Write down the structure and IUPAC name for neo-pentylbromide.

NCERT Exemplar · Q59

Q. Write down the structure and IUPAC name for neo-pentylbromide.

CBSEShort· 2mImportance★★★★★

67% · 99/147 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

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 …

Start your 14-day free trial to unlock the full solution →

Neo-pentylbromide is a primary alkyl halide with a neopentane skeleton. Its structure is $(CH_3)_3C-CH_2Br$ and its IUPAC name is 1-bromo-2,2-dimethylpropane.

Why this approach works

The name "neo-pentylbromide" is a common (trivial) name that tells you two things: the carbon skeleton is "neo-pentyl" (meaning a five-carbon chain with a quaternary carbon at one end) and the functional group is bromine attached to the "pentyl" chain. To convert this to IUPAC, you need to identify the longest continuous carbon chain that includes the bromine, then number it to give the bromine the lowest possible locant.

The key insight: "neo" in organic nomenclature specifically refers to a structure where a carbon atom is bonded to four other carbons — a quaternary carbon. For a five-carbon system, neo-pentane is 2,2-dimethylpropane. The "bromide" tells us bromine replaces one hydrogen.

Step-by-step reasoning

Decode the trivial name.

"Neo-pentyl" means the carbon skeleton is derived from neopentane. Neopentane itself is $C(CH_3)_4$ — a central carbon bonded to four methyl groups. In neo-pentyl, one of those methyl groups loses a hydrogen to become a $CH_2$ group that attaches to the bromine. So the structure is: a central carbon (call it C2) bonded to three methyl groups and one $CH_2Br$ group.
Write the condensed structure.

The central carbon has four single bonds: three to $CH_3$ and one to $CH_2Br$ . This gives:

(CH_3)_3C-CH_2Br

Identify the longest carbon chain for IUPAC.

Starting at the $CH_2Br$ carbon, the chain runs to the central carbon, then to any ONE of its three methyl groups — that's 3 carbons in a row ( $C{-}C{-}C$ ), with the other two methyls left as branches on the middle carbon. Since the central carbon is bonded to three methyls (all equivalent), no path through it can be longer than 3 carbons: the longest chain containing the bromine is exactly propane.

Tip

A common mistake is to think the chain is 5 carbons long. In neopentane, the longest chain is only 3 carbons because the central carbon is quaternary. The "pentyl" in the trivial name refers to the total number of carbons (5), not the chain length.

The structure $(CH_3)_3C-CH_2Br$ :
- Carbon 1: $CH_2Br$ (the bromine-bearing carbon)
- Carbon 2: the central carbon $C$
- Carbon 3, 4, 5: the three methyl groups attached to carbon 2. …

Start your 14-day free trial to unlock the full solution →

Common Mistakes: Neo-pentylbromide Structure & IUPAC Name

The Concept at a Glance

Neo-pentylbromide is an alkyl halide. The key is understanding the "neo" prefix — it indicates a specific branched alkyl group: a quaternary carbon bearing three methyl groups, plus the CH₂ that carries the free bond.

Correct structure: $(CH_3)_3C{-}CH_2{-}Br$

Correct IUPAC name: 1-bromo-2,2-dimethylpropane

Mistake #1: Confusing "neo" with "tert" or "iso"

The error: Students often draw the "neo" group as a tert-butyl group (three methyls on a central carbon) attached directly to bromine — giving (CH₃)₃C—Br (tert-butyl bromide).

Why it happens: The prefixes iso, sec, tert, and neo all describe branching patterns, but they are not interchangeable. "Neo" specifically means the carbon chain has five carbons with a quaternary carbon at one end.

How to avoid: Memorise the pattern:

Iso: One methyl branch at the end → (CH₃)₂CH—
Neo: Two methyl branches at the end → (CH₃)₃C—CH₂—
Tert: Three methyls on a single carbon → (CH₃)₃C—

For neo-pentylbromide, the group is neo-pentyl = (CH₃)₃C—CH₂—, not tert-butyl.

Mistake #2: Incorrect IUPAC numbering

The error: Naming it as 2-bromo-2-methylbutane or 2-bromo-2,2-dimethylpropane (wrong locant for bromine).

Why it happens: Students forget that the longest continuous carbon chain must be identified first. In neo-pentylbromide, the longest chain is 3 carbons (propane), not 4 or 5.

How to avoid: Always follow the IUPAC priority:

Find the longest carbon chain — here it's propane (3 carbons).
Number from the end closest to the first substituent (bromine gets lowest number).
Bromine is on carbon-1 → 1-bromo.
Two methyl groups are on carbon-2 → 2,2-dimethyl.

So: 1-bromo-2,2-dimethylpropane — not 2-bromo-anything.

Mistake #3: Forgetting the "2,2-dimethyl" part

The error: Writing the name as simply 1-bromopentane or 1-bromo-2-methylbutane.

Why it happens: Students see five carbons total and assume it's a straight chain (pentane) or a simple branched chain.

How to avoid: Count the carbons in the neo group:

Neo-pentyl = (CH₃)₃C—CH₂— = 5 carbons total.
But the longest continuous chain is only 3 carbons (propane) because the quaternary carbon forces branching.
The two extra carbons are methyl substituents on carbon-2.

Always draw the structure first, then identify the parent chain — never guess from the total carbon count.

Mistake #4: Writing the structure with bromine on the wrong carbon

The error: Drawing Br—CH₂—C(CH₃)₃ as (CH₃)₃C—Br (bromine directly on the quaternary carbon). …

Start your 14-day free trial to unlock the full solution →

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) ✓ …
Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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°) …
Start your 14-day free trial to see the full solution →
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}.

…

Start your 14-day free trial to see the full solution →
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 (C4H1 …

Start your 14-day free trial to see the full solution →
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).

…

Start your 14-day free trial to see the full solution →
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), $(CH_3)_3CCl$ . …
Start your 14-day free trial to see the full solution →
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).

…

Start your 14-day free trial to see the full solution →

←Previous questionQ58 — Which of the following compounds will have the highest melting point a…→Next questionQ60 — A hydrocarbon of molecular mass 72 g mol⁻¹ gives a single monochloro d…

Q.Write down the structure and IUPAC name for neo-pentylbromide.

Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

The Precise Statement

A Concrete Example: C₄H₁₀

The Three Main Types

Why This Matters …

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

The Core Principle: Connectivity ≠ Composition

Why This Happens: The Valency Constraint

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

Why No Simple Formula Exists

Key Reasoning: The Branching Principle

The Functional Group Position Rule

Why this approach works

Step-by-step reasoning

Concept: IUPAC Nomenclature of Alkyl Halides (Common vs. IUPAC Names)

Method: Decoding Common Alkyl Group Names

Common Mistakes: Neo-pentylbromide Structure & IUPAC Name

The Concept at a Glance

Mistake #1: Confusing "neo" with "tert" or "iso"

Mistake #2: Incorrect IUPAC numbering

Mistake #3: Forgetting the "2,2-dimethyl" part

Mistake #4: Writing the structure with bromine on the wrong carbon

Q.Write down the structure and IUPAC name for neo-pentylbromide.

Why this approach works

Step-by-step reasoning

🎓Unlock everything free for 14 days

Unlock everything free for 14 days