the principal group/unsaturation (a genuine tie), the tie-break is the lowest locant SET for the substituents as a group — compare the two sets at their first point of difference. Only if the sets are themselves tied does the alphabetically-first substituent get the lower number.
Name and cite the substituents as prefixes, in alphabetical order (ignoring multiplying prefixes like di-/tri- but not ignoring structural prefixes like iso-/cyclo-), each with its own locant.
Assemble the name: locants + substituent prefixes (alphabetical) + parent chain name + suffix (if any).

Worked Example

$CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3$ : the longest chain is 5 carbons (pentane), no principal characteristic group (just a halogen substituent), so number for the lowest locant set. From the left: Cl at C2, methyl at C3 → set {2,3}. From the right: methyl at C3, Cl at C4 → set {3,4}. {2,3} is lower, so numbering from the left wins: 2-chloro-3-methylpentane. …

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

IUPAC nomenclature is not a single formula, but a system of rules designed to give every organic compound a unique, unambiguous name. The "why" behind these rules lies in clarity, consistency, and communication — ensuring that a chemist in Tokyo and one in Toronto draw the same structure from the same name.

1. The Core Principle: The Longest Carbon Chain

Rule: Identify the longest continuous chain of carbon atoms. This becomes the parent chain (e.g., pentane, hexane).

Why?

The longest chain represents the backbone of the molecule.
It gives the most stable, fundamental name — shorter chains would be branches, not the main structure.
Example: In a molecule with 5 carbons in a row and a 2-carbon branch, calling it "pentane" (not "ethane") tells you the core skeleton is 5 carbons long.

Key idea: The parent chain is the maximum continuous path — not necessarily the one that looks "straight" on paper.

2. Numbering: Lowest Locants (The "First Point of Difference" Rule)

Rule: Number the parent chain so that substituents get the smallest possible numbers. When there's a tie, compare the first point of difference.

Why?

This ensures reproducibility — two chemists will always number the same way.
It avoids ambiguity: "2-methylpentane" is unambiguous; "3-methylpentane" would be a different compound.
The first point of difference rule: If you have substituents at positions 2,4 and 3,5, choose 2,4 because 2 < 3 (the first number is smaller).

Example:

For a methyl group on carbon 2 vs. carbon 4 of a 5-carbon chain:
- 2-methylpentane (correct)
- 4-methylpentane (wrong — higher number)

3. Alphabetical Order of Substituents

Rule: List substituents in alphabetical order (ignoring prefixes like di-, tri-, sec-, tert- but not iso-).

Why?

Alphabetical order is a universal sorting convention — no need to remember priority based on size or complexity.
It makes names searchable and predictable.
Example: "3-ethyl-2-methylpentane" (e before m) — not "2-methyl-3-ethylpentane".

Exception: Prefixes like iso- and neo- are considered part of the name (e.g., isopropyl comes before methyl because "i" < "m").

4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes

Rule: Number the chain so that the double or triple bond gets the lowest possible number, even if it means giving a substituent a higher number.

Why?

The functional group (alkene/alkyne) is more important than alkyl substituents.
The bond position defines the compound's reactivity and geometry.
Example: In pent-2-ene (not pent-3-ene), the double bond is between carbons 2 and 3 — the lower number (2) is used.

Priority order:

Principal functional group (e.g., -OH, -COOH, C=C)
Multiple bonds
Substituents (alkyl, halo, etc.)

5. The "Suffix" and "Prefix" System …

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Alcohols with formula $\mathrm{C_4H_{10}O}$ are saturated monohydric alcohols — four structural isomers exist (two primary, one secondary, one tertiary). Only butan-2-ol has a chiral carbon and therefore exhibits optical activity.

The molecular formula $\mathrm{C_4H_{10}O}$ fits the general formula $\mathrm{C_nH_{2n+2}O}$ for a saturated alcohol or ether. Since we are asked for alcohols, the functional group is $-\mathrm{OH}$ attached to a carbon chain. The key to finding all isomers is to vary the carbon skeleton (straight vs. branched) and the position of the $-\mathrm{OH}$ group.

Optical activity arises when a molecule has no plane of symmetry — most commonly because it contains a carbon atom bonded to four different groups. That carbon is called a chiral centre (or stereocentre). For a molecule to be optically active, it must exist as non-superimposable mirror images (enantiomers). So among the isomers, we look for one with a chiral carbon.

Let’s build the isomers systematically.

Straight-chain (n-butane) skeleton: $\mathrm{C-C-C-C}$

Place the $-\mathrm{OH}$ at the end of the chain: $\mathrm{CH_3CH_2CH_2CH_2OH}$ — this is butan-1-ol (a primary alcohol).

Place the $-\mathrm{OH}$ on the second carbon: $\mathrm{CH_3CH_2CH(OH)CH_3}$ — this is butan-2-ol (a secondary alcohol).

No other positions are possible on a four-carbon straight chain (positions 3 and 4 are identical to 2 and 1 by symmetry).
Branched skeleton (isobutane): $\mathrm{C-C(C)-C}$

The carbon backbone is $\mathrm{CH_3CH(CH_3)CH_3}$ (2-methylpropane).

Place $-\mathrm{OH}$ on a terminal carbon: $\mathrm{(CH_3)_2CHCH_2OH}$ — this is 2-methylpropan-1-ol (a primary alcohol).

Place $-\mathrm{OH}$ on the central (tertiary) carbon: $\mathrm{(CH_3)_3COH}$ — this is 2-methylpropan-2-ol (a tertiary alcohol).

No other distinct positions exist.

So we have exactly four structural isomers:

IUPAC Name	Structure	Type
Butan-1-ol	$\mathrm{CH_3CH_2CH_2CH_2OH}$	Primary
Butan-2-ol	$\mathrm{CH_3CH_2CH(OH)CH_3}$	Secondary
2-Methylpropan-1-ol	$\mathrm{(CH_3)_2CHCH_2OH}$	Primary
2-Methylpropan-2-ol	$\mathrm{(CH_3)_3COH}$	Tertiary

Now, which one is optically active? Check each for a chiral carbon.

Butan-1-ol: Carbon-2 has two H atoms, carbon-3 has two H atoms — no carbon is bonded to four different groups. No chirality.
2-Methylpropan-1-ol: The carbon bearing $-\mathrm{OH}$ is $\mathrm{CH_2OH}$ (two H atoms). The central carbon is $\mathrm{CH}$ bonded to two identical methyl groups — not chiral. …

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Step 1: Understand the formula

The molecular formula is $C_4H_{10}O$ .

For an alcohol ( $R-OH$ ), the general formula is $C_nH_{2n+2}O$ .

Here $n=4$ , so $C_4H_{10}O$ fits perfectly — no unsaturation, only saturated alcohols (and ethers, but we focus on alcohols here).

Step 2: Draw all alcohol isomers

We need to arrange 4 carbons in a chain with an –OH group attached. The key: change the carbon skeleton and position of –OH.

1. Butan-1-ol (primary alcohol)

CH3-CH2-CH2-CH2-OH

Straight chain, –OH at terminal carbon.

2. Butan-2-ol (secondary alcohol)

CH3-CH2-CH(OH)-CH3

Straight chain, –OH at carbon #2.

3. 2-Methylpropan-1-ol (primary alcohol)

CH3-CH(CH3)-CH2-OH

Branched chain (isobutyl group), –OH at terminal carbon.

4. 2-Methylpropan-2-ol (tertiary alcohol)

CH3-C(OH)(CH3)-CH3

Branched chain, –OH at the central (tertiary) carbon.

Total alcohol isomers = 4

Step 3: Which one exhibits optical activity?

Optical activity requires a chiral carbon — a carbon with four different substituents.

Butan-1-ol: No chiral carbon (all carbons have at least two identical H’s or groups).
Butan-2-ol: Carbon #2 has –OH, –H, –CH $_3$ , –CH $_2$ CH $_3$ → four different groups → chiral → exhibits optical activity.
2-Methylpropan-1-ol: No chiral carbon.
2-Methylpropan-2-ol: Central carbon has two identical –CH $_3$ groups → not chiral.

Answer: Butan-2-ol shows optical activity.

Common Mistakes & How to Avoid Them

Mistake	Why it happens	How to avoid
Forgetting branched isomers	Students only draw straight-chain alcohols.	Always consider carbon skeletons: straight, branched once, branched twice (if possible). For $C_4$ , draw all possible carbon backbones first.
Counting ethers as alcohols	$C_4H_{10}O$ also includes ethers (e.g., diethyl ether).	The question specifically asks for alcohols — check functional group: –OH must be present.

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Q.Write the structures of the isomers of alcohols with molecular formula $C_4H_{10}O$ . Which one of these exhibits optical activity?

Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

The Recipe

Worked Example

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

1. The Core Principle: The Longest Carbon Chain

2. Numbering: Lowest Locants (The "First Point of Difference" Rule)

3. Alphabetical Order of Substituents

4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes

5. The "Suffix" and "Prefix" System …

Method: Systematic Isomer Enumeration + Optical Activity Check

Step 1 – Identify the functional group and degree of unsaturation

Step 2 – Draw all carbon skeletons for 4 carbons

Step 3 – Place the $-OH$ group on distinct carbons

Step 4 – Check for optical activity

Step 1: Understand the formula

Step 2: Draw all alcohol isomers

1. Butan-1-ol (primary alcohol)

2. Butan-2-ol (secondary alcohol)

3. 2-Methylpropan-1-ol (primary alcohol)

4. 2-Methylpropan-2-ol (tertiary alcohol)

Step 3: Which one exhibits optical activity?

Common Mistakes & How to Avoid Them

Q.Write the structures of the isomers of alcohols with molecular formula C4H10OC_4H_{10}OC4​H10​O. Which one of these exhibits optical activity?

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Q.Write the structures of the isomers of alcohols with molecular formula $C_4H_{10}O$ . Which one of these exhibits optical activity?

Unlock everything free for 14 days