Exam-Style Mock Test | Class 11 Chemistry: Hydrocarbons
GKaim: Measure | Improve | Achieve

Hydrocarbons Mock Test – Class 11 Chemistry

Progressive Test — Guest First Round

0%

Hydrocarbons – Progressive Test

Welcome to the Progressive Test.

Click Start Test to begin the loaded practice round.

Good luck!

1 / 50

1. A benzene ring carrying a group is commonly named

2 / 50

2. In electrophilic substitution of nitrobenzene, the meta product is favoured because ortho and para attack would produce sigma-complex forms with

3 / 50

3. A bond-line formula for a hydrocarbon omits most symbols for carbon and hydrogen. In such a representation, each line end or line angle usually represents

4 / 50

4. A synthesis route must convert into without changing the carbon count. The suitable route is

5 / 50

5. Chloromethane reacts with sodium in dry ether. The main alkane product expected from the Wurtz reaction is

6 / 50

6. Friedel-Crafts acylation differs from Friedel-Crafts alkylation because acylation introduces

7 / 50

7. A reaction sequence is written as


The products and are respectively

8 / 50

8. A hydrocarbon gives and in equimolar amounts on reductive ozonolysis. If of is ozonolysed completely, the total moles of carbonyl products formed are

9 / 50

9. A Newman projection of ethane is drawn by looking

10 / 50

10. In naming an alkene, the parent chain must be chosen so that it

11 / 50

11. The formula-test record for gives , rapid bromine-water decolourisation, and plus on reductive ozonolysis. The best identity of is

12 / 50

12. A simple orientation of hydrocarbon reactions pairs the family and common reaction tendency as

13 / 50

13. In an ethane Newman projection, the conformation with maximum torsional strain has

14 / 50

14. A table compares local bonding in hydrocarbons.

Hydrocarbon site Hybridisation at the carbon site Geometry
P. Alkane carbon with four single bonds tetrahedral
Q. Alkene carbon of trigonal planar
R. Alkyne carbon of linear
S. Alkene carbon of tetrahedral

The row that needs correction is

15 / 50

15. For , of propene reacts completely with excess . The total amount of monobromopropane formed is

16 / 50

16. A hydrocarbon has the molecular formula . Using only the open-chain general formulas, it is best identified as a member of the

17 / 50

17. A butane Newman projection shows the two groups directly eclipsing each other. This conformation is best described as

18 / 50

18. A benzene ring already carrying one group undergoes further substitution. The group is generally

19 / 50

19. Assertion: A methyl group directs incoming electrophiles mainly to ortho and para positions in benzene substitution.
Reason: The methyl group increases electron density at ortho and para positions through electron-releasing effects.

20 / 50

20. A hydrocarbon has one tertiary hydrogen and nine primary hydrogens. On radical bromination, the major monobromo product is expected from replacement of

21 / 50

21. A four-carbon chain contains a double bond between carbon and carbon . Its suitable IUPAC name is

22 / 50

22. Acid-catalysed hydration of propyne mainly gives

23 / 50

23. Catalytic hydrogenation of ethene produces ethane because

24 / 50

24. The condensed formula represents a hydrocarbon in which the carbon skeleton is

25 / 50

25. Assertion: Soda-lime decarboxylation of sodium acetate gives methane.
Reason: In soda-lime decarboxylation, the carbon atom of the group is not retained in the alkane.

26 / 50

26. The number of -electrons in benzene is

27 / 50

27. The electrophile in nitration of benzene is

28 / 50

28. Open-chain alkyne produces of on complete combustion of . Complete addition of bromine to the same amount of requires

29 / 50

29. A formula card has three entries:
I. for open-chain alkanes
II. for open-chain monoalkenes
III. for cycloalkanes
The best judgement is that

30 / 50

30. A structure can be numbered from either end, giving possible double-bond locants and . The proper numbering choice gives

31 / 50

31. Alkenes are generally insoluble in water mainly because they are

32 / 50

32. A potential-energy graph for rotation about the central bond of butane has the highest peak when

33 / 50

33. Bromination of alkanes is generally more selective than chlorination because bromine radicals are

34 / 50

34. A data note compares physical states of straight-chain alkanes at ordinary conditions.

Alkane range Typical physical state trend
P. Lower members gases
Q. Middle members liquids
R. Higher members waxy solids

This trend is best explained by the increase in

35 / 50

35. The initiation step in chlorination of methane is

36 / 50

36. Ethene, , does not show geometrical isomerism because

37 / 50

37. Ozonolysis of , followed by reductive work-up, gives

38 / 50

38. A liquid alkene is shaken with water and then with a non-polar organic solvent. It is expected to dissolve better in the non-polar solvent because

39 / 50

39. A compact case is given below.

An unknown alkene undergoes ozonolysis followed by reductive work-up. The only carbonyl product detected is ethanal, .

The alkene is most likely

40 / 50

40. The compound that contains an acidic terminal alkyne hydrogen is

41 / 50

41. An open-chain hydrocarbon record shows that of gives of and of on combustion. Complete hydrogenation of consumes of . The most suitable formula and unsaturation pattern of are

42 / 50

42. The most stable conformation of ethane is the

43 / 50

43. A saw-horse representation of ethane is most useful because it

44 / 50

44. Formula-and-hydrogenation clues show that hydrocarbon has molecular formula , consumes of per , and has no ring. The most suitable structural class of is

45 / 50

45. An alkene can show geometrical isomerism when each carbon of the bond has

46 / 50

46. Dehydration of an alcohol prepares an alkene by removing

47 / 50

47. A hydrocarbon molecule containing a or bond is described as

48 / 50

48. A structural record contains the following entries.

Row Pair of hydrocarbons Claimed relation
P and Chain isomers
Q and Position isomers
R and Same molecular formula
S with - and -substitution Positional isomers

The row that needs correction is

49 / 50

49. A set of three hydrocarbons is listed as , , and . The feature that most strongly supports placing them in one homologous series is

50 / 50

50. A mixture contains , , and . A sample consumes on complete hydrogenation. After hydrogenation, complete combustion gives how many moles of ?

Your score is

Share your achievement!

LinkedIn Facebook
0%

Complete at least one Progressive Test round with incorrect or unanswered questions to unlock Mistake Review.
Complete at least 25% of the Progressive Test to unlock the Certificate Challenge Section.

Subscribe
Notify of
guest
0 Comments
Scroll to Top