Education
Class 10 Chemistry Chapter 21 Organic Chemistry Test 4
Grade 10 Chemistry chapter 21 Organic Chemistry Test 4
Prepare for the Punjab Board Class 10 Chemistry Examination 2027 with Chapter 21: Organic Chemistry – Class Test No. 4. This full chapter test covers the complete chapter, including exercise questions, to help students revise important concepts and assess their exam preparation. Designed according to the new 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 board-standard questions, strengthen your understanding of Organic Chemistry, and improve your preparation for the Class 10 annual examinations in 2027.
786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times
Chapter: 21
Total Time: 35 min
CHAPTER WISE CLASS TEST(Test # 4)
Chemistry
Total Marks: 25
ABCD
1ABCD
2ABCD
3ABCD
ABCD
4ABCD
5ABCD
6ABCD
ABCD
7ABCD
8ABCD
NOTE: Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle in front of that question with Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
TIME: 10 MinOBJECTIVEMARKS: 8
Q.1:(08)
1.Select the general formula of alkene family.
(A) CnH2n+1(B) CnH2n+2(C) CnH2n(D) CnH2n-1
Alkenes: CnH2n.
2.As the number of carbon atoms increases, the number of possible isomers:
(A) Decreases(B) Remains same(C) Increases very rapidly(D) Becomes zero
The number of isomers increases very rapidly.
3.How many structural isomers will be there for a saturated hydrocarbon having five carbon atoms?
(A) 2(B) 3(C) 4(D) 5
Pentane C5H12 has 3 isomers: pentane, 2-methylbutane, 2,2-dimethylpropane.
4.The functional group of amines is:
(A) –OH(B) –COOH(C) –CHO(D) –NH2
Amines contain the –NH2 group.
5.Which ester among the following will be called ethylpropanoate?
(A) (B) (C) (D)
Propanoate part: CH3CH2COO–, ethyl part: –C2H5, so (D).
6.The suffix of an organic compound’s name tells us:
(A) Number of carbon atoms(B) Number of hydrogen atoms(C) Class of organic compound(D) Molecular mass
The suffix shows the class (functional group) of the compound.
7.Which functional group is present in alkenes?
(A) No Functional group(B) Carbon – carbon triple bond(C) Carbon – carbon double bond(D) Carbon–carbon single bond
Alkenes contain a C=C double bond.
8.The suffix used for carboxylic acids is:
(A) –ane(B) –ene(C) –ol(D) –oic acid
Carboxylic acids end with “–oic acid”, e.g. ethanoic acid.
TIME: 25 MinSUBJECTIVEMARKS: 17
(PART – I)
Q.2 Write short answers to following questions.(6×2=12)
i.Define structural isomerism.
Compounds having the same molecular formula but different structural formulae (different arrangement of atoms) show structural isomerism, e.g. butane and 2-methylpropane (C4H10).
ii.Write down the structural formulae of isomers shown by butene.
Butene (C4H8):
1) But-1-ene: CH2 = CH – CH2 – CH3
2) But-2-ene: CH3 – CH = CH – CH3
3) 2-Methylpropene: CH2 = C(CH3) – CH3
1) But-1-ene: CH2 = CH – CH2 – CH3
2) But-2-ene: CH3 – CH = CH – CH3
3) 2-Methylpropene: CH2 = C(CH3) – CH3
iii.What is common in carboxyl group, amide functional group and ester functional group?
All three contain the carbonyl group (C=O) joined to a second atom or group (–OH in carboxyl, –NH2 in amide, –OR in ester).
iv.What is the root in IUPAC nomenclature?
The root is the part of the name that shows the number of carbon atoms in the longest chain (meth-, eth-, prop-, but-…).
v.Why the members of a homologous series have similar chemical properties?
Because all members have the same functional group, which decides the chemical properties.
vi.Give IUPAC name of CH3COOH.
Ethanoic acid
(PART – II)
Note: Attempt the following question.
Q.3: Give IUPAC rules for naming alcohols.(5)
1) Select the longest carbon chain that contains the carbon bearing the –OH group.
2) Name the chain as the alkane and replace the final “e” with “ol” (ethane → ethanol).
3) Number the chain from the end that gives the –OH carbon the lowest number.
4) Write the position number of –OH before “ol”, e.g. propan-2-ol.
5) Name and number any side groups as prefixes in alphabetical order, e.g. 2-methylpropan-1-ol.
2) Name the chain as the alkane and replace the final “e” with “ol” (ethane → ethanol).
3) Number the chain from the end that gives the –OH carbon the lowest number.
4) Write the position number of –OH before “ol”, e.g. propan-2-ol.
5) Name and number any side groups as prefixes in alphabetical order, e.g. 2-methylpropan-1-ol.
