How many alcohols with molecular formula $C_4H_{10}O$ are chiral in nature? (i) 1 (ii) 2 (iii) 3 (iv) 4

> [!TLDR] > Chiral alcohols require a carbon atom bonded to four different groups. For $C_4H_{10}O$, only one structural isomer — butan-2-ol — has such a chiral centre, so the answer is **1**. ## Why chirality matters here Chirality in organic molecules arises when a carbon atom is *sp3-hybridised* and carries four *different* substituents. That carbon is called a chiral centre (or stereocentre). For an alcohol with formula $C_4H_{10}O$, we first need to list all possible structural isomers (different connectivity of atoms), then check each for a carbon with four distinct groups. A common mistake is to count all alcohols that *look* like they might be chiral, forgetting that symmetry can destroy chirality. Let's be systematic. ## Step-by-step reasoning 1. **Draw all structural isomers of $C_4H_{10}O$ that are alcohols** The formula $C_4H_{10}O$ fits the general formula $C_nH_{2n+2}O$, so it's a saturated alcohol (no rings, no double bonds). The –OH group can be placed on different carbons of the butane chain. There are four structural isomers: - Butan-1-ol: $CH_3CH_2CH_2CH_2OH$ - Butan-2-ol: $CH_3CH(OH)CH_2CH_3$ - 2-Methylpropan-1-ol: $(CH_3)_2CHCH_2OH$ - 2-Methylpropan-2-ol: $(CH_3)_3COH$ > [!TIP] > Always check for branching — the '2-methyl' isomers come from the isobutane skeleton. 2. **Identify chiral centres in each isomer** A chiral centre must be an *sp3 carbon* with four *different* groups attached. Let's examine each: - **Butan-1-ol**: The carbon bearing –OH is $C_1$. Its four groups are: –H, –OH, –H, and –CH2CH2CH3. Two hydrogens are identical → not chiral. No other carbon has four different groups either. - **Butan-2-ol**: The carbon bearing –OH is $C_2$. Its groups are: –H, –OH, –CH3, and –CH2CH3. All four are different → **this is a chiral centre**. - **2-Methylpropan-1-ol**: The –OH is on $C_1$, which has two identical –H atoms. The central carbon ($C_2$) has two identical –CH3 groups. No chiral centre.

How many alcohols with molecular formula C₄H₁₀O are chiral in nature? (i) 1 (ii) 2 (iii) 3 (iv) 4

NCERT Exemplar · Q2

C_4H_{10}O

(i) 1

(ii) 2

(iii) 3

(iv) 4

CBSEMCQ· 1mImportance★★★★★

40% · 54/135 Questions

✓ Free question

Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

Imagine you have a box of identical Lego bricks — four red, ten blue, and six yellow. You build two different models: a car and a house. Both use exactly the same number of each colour brick, but the structures are completely different. That is the core idea of isomerism: same atoms, different arrangement.

In chemistry, molecules are not just a list of atoms. How those atoms are connected matters enormously. Two molecules can have the exact same molecular formula (same number of each atom) but be connected in different ways. Those are structural isomers (also called constitutional isomers).

The Precise Statement

Structural isomers are compounds that have the same molecular formula but different connectivity of atoms — that is, different structural formulas.

The key word is connectivity. Which atom is bonded to which? If you change that, you get a different substance with different physical and chemical properties.

A Concrete Example: C₄H₁₀

Take butane, C₄H₁₀. There are exactly two ways to connect four carbon atoms and ten hydrogen atoms:

n-Butane — a straight chain: C–C–C–C
Isobutane (2-methylpropane) — a branched chain: a central carbon bonded to three methyl groups

Both have formula C₄H₁₀. But n-butane boils at –0.5 °C, while isobutane boils at –11.7 °C. Same atoms, different connectivity → different substance.

Note

Structural isomers are not the same molecule. They are distinct compounds that happen to share a molecular formula. You cannot rotate or flip one to get the other — you must break and reform bonds.

The Three Main Types

Structural isomerism comes in three flavours:

Type	What changes	Example (C₃H₆O)
Chain isomerism	The carbon skeleton (straight vs. branched)	Butane vs. isobutane
Position isomerism	The location of a functional group or substituent	Propan-1-ol vs. propan-2-ol (OH on carbon 1 vs. carbon 2)
Functional group isomerism	The atoms are rearranged into a different functional group	Propanal (aldehyde) vs. propanone (ketone) — both C₃H₆O

Watch out

Do not confuse structural isomers with stereoisomers. Stereoisomers have the same connectivity but differ in spatial arrangement (like left and right hands). That is a completely different chapter. For now: structural isomers = different bond connections.

Why This Matters

Structural isomers can have wildly different properties. Ethanol (C₂H₆O) is a drinkable alcohol; its isomer dimethyl ether is a gas used as a refrigerant. Same atoms, but one is a liquid you can consume, the other is a gas that would kill you. That is why chemists care so much about connectivity — it determines everything.

Quick Check

Question: Are these structural isomers?

Molecule A: CH₃–CH₂–CH₂–CH₃

Molecule B: CH₃–CH(CH₃)–CH₃

Answer: Yes. Both are C₄H₁₀. A is n-butane (straight chain), B is isobutane (branched). Different connectivity → structural isomers.

Important

The molecular formula must be identical. If the formulas differ, they are not isomers at all — just different compounds.

Structural isomerism is introduced in the NCERT/CBSE Class 11 Chemistry chapter on Organic Chemistry: Some Basic Principles and Techniques, and ‘structural isomerism examples class 11’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Correctly distinguishing structural isomers by connectivity, rather than just matching molecular formulas, is a skill tested throughout competitive organic chemistry exams.

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

Structural isomerism arises when molecules share the same molecular formula but differ in the connectivity of atoms. There is no single "formula" for structural isomerism — instead, the key is understanding why different arrangements are possible.

The Core Principle: Connectivity ≠ Composition

A molecular formula tells you how many of each atom are present, but not how they are joined. Structural isomers exist because atoms can form bonds in multiple distinct sequences while satisfying valency rules.

Why This Happens: The Valency Constraint

Each atom has a fixed bonding capacity (valency):

Carbon: 4 bonds
Hydrogen: 1 bond
Oxygen: 2 bonds
Nitrogen: 3 bonds

Example: For $C_4H_{10}$ , the formula satisfies $4(4) + 10(1) = 26$ valence electrons. But the carbon atoms can be arranged as:

A straight chain: $CH_3-CH_2-CH_2-CH_3$ (n-butane)
A branched chain: $CH_3-CH(CH_3)-CH_3$ (isobutane)

Both satisfy valency, but the connectivity differs.

The "Formula" for Counting Isomers: Why It's Not Simple

There is no closed-form formula to count structural isomers for a given molecular formula. The number grows rapidly and depends on:

Carbon skeleton branching possibilities
Functional group positions
Ring formation possibilities

Why No Simple Formula Exists

The problem is combinatorial — the number of possible trees (acyclic graphs) with $n$ carbon atoms grows exponentially. For example:

$C_4H_{10}$ : 2 structural isomers
$C_5H_{12}$ : 3 structural isomers
$C_6H_{14}$ : 5 structural isomers
$C_{10}H_{22}$ : 75 structural isomers

The pattern follows Cayley's formula for trees, but even that counts only carbon skeletons — not functional group positions.

Key Reasoning: The Branching Principle

The fundamental reason structural isomers exist is that carbon chains can branch. Consider $C_5H_{12}$ :

Straight chain: $C-C-C-C-C$ (n-pentane)
One branch: $C-C-C(C)-C$ (isopentane) — the branch can be at position 2 or 3, but these are identical due to symmetry
Two branches: $C-C(C)(C)-C$ (neopentane) — a quaternary carbon

Why position matters: The branch location changes the carbon's environment, altering physical and chemical properties.

The Functional Group Position Rule

For compounds with functional groups (e.g., alcohols $C_nH_{2n+2}O$ ), the position of the -OH group creates isomers:

$CH_3CH_2CH_2OH$ (propan-1-ol) — OH at end
$CH_3CH(OH)CH_3$ (propan-2-ol) — OH in middle

Why these are distinct: The OH group's position changes the carbon's hybridization environment and the molecule's polarity.

The Ring-Chain Isomerism Reason

For unsaturated formulas like $C_4H_8$ , the same formula can represent:

A straight alkene: $CH_2=CH-CH_2-CH_3$
A branched alkene: $CH_3-C(=CH_2)-CH_3$
A cycloalkane: cyclobutane (ring)

Why rings form: Carbon atoms can bond to form closed loops, reducing the number of hydrogen atoms needed. The formula $C_nH_{2n}$ can be either an alkene (one double bond) or a cycloalkane (one ring).

Summary: The Takeaway

Aspect	Why It Holds
Different connectivity	Atoms can bond in multiple sequences while satisfying valency
No simple counting formula	The number of possible trees grows combinatorially
Branching creates isomers	Carbon chains can have branches at different positions
Position matters	Functional groups at different locations change properties
Rings vs. chains	Same formula can represent open chains or closed rings

The key insight: Structural isomerism exists because molecular formula is a constraint, not a blueprint — it tells you the ingredients, not the recipe.

Chiral alcohols require a carbon atom bonded to four different groups. For $C_4H_{10}O$ , only one structural isomer — butan-2-ol — has such a chiral centre, so the answer is 1.

Why chirality matters here

Chirality in organic molecules arises when a carbon atom is sp3-hybridised and carries four different substituents. That carbon is called a chiral centre (or stereocentre). For an alcohol with formula $C_4H_{10}O$ , we first need to list all possible structural isomers (different connectivity of atoms), then check each for a carbon with four distinct groups.

A common mistake is to count all alcohols that look like they might be chiral, forgetting that symmetry can destroy chirality. Let's be systematic.

Step-by-step reasoning

Draw all structural isomers of $C_4H_{10}O$ that are alcohols

The formula $C_4H_{10}O$ fits the general formula $C_nH_{2n+2}O$ , so it's a saturated alcohol (no rings, no double bonds). The –OH group can be placed on different carbons of the butane chain. There are four structural isomers:
- Butan-1-ol: $CH_3CH_2CH_2CH_2OH$
- Butan-2-ol: $CH_3CH(OH)CH_2CH_3$
- 2-Methylpropan-1-ol: $(CH_3)_2CHCH_2OH$
- 2-Methylpropan-2-ol: $(CH_3)_3COH$
Tip

Always check for branching — the '2-methyl' isomers come from the isobutane skeleton.
Identify chiral centres in each isomer

A chiral centre must be an sp3 carbon with four different groups attached. Let's examine each:
- Butan-1-ol: The carbon bearing –OH is $C_1$ . Its four groups are: –H, –OH, –H, and –CH2CH2CH3. Two hydrogens are identical → not chiral. No other carbon has four different groups either.
- Butan-2-ol: The carbon bearing –OH is $C_2$ . Its groups are: –H, –OH, –CH3, and –CH2CH3. All four are different → this is a chiral centre.
- 2-Methylpropan-1-ol: The –OH is on $C_1$ , which has two identical –H atoms. The central carbon ( $C_2$ ) has two identical –CH3 groups. No chiral centre.
- 2-Methylpropan-2-ol: The –OH carbon ( $C_2$ ) has three identical –CH3 groups — clearly not chiral.
Watch out

Do not confuse 'having a carbon with four different groups' with 'the molecule has no plane of symmetry'. For butan-2-ol, the molecule itself is asymmetric, but the only requirement for chirality is the presence of at least one chiral centre. Here, butan-2-ol has exactly one.
Count the chiral alcohols

Only butan-2-ol qualifies. So the number of chiral alcohols among $C_4H_{10}O$ isomers is 1.

C_nH_{2n+2}O

✓Final answer

The correct option is (i) 1 .

Method: Identify Chiral Centres in Skeletal Isomers

Step 1 — Write all structural isomers of $C_4H_{10}O$ that are alcohols

Alcohols have the –OH group. For $C_4H_{10}O$ , the possible carbon skeletons are:

Butan-1-ol (straight chain, OH at end)

$\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{OH}$
Butan-2-ol (straight chain, OH at C2)

$\text{CH}_3\text{–CH(OH)–CH}_2\text{–CH}_3$
2-Methylpropan-1-ol (branched, OH at end)

$\text{(CH}_3)_2\text{CH–CH}_2\text{OH}$
2-Methylpropan-2-ol (branched, OH at central C)

$\text{(CH}_3)_3\text{C–OH}$

Step 2 — Check each for a chiral carbon (carbon with 4 different groups)

Alcohol	Structure	Chiral carbon?
Butan-1-ol	$\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{OH}$	No — no carbon has 4 different groups
Butan-2-ol	$\text{CH}_3\text{–CH(OH)–CH}_2\text{–CH}_3$	Yes — C2 has H, OH, CH $_3$ , CH $_2$ CH $_3$ (all different)
2-Methylpropan-1-ol	$\text{(CH}_3)_2\text{CH–CH}_2\text{OH}$	No — the central CH has two identical CH $_3$ groups
2-Methylpropan-2-ol	$\text{(CH}_3)_3\text{C–OH}$	No — central C has three identical CH $_3$ groups

Step 3 — Count the chiral alcohols

Only Butan-2-ol has a chiral centre.

Answer: $\boxed{1}$ chiral alcohol → Option (A)

Quick Tip for Exams

A carbon with four different substituents is chiral.
In $C_4H_{10}O$ alcohols, only butan-2-ol satisfies this — the others have symmetry or identical groups.

Common Mistakes & How to Avoid Them

Mistake 1: Counting all isomers instead of only chiral ones

Students often list all structural isomers of $C_4H_{10}O$ (there are 7 alcohols + 3 ethers) and then guess a number like 3 or 4.

How to avoid:

First, identify only the alcohol isomers (—OH group attached to carbon).
Then, check each for a chiral centre — a carbon with four different substituents.
For $C_4H_{10}O$ $C_{4} H_{10} O$ , the alcohols are:
- Butan-1-ol
- Butan-2-ol
- 2-Methylpropan-1-ol
- 2-Methylpropan-2-ol

Only butan-2-ol has a chiral carbon (the second carbon has —H, —OH, —CH $_3$ , and —C $_2$ H $_5$ ).

Answer: (A) 1

Mistake 2: Thinking 2-methylpropan-1-ol is chiral

Some students see a branched carbon and assume chirality.

How to avoid:

Draw the structure:
```
    CH3
     |
HO—CH2—CH—CH3
```
The carbon attached to —OH has two identical —CH $_3$ $_{3}$ groups? No — actually it has:
- 2 H atoms (on CH $_2$ )
- 1 —OH
- 1 —CH(CH $_3$ ) $_2$ group Two hydrogens → not four different groups → not chiral.

Mistake 3: Forgetting that a chiral carbon must have 4 different groups

Students sometimes count carbons with 3 different groups as chiral.

How to avoid:

Rule: A chiral centre must have all four substituents distinct.
Example: In butan-2-ol, the chiral carbon has:
- H
- OH
- CH $_3$
- CH $_2$ CH $_3$ All four are different → chiral.

Mistake 4: Including ethers in the count

The question asks for alcohols only, but some students count all $C_4H_{10}O$ isomers.

How to avoid:

Read carefully: "alcohols with molecular formula $C_4H_{10}O$ "
Ethers (like methoxypropane, ethoxyethane) are not alcohols — they lack the —OH group.

Quick Checklist to Avoid Errors

Step	Action
1	List only alcohol isomers (—OH group)
2	For each, identify carbons with 4 different substituents
3	Count only those carbons
4	Verify no two substituents are identical

Final answer: $\boxed{1}$ (Option A)

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set DZ1 markMCQ
$Q. Which of the following is an aldehyde?$
(a) $CH_3-\overset{O}{\overset{||}{C}}-H$
(b) $CH_3-CH_2-\overset{O}{\overset{||}{C}}-CH_3$
(c) $CH_3-CH_2-\overset{O}{\overset{||}{C}}-CH_2-CH_3$
(d) $CH_3-\overset{O}{\overset{||}{C}}-CH_3$
›Reveal solutionSolution
An aldehyde has the $-CHO$ group (a carbonyl carbon bonded to at least one H). Only option (a) has this; the rest are ketones.

The functional-group test: in an aldehyde the carbonyl carbon $(>\!C=O)$ carries at least one hydrogen ( $R-CHO$ ). In a ketone the carbonyl carbon is bonded to two carbon atoms ( $R-CO-R'$ ).

(a) $CH_3-CHO$ → carbonyl C bonded to one H → aldehyde (ethanal) ✓

(b) $CH_3CH_2-CO-CH_3$ → butan-2-one → ketone

(c) $CH_3CH_2-CO-CH_2CH_3$ → pentan-3-one → ketone

(d) $CH_3-CO-CH_3$ → propanone (acetone) → ketone

✓Final answer
$CH_3CHO$
CBSE 2026Set A1 markMCQ
$Q. Which of the following is Isopropyl amine ?$
(a) CH3-CH2-CH2-NH2
(b) CH3-NH-C2H5
(c) CH3-CH(NH2)-CH3
(d) CH3-CH(CH3)-CH2-NH2
›Reveal solutionSolution
Isopropyl amine = propan-2-amine, CH3-CH(NH2)-CH3.

The isopropyl group is (CH3)2CH-, so isopropylamine has the amino group attached to the central carbon of a propane chain: CH3-CH(NH2)-CH3 (propan-2-amine). For reference:

(a) CH3-CH2-CH2-NH2 is n-propylamine (propan-1-amine)

(b) CH3-NH-C2H5 is N-methylethylamine (a secondary amine)

(d) is isobutylamine

✓Final answer
$(c) CH3-CH(NH2)-CH3.$
CBSE 2026Set ANNUAL1 mark
$Q. How many structural isomers of C5H11Br are possible?$
›Reveal solutionSolution
$C_5H_{11}Br$ has 8 possible structural isomers, arising from bromine substitution at different positions on the three possible pentane carbon skeletons.

$C_5H_{11}Br$ is derived from pentane ( $C_5H_{12}$ ) by replacing one H with Br. Pentane itself has 3 carbon-skeleton isomers, and Br can go on different, non-equivalent carbon positions of each:

From n-pentane skeleton ( $CH_3CH_2CH_2CH_2CH_3$ ):

1-bromopentane

2-bromopentane

3-bromopentane (C4 and C5 positions are equivalent to C2 and C1 by the molecule's symmetry)

From isopentane / 2-methylbutane skeleton ( $(CH_3)_2CHCH_2CH_3$ ):

4. 1-bromo-2-methylbutane

5. 2-bromo-2-methylbutane

6. 3-bromo-2-methylbutane

7. 1-bromo-3-methylbutane (isoamyl bromide)

From neopentane / 2,2-dimethylpropane skeleton ( $(CH_3)_3CCH_3$ , i.e. $(CH_3)_4C$ ):

8. Neopentyl bromide (1-bromo-2,2-dimethylpropane)

Of these 8: 4 are primary bromides, 3 are secondary, and 1 is tertiary.

✓Final answer
$8 structural isomers.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Ethylidine dichloride is a:$
(a) vic-dihalide
(b) gem-dihalide
(c) allylic dihalide
(d) vinylic halide
›Reveal solutionSolution

Ethylidene dichloride, CH₃CHCl₂, has both chlorine atoms on the same carbon — a gem-dihalide.

Ethylidene dichloride has the structure CH₃–CHCl₂. Both chlorine atoms are attached to the same carbon atom (the second carbon). Dihalides in which both halogens sit on one carbon are called geminal (gem) dihalides; when they are on adjacent carbons they are called vicinal (vic) dihalides. Since both Cl atoms here are on one carbon, this is a gem-dihalide.

✓Final answer
$(b) gem-dihalide.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The correct IUPAC name of the organic compound CH₃—CH(C₂H₅)—CH₂Br is-$
(a)
(i) 1-Bromo-2-ethyl-2 methyl ethane
(b)
(ii) 1-Bromo-2-ethyl propane
(c)
(iii) 1-Bromo-2-methyl butane
(d)
(iv) 2-Methyl-1-bromo butane
›Reveal solutionSolution
$\text{CH}_3\text{–CH}(\text{C}_2\text{H}_5)\text{–CH}_2\text{Br}$ names as 1-bromo-2-methylbutane. Correct option: (iii).

Concept. IUPAC naming of a haloalkane: (1) pick the longest carbon chain that contains the carbon bearing the halogen; (2) number so the substituents get the lowest set of locants; (3) cite substituents alphabetically as prefixes.

Steps.

The structure is $\text{BrCH}_2\text{–CH(CH}_3\text{)–CH}_2\text{–CH}_3$ .

Longest chain through the C–Br carbon = 4 carbons (butane): $\text{C1(CH}_2\text{Br)} - \text{C2(CH)} - \text{C3(CH}_2\text{)} - \text{C4(CH}_3\text{)}$ .

Number from the Br end: bromo at C-1, methyl branch at C-2.

Name: 1-bromo-2-methylbutane.

Why other options are wrong. (i) and (ii) use "ethane/propane" as the parent — they fail to select the longest chain (butane). (iv) "2-methyl-1-bromobutane" mis-orders the substituents (alphabetical order requires bromo before methyl).

✓Final answer
$(iii) 1-Bromo-2-methylbutane. This IUPAC nomenclature is common to the NCERT/CBSE-aligned UBSE Class-12 Chemistry course.$
CBSE 2025Set JZ1 markMCQ
$CH_2=CHCH_2NHCH_3$
(a) Allylmethylamine
(b) 1-amine-4-pentene
(c) 4-aminopent-1-ene
(d) N-methylprop-2-ene-1-amine
›Reveal solutionSolution
Naming $CH_2=CHCH_2NHCH_3$ as a substituted amine gives N-methylprop-2-en-1-amine — option (d).

Concept. For a secondary amine, choose the longest carbon chain attached to nitrogen as the parent amine; the smaller alkyl group on N is named as an N-substituent.

Working:

Parent chain: $CH_2=CH-CH_2-$ is a 3-carbon chain with a double bond → prop-2-ene; the amino group is on C-1 → prop-2-en-1-amine.

The nitrogen also carries a $-CH_3$ , written as the prefix N-methyl.

Combining: N-methylprop-2-en-1-amine (the common name is allylmethylamine, but the IUPAC name is required).

✓Final answer
$(d) N-methylprop-2-ene-1-amine.$
CBSE 2025Set D1 markMCQ
$Q. The number of isomeric alcohols of molecular formula C4H10O is$
(a) 2
(b) 4
(c) 7
(d) 8
›Reveal solutionSolution
The four alcohols of formula C4H10O are 1-butanol, 2-methyl-1-propanol, 2-butanol and 2-methyl-2-propanol.

For the molecular formula C4H10O with an -OH group, the possible alcohols are:

n-Butanol (butan-1-ol): CH3CH2CH2CH2OH (1°)

Isobutyl alcohol (2-methylpropan-1-ol): (CH3)2CHCH2OH (1°)

sec-Butyl alcohol (butan-2-ol): CH3CH2CH(OH)CH3 (2°)

tert-Butyl alcohol (2-methylpropan-2-ol): (CH3)3COH (3°)

That gives 4 isomeric alcohols. (The remaining C4H10O isomers such as diethyl ether are ethers, not alcohols.)

✓Final answer
$(B) 4 isomeric alcohols of formula C4H10O.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of isomers in C2BrClFI is$
(a) 3
(b) 4
(c) 5
(d) 6
›Reveal solutionSolution

C2BrClFI, an ethylene bearing all four halogens (Br, Cl, F, I) with no hydrogens, has 6 possible isomers.

Since the formula has 2 carbons and exactly 4 substituents (Br, Cl, F, I) with no hydrogens, this corresponds to a fully-substituted ethylene, i.e. an alkene of type (X)(Y)C=C(Z)(W), where each carbon bears 2 of the 4 different halogens.

Step 1 — constitutional (positional) isomers: choose which 2 of the 4 halogens sit on one carbon (the other 2 automatically go on the other carbon). The number of distinct ways to split 4 different halogens into two unordered pairs is 3: {Br,Cl}|{F,I}, {Br,F}|{Cl,I}, {Br,I}|{Cl,F}.

Step 2 — geometric (cis/trans) isomers: for each of these 3 constitutional arrangements, since each carbon carries 2 different substituents, the molecule can exist as 2 geometric isomers (cis and trans, i.e. Z and E).

Total isomers = 3 (constitutional arrangements) x 2 (geometric isomers each) = 6.

✓Final answer
$(D) 6.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. An isomer of ethanol is$
(a) Methanol
(b) Dimethyl ether
(c) Diethyl ether
(d) Ethylene glycol
›Reveal solutionSolution

Dimethyl ether (CH3–O–CH3) is the classic functional isomer of ethanol (C2H6O).

Ethanol (CH3CH2OH) has molecular formula C2H6O. Dimethyl ether (CH3–O–CH3) has the identical molecular formula C2H6O but a completely different functional group (ether linkage instead of an –OH group) — this is an example of functional group isomerism. Methanol (CH4O) and diethyl ether (C4H10O) have different molecular formulas, so they are not isomers of ethanol; ethylene glycol (C2H6O2) also has a different formula.

✓Final answer
$(B) Dimethyl ether.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The total number of isomers for the compounds having molecular formula C4H10O is$
(a) 7
(b) 6
(c) 3
(d) 4
›Reveal solutionSolution

C4H10O has 7 total structural isomers: 4 alcohols + 3 ethers.

Alcohols (C4H9OH, 4 isomers): butan-1-ol, butan-2-ol, 2-methylpropan-1-ol (isobutanol), 2-methylpropan-2-ol (tert-butanol).

Ethers (C4H10O, 3 isomers): diethyl ether (C2H5–O–C2H5), methyl n-propyl ether (CH3–O–CH2CH2CH3), methyl isopropyl ether (CH3–O–CH(CH3)2).

Total = 4 + 3 = 7 constitutional isomers.

✓Final answer
$(A) 7.$
CBSE 2023Set ANNUAL1 mark
$Q. How many isomeric monochloro derivatives will be formed when 2-methylpropane is subjected to photochlorination ?$
›Reveal solutionSolution
2-Methylpropane has only two chemically distinct kinds of hydrogen (9 equivalent primary H's and 1 tertiary H), so free-radical photochlorination gives exactly two monochloro isomers.

2-Methylpropane (isobutane), $(CH_3)_3CH$ , has two types of hydrogen atoms:

9 primary hydrogens (three equivalent $CH_3$ groups), all chemically equivalent by symmetry — substitution at any of these gives the same product: 1-chloro-2-methylpropane, $(CH_3)_2CHCH_2Cl$ .

1 tertiary hydrogen (on the central carbon) — substitution here gives: 2-chloro-2-methylpropane (tert-butyl chloride), $(CH_3)_3CCl$ .

Since photochlorination is a free-radical substitution that can occur at any C–H bond, but chemically equivalent hydrogens always give the identical product, the number of distinct isomeric monochloro products depends only on the number of structurally different types of hydrogen, not the total count of hydrogens.

✓Final answer
$2 isomeric monochloro derivatives : 1-chloro-2-methylpropane (from the primary H's) and 2-chloro-2-methylpropane / tert-butyl chloride (from the tertiary H).$
CBSE 2023Set ANNUAL1 markMCQ
$Q. CH2=CH-CH2-CH3 and CH3-CH=CH-CH3 are:$
(a) Chain isomers
(b) Position isomers
(c) Functional isomers
(d) Metamers
›Reveal solutionSolution

CH2=CH-CH2-CH3 (1-butene) and CH3-CH=CH-CH3 (2-butene) share the same carbon skeleton and functional group but differ only in where the double bond sits — that is positional isomerism.

Both molecules have molecular formula $C_4H_8$ and the same unbranched 4-carbon chain, and both are alkenes (same functional group, so not functional isomers; same chain, so not chain isomers).

The only difference is the location of the C=C double bond: between C1–C2 in the first compound (1-butene) versus between C2–C3 in the second (2-butene). This difference in position of the double bond (or functional group) along an otherwise identical skeleton defines position isomers.

✓Final answer
$Position isomers (option b).$

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Q.How many alcohols with molecular formula C4H10OC_4H_{10}OC4​H10​O are chiral in nature?

Why chirality matters here

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Q.How many alcohols with molecular formula $C_4H_{10}O$ are chiral in nature?

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