Organic Chemistry : Some Basic Principles - Practice Test (1)


25 Questions MCQ Test Additional Question Bank | Organic Chemistry : Some Basic Principles - Practice Test (1)


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This mock test of Organic Chemistry : Some Basic Principles - Practice Test (1) for Class 9 helps you for every Class 9 entrance exam. This contains 25 Multiple Choice Questions for Class 9 Organic Chemistry : Some Basic Principles - Practice Test (1) (mcq) to study with solutions a complete question bank. The solved questions answers in this Organic Chemistry : Some Basic Principles - Practice Test (1) quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this Organic Chemistry : Some Basic Principles - Practice Test (1) exercise for a better result in the exam. You can find other Organic Chemistry : Some Basic Principles - Practice Test (1) extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
QUESTION: 1

Hybridisation in methane (CH4), ethene (C2H4), ethyne (C2H2) involves s and p orbitals. Choose the correct hybrid orbitals in the options given below for CH4, C2H4CH4, C2H4 and, C2H4 respectively

Solution:

In methane C is sp3 hybridised because of 3 sigma bond formation. While in ethane C is sp2 hybridised and in ethyne it is sp hybridised

QUESTION: 2

Correct IUPAC name the following compound is

Solution:

3,4- dimethylhexane will be correct iupac name as 6 membered carbon chain will be the main chain.

QUESTION: 3

In which of the following, functional group isomerism is not possible?

Solution:

alkyl halides donot give alkyl halides.

QUESTION: 4

The technique that is extensively used to separate mixtures into their components, purify compounds and also to test the purity of compounds is

Solution:

Chromatography is used to separate mixtures of substances into their components. All forms of chromatography work on the same principle. They all have a stationary phase (a solid, or a liquid supported on a solid) and a mobile phase (a liquid or a gas).

QUESTION: 5

Electrophiles are electron seeking species. Which of the following groups contain only electrophiles?

(i) 

(ii)

(iii)

(iv) 

Solution:

(ii) and (iii) contains all the electrophiles.

QUESTION: 6

Change in hybridisation affects the carbon’s and the organic compound’s

Solution:

in hybridization as % s character increases electronegativity also increases.

QUESTION: 7

Benzene and other related ring compounds (benzenoid) such as given below are called:

Solution:

they are aromatic system as they have benzene ring intact.

QUESTION: 8

The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is:

Solution:

A suitable method for the extraction of these oils from the flowers is Steam distillation

QUESTION: 9

Aniline is separated by technique from aniline . water mixture

Solution:

They are separated by steam distillation.

QUESTION: 10

In the four compounds below, which of the following pairs are position isomers? (i) I and II (ii) II and III(iii) II and IV(iv) III and IV

Solution:

(I) and (II) will be position isomer as they have same chain legth while they only differs in the position of carbonyl functional group.

QUESTION: 11

Condense the following complete structural formula given below. Choose the appropriate answer given below

Solution:

All the single bonds are broken and atoms are written together.

QUESTION: 12

The following compound is called 

Solution:

this is known as tropolone. It is polar and aromatic.

QUESTION: 13

During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results?

Solution:

TLC will give the best results.

QUESTION: 14

Sodium cyanide, sulphide and halide,( -CN, -S and -H, coming from organic compound) so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This is called

Solution:

This is lassaignes test.

QUESTION: 15

In Carius method, the organic compound is heated with

Solution:

fuming nitric acid is used.

QUESTION: 16

2-bromo butane can be represented in various ways. Choose the incorrect one from the following

Solution:

A will be wrong as terminal group represent methyl group not bromine.

QUESTION: 17

Compounds contain carbon atoms joined in the form of a ring are called

Solution:

they are known as alicyclic compounds.

QUESTION: 18

The principle involved in paper chromatography is

Solution:

Paper chromatography is based on the principle of partition

QUESTION: 19

Geometrical isomers and Optical isomers are:

Solution:

They are stereoisomers as they differ in the orientation of atoms in space.

QUESTION: 20

Write the state of hybridisation of carbon in the compound, HC≡≡N and shapes of each of the molecules

Solution:

C is sp hybridized and is linear in shape.

QUESTION: 21

Which of the following is the correct IUPAC name?

Solution:

4, 4-Dimethyl-3-ethylheptane is correct iupac name.

QUESTION: 22

The molecular formula C3H8O represents two alcohols: propan-1-ol and propan-2-ol. This property is called

Solution:

propan-1-ol and propan-2-ol are positional isomers as they differ ony in the position of functional group.

QUESTION: 23

In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?

Solution:

Cl is more electronegative.

QUESTION: 24

When an organic compound is present in an aqueous medium, it is separated by

Solution:

They are separated by differential extraction.

QUESTION: 25

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Solution:

moles of BaSO4 = moles of S = 0.418/(137+32+64) = 0.002065 moles % of S = 0.002065×32 × 100/0.157 = 42.10%