Q.Draw the structures of all the eight structural isomers that have the molecular formula $C_5H_{11}Br$. Name each isomer according to IUPAC system and classify them as primary, secondary or tertiary bromide.

Showing the 12 most recent of 17 on this concept.

1. 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

   (a) $CH_3CHO$ (acetaldehyde/ethanal).

2. 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.

3. 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. 1-bromopentane
   2. 2-bromopentane
   3. 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.

4. 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.

5. 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.

6. CBSE 2025Set JZ1 markMCQ
   Q.The correct IUPAC name for $CH_2=CHCH_2NHCH_3$ is :

   (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.

7. 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:

   1. n-Butanol (butan-1-ol): CH3CH2CH2CH2OH (1°)
   2. Isobutyl alcohol (2-methylpropan-1-ol): (CH3)2CHCH2OH (1°)
   3. sec-Butyl alcohol (butan-2-ol): CH3CH2CH(OH)CH3 (2°)
   4. 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.

8. 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.

9. 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.

10. 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.

11. 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).

12. 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).