NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | CHEMISTRY IMPORTANT QUESTIONS PART-2 | CHAPTER -5| ORGANIC CHEMISTRY: SOME BASIC PRINCIPLE AND TECHNIQUE | EDUGROWN |

In This Post we are  providing Chapter-5 ORGANIC CHEMISTRY: SOME BASIC PRINCIPLE AND TECHNIQUE NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON ORGANIC CHEMISTRY: SOME BASIC PRINCIPLE AND TECHNIQUE

Question 1.
Without using column chromatography, how will you separate a mixture of camphor and benzoic acid?

Answer:
Sublimation cannot be used as both camphor and benzoic acid sublime on heating. Therefore, a chemical method using NaHCO₃ solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is then cooled with dilute HCl to get benzoic acid.

Question 2.
Compare inductive & mesomeric effects.

Answer:

Inductive effect	Mesomeric effect
1. It operates in saturated gp. of compounds.	1. It occurs in unsaturated & especially in conjugated compounds.
2. It involves electrons in σ – bonds.	2. It involves electrons in π – bonds.
3. Electron pair is slightly displaced & there only partial charges are developed.	3. The electron pair is transferred completely with the result full positive & negative charges are created.
4. It is transmitted over only a quite short distance.	4. It is transmitted from one end to the other of quite large molecules provided conjugation (i.e. delocalised orbitals) is present through which it can proceed.

Question 3.
What is the difference between distillation, distillation under reduced pressure & steam distillation?

Answer:

Distillation	Distillation under reduced pressure	Steam distillation
This is used to separate volatile liquid from non-volatile liquid or solid separately.	This is used to purify liquids that decompose at or below their boiling points.	This is used for purifying substances that are steam volatile & immiscible with water.

Question 4.
How will you purify sugar which has impurities of sodium chloride?

Answer:
Sugar may be purified by the crystallization method. This can be purified by shaking the impure solid with hot ethanol at 345K. The sugar will dissolve whereas common salt remains insoluble. The hot solution is filtered, concentrated & allowed to cool when crystals of sugar will separate out. In this case, hot water has been used as a solvent. The purification of sugar would not have been possible since both sugar’& common salt are soluble in water.

Question 5.
Differentiate between Ionic & free radical reactions.

Answer:

Ionic reactions	Free radical reactions
1. These occur only rarely in the gas phase but mainly in a solution of polar solvents; the reaction is influenced by the polarity of the solvent.	1. These occur in gas phases or in non-polar solvents.

Question 6.
For each of the following compounds, write a more condensed formula & also their bond-line formula.

(b) HOCH2CH2CH2CHCH3CHCH3CH3

Answer:

Question 7.
Expand each of the following bond line formulae to show all the atoms including carbon & hydrogen.

Answer:

Answer:

Answer:

Question 8.
Explain why is (CH₃) C+ more stable than CH₃ C⁺ H₂ & C⁺ H₃ is the least stable cation?

Answer:
Hyperconjugation interaction in (CH₃)₃ C⁺ is greater than in C+⁺ H₃ C⁺ H₂ has 9 C -H bonds. In C H₃, the C -H bonds are in the nodal plane of the vacant 2p-orbital & hence cannot overlap with it.
Thus C⁺ H₃ is least stable.

Question 7.
The choice of the solvent is of great importance in crystallizing organic substances. What are the characteristics of a suitable solvent?

Answer:
A suitable solvent must have the following characteristics;

1. The impurities & pure compound must have a large difference in their solubilities.
2. The pure compound must have low solubility at room temperature but high solubility at its boiling point.
3. The impurity should either be insoluble at room temperature or must have high solubility so that crystallization may give a high yield.
4. The solvent should have an average boiling point.
5. The solvent should neither react with the compound nor with impurities.
6. The solvent should not be highly inflammable.

Question 8.
Explain the principle of steam distillation.

Answer:
Steam distillation: The process of steam distillation is employed in the purification of substance from non-volatile impurities provided the substance itself is volatile in steam and insoluble in water.

This method is based on the facts that

1. A liquid boils at a temperature when its vapour pressure becomes equal to the atmospheric pressure.
2. The vapour pressure of a mixture of two immiscible liquids is equal to the sum of the vapour pressures of the individual liquids.

In the actual process, steam is continuously passed through the impure organic liquid. Steam heats the liquid and it gets practically condensed to water. After some time mixture of the liquid and water begins to boil, because the vapour pressure of the mixture becomes equal to the atmospheric pressure.

Obviously, this happens at a temperature that is lower than the boiling point of the substance or that of water. Thus an organic compound boils below its boiling points and chances of decomposition avoided. For example, a mixture of aniline (b.p 453 K) with decomposition and water (b.p. 373 K) under normal atmospheric pressure boils at 371K. At this temperature the

Steam Distillation

water boils at 371 K. At this temperature, the vapour pressure of water is 717 mm and that of aniline is 43 mm and therefore the total pressure is equal, to 760 mm. Thus in steam distillation, the liquid gets distilled at a temperature lower than its boiling point and chances of decomposition avoided. The proportion of water and liquid in the mixture that distils over is given by the relation.

w1w2=P1×18P2×M
where w₁ and w₂ stand for the masses of water and liquid that distils over. P₁ and P₂ are vapour pressure of water and of liquid at the distillation temperature and M is the molecular mass of the liquid.

Question 9.
Dehydrobromination of compounds (A) and (B) yield the same alkene (c) Alkene (c) Can regenerate (A) and (B) by the addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of A and B give isomeric products (D) and (E) respectively. 1, 1-Diphenyl ethane is obtained on the reaction of (C) of benzene in the presence of H⁺ ions. Give structures of A to E with reactions.

Answer:
Alkene (C) on reaction with benzene in the presence of H+ ions gives 1, 1-Diphenyl ethane. Therefore C must be styrene as depicted below

Now dehydrobromination of A and B give the same alkene C, i.e.,
styrene.
∴ A and B must be isomeric alkyl bromide.

A and B can be obtained by the addition of HBr in the presence and absence of peroxide to styrene.

Hydrolysis of A and B give isomeric alcohols (D) & (E) as

Question 10.
What are reaction intermediates? How are they generated by bond fission?

Answer:
The species which are generated as a result of bond fission are called reaction intermediates. The important reaction intermediates are:
1. Free Radicals: A free radical may be defined as an atom or group of atoms having an impaired electron. These are obtained as a result of homolytic fission of covalent bonds.

These free radicals are neutral particles, extremely transient, (short-lived) and highly reactive. They get consumed as soon as they are formed. They pair up their electron with another electron from wherever it is available. They occur only as a reaction intermediate. Their presence is felt in reactions, but cannot be isolated in a free state. For example dissociation of Cl2 gas in the presence of Ultraviolet light produces free radicals.

The alkyl free radicals are obtained when free radical: Cl reacts with alkanes.

Free radical may be primary, secondary, tertiary depending upon whether, one, two or three carbon atom attached to the carbon atoms carrying the odd electron.

The stability is CH₃ < 1° < 2° < 3°.

2. Carbocation or carbonium ion: It is defined as a group of atoms that contain positively charged carbon having only six electrons. It is obtained by heterolytic fission of a covalent bond involving a carbon atom.

They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon bearing the positive charge as:

Thus the order of stability if CH₃⁺ < 1° < 2° < 3°.

3. Carbanion: A carbanion may be defined as a species containing a carbon atom carrying a negative charge. These are generated by the atom in which the atom linked to carbon goes without the bonding electrons. As a result of this carbon acquires a negative charge. For example, the removal of hydrogen of methyl part of acetaldehyde molecule as H+ ion leaving both the electron on carbon.

They are also very reactive species. They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon atom bearing negative
charge.

The order of stability is the reverse of free radicals and carbocations
CH₃ ^– > 1° > 2° > 3°.

(iv) Carbenes: The carbenes are reactive neutral species in which carbon atom has six electrons in the valency shell out of which two are shared. The simplest carbene is methylene (CH₂). It is formed when diazomethane is decomposed by the action of light.

It is very reactive. It reacts with alkenes by adding to the double bond forming cyclopropane.