Prepare for the Punjab Board Class 10 Chemistry Examination 2027 with Chapter 21: Organic Chemistry – Class Test No. 6. This test covers textbook topics 21.4 to 21.6 and includes a full chapter test with exercise questions to help students revise important concepts and evaluate their exam preparation. Designed for students following the new Class 10 Chemistry syllabus issued by PECTAA Punjab, this test is part of our Punjab Board 2027 Model Papers, Full Unit Model Papers, and Chapter-Wise Test Series. Practice these Chemistry questions to strengthen your understanding of Organic Chemistry and prepare effectively for the 10th Class annual board examinations.
Chapter Wise Class Test
T-6
Chapter-21
Organic Chemistry
1.
The compounds having the same molecular formula but different structural formulae are called:
(a) Homologues(b) Isomerism(c) Allotropes(d) Isotopes
Answer: (b) Isomerism
2.
The functional group -OH (hydroxyl group) is characteristic of which family?
(a) Ethers(b) Alcohols(c) Aldehydes(d) Ketones
Answer: (b) Alcohols
3.
Which of the following functional groups contains Nitrogen?
(a) Ester group(b) Ether group(c) Amino group(d) Alcohol group
Answer: (c) Amino group (–NH2)
4.
The simplest member of saturated hydrocarbons is:
(a) Ethane(b) Propane(c) Methane(d) Butane
Answer: (c) Methane (CH4)
5.
Hydrocarbons containing at least one carbon-carbon double bond are called:
(a) Alkanes(b) Alkenes(c) Alkynes(d) Alcohols
Answer: (b) Alkenes
6.
The general formula of alcohol is:
(a) RCOOH(b) RCHO(c) ROH(d) RH
Answer: (c) ROH
7.
The IUPAC name of propyl alcohol is:
(a) Propan-2-ol(b) Propan-1-ol(c) Butan-1-ol(d) Butan-2-ol
Answer: (b) Propan-1-ol (CH3–CH2–CH2–OH)
8.
Formic acid was first isolated from:
(a) Honey bees(b) Red ants(c) Vinegar(d) Plants
Answer: (b) Red ants (Latin formica = ant)
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Chapter Wise Class Test
T-6
Chapter-21
Organic Chemistry
2. Write short answers to any six (6) questions.[6 × 2 = 12]
(i)
Why do Methane, Ethane, and Propane not show isomerism?
Because with only one, two or three carbon atoms there is only one possible way to arrange the carbon atoms (no branching is possible), so each has just one structure.
(ii)
How does the number of isomers change with the increase in carbon atoms?
The number of isomers increases rapidly as the number of carbon atoms increases. For alkanes: C4H10 has 2, C5H12 has 3, C6H14 has 5, C7H16 has 9 and C10H22 has 75 isomers.
(iii)
Define functional Group. Give example.
A functional group is an atom or group of atoms in an organic molecule that is responsible for its characteristic chemical properties. Example: hydroxyl group (–OH) in alcohols, carboxyl group (–COOH) in carboxylic acids.
(iv)
Why is the study of functional groups important in Organic Chemistry?
Compounds with the same functional group behave in a similar way, so studying functional groups lets us classify millions of compounds into families, predict their properties and reactions, and name them systematically.
(v)
What is the “Ester Linkage”?
The ester linkage is the functional group –COO– (a carbonyl carbon joined to an oxygen that is attached to an alkyl group). General formula of an ester: R–COO–R′. Example: ethyl ethanoate, CH3COOC2H5.
(vi)
How are common names of carboxylic acid usually derived?
Common names are derived from the natural source from which the acid was first obtained (usually from Latin or Greek names). Examples: formic acid (from red ants, formica), acetic acid (from vinegar, acetum), butyric acid (from butter, butyrum).
(vii)
Why was the IUPAC system of nomenclature introduced?
Common names do not follow any rule and cannot describe the structure of the many organic compounds. The IUPAC system gives each compound a systematic, unique name that shows its structure and is accepted worldwide.
(viii)
Write down the molecular formula of a carboxylic acid having three carbon atoms.
Propanoic acid: C3H6O2 (structural: CH3–CH2–COOH).
(ix)
What is common between the formula of aldehydic and ketonic functional groups?
Both contain the carbonyl group (>C=O). Aldehyde: R–CHO (carbonyl at the end of the chain). Ketone: R–CO–R′ (carbonyl between two alkyl groups).
Note: Attempt any one question.[5 × 1 = 5]
3.
Explain the three parts of an IUPAC name (Root, Prefix and Suffix) used in naming organic compounds.
1. Root word: shows the number of carbon atoms in the longest continuous chain: meth (1), eth (2), prop (3), but (4), pent (5), hex (6).
2. Prefix: comes before the root and names the side chains or substituents (e.g. methyl, ethyl, chloro, bromo) with their position numbers.
3. Suffix: comes after the root and shows the functional group or the type of bond: –ane (single bonds), –ene (double bond), –yne (triple bond), –ol (alcohol), –al (aldehyde), –one (ketone), –oic acid (carboxylic acid).
Example: 2-methylpropan-1-ol → prefix = 2-methyl, root = prop, suffix = an-1-ol.
2. Prefix: comes before the root and names the side chains or substituents (e.g. methyl, ethyl, chloro, bromo) with their position numbers.
3. Suffix: comes after the root and shows the functional group or the type of bond: –ane (single bonds), –ene (double bond), –yne (triple bond), –ol (alcohol), –al (aldehyde), –one (ketone), –oic acid (carboxylic acid).
Example: 2-methylpropan-1-ol → prefix = 2-methyl, root = prop, suffix = an-1-ol.
4.
Give IUPAC rules for naming alcohols.
Rules:
1. Select the longest continuous carbon chain that contains the –OH group.
2. Name the chain as the alkane and replace the last “e” with “ol” (alkane → alkanol).
3. Number the chain from the end that gives the –OH group the lowest number.
4. Write the position number of –OH just before “ol”.
5. Name side chains as prefixes with their positions, in alphabetical order.
Examples:
CH3–OH → Methanol
CH3–CH2–CH2–OH → Propan-1-ol
1. Select the longest continuous carbon chain that contains the –OH group.
2. Name the chain as the alkane and replace the last “e” with “ol” (alkane → alkanol).
3. Number the chain from the end that gives the –OH group the lowest number.
4. Write the position number of –OH just before “ol”.
5. Name side chains as prefixes with their positions, in alphabetical order.
Examples:
CH3–OH → Methanol
CH3–CH2–CH2–OH → Propan-1-ol
OH
|
CH3 – CH – CH3 → Propan-2-ol