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Students get through the MP Board Class 11th Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques which are most likely to be asked in the exam.

MP Board Class 11th Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques Class 11 Important Questions Very Short Answer Type

Question 1.
What are the hybridisation state of each carbon atom in the following com-pounds :
(i) CH₂ = C = O,
(ii) CH₃CH = CH₂,
(iii) (CH₃)₂CO,
(iv) CH₂ = CHCN,
(V) C₆H₆.
Answer:
(i)
(ii) 
(iii) 
(iv)
(v)

Question 2.
Identify the reagents shown in underline in the following equations as nu-cleophiles or electrophiles:
(a) CH₃COOH + HO^– → CH₃COO^–+H₂O
(b) CH₃COCH₃ + CN^– → (CH₃)₂C(CN)(OH)
(c) C₆H₅^–+CH₃CO → C₆H₅COCH₃.
Answer:
(a) HO^– is nucleophile,
(b) CN^– is nucleophile,
(c) CH₃ img is electrophile.

Question 3.
Classify the following reactions :
(a) CH₃CH₂Br + HS^– → CH₃CH₂SH+ Br^–
(b) (CH₃)₂C = CH₂ +HCl → (CH₃)₂CCl-CH₃
(c) CH₃CH₂Br + HO_– → CH₂ = CH₂+H₂O + Br^–
(d) (CH₃)₃ C- CH₂OH + HBr → (CH₃)₂CBrCH₂CH₃ +H₂O.
Answer:
(a) Nucleophilic substitution reaction. Nucleophile (Br^–) is substituted by other nucleophiles (HS^–).
(b) Electrophilic addition reaction.
(c) β -Elimination reaction.
(d) Nucleophilic substitution reaction along with rearrangement.

Question 4.
What is Isomerism? Explain with example.
Answer:
Compound with same molecular formula but different structural formula are called isomers and this property is known as isomerism.
Example: Ethyl alcohol (C₂H₅OH) and dimethyl ether have same molecular formula C₂H₆O but their structural formulae are different, so they are isomers.

Question 5.
Explain position isomerism with example.
Answer:
Position isomerism: It includes those compounds in which there is a difference in the position of double or triple bond or in the linkage of the functional group with different carbon atoms.
Example: C₃H₈O:

Question 6.
Write structural formulae of two functional isomers of molecular formula C₃H₆O.
Answer:

Question 7.
Write the isomers of pentane and write name of the isomerism present.
Answer:
C₅H₁₂ exhibits chain isomerism. It has three isomers :
1. CH₃ – CH₂ – CH₂ – CH₂ – CH₃ w-pentane
2. 
3.

Question 8.
Explain chain and position isomerism in Alkyne.
Answer:
Alkynes represent chain and position isomerism.
1. Chain isomerism:

2. Position isomerism:

Question 9.
Which isomerism is found in Arenes? Give example.
Answer:
Arenes exhibit position isomerism.
Example: C₈H₁₀:

Question 10.
What type of isomerism exist in Alkane? Explain with example.
Answer:
Alkanes exhibit chain isomerism.
Examples : C₅H₁₂: CH₃– CH₂– CH₂– CH₂– CH₃ n-pentane

Question 11.
Give name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Answer:
Mixture of CaSO₄ and camphor is separated by sublimation because camphor is sublime and CaSO₄ is not. Thus, on heating camphor will deposit on the walls of the funnel and CaSO₄ will remain in the crucible.

Question 12.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
Nitrogen, sulphur and halogens present in organic compounds are in covalent state, therefore their identification is not easy. On fusing with Na metal, they get converted to NaCN, Na₂S or NaX, that is they get converted to ionic form. Thus, in ionic form, these elements can be easily identified.

Question 13.
Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Answer:
In steam distillation method, mixture of organic liquid and water boils when the sum of vapour pressure of the organic liquid (P₁) and vapour pressure of water (P₂) becomes equal to the atmospheric vapour pressure (P).
Thus, P = P₁ + P₂
Since, P₁ is less than P, thus organic liquid starts boiling at a temperature below its boiling point.

Question 14.
Structural formula of the following compounds are given. Write the name of isomerism found in them.
1. CH₃ – CH₂ -CHO and
2. CH₃– O- C₃H₇ and C₂H₅ – O -C₂HS
3.
Answer:
1. Functional isomerism,
2. Metamerism,
3. Tautomerism.

Question 15.
What type of isomerism is shown by Alkenes? Explain with example.
Answer:
Alkenes show Chain, Position and Geometrical isomerism.
1. Chain isomerism: C4H8 :

2. Position isomerism: C4H8

3. Geometrical isomerism : C4H8:

Question 16.
Explain ring chain isomerism with example.
Answer:
When molecular formula of two or more than two compounds is same but carbon in the compound may be arranged as an open chain or closed chain, then this type of isomerism are known as ring chain isomerism.
Example: C₆H₁₂:

Question 17.
Write the method for the estimation of C and H in organic compounds.
Answer:
Liebig method is used for the estimation of C and H in organic compounds.

Question 18.
Write the name of method to purify the following mixtures :

1. Impure Naphthalene,
2. Two volatile liquids,
3. Iodine and NaCI,
4. Benzene and Toluene.

Answer:

1.  Impure Naphthalene: Sublimation.
2. Two volatile liquids: Fractional distillation.
3. Iodine and NaCl: Sublimation.
4. Benzene and Toluene: Fractional distillation.

Question 19.
Write names of different methods used for purification and separation of organic compounds.
Answer:
For purification and separation of organic compounds different methods are following :
(A) For solid substances :
1. Crystallisation :

• Simple crystallisation,
• Fractional crystallisation.

2. Sublimation.

(B) For liquid substances :

• Simple distillation,
• Fractional distillation,
• Steam distillation,
• Vacuum distillation or reduced pressure distillation.

Question 20.
What is chromatography?
Answer:
The process in which different components of a mixture are separated by distributing in stationary or mobile phases on the basis of difference in adsorption abilities on any adsorbent is called chromatography.

Question 21.
If an organic compound is given in very small amount, then how is its purity identified?
Answer:
If an organic compound is given in very small amount, then its purity is identified by determining its melting or boiling point or it is done by column chromatography if one band is obtained, then the compound is pure, if more than one band is obtained, then compound is impure.

Question 22.
What is the use of adsorption coloured band process in the purification of organic compounds?
Answer:
Adsorption coloured band process is used in the separation of complex compounds like vitamins and hormones and in the identification of purity of substances.

Question 23.
Small quantity (1 or 2 drops) of cone. HNO₃ is added when halogens are tested in presence of nitrogen. Why?
Answer:
When halogens are tested in presence of nitrogen, it creates problem. Therefore, one or two drops of cone. HNO₃ is added previously which decomposes N and S. Both are liberated in the form of HCN and H₂S gas.
NaCN + HNO₃→NaNO₃ +HCN ↑
Na₂S + 2HNO₃ → 2NaNO₃ +H₂S↑.

Question 24.
While testing for nitrogen in organic compounds sometimes blood red colour is obtained on adding FeCl3 solution. Why?
Answer:
If blood red colour is obtained while testing for nitrogen in the given organic compound, it means both N and S are present in the organic compound. If both N and S are present NaCNS is formed in the Na-extract which gives blood red colour with FeCl₃.
Na + C + S + N → NaCNS
3NaCNS+FeCl₃ →Fe(CNS)₃ ↓ +3NaCl Ferric sulphocyanide (Blood red).

Question 25.
How can molecular mass of compounds be determined by volumetric analysis?
Answer:
By volumetric analysis, compounds of acidic or basic property is titrated in presence of an indicator by a standard base or acid, and their equivalent masses are determined. Standard acid or base is prepared by dissolving its equivalent mass in one litre.
Molecular mass of acid or base = Equivalent mass × Acidity or Basicity.

Question 26.
Why sodium extract is prepared for testing the elements of organic compounds? Give reason.
Answer:
The nature of organic compounds is covalent, as a result they do not ionise and thus it is difficult to test the presence of elements in it. Sodium is active metal which reacts with the elements present in the organic compound (N, S, Cl, Br, I) and forms sodium compounds which are ionic in nature and thus test of elements becomes easy.

Question 27.
Why is chloroform or CCl₄ added in sodium extract for the identification of bromine or iodine?
Answer:
Sodium extract is neutralised by dilute HNO₃ and CHCl₃ or CCl₄ and chlorine water is added. Chlorine displaces the bromine or iodine present in sodium extract which dissolves in CHCl₃ or CCl₄ and provides brown or violet colour.

Question 28.
How are carbon and hydrogen detected in an organic compound?
Answer:
In a test tube organic compound is heated with CuO, the CO₂ gas evolved turns lime water milky whereas water vapour gets condensed in the form of drops, which give blue colour on being absorbed by CuSO₄.
C + 2CuO → CO₂ + 2Cu
2H + CuO → H₂O + Cu
CuSO₄ White + 5H₂O → CUSO₄ Blue .5H₂O
CO₂+Ca(OH)₂ → CaCO₃ Milky+ H₂O

Question 29.
What do you underštand by mixed melting point process of an organic compound?
Answer:
Mixture of two completely, mixed substances is called mixed melting point. In this method melting point of mixture is determined. If it is same as the melting point of pure substance, the substance is pure.

Question 30.
How are two liquids with different boiling points separated?
Answer:
Mixture of two liquids which differ in their boiling points can be separated by fractional distillation. The more volatile liquid evaporates first passes through th condenser get condensed and collected in the receiver.

Vapours of less volatile liquid condense in the fractionating column and returns back. After the removal of more vohitile liquid at a particu lar temperature then the less volatile liquid gets distilled at a higher temperature.

Question 31.
What is empirical formula? Give its importance.
Answer:
Formula showing simplest whole number ratio of different constituents of compound is called empirical formula.
Example: Empirical formula of glucose is CH₂O.

Question 32.
What Is molecular formula ? Explain with example.
Answer:
Chemical formula which expresses the actual number of atoms of various elements in a compound is called molecular formula.
Example: Molecular formula of glucose is C₆ H₁₂O₆.
Molecular formula = Empincal formula × n, [ n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }} \) ]

Question 33.
What are heterocyclic compounds? Explain with example.
Answer:
This series includes those cyclic compounds in which apart from carbon one or more polyvalent atoms are present like, Nitrogen, Sulphur, Oxygen etc
Example : (i) Pyridine, C₅H₅N
(ii) Furan C₄H₄O,
(iii) Thiophene C₄H₄S.

Question 34.
Will CCl₄ give white precipitate of AgCI on heating with silver nitrate? Explain your answer with reason.
Answer:
CCl₄ will not give white precipitate with AgNO₃ solution because CCl₄ is a covalent compound. It doés not ionise due to which Cl^– ions are not available for the formation of precipitate of AgCl.

Question 35.
What are homocyclic compounds? Explain with example.
Answer:
This series include organic compounds which contain only carbon atoms in the cycle. Homocyclic compounds are of two types:

1. Aromatic compounds
2. Alicyclic compounds

Examples of Aromatic compound:
1. Benzene: C₆H₆ or
2. Toluene: C₆H₅CH₃ or
Examples of Alicyclic compound:
1. Cyclohexane: C₆H₁₂ or
2. Cyclopentane : C₅ H₁₀ or

Question 36.
What are functional groups?
Answer:
Every organic compound is made up of two parts. The part which determines the common properties or chemical properties of a particular homologous series is called a functional group. For example, -OH is the functional group present in alcohols C₂H₅OH.

Question 37.
Write general formula and name of first two members of Aldehyde series.
Answer:
General formula of Aldehyde series is R-CHO or C_nH_2n+1-CHO
Chemical formula Common name IUPAC name
HCHO Formaldehyde Methanal
CH₃CHO Acetaldehyde Ethanal.

Question 38.
Write names and formulae of 5 carbon alkane, alkene and alkyne.
Answer:
1. Alkane – C₅H₁₂ – Pentane
2. Alkene – C₅H₁₀ – Pentene
3. Alkyne – C₅H₈ – Pentyne.

Question 39.
What is a radical?
Answer:
The part left after the removal of a hydrogen atom from a hydrocarbon is called radical. The part left after the removal of a hydrogen atom from alkane is called alkyl radical. It is represented by – R.
Example: CH₃ – methyl, C₂H₅ – ethyl.
The part left after the removal of a hydrogen atom from an aromatic hydrocarbon is called aryl radical. It is represented by Ar.
Example: C₆H₅ – Phenyl.

Question 40.
Write name and formulae of first two members of Ester series.
Answer:
General formula of Ester series is RCOOR or C_nH₂O₂

Question 41.
Give examples of Nucleophille and Electrophilic reagents.
Answer:
Nucleophilic reagents:

1. Negative nucleophilic reagent: Cl^–, Br^–, SO₄^-2, OH^–
2. Neutral nucleophilic reagent: NH₃, H₂O, R – OH.

Electrophilic reagents:

1. Positive electrophilic reagent NH_4⁺, H₃O⁺, Cl⁺ NO₂⁺
2. Neutral electrophilic reagent: BF₃, AlCl₃, FeCl₃.

Organic Chemistry: Some Basic Principles and Techniques Class 11 Important Questions Short Answer Type 

Question 1.
Why is it necessary to use acetic acid and not sulphuric acid for acidifica tion of sodium extract for testing sulphur by lead acetate test?
Answer:
If H₂SO₄ is used, then lead acetate itself reacts with H₂SO₄ to give white precipi
late of lead sulphate.
Pb(CH₃COO)₂ + H₂SO₄ → PbSO₄ White ppt ↓+ 2CH₃COOH .

Thus, white precipitate of PbSO₄ will affect the following test of sulphur.

But, if acetic acid is used, then it does not react with lead acetate due to which no obstruction is produced in the test.

Question 2.
Why is a solution of Potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in organic compound?
Answer:
CO₂ is slightly acidic in nature. Therefore, it reacts with strong base KOH to form K₂CO₃ by the help of which, percent amount of carbon in organic compound can be determined by the CO₂ obtained.
2KOH + CO₂ → K₂CO₃+H₂O

By the increase in mass of U-tube of KOH, mass of CO₂ can be determined. By the mass of CO₂ obtained carbon present in organic compound can be determined as follows :
%C = \(\frac{12}{44}\) × \(\frac{\text { Mass of } \mathrm{CO}_{2} \text { produced }}{\text { Mass of organic compound }} \) × 100.

Question 3.
What are Nucleophiles?
Answer:
Nucleophiles or Nucleophilic reagents: Reagents which donate an electron pair in chemical reactions are called nucleophiles.
These are negative ions including carb-anion or neutral molecules containing ffee electron pairs.
Example: Negative nucleophiles: Cl^–, Br^–, I^–, CN^–, HO^–, R-CH₂.
Neutral nucleophiles: NH₃, RNH₂, H₂O, ROH.
Nucleophilic reagents are generally represented by Nu.

Question 4.
What are electrophiles?
Answer:
Electrophiles or Electrophilic reagents: Reagents which can accept an electron pair in chemical reaction are called electrophiles.
These are positive ions including carbonium ion or neutral molecules in which electron deficient centres are present.
Example : Positive electrophiles : H⁺, Cl⁺, Br⁺, I⁺, NO², R₃C+, RN₂.
Neutral electrophiles : SO₃H, AlCl₃, BF₃.
Electrophilic reagents are generally represented by E⁺.

Question 5.
Write difference between Electrophilic and Nucleophilic reagents.
Answer:
Differences between Electrophilic and Nucleophilic reagent

Question 6.
Explain Can’t Hoff and Le Bell rule of valency of carbon atom.
Answer:
Atomic number of carbon is 6. Its electronic configuration is 2,4. In this way its valence shell contains 4 electrons and its valency is 4.
According to Can’t Hoff and Le Bell, carbon is sp³ hybridised, therefore its structure is tetrahedral in which carbon atom is at the centre and the four valencies are situated at the four comers and bond angle between two valencies is 109° 28′.

Question 7.
Carbon forms only covalent compounds. Why?
Answer:
Atomic number of carbon is 6 and its valence shell contains 4 electrons. It requires 4 extra electrons to complete its octet. Thus carbon can either donate 4 electrons or can accept 4 electrons or can share 4 electrons. But due to its small size its ionisation energy and electron affinity is very high.

Thus, it is not possible for carbon to donate or accept 4 electrons. Thus, to complete its octet carbon can only share electrons, therefore it forms only covalent compounds.

Question 8.
What is the reason of formation of large number of compounds by carbon? Or, Explain peculiar behaviour of carbon.
Answer:
Causes of existence of large number of carbon compounds are as follows :
1. Catenation: The tendency of atoms of an element to link with atoms of its own kind by covalent bond is called catenation. Tendency of catenation is maximum in carbon.

2. Strong bond between carbon-carbon atom: Size of carbon atom is very small. Due to it, overlapping between half filled orbitals of carbon is easier and more, as a result bond formed between carbon-carbon is very strong.

3. Tendency to form multiple bonds: Due to its small size, carbon can easily combine and form double and triple bond with other atoms. Like C, H, O, N.

4. Isomerism: When two or more than two compounds possess the same molecular formula but different structural formula, then such compounds are called isomers and the phenomenon is called isomerism. This property is maximum in carbon.
Due to these properties, carbon forms a large number of compounds.

Question 9.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:
Sodium extract is boiled with concentrated nitric acid by which if NaCN and Na₂S are present they dissociate.
NaCN + HNO₃ → NaNO₃ + HCN ↑
Na₂S + 2HNO₃→ 2NaNO₃ + H₂S ↑

If cyanide and sulphide are not removed, then they will react with AgN03 and obstruct the silver nitrate test of halogens.

Question 10.
Write the difference between Organic and Inorganic compound.
Answer:
Differences between Organic and Inorganic compound

Question 11.
What is homologous series? Write its characteristics.
Answer:
Homologous series: A homologous series is a group of organic compounds with similar chemical constitutions and having a common functional group. Each member of series is called homologue. They have same chemical properties. The physical properties show a gradation.

Characteristics of homologous series :

• All the members of a particular series can be represented by a general molecular formula. For example, the alkanes or saturated hydrocarbons can be represented by a general formula C_nH_2n+2, where n is the number of carbon atoms.
• All the homologous of a particular series can be prepared by certain general methods.
• There is a common difference of one -CH₂ group between any two consecutive members of series.
• All the members of a series have same chemical properties due to the presence of same functional group, e.g., all alcohols react with sodium metals liberating hydrogen gas.
• There is a regular gradation in the physical properties of the successive members of a particular series.

Question 12.
What are primary (1°), secondary (2°), tertiary (3°) and quaternary (4°) carbon atoms in the organic compounds?
Answer:
The classification of carbon atoms in alkane is done in the following manner:

• Primary carbon (1°) atom is that which is linked with one carbon atom.
• Secondary carbon (2°) atom is that which is linked with 2 carbon atoms.
• Tertiary carbon (3°) atom is that which is linked with 3 carbon atoms.
• Quaternary carbon (4°) atom is that which is linked with 4 carbon atoms.

In the following example primary (p), secondary (s), tertiary (t) and quaternary (q) carbon atoms are shown.

Question 13.
For the following bond cleavages, use curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
(a) CH₃O – OCH₃ → CH₃O + ÒCH₃
(b) >=O+^–OH → > = O+H₂O
(c)
(d)
Answer:
(a)
(b)
(c)
(d)

Question 14.
Explain the method of separation of a mixture of ammonium chloride and sodium chloride. Or, Explain the method of separation of a mixture of naphthalene and common salt.
Answer:
Mixture of ammonium chloride or naphthalene and NaCl is separated by sublimation. The mixture is placed in a porcelain dish covered with porous filter paper and a funnel is kept inverted over it. Mouth of the funnel is closed by cotton plugs. This is then heated slowly on a sand bath. NH₄Cl or naphthalene sublimes pass through the pores of filter paper and get collected over the walls of the funnel from where it is collected and NaCl remains behind in the dish.

Question 15.
How nitrogen is tested in any organic compound by Lassaigne’s method?
Answer:
Test of nitrogen (Lassaigne’s test): In 2 ml of sodium extract, 2 ml of freshly prepared solution of FeSO₄ is added along with little NaOH, warm it. Green precipitate of Fe(OH)₂ is obtained. The solution is cooled and small quantity of cone. HC1 is added in which green ppt. dissolves. Then little amount of FeCl₃ solution is added. If green or blue colour precipitate is obtained, then the substance contains nitrogen.
Na + C + N → NaCN
FeSO₄ + 2NaOH → Na₂SO₄ + Fe(OH )₂

Question 16.
How is sulphur tested in an organic compound?
Answer:
Test of sulphur : (i) Sodium nitroprusside solution is added to the sodium extract of organic substance. If violet colour appears, it confirms the presence of sulphur in the given organic compound.
2Na+S → Na₂S

(ii) If acetic acid and lead acetate solution is added to the sodium extract then a black precipitate is obtained, presence of sulphur is confirmed in it.
Na₂S + (CH₃COO)₂ Pb → PbS ↓ Black + 2CH₃COONa.

Question 17.
When is steam distillation more advantageous? Explain this method with diagram.
Answer:
Steam distillation: There are many organic compounds which are insoluble in water but are soluble in steam (water vapour). The distillation of such substances is called steam distillation. In this method impurities of non-volatile substance remain behind if the flask.

The impure mixture is taken in the flask and heated over sand bath and steam is passed through it. The steam along with vapour of substance passes through delivery tube and condenses. In receiver mixture of substance and water is collected. If organic substance is
solid, it is separated by filtration and if it is liquid then separation is done by separating funnel.

Question 18.
What is the relationship between the members of following pairs of struc tures? Are they structural or geometrical isomers or resonance contributors:
(a)
(b)
(c)
Answer:
(a) Structural isomers (Difference in position of functional group),

(b) Geometrical isomers.

(c) Resonance structures (position of electrons is different, but not of atoms.)

Question 19.
What is the principle of vacuum distillation or low pressure distillation? Explain with diagram.
Answer:
Vacuum distillation or Distillation under reduced pressure: Those organic compounds which decompose at their boiling point or even below the boiling point are purified by this method. As shown in the figure, partial vacuum is created in the receiver flask. This results in lowering of pressure in the distillation flask.

The liquid starts boiling at low temperature. Glycerol is purified by this method. At atmospheric pressure, its boiling point is 563 K but at a pressure of 12 mm, it boils at 453 K only.

Question 20.
How are halogens present in organic compounds detected by silver nitrate test? Explain giving equation.
Answer:
Test of halogens by Silver nitrate test: To a part of sodium extract add cone, nitric acid and the silver nitrate solution. A curdy white precipitate shows the presence of chlorine, yellowish-white and lemon yellowish precipitates show the presence of bromine and iodine respectively.

NaCl + AgNO₃ → NaNO₃ + AgCl ↓ (curdy white precipitate)
NaBr + AgNO₃ → NaNO₃ + AgBr ↓ (yellowish white precipitate)
Nal + AgNO₃→ NaNO₃ + Agl↓ (lemon yellowish precipitate)
White precipitate of AgCl dissolves easily in dilute NH4OH to form a soluble complex.
AgCl + 2NH₄OH → [Ag(NH₃)₂]Cl + 2H₂O

Yellowish white precipitate of AgBr is sparingly soluble in NH₄OH, hence dissolves with difficulty to form a soluble complex.
AgBr + 2NH₄OH → [Ag(NH₃)₂ ]Br + 2H₂O
Lemon yellowish precipitate of Agl is insoluble in NH₄OH as well as in cone. HNO₃.

Question 21.
Which of the following represents the correct IUPAC name for the compounds concerned:
(a) 2,2-Dimethylpentane or 2-Dimethylpentane.
(b) 2,4,7-Trimethyloctane or 2,5, 7-TVimethyloctane.
(c) 2-Chioro-4-MethyIpentane or 4-Chloro-2-MethyIpentane.
(d) But-3-yn-l-ol or But-4-oI-l-yn.
Answer:
(a) 2, 2-Dimethylpentane (Because for two alkyl groups on the same carbon atom, numbering should be twice).
(b) 2,4,7-Trimethyloctane (Because value of 2,4,7 numeral set is lesser than 2,5,7 numeral set.)
(c) 2-Chloro-4-Methylpentane (Because substituted groups are written in alphabetical order).
(d) But -3-yn-l-ol (Because functional group or alcohol group possess the lowest number).

Question 22.
How is sulphur present in organic compounds estimated?
Answer:
Estimation of sulphur: Sulphur is also estimated by Carius method. In this case, organic substance is heated with only HNO₃ (AgNO₃ is not used). Carbon, hydrogen and sulphur present in the organic compound gets oxidized to CO₂, H₂O and H₂SO₄ respectively on account of oxidation.

Carius tube is cut and contents removed by washing. Solution is filtered and washed well. Solution is warmed and treated with excess of BaCl₂ solution, when a white precipitate of BaSO₄ is obtained. Heat the ppt. at low temperature for sometime for aglomerization. Precipitate of BaSO₄ is filtered, washed, dried and weighed. Percentage of sulphur is then calculated.

H₂SO₄ + BaCl₂ → BaSO₄↓ White precipitate + 2HCL
% of sulphar = \(\frac{32}{233} \) × \(\frac{\text { Weight of } \mathrm{BaSO}_{4}}{\text { Weight of org. compound }}\) × 100

Question 23.
How are halogens present in organic compound estimated?
Answer:
Estimation of halogens: Canus method: In this method estimation of halogens (Cl, Br and I) is done. A known weight of organic compound containing halogen is heated with AgNO₃ and fuming nitric acid. Carbon, hydrogen and sulphur present in the compound gets oxidized and halogens form AgX (Silver halide).


Observation: Mass of Silver halide = Mgm
Mass of organic compound = Wgm
% amount of Cl = \(\frac{35 \cdot 5}{143 \cdot 5}\) × \(\frac{\text { Mass of } \mathrm{AgCl}}{\text { Mass of organic compound }} \) × 100
%amount of Br = \(\frac{80}{188}\) × \(\frac{\text { Mass of } \mathrm{AgBr}}{\text { Mass of organic compound }}\) × 100
% amount of l =\(\frac{127}{235}\) × \(\frac{\text { Mass of AgI }}{\text { Mass of organic compound }}\) × 100

Question 24.
Write the principle of Kjeldahl’s method.
Answer:
Principle: Nitrogen-containing organic compound is heated with concentrated sulphuric acid in presence of K₂SO₄ and CuSO₄. The nitrogen present is converted into ammonium sulphate. On heating ammonium sulphate with NaOH, the whole of N₂ is changed into ammonia gas.

The NH₃gas is passed into a known volume of standard solution of sulphuric acid. NH₃ neutralises sulphuric acid. The amount of unreacted sulphuric acid is determined by titrating it with a standard alkali. From this percentage of N₂ is calculated.

(NH₄)₂ SO₄ +2NaOH → Na₂SO₄ +2NH₃ + 2H₂O
2NH₃+H₂SO₄ → (NH₄)₂SO₄.

Question 25.
Explain distillation method of purification of compounds.
Answer:
The various types of distillation are :

• Simple distillation,
• Fractional distillation,
• Vacuum distillation,
• Steam distillation.

Simple distillation: This method is employed for the purification of those liquids which, boil without decomposition and are associated with non-volatile impurities.
Liquids which have a difference of 30-40°C in their boiling points are purified by this method. On heating the mixture, vapours of pure substance are formed which condenses as they pass through the air or water condenser.

The pure liquid collects in the receiver while the non-volatile impurities are left behind in the flask. Some glass beads are also added to the distillation flask to avoid bumping.

Question 26.
What is the principle of adsorption chromatography? Explain with diagram.
Answer:
Adsorption chromatography: It is also known as column chromatography. It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

The mixture to be separated and purified is first dissolved in a suitable non-polar organic solvent like petroleum ether, benzene, chloroform, alcohol, etc. This solution is allowed to flow down an absorption column.
For using this technique there should be proper adsorption column. Adsorbents like activated magnesia, alumina, calcium carbonate, gypsum, etc. is filled in a hard vertical tube (adsorbent column.).

Question 27.
Explain chain isomerism and functional isomerism with example.
Answer:
1. Chain isomerism: When the same molecular formula represents two or more compounds which differ in the nature of carbon chain (carbon skeleton) then such compounds are called chain isomers and the phenomenon is called chain isomerism.
Example : (a) C₄H₁₀ (butane) has two chain isomers.

(b) C₅H₁₂ (pentane) has three chain isomers.

2. Functional group isomerism: When the molecular formula represents two or more compounds which differ in the nature of the functional group, then such compounds are called the functional isomers and the phenomenon is called functional group isomerism. Example :
(a) C₃H₈O : CH₃ – CH₂ – CH₂ – OH
(i) Propanol
CH₃ -O – C₂H₅
(ii) Ethyl methyl ether

(b) C₄H₈O:

Question 28.
Write a note on carbocation and carbanion.
Answer:
Carbocations or Carbonium ions: Carbocations are positively charged carbon atoms having only six electrons in its valence shell. They are also highly reactive and act as reaction intermediate. They are also called carbonium ions. They are produced by heterolytic cleavage. Some examples are :

Types of carbocations: Carbocations are classified as primary, secondary or tertiary depending upon the nature of carbon atom carrying the positive charge.

Stability of Carbonium ion: Larger the dispersal of positive ion on other groups or atoms, larger is the stability of carbonium ion. With the increase in the number of alkyl group its stability increases.
Structure: Positively charged carbon atom is sp² hybridised and one p-orbital is vacant.

Question 29.
Give condensed and bond line structural formulae and identify the functional groups present if any, for :
(a) 2, 2,4-Trimethylpentane.
(b) 2-Hydroxy-1, 2,3 propane tricarboxylic acid.
(c) Hexanedial.
Answer:
To write the condensed and bond line formulae first let us write the structural formulae of the compounds :
(a)

(b)

(c)

Question 30.
Write a note on carbanion or cabon anion.
Answer:
Carbanions: Carbanions are negatively charged carbon surrounded by eight electrons in the valence shell. They are also short-lived and highly reactive. They also act as reaction intermediates. They are also called carbon anions.
Methyl carbanion: H₃C^– :
Ethyl carbanion :

Types of carbanions: Carbanions are classified as primary, secondary and tertiary depending upon the nature of carbon carrying negative charge.

Relative stability of carbanion: Carbanions are electron rich species and hence their stability depend on dispersal of electrons away from the carbon carrying negative charge. The order of stability of alkyl carbanion is as follows :
Methyl carbanion > Primary carbanion > Secondary carbanion > Tertiary carbanion

Structure: In it, carbon is in sp³ hybridised state, three bond paired electrons and one lone pair electron is present.
Lone pair of electron

Question 31.
Explain Metamerism and Tautomerism with example.
Answer:
Metamerism: When the same molecular formula represents two or more compounds which only differ in the nature of the alkyl groups attached to the same functional group, then such compounds are called metamers and the phenomenon is called metamerism.
Example: Molecular formula, C₄H₁₀O (ether) has three metamers :

Tautomerism: This is a special type of functional isomerism in which one compound
exhibits two isomers. Both of the isomers are present in equilibrium. It is a dynamic state in which both isomers change into each other.
These are called tautomers and the phenomenon is known as tautomerism.
Example : (i) Diad system: Hydrogen vibrates between two multivalent atoms.
H – C ≡ N

(ii) Triad system: Hydrogen vibrates between three multivalent atoms.

Question 32.
What Is Fractional crystallisation?
Answer:
When two or more constituents of an impure mixture differ in the amount of their solubility in the same solvent, they are separated by fractional crystallisation process. Let the two constituents of the mixture are A and B but solubility of A is less than B. Saturated solution of this mixture is prepared in a suitable solvent like alcohol any filtered. The solution is then cooled.

A being less soluble forms crystals first which are separated by filtration. Then the solution is cooled. Crystals of B are obtained which are separated by filtration. This process is repeated several times because crystals obtained in the first time still contain impurities.

Question 33.
What is inductive effect? What is its use?
Answer:
Inductive effect: In a covalent bond between two dissimilar atoms having different electronegativities, the electron pair does not remain in the centre but get attracted towards the more electronegative atoms. The bond becomes somewhat polar due to unequal

sharing of the electron pair. For example, in the bond if X is more electronegative than C, the electron pair get attracted towards X. This shifting of electrons develops a partial negative charge denoted by δ^– on X and C attains a partial positive charge denoted by δ⁺. Thus,

Now consider a long chain of carbon atoms with a more electronegative element say chlorine attached at one end.

The electron pair of the bond between C₁ and X gets displaced towards electronegative chlorine atom. This results in developing of partial negative charge on chlorine and partial positive charge on carbon. This displacement is further transmitted to other carbon atoms of the chain but the magnitude of displacement goes on decreasing with the increase in the distance of the carbon atoms from the chlorine atom as shown below :

(Here δδδ⁺ < δδ⁺ < δ⁺ )
Thus, it can be concluded that a polar bond induces polarity in the other covalent bonds in a chain. This type of displacement of electrons is referred to as inductive effect (or I effect) or transmission effect. Thus, inductive effect may be defined as, the permanent displacement of electrons along the chain of carbon atoms due to the presence of a polar covalent bond in the chain.

Types of inductive effect: There are two types of inductive effect:
(a) Electron withdrawing inductive effect (-1 effect): If the substance attached to the end of the carbon chain is electron-withdrawing, the effect is called -I effect. The de¬creasing order of-I effect of some atoms or groups is as follows :
– NO₂ > -C ≡ N> -F> -COOH> -Cl> -Br>-I> -OCH₃> – C₆H₅ > – H

(b) Electron releasing inductive effect (+1 effect): If the substance attached to the end of the carbon chain is electron releasing, the effect is called +1 effect. Alkyl groups are electron releasing in nature. Thus, the decreasing order of+1 effect is as follows :

Uses :

1. Comparison of strength of fatty acids
2. Basic character of amines
3. Presence of dipole moment.

Question 34.
Write IUPAC name of the following compounds :
(a)
(b)
(c)
(d)
(e)
(f) Cl₂CHCH₂OH
Answer:
(a)
(b)
(c)
(d)
(e)

(f)

Question 35.
What do you understand by Electromeric effect? Write a note on it.
Answer:
Electromeric effect: It is a temporary effect which takes place in compounds containing multiple covalent bonds (e.g., C = C, C = O, C = N, etc.) or an atom with a lone pair of electrons adjacent to a covalent bond.
It is known to us that a multiple bond consist of σ – and π – bonds and the π -bonding electrons in π -bonds are loosely held.

Therefore, when a charged species approaches compound having double or triple bond, the electrons of the π – bond due to electrostatic attraction are completely polarized or displaced towards one of the constituent atoms. This transference of electron pair is shown by a curved arrow, starting at the original position of the electron pair and ending to the new position of the electron pair as shown below :

Types of Electro meric effect: There are two types of electromeric effects :

1. +E effect: When the transfer of electron takes place towards the atom of the double bond to which the attacking species gets finally attached, the effect is called +E effect. Like -F, -NR₂, -OR etc.
2. -E effect: When the transfer of electron takes place away from the atom of the double bond to which the attacking species finally gets attached, the effect is called -Eeffect.

Uses:

• Addition reaction of halogen on unsaturated hydrocarbons (Alkene and Alkyne)
• Reaction of halogen acid on alkene and alkyne containing carbonyl group.

Question 36.
Write the difference between Inductive effect and Electromeric effect.
Answer:
Differences between Inductive and Electromeric Effect

Question 37.
Explain the rules of nomenclature of compounds containing functional group with example.
Answer:
Rules of Nomenclature:

• The carbon chain containing the functional group is considered to be the main chain.
• If functional group contains carbon then it is counted in the main chain.
• Numbering of chain is started from the side which the functional group is nearer.
• During naming the substituents are named first and then the main functional group.
• If more than one functional group is present in the chain then on the basis of priority the functional group is selected.
• Then before naming the substituents, leaving the main functional group, position of other functional group and their names are written.

Question 38.
Represent σ – and π -bond in the following :
(i) C₆H₆,
(ii) C₆H₁₂,
(iii) CH₂C₁₂,
(iv) CH₂ =C = CH₂,
(V) CH₃ NO₂,
(vi) HCO – NHCH₃
Answer:
(i)

(ii)

(iii)
(iv)
(v)

(vi)

Question 39.
Write IUPAC name and formula of the following compounds :
1. Iodoform
2. Vinegar
3. Formic acid
4. Marsh gas
5. Acetone
6. Glycol
Answer:

Question 40.
Write structural formulae of the following compounds :
1. 3, 3-Dimethylpentene,
2. 4-Methylhexan-2-ol,
3. 1, 2-Dibromoethane,
4. Nitromethane,
5. 4-Methyl-3-hydroxypentanal.
Answer:



4. CH₃ – NO₂

Question 41.
Write the difference between Aliphatic compound and Aromatic compound.
Answer:
Differences between Aliphatic compound and Aromatic compound

Question 42.
What is resonance effect? Write its application.
Answer:
Resonance effect is defined as the polarity produced in the molecule by the interaction of two π -bonds or between a π -bond and lone pair of electrons present on an adjacent atom. The effect is transmitted through the chain. It is also known as mesomerie effect. It is a permanent effect.

Negative mesomerie effect: Electron pair is transferred towards the atom or substituent group. It is known as -M effect.
Example : – NO₂,-CHO, – SO₃H, > C = O.
Positive mesomeric effect: Electron pair is transferred away from an atom or substituent group. It is known as +M effect.
Example : – OH – OR, – SH, – SR, – NH₂, – Cl, – Br.

Applications :

• In the determination of structure of benzene.
• In description of dipole moment.
• Capability of acid and base.
• In conjugated addition reactions.

Question 43.
What are substitution reactions? Explain with example.
Answer:
Reactions in which an atom or group of atoms of a reactant molecule displace atoms or group of atoms of the reagent are called substitution reactions.
Example :
(i)
Here, H atom of methane is substituted by Cl atom.
(ii) CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl
In this reaction – OH group is substituted by Cl.

Organic Chemistry: Some Basic Principles and Techniques Class 11 Important Questions Long Answer Type

Question 1.
Identify the functional group of the following compounds :
(a)

(b)

(c)

Answer:
(a)

(b)

(c)

Question 2.
Explain Kjeldahl’s method of estimation of nitrogen under the following points :
1. Principle,
2. Chemical equation,
3. Diagram,
4. Observation and calculation.
Answer:
1. Principle: When nitrogenous organic substance is heated with cone. H₂SO₄, then nitrogen is completely converted to ammonium sulphate. The reaction mixture when heated with caustic alkali evolves ammonia which is passed in known volume of a standard acid solution.
After the neutralization the volume of remaining acid is determined by titration with a standard solution of an alkali.

2. Chemical equation:
N from organic compound
(NH₄ )₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O + 2NH₃
2NH₃ + H₂SO₄ → (NH₄)₂ SO₄

3. Diagram:

4. Observation and calculation :
Suppose,

•  Mass of organic substance =Wgm
• Normality of acid taken initially = N
• Volume of N normal acid taken initially = V₁ ml
• Volume of N normal alkali required to neutralize residual acid =V₂ml
  (This is equivalent to the amount of acid left after neutralization with NH₃) Therefore, volume of N normal acid neutralized by NH₃
  = V₁ – V₂ = V ml

Because V ml of N normal acid = V ml of N normal alkali = V ml of N normal of NH₃ (Acid and base always react in equivalent proportion)

1000 ml of normal solution of NH₃ contains =17 gm NH₃
= 14 gm nitrogen
∴ Mass of V ml of 1 N NH₃ = \(\frac{14}{1000} \) × V gm
∴ Mass of N₂ in V ml of N normal NH₃ = \(\frac{14}{1000}\) × V × N gm

This weight of nitrogen is present in W gm of substance W gram of substance contains = \(\frac{14 \times \mathrm{V}}{1000} \)
∴ 100 gm substance will contain = \(\frac{14 \times \mathrm{V}}{1000}\) × \( \frac{\mathrm{N} \times 100}{\mathrm{~W}} \) gm nitrogen
= \(\frac{1 \cdot 4 \times V \times N}{W} \)
∴ % of N = \(\frac{1 \cdot 4 \times V \times N}{W}\)

Question 3.
Which of the two: O₂NCH₂CH₂O or CH₃CH₂O is expected to be more stable and why?
Answer:
O₂NCH₂CH₂CT is more stable than CH₃CH₂O“ because (-) NO₂ group exhibit -1 effect due to which negative charge is rare. On the other side -CH₃ group exhibits +1 effect due to which negative charge becomes dense. Due to less of negative charge stability of ion increases whereas due to dense negative charge stability of ion decreases.

Question 4.
Give a brief description of the principles of the following techniques taking an example in each case :
(a) Crystallisation,
(b) Distillation,
(c) Chromatography.
Answer:
(a) Crystallisation: In this method, impure compound is converted to pure crystals. It is based on the difference in solubilities of compound and impurities in a suitable solvent. Impure compound is dissolved in such a solvent in which compound is partially soluble at normal temperature, but more soluble at high temperature.

Then the solution is concentrated till saturation. Solution is then cooled when pure compound crystallises. Example: Iodoform crystallises with alcohol. Benzoic acid mixed with naphthalene is purified by hot water.

(b) Distillation: In this method, impure liquid is converted to vapours by heating and again vapours are cooled to convert into liquid. This method is suitable for such liquids which do not decompose on heating and which contain non-volatile impurities. Two such liquids which differ sufficiently in the boiling points can be separated and purified by this method.

Example: Chloroform (b.p. 334K) and aniline (b.p. 457 K) can be easily separated by this method on boiling, liquid of lower b.p. evaporates first which is cooled and collected in the receiver.

(c) Chromatography: It is a technique of separation, purification and identification of components of a mixture. It is based on the specific adsorption of mobile phase of component on stationary phase. Stationary phase can be solid or liquid whereas mobile phase is liquid or gas.

Types of chromatography :

1. Column chromatography,
2. Thin layer chromatography,
3. Paper chromatography.

In column chromatography, stationary phase is solid and mobile phase is a mixture of solvents of variable polarity. Solid adsorbent is taken in a long glass tube with a suitable non-polar solvent. Concentrated solution of the mixture to be separated and purified is passed through upper part of the column. Due to difference in adsorption, the components of the mixture are separated as separated layers known as band (chromatogram).

Question 5.
Explain Victor Meyer method and volumetric method of determination molecular mass of organic compounds.
Answer:
Principle: This method is applicable for determination of mole-cular mass of a volatile substance. A definite mass of organic substance is heated in Victor Meyer’s tube to convert it into vapour state and volume of moist air displaced by vapours is noted. This volume is converted to volume at NTP by further calculation.

Mass of substance which gives 22-4 litre of vapours at NTP is determined. The weight of substance expressed in gram is the molecular mass of substance.

Apparatus: The apparatus used consists of the following parts :

• Copper jacket (J)
• Victor Meyer’s tube (A)
• Hofmann’s bottle.

Calculation: Suppose mass of organic compound = W gram
Volume of dry air displaced at S.T.P. = V ml
Atmospheric pressure = Pmm
Room temperature = t°C or t + 273 K
Aqueous tension at t° C =f
Pressure of dry air = P —f
Volume of vapour at S.T.P. :
In Experimental state At S.T.P.

P₁ = P -f
V₁ = V ml
T₁ = t + 273
P₂ = 760
V₂ = ?
T₂ = 273K

Using gas equation,
\(\frac{P_{1} V_{1}}{T_{1}} \) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}} \)
⇒ V₂ = \(\frac{P_{1} \times V_{1} \times T_{2}}{T_{1} \times P_{2}}\)
⇒ V ₂ = \(\frac{(P-f) \times V \times 273}{(t+273) \times 760}\)
At S.T.P. V₂ ml vapours are obtained from W gm substance.
∴ At S.T.P. 22,400 ml vapours obtained from = \(\frac{\mathrm{W}}{\mathrm{V}_{2}} \) × 22,400 gm.
Molecular mass = \(\frac{\mathrm{W}}{\mathrm{V}_{2}} \) × 22,400 gm.

Question 6.
Describe silver salt method of determination of molecular mass of organic acids.
Answer:
Principle: Organicacid is treated with silver nitrate to form silver salt. These silver salts are heated to obtain silvers. By the help of weight of silver salt and weight of silver, equivalent mass of silver salt is determined. By the equivalent mass of silver salt, equivalent mass and molecular mass of acid can be determined.

Method: Acid is first reacted with excess of NH₄OH by which ammonium salt is obtained.
R-COOH+NH₄OH → RCOONH₄+H₂O
It is heated to remove ammonia and neutral solution is treated with silver nitrate to obtain precipitate of silver salt which is washed, dried and its weight is determined.
R- COONH₄ + AgNO₃ → RCOOAg + NH₄NO₃
On heating the known weight of silver salt in platinum crucible, silver is obtained whose weight is determined.
Observation and calculation:
Weight of silver salt = Wgm
Weight of silver =x gin

\(\frac{\text { Equivalent weight of silver salt }}{\text { Equivalent weight of silver }}\) = \(\frac{\text { Weight of silver salt }}{\text { Weight of silver }} \)
\(\frac{\text { Equivalent weight of silver salt }}{108}\) = \(\frac{\mathrm{W}}{x}\)
∴ Equivalent weight of silver salt = \(\frac{\mathrm{W}}{x}\) × 108
If E is equivalent weight of acid then on substituting the H by silver, then equivalent weight of silver salt
(E-1) + 108 = E+107
⇒ E + 107 = \(\frac{\mathrm{W}}{x}\) × 108
⇒ E = \(\frac{\mathrm{W}}{x}\) × 108 – 107
If n is the basicity of acid
Molecular mass = [ \(\frac{\mathrm{W}}{x}\) × 108 – 107 ] × n.

Question 7.
Write principle of the chemistry of Lassaigne’s test.
Answer:
Sodium extract: Nitrogen, sulphur and halogens present in organic compound can be detected by Lassaigne’s test. Due to covalent nature of organic compounds, their reaction proceed with slow rate, thus it is impossible to test them directly in the laboratory.

Therefore, they are fused with sodium metal to convert them to ionic compounds and the solution formed is called sodium extract.

Test of Nitrogen:
Test of nitrogen (Lassaigne’s test): In 2 ml of sodium extract, 2 ml of freshly prepared solution of FeSO₄ is added along with little NaOH, warm it. Green precipitate of Fe(OH)₂ is obtained. The solution is cooled and small quantity of cone. HC1 is added in which green ppt. dissolves. Then little amount of FeCl3 solution is added. If green or blue colour precipitate is obtained, then the substance contains nitrogen.
Na + C + N → NaCN
FeSO₄ + 2NaOH → Na₂SO₄ + Fe(OH )₂

Test of sulphur :
(i) Sodium nitroprusside solution is added to the sodium extract of organic substance. If violet colour appears, it confirms the presence of sulphur in the given organic compound.
2Na+S → Na₂S

(ii) If acetic acid and lead acetate solution is added to the sodium extract then a black precipitate is obtained, presence of sulphur is confirmed in it.
Na₂S + (CH₃COO)₂ Pb → PbS ↓ Black + 2CH₃COONa.

Combined test of Nitrogen and Sulphur :

Neutralize the solution with HCl and add FeCl₃ solution. Sodium thiocyanate gives blood red colour and does not produce Prussian blue colour like in the test of nitrogen because in this reaction free cyanide ions are not present.

Estimation of halogens: Canus method: In this method estimation of halogens (Cl, Br and I) is done. A known weight of organic compound containing halogen is heated with AgNO₃ and fuming nitric acid. Carbon, hydrogen and sulphur present in the compound gets oxidized and halogens form AgX (Silver halide).


Observation:
Mass of Silver halide = Mgm
Mass of organic compound = Wgm
%amount of Cl = \(\frac{35 \cdot 5}{143 \cdot 5}\) × \(\frac{\text { Mass of } \mathrm{AgCl}}{\text { Mass of organic compound }} \) × 100
%amount of Br = \(\frac{80}{188}\) × \(\frac{\text { Mass of } \mathrm{AgBr}}{\text { Mass of organic compound }}\) × 100
% amount of l =\(\frac{127}{235}\) × \(\frac{\text { Mass of AgI }}{\text { Mass of organic compound }}\) × 100.

Question 8.
Describe chloroplatinate method of determination of molecular mass of organic base.
Answer:
Principle: Organic base reacts with chloroplatinic acid to form insoluble substance of general formula B₂H₂PtCl₆.
B represents equivalent mass of base. By the combustion of platinum salt, metallic platinum is obtained.

With different masses of platinum salt and platinum, molecular mass of base is determined.

Method: Organic base is first reacted with dilute HCl to obtain soluble hydrochloride. Platinic chloride is mixed in it to obtain salt of the base in the form of chloroplatinate. It is washed and dried. Known weight of it, is heated in platinum crucible to obtain platinum in the crucible as residue. By it equivalent mass of base is determined.

Calculation : Equivalent weight of base = E
Weight of platinum salt = Wgm Weight of platinum = xgm
Molecular mass of B₂ H₂PtCl₆ = 2E + 2 + 195 + 213 = 2 E + 410
\(\frac{\text { Molecular mass of salt }}{\text { Molecular mass of platinum }}\) =\(\frac{\text { Weight of platinum salt }}{\text { Weight of platinum }}\)
or \(\frac{2 \mathrm{E}+410}{195}\) = \(\frac{\mathrm{W}}{x}\)
or 2E = [\(\frac{\mathrm{W}}{x}\) × 195 – 410]
or = \(\frac{1}{2}\) [\(\frac{\mathrm{W}}{x}\) × 195 – 410]
If acidity of base in then,
Molecular mass of base = \(\frac{1}{2}\) [\(\frac{\mathrm{W}}{x}\) × 195 – 410] × n.

Question 9.
Explain Duma’s method of estimation of nitrogen under the following heads:
(i) Principle and method,
(ii) Labelled diagram,
(iii) Observation and calculation.
Answer:
(i) Principle and method: Organic substance and cupric oxide is ignited in hard glass combustion tube in atmosphere of carbon dioxide. Carbon of compound changes to CO₂ and hydrogen to H₂O. By oxidation nitrogen is obtained in free state. Some oxide of nitrogen are also formed.

Nitrogen+CuO → N₂ + Some amount of some oxides of N₂.
Oxides of nitrogen when passed over hot reduced copper gauze change to nitrogen by losing their oxygen.
Oxides of nitrogen + Cu → N₂ + CuO

The nitrogen set free is passed through Schiff’s nitrometer filled with 40% solution of caustic potash. Caustic potash solution absorb CO₂ while water formed gets condensed. Again CO₂ is passed through combustion tube.to drive out all the remaining nitrogen gas to nitrometer. After cooling nitrometer, the volume of nitrogen in nitrometer is noted.

(ii) Labelled diagram :

(iii) Observation and calculation
Let,

• Weight of organic compound = W gm
• Volume of moist N = V ml
• Temperature = t°C
• Atmospheric pressure = P mm
• Aqueous tension at t°C = p mm.
  Calculation of volume of nitrogen at N.T.P.:

By experiment At N.T.P.
P1 = (P – p) mm
P2 = 760 mm
(Pressure of dry gas)
V, = V ml
V2 = ?
T1 = (t + 273)K
T2 = 0 + 273 = 273K

Using gas equation,
\(\frac{P_{1} V_{1}}{T_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
\(\frac{(P-p) \times V}{(t+273)}\) = \(\frac{760 \times V_{2}}{273}\)
∴ V₂ = \(\frac{(P-p) \times V}{(t+273)}\) × \(\frac{273}{760} \)
∵ Mass of 22400 ml of N₂ at N.T.P = 28 gm
∴ Mass of V₂ ml of N₂ at N.T.P. = \(\frac{28 \times \mathrm{V}_{2}}{22,400}\)
When in W gm of substance N₂ is = \(\frac{28 \times V_{2}}{22,400} \times \frac{100}{W}\)
∴ Percentage of N₂ in organic compound = \(\frac{28}{22,400} \times \frac{\text { Volume of } \mathrm{N}_{2} \text { at N.T.P. }}{\text { Weight of organic substance }} \times 100\)

Organic Chemistry: Some Basic Principles and Techniques Class 11 Important Numerical Questions

Question 1.
By the combustion of 0.1877gm sample of an organic compound at 14°C and 758mm pressure 31.7ml moist N₂ gas is obtained. Calculate the % amount of nitrogen in the compound. (Aqueous tension at 14°C = 12 mm) (NCERT)
Solution: \(\frac{P_{1} V_{1}}{T_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
P₁ = 758 -12mm,
P₂ = 760
V₁ = 31.7ml
V₂ = ?
T1 =14 + 273
T₂ = 273

Volume of nitrogen at N.T.P. V₂ = \(\frac{(758-12) \times 31 \cdot 7 \times 273}{(14+273) \times 760}\)= 29.6 ml
Percentage amount of N₂ =\(\frac{28 \times \mathrm{V}_{2}}{22,400} \times \frac{100}{\mathrm{~W}} \)
= \(\frac{28 \times 29 \cdot 6}{22,400} \times \frac{100}{0 \cdot 1877}\) = 19.72%

Question 2.
0.2870 gram of AgCl is obtained by Carius method by 0.2 gm of an organic compound containing chlorine. Calculate the percentage of chlorine in the compound. (NCERT)
Solution:
Weight of organic compound = 0.2 gm
Weight of AgCl = 0.2870 gm.

∴ Percentage of chlorine = \(\frac{35 \cdot 5}{143 \cdot 5}\) × \(\frac{\text { Weight of } \mathrm{AgCl}}{\text { Weight of organic compound }} \) × 100
= \(\frac{35 \cdot 5}{143 \cdot 5} \times \frac{0 \cdot 2870}{0 \cdot 2} \times 100\) = 35.5%

Question 3.
In Victor Meyer method 20ml air was displaced over water by the vapours of 0.1gm liquid at 15°C and 765mm pressure. Determine the molecular mass of the liquid (Aqueous tension at 15°C = 13mm).
Solution:
\(\frac{P_{1} V_{1}}{T_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)

P₁ = 765 – 13
V₁ = 20
T₁ = 25 + 273
P₂ = 760
V₂ = ?
T₂ = 273

∴ V₂ = \(\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}\)
V₂ = \(\frac{752 \times 20 \times 273}{288 \times 760}\) = 18.76 ml
∵ Mass of 18.76 ml vapours = 0.1 gm
∴ Mass of 22,480 ml vapours = \(\frac{0 \cdot 1 \times 22,400}{18 \cdot 76}\) = 119.4 gm
Molecular Mass = 119.4 gm

Question 4.
Ammonia produced by the analysis of 06 gm organic compound was absorbed in 90m1 of \(\frac{\mathbf{N}}{\mathbf{9}} \) H₂SO₄. 20m1 of 01 N caustic soda was required to neutralize the remaining acid. Determine the percent amount of nitrogen in compound. (Atomic mass of H = 1,O = 16, N =14, Na = 23, S = 32) .
Solution:
Acid taken =90 ml \(\frac{\mathbf{N}}{\mathbf{9}} \) H₂SO4
= 10 ml NH₂SO₄
Caustic soda used to neutralize the remaining acid = 20ml 0.1 N NaOH = 2 ml N NaOH
Acid neutralised by ammonia = 10 – 2 = 8 ml (NH₄)₂SO₄
Percent amount of nitrogen = \(\frac{1 \cdot 4 \times N \times V}{W}\)
= \(\frac{1 \cdot 4 \times 1 \times 8}{0 \cdot 6}\)
= 18.66%

Question 5.
In the estimation of sulphur by Carius method, 0.468gm of an organic sulphur compound afforded 0.668gm of barium sulphate. Find out the percentage of sulphur in the given compound. (NCERT)
Solution:
Given, weight of organic compound = 0.468gm.
Weight of BaS04 produced = 0.668gm
Percentage amount of sulphar = \(\frac{32}{233} \) × \(\frac{\text { Weight of } \mathrm{BaSO}_{4} \text { produced }}{\text { Weight of organic compound }} \times 100 \)
= \(\frac{32}{233} \) × \(\frac{0 \cdot 668}{0.468} \) × 100
= 19.60%

Question 6.
An organic substance, A contains C = 40%, H = 6.66% and the rest is oxygen. Vapour density of A is 30. Calculate empirical formula and molecular formula. Or, An organic substance A contains C = 40% and H = 6.66% vapour density of A is 30. It turns blue litmus red and reacts with bases. On heating it, sodium salt with soda lime, first member of paraffin series is obtained, what is it?
Solution:
C = 40%, H = 6.66%
O = 100 -(40 + 6.66)
= 100-46.66 = 53.34%.

The empirical formula of A = CH₂O.
Empirical formula mass = 12 + 2 + 16 = 30
∵ Vapour density of compound = 30
∴ Molecular mass = 2 × V.D. = 2 × 30 = 60

Now, n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\) = \(\frac{60}{30}\) = 2
∴ Molecular formula of compound = [CH₂O]₂
= C₂H₄O₂
Substance A turns blue litmus red, therefore it is an acid. On heating its sodium salt with soda lime first member of paraffin series (methane) is obtained therefore the acid will be CH₃COOH.

Organic Chemistry: Some Basic Principles and Techniques Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Alicyclic compound is:
(a) Aromatic compound
(b) Aliphatic compound
(c) Heterocyclic compound
(d) Aliphatic cyclic compound.
Answer:
(d) Aliphatic cyclic compound.

Question 2.
Chemical formula of dry ice is:
(a) H₂O
(b) (H₂O)
(c) CO₂
(d) H₂CO₃.
Answer:
(c) CO₂

Question 3.
It mainly present in marsh gas:
(a) C₂H₂
(b) CH₄
(c) H₂S
(d) CO₂.
Answer:
(b) CH₄

Question 4.
General formula of alkyne Is:
(a) C_nH_2n+2
(b) C_nH_2n+1
(c) C_nH₂
(d) C_nH_2n-2
Answer:
(d) C_nH_{2n -2}

Question 5.
IUPAC name of (CH₃)₂ CH-CH = CH – CH₃ is:
(a) 2-methylpent-3-ene
(b) 4-methylpent-2-ene
(c) 2-isopropyiprop- 1-ene
(d) 3-isopropylprop-2-ene.
Answer:
(b) 4-methylpent-2-ene

Question 6.
Chief source of organic compounds is :
(a) Coaltar
(b) Petroleum
(c) Both (a) and (b)
(d) None of these.
Answer:
(c) Both (a) and (b)

Question 7.
What type of isomers are alcohol and ether:
(a) Chain
(b) Position
(c) Functional
(d) Metamerism.
Answer:
(c) Functional

Question 8.
General formula of alcohol is:
(a) C_nH_2n+1
(b) C_nH_2n+1.OH
(c) C_nH_2n-2
(d) C_nH_2n.
Answer:
(b) C_nH_2n+1.OH

Question 9.
Which type of isomerism is shown by the following compounds:

(a) Only functional
(b) Only chain
(c) Position and chain
(d) Only position.
Answer:
(b) Only chain

Question 10.
Which of the following is an example of carbanion:
(a) CH₃^–
(b)
(c)
(d) CH₃.
Answer:
(b)

Question 11.
Are known as chiral molecule:
(a) Which do not superimpose on their mirror image
(b) Which superimpose on their mirror image
(c) Which show geometrical isomerism
(d) Which are stable molecules.
Answer:
(a) Which do not superimpose on their mirror image

Question 12.
Isomers are similar in:
(a) Structural formula
(b) Molecular mass
(c) Chemical properties
(d) Physical properties.
Answer:
(b) Molecular mass

Question 13.
In which of the following all the three groups represent – 1 effect:
(a) -NO₂, -Br, -CH₃
(b) -I, -NO₂, -C₂H₅
(c) -Cl, -C₂H₅, -CH₃
(d) -F, -NO₂, -C₆H₅.
Answer:
(d) -F, -NO₂, -C₆H₅.

Question 14.
Which of the following compound can exist In geometrical isomer form:
(a)
(b)
(c)
(d)
Answer:
(b)

Question
Example of nucleophile is:
(a) F ion
(b) H₃O₄ion
(c) Cl atom
(d) Aniline hydrochloride.
Answer:
(a) F ion

Question 16.
ExIstence of a substance In two or more solid forms is known as:
(a) Allotropy
(b) Polymensm
(c) Isomerism
(d) Enantiomerism.
Answer:
(a) Allotropy

Question 17.
Name of compound of structure CI. CH₂CH₂COOH is:
(a) 3-chioropropanoic acid
(b) 2-chloropropanoic acid
(c) 2-chloroethanoic acid
(d) Chiorosuccinic acid.
Answer:
(a) 3-chioropropanoic acid

Question 18.
Structure of isobutyl chloride is:
(a) CH₃CH₂CH₂CH₂Cl
(b) (CH₃)₂CH.CH₂Cl
(c) CH₃CH₂CHCI.CH₃
(d) (CH₃)₃C-Cl.
Answer:
(c) CH₃CH₂CHCI.CH₃

Question 19.
R-CONH₂ is :
(a) An amide
(b) An alcohol
(c) An acid
(d) A cyanide.
Answer:
(a) An amide

Question 20.
C_nH_2n is the general formula of:
(a) Alkanes
(b) Alkenes
(c) Alkynes
(d) Arenes.
Answer:
(b) Alkenes

Question 21.
Bond angle in carbon tetrachloride is about:
(a) 90°
(b) 109°
(c) 120°
(d) 180°.
Answer:
(b) 109°

Question 22.
Name of (CH₃)₂CH-O-CH₂-CH₂ -CH₃ is :
(a) Isopropyl propyl ether
(b) Dipropyl ether
(c) Di-isopropyl ether
(d) Isopropyl propyl ketone.
Answer:
(a) Isopropyl propyl ether

Question 23.
IUPAC name of compound is :
(a) 4-bromo-2-hydroxypentanoic acid
(b) 2-bromo-4-hydroxypentanoic acid
(c) 2-bromo-4-hydroxypentanal
(d) 2-bromo-4-hydroxypentanone.
Answer:
(b) 2-bromo-4-hydroxypentanoic acid

Question 24.
CH₃CH₂I + KOH(aq) → CH₃CH₂OH + Kl. Classify this reaction into the following type:
(a) Electrophilic substitution
(b) Nucleophilic substitution
(c) Elimination
(d) Combination.
Answer:
(b) Nucleophilic substitution

Question 25.
General formula of Alkenes Is:
(a) C_nH_2n+2
(b) C_n H_2n
(c) C_n H_2n-2
(d) C_2n+2H_n.
Answer:
(b) C_n H_2n

Question 26.
IUPAC name of the following compound is:

(a) 3, 4-dimethyl-3-n-propylnonane
(b) 6, 7-dimethyl-7-ethyldecane
(c) 6, 7-dimethyl-7-propylnonane
(d) 4, 5-dimethyl-4-ethyLnonane.
Answer:
(d) 4, 5-dimethyl-4-ethyLnonane.

Question 27.
The gas formed by the action of water on calcium carbide is:
(a) Methane
(b) Ethane
(c) Ethylene
(d) Acetylene.
Answer:
(d) Acetylene.

Question 28.
Baeyer’s reagent is:
(a) Alkaline KMnO₄
(b) Ammoniacal AgNO₃
(c) Ammoniacal CuSO₄
(d) Acidic CaSO₄.
Answer:
(a) Alkaline KMnO₄

Question 29.
Butane-I, 2-diene contains:
(a) Only sp hybridized carbon atoms
(b) Only sp² hybridized carbon atoms
(c) sp and sp² hybridized carbon atoms
(d) sp, sp² and sp³ hybridized carbon atoms.
Answer:
(d) sp, sp² and sp³ hybridized carbon atoms.

Question 30.
IUPAC name of compound
(a) 2-ethylbut-2-ene
(b) 3-ethylbut-2-ene
(c) 3-methylpen-2-ene
(d) 3-methylpent-2-ene.
Answer:
(d) 3-methylpent-2-ene.

Question 31.
IUPAC name of Cl₃C.CH₂CHO is :
(a) 3, 3, 3-trichloropropanal
(b) 1, 1, l-trichloropropanal
(c) 2, 2, 2-trichloropropanal
(d) Chloral.
Answer:
(a) 3, 3, 3-trichloropropanal

Question 32.
Three-dimensional isomers differ in :
(a) Configuration
(b) Conformation
(c) They are not different
(d) None of these.
Answer:
(a) Configuration

Question 33.
CH₃ -CH(OH) – COOH represent:
(a) Geometrical isomerism
(b) Optical isomerism
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Optical isomerism

Question 34.
Nitroethane is one of the type of the following isomerism :
(a) Metamerism
(b) Optical activity
(c) Tautomerism
(d) Position isomerism.
Answer:
(c) Tautomerism

Question 35.
Aniline is generally purified by :
(a) Steam distillation
(b) Simple distillation
(c) Distillation under reduced pressure
(d) Sublimation.
Answer:
(a) Steam distillation

Question 36.
Glycerol boils at 290°C with slight decomposition. Impure glycerol is purified by:
(a) Steam distillation
(b) Simple distillation
(c) Vacuum distillation
(d) Extraction with solvent.
Answer:
(c) Vacuum distillation

Question 37.
The blue or green colour obtained in Lassaigne’s test is due to the formation of:
(a) NaCN
(b) Na₄[Fe(CN)₆]
(c) Fe₃[Fe(CN)₆]₄
(d) Fe₄[Fe(CN)₆]₃.
Answer:
(d) Fe₄[Fe(CN)₆]₃.

Question 38.
Formation of blood-red colouration in Lassaigne’s test indicates the presence of:
(a) Nitrogen
(b) Sulphur
(c) Nitrogen and sulphur
(d) Sulphur and iodine.
Answer:
(b) Sulphur

Question 39.
The blood-red coloured compound formed during the qualitative analysis of nitrogen and sulphur together is :
(a) Fe₄[Fe(CN)₆]₂
(b) Fe(SCN)₃
(c)KSCN
(d) Na₂S.NaCN.
Answer:
(b) Fe(SCN)₃

Question 40.
Kjeldahl method is used for estimation of:
(a) Sulphur
(b) Nitrogen
(c) Halogens
(d) Oxygen.
Answer:
(c) Halogens

Question 41.
A compound with empirical formula CH₂O has molecular mass 90. The formula of compound is:
(a) C₄H₁₀O₂
(b) C₂H₅O
(c) C₃H₆O₃
(d) C₅H₁₄O.
Answer:
(c) C₃H₆O₃

Question 42.
The amount of sulphur present in an organic compound is estimated by changing it into:
(a) H₂S
(b) SO₂
(c) H₂SO₃
(d) H₂SO₄.
Answer:
(d) H₂SO₄.

Question 43.
The reagent used in Carius method to estimate halogens is :
(a) HNO₃ and HCl
(b) HNO₃ and H₂SO₄
(c) Fuming HNO₃ and BaCl₂
(d) Fuming HNO₃ and AgNO₃.
Answer:
(d) Fuming HNO₃ and AgNO₃.

Question 44.
The gas collected in Duma’s method of estimation of nitrogen in organic com-pound is:
(a) N₂
(b) NO
(c) NH₃
(d) None of these.
Answer:
(a) N₂

Question 45.
An organic compound contains C = 80% and H = 20%. The compound shall be:
(a) C₆H₆
(b) C₂H₅OH
(c) C₂H₆
(d) CHCl₃.
Answer:
(c) C₂H₆

Question 46.
An organic compound contains C = 39-9%, H = 6-7%, O = 53*4%. Its empirical formula will be:
(a) CHO
(b) CHO₂
(c) CH₂O₂
(d) CH₂O.
Answer:
(d) CH₂O.

Question 47.
In an organic compound, the ratio of mass is C: H: O = 4: 1: 5. Its empirical formula shall be:
(a) C₂HO
(b) C₅H₄O₄
(c) CH₄O₂
(d) CH₃O.
Answer:
(d) CH₃O.

Question 48.
The best modern technique of separation and purification of organic compounds is :
(a) Crystallization
(b) Distillation
(c) Sublimation
(d) Chromatography.
Answer:
(d) Chromatography.

Question 49.
In the organic compound CH₂ = CH – CH₂ – CH₂ – C = CH, by which hybrid orbital C₂– C₃ bond is formed :
(a) sp – sp²
(b) sp – sp³
(c) sp² – sp³
(d) sp³ – sp³.
Answer:
(c) sp² – sp³

Question 50.
Formation of Prussian blue colour for the identification of nitrogen by Lassaigne’s test in an organic compound is obtained due to :
(a) Na₄[Fe(CN)₆]
(b) Fe₄[Fe(CN)₆]₃
(c) Fe₂[Fe(CN)₆]
(d) Fe₃[Fe(CN)₆]_4.
Answer:
(b) Fe₄[Fe(CN)₆]₃

Question 51.
Which of the following carbocation is most stable :
(a)
(b)
(C)
(d)
Answer:
(b)

2. Fill in the blanks:

1. IUPAC name of is ………………….. .
Answer:
3-methylpent-2-ene

2. Chemical name of freon organic compound which is used in air conditioner and refrigerators is………………….. and its chemical formula is ………………….. .
Answer:
Chlorofluoro carbon,CF₂Cl₂

3. ………………….. ratio of elements in a compound is called its empirical formula.
Answer:
Simplest

4. The process of fractional crystallization of separation of two substances depend on the difference of ………………….. .
Answer:
Solvent

5. In organic compound, presence of amount of halogen can be detected by converting it into ………………….. .
Answer:
Silver halide

6. A compound contains 80% carbon and 20% hydrogen, its formula will be ………………….. .
Answer:
C₂H₆

7. Marsh gas mainly contains ………………….. .
Answer:
Methane

8. Process of separation of substances at different rates of adsorption is known as ………………….. .
Answer:
Elution

9. Kjeldahl method is used for the estimation of …………………… element.
Answer:
Nitrogen

10. Separation of mixture of two substances is based on ………………….. .
Answer:
Solubility

11. Separation of mixture of o-nitrophenol and p-nitrophenol is based on ………………….. .
Answer:
Steam distillation

12. Glycerine decomposes at its boiling point, it is purified by ………………….. .
Answer:
Distillation at low pressure.

3. Match the following:

‘A’	‘B’
1. Marsh gas	(a) (CH3)-C-OH
2. Vinegar	(b) CH3-CH2-OH
3. Absolute alcohol	(c) HCHO
4. Formalin	(d) CH3NC
5. lodoform	(e)
6. Ethane nitrile	(f)
7. Carbylamine methane	(g) CH3-COOH
8. Formic acid	(h) CH4
9. Gem dihalide	(i) CHl3
1o. Tertiary butyl alcohol	(j)CH3CN.

Answer:
1. (h) CH₄
2. (g) CH₃-COOH
3. (b) CH₃-CH₂-OH
4. (c) HCHO
5. (j) CH₃CN
6. (i) CHl₃
7. (d) CH₃NC
8. (e)
9.  (f)
10.  (a) (CH₃)-C-OH.

4. Answer in one word/sentence:

1. IUPAC name of CH₃ – CO – CH₃ is.
Answer:
Propanone-2

2. Chemical name of CH₃ – CH₂ – CHCl – CH₃ is.
Answer:
Iso butyl chloride

3. Mixture of KMnO₄ and KOH is known as.
Answer:
Baeyer’s reagent

4. On adding femc chloride in sodium extract red colour is obtained which prove the presence of which element?
Answer:
Nitrogen and sulphur

5. In organic compounds amount of S is determined by converting it to which compound?
Answer:
H₂SO

6. How many moles of H₂O are formed by the burning of one mole C₃H₆ in air?
Answer:
6

7. IUPAC name of Vinegar is.
Answer:
Ethanoic acid

8. IUPAC name of çthyl alcohol is.
Answer:
Ethanol

9. Camphor is separated from a mixttire of NaCl is?
Answer:
Sublimation

10. Naphthalene is purified by.
Answer:
Sublimation

11. Separation by coLumn chromatography is due to.
Answer:
Difference in adsorption

12. Petroleum is refined by.
Answer:
Fractional distillation

13. Beilstein Test is used.
Answer:
In detection of halogens

14. Free radicals are formed due to which type of fission?
answer:
Homolytic fission.

These MP Board Class 11th Chemistry Notes for Chapter 6 Thermodynamics help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 6 Thermodynamics

→ Energy : Capacity to do work is Energy.

→ Chemical energetics : Branch of chemistry which deals with the energy changes associated with chemical reaction is called chemical energetics.

→ In chemical reaction energy is used in breaking the bonds of reactant compounds whereas energy is released in the making of new bonds in product compounds.

→ Exothermic reactions : Reactions in which energy is released.
A + B → C + D,+ Q or – ΔH

→ Endothermic reactions : Reactions in which energy is absorbed.
A + B → C + D,- Q or + Δ H

→ Internal energy is the sum of various types of energies of molecules of a substance. It is a state function.

→ Internal energy change (Δ E): Heat given to a system at constant temperature and constant volume is equal to increase in its Internal Energy (ΔE = q_v).

→ Enthalpy (H) : Enthalpy of any system is the sum of internal energy and product of pressure and volume, i.e., H = E + PV.

→ Enthalpy change (ΔH) : Enthalpy change is the difference of enthalpies of products and reactants in a chemical reaction, i.e.,
ΔH = H_(products) – H_(reactants)

Enthalpy change is also related with change in internal energy and work done, i.e.,
ΔH = ΔE + PΔV

Heat (q_p) given to the system at constant pressure is equal to increase in enthalpy of the system ΔH = q_p

→ First law of thermodynamics : According to it, Total energy of the universe is always con¬served. Mathematically AE = q + W

→ Heat of reaction : Is the enthalpy change in a chemical reaction (Difference in Enthalpy of products and reactants).

→ Heat of neutralization : Energy produced in neutralization of one gram equivalent of acid by one gram equivalent of base at constant temperature.
For a strong acid and strong base value of ΔH = -57.1 kJ.

→ Heat of combustion: At constant pressure, when one mole of any substance completely reacts with excess of oxygen, energy released in this process is called heat of combustion.

→ Heat of formation: When 1 mole of substance is formed from its constituent elements in their stable states, the energy absorbed or released is called heat of formation.

→ Heat of fusion : Enthalpy change when one mole of solid substance is liquified, is called heat
of fusion (H_fussion)

→ Heat of vaporization : When one mole liquid substance changes into gaseous state, enthalpy change in this process is called heat of vaporization (H_vap)

→ Hess’s law: The enthalpy change in a chemical process is always same whether the process is carried out in one or in several steps.

→ Bond energy: The average energy required to break the bonds present in 1 mole of a gaseous molecule is called bond energy.

→ Endothermic Reaction : Absorption of heat in a reaction ΔH = +Ve .
→ Enthalpy (H) is a state function :
H = E + PV
ΔH = ΔE + PΔV .
q_p, =ΔH
ΔH = ΔE + ΔnRT
g_v = ΔE
4.18 x 10⁷ Erg = 1 calorie

→ Entropy change (ΔS) = \(\frac{q_{\mathrm{rev}}}{\mathrm{T}}\)

→ For spontaneous process, ΔG is negative.

→ If the process is in equilibrium, ΔS is zero.

→ Free energy, G = H-TS and AG = ΔH-TΔS.

→ Heat of reaction at constant volume : Heat of reaction at constant volume is equal to the difference of internal energy of products (EP) and internal energy of reactants (ER). /.e.,
ΔE = E_p – E_R = q_v

→ Heat of reaction at constant pressure (q_p) : Heat of reaction at constant pressure is equal to the heat absorbed or evolved when all moles of reactants react completely at constant pressure.

These MP Board Class 11th Chemistry Notes for Chapter 5 States of Matter help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 5 States of Matter

→ Three states of matter are solid, liquid and gas. These states are interconvertible by the effect of temperature and pressure.

→ Attractive force between the molecules is known as intermolecular force.

→ In crystalline solid, molecules are arranged in a definite order.

→ Molecules in gas are arranged in an indefinite order.

→ At STP/NTP, T = 273.15 K, P = 1 atm = 101.325 kP_a, V = 22.4 L mol^-1.

→ 1 Pascal (pa) = 1 Nm^-2, 1 atmospheres (atm) = 101325 x 10⁵ Nm^-2.

→ Boyle’s law : At constant tçmperature, the volume of definite mass of gas is inversely proportional to pressure (P).
V ∝ \(\frac{1}{\mathrm{P}}\) or PV = Constant (At constant T) or P₁V1 = P₂V2

→ Charles’ law: At constant pressure, the volume of a given mass of a gas is directly proportional to absolute temperature.
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (At constant P)

→ Gay Lussac’s law : Whenever gases combine, they combine in simple ratio of their volumes, if the products are in gaseous state, then their volume is also in simple ratio of the reacting gases.
P ∝ T (At Constant V)

→ Avogadro’s law : Equal volume of aH gases, under similar conditions of temperature and pressure contain equal number of molecules.
V ∝ n (At constant temperature and pressure)

→ Ideal gas equation: PV = nRT or P = \(\frac{n \mathrm{RT}}{\mathrm{V}}=\frac{m \mathrm{RT}}{\mathrm{MV}}=\frac{d \mathrm{RT}}{\mathrm{M}}\)
Where d= Density = gL¹
PV = RT if n = 1
Where R = Gas constant.

→ Graham’s law of diffusion : At constant temperature and pressure, rate of diffusion of gases is inversely proportional to the square root of their densities.
\(r \propto \sqrt{\frac{1}{d}} \text { or } \frac{d_{2}}{d_{1}}\)

(i) Under similar pressure
\(\frac{r_{1}}{r_{2}}=\sqrt{\frac{d_{1}}{d_{2}}}=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}=\frac{\mathrm{V}_{\mathrm{l}} t_{2}}{\mathrm{~V}_{2} t_{1}}\)

Where, d = Vapour density, M = molar mass of gas, V = volume of gas diffused, t = time
required in difusion.

(ii) Diffusion at different pressure:

→ Kinetic theory: PV = \(\frac{1}{2}\)mu²
Where, u = square root velocity,
KE = \(\frac{3}{2}\)nRT

Average kinetic energy KE. =\(\frac{3}{2} \frac{\mathrm{RT}}{\mathrm{N}_{0}}=\frac{3}{2}\)KT where, K = \(\frac{\mathbf{R}}{\mathrm{N}_{0}}\) = Boltzmann constant.

→ Dalton’s law of partial pressure: At constant temperature the total pressure of a mixture of
gases is equal to the sum of their partial pressures (P = p₁ + p₂ + p₃ +…………………. ).
P₁ = P_total. X₁
Where, P₁, P₂ are partial pressure,
X₁ = Mole fraction
P_{dry gas} = P_total = Aqueous tension.

→ Kinetic gas equation : PV = \(\frac{1}{3}\)mnv²

→ Root Mean Square velocity : R.M.S. velocity is the square root of the average mean of velocities of n moles present in a gas.

→ Average velocity : It is the mathematical mean of velocities of all gas molecules.

→ Ideal gas : Gas which obey gas equation at all temperature and pressure.

→ Compressibility factor : Ratio of observed volume to calculated volume is called Com-pressibility factor.
Z = \(\left[\mathrm{P}+\frac{a n^{2}}{\mathrm{~V}^{2}}\right]\)

→ vander Waals equation :
\(\left[\mathrm{P}+\frac{a n^{2}}{\mathrm{~V}^{2}}\right]\) [V – nb] = nRT

a = unit = atmL² mol²
b = unit = L mol^-1

→ Boiling point: Temperature at which vapour pressure of a liquid becomes equal to atmospheric pressure.

→ Surface tension : It is an important property related to inter-molecular force of liquids. It is the force acting perpendicular to the surface having unit centimetre length. Due to this property, small drops of liquids are spherical in shape. It decreases with increase in temperature.

→ Viscosity : It is the resistance to the flow of liquid or friction in surface of liquids.

F = ηA \(\frac{d u}{d z}\)
Where, \(\frac{d u}{d z}\) = velocity gradient
η is expressed by the unit poise.
1 poise = dynes/cm²s = 1 g cm^-1s^-1 = 10^-1 kg m^-1s^-1.

→ Critical temperature : Temperature above which a gas cannot be liquefied.

→ Critical pressure : The minimum pressure by which gas can be liquefied is called Critical pressure.

→ Critical volume : Volume of 1 mole gas at the critical temperature and critical pressure is called Critical volume.

These MP Board Class 11th Chemistry Notes for Chapter 4 Chemical Bonding and Molecular Structure help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 4 Chemical Bonding and Molecular Structure

→ Chemical bond : Electrons of outermost shell participate in combination.

→ Lewis structure : Representation of electrons of outermost shell by dots. Lewis symbol is an easy way of representation of valence electrons.

→ s-orbital is spherical, p-orbitals are of three types : p_x, p_y and p_z. Their shape is like dumbbell.

→ d-orbitals are situated on two axis, are double dumbbell in shape and are of five types : d_xy,d_yz, d_xz,d_{x² – y²}, d_z²

→ Valence bond theory was proposed by Linus Pauling.

→ Molecular orbital theory was proposed by Robert Mullican.

→ Bond order = \(\frac { 1 }{ 2 }\) (N_b – N_a).

→ Higher the bond order, higher is the dissociation energy.

→ Paramagnetic molecule: If unpaired electrons are present in a molecule, then the molecule is paramagnetic.

→ Molecular orbital: These are formed by the combination of atomic orbitals.

→ Atomic orbitals are of three types :

• Bonding molecular orbital : Their energy is minimum, lesser than the energy of the component atomic orbitals.
• Non-bonding molecular orbital: Molecular orbitals whose energy is equal to the energy of the component atomic orbitals.
• Antibonding molecular orbital: These are the molecular orbitals, whose energy is maximum and is higher than the energy of the component atomic orbitals.

→ In molecular orbitals, electrons are filled in the increasing order of their energy.

→ With the increase in bond order, bond length decreases.

→ Ionic or Electrovalent bond: Electrostatic attraction between cation formed by loss of electrons and anioi formed by the gain of electrons.

→ Chemical bond : Sharing of equal numbers of electrons between the two combining atoms.

→ Co-ordinate bond : Donation of an electron pair from one atom to another atom. It contains both ionic and polar character.

→ Orbital theory oi covalent bond : Formation of covalent bond between two atoms is due to overlapping of orbitals with unpaired electrons.

→ VSEPR theory : Any molecule orient in such a way that there should be maximum distance between electron pairs of valence shell and minimum repulsion. Order of repulsion between bond pair and lone pair is following :
bond pair – bond pair < lone pair – bond pair < lone pair – lone pair

→ Quantum theory of covalent bond : By overlapping of orbitals or bonding electrons 5-bonds and p-bonds are formed; 5-bonds are stronger than p-bonds.

→ a-bond: Bond formed between axial overlapping of atomic orbitals (s -s, s-p and p-p). a-bond can exist alone.

→ 7t-bond : Bond formed between lateral overlapping of (p-p) atomic orbitals, jr-bonds are formed only when 5-bonds are formed or existing.

→ Hybridization: Atomic orbitals of similar energies mixed and same number of orbitals of same energy and shape are formed. This process is called hybridization.

→ Hydrogen bond : The electrostatic force which exist between H-atom of one molecule with highly electro-negative element of the other molecule is called Hydrogen bond.

→ Intra-molecular hydrogen bond: When hydrogen atom and electro-negative element both are present in the same molecule, then this type of hydrogen bond formed is called intra molecular hydrogen bond.

→ Inter-molecular hydrogen bond: When hydrogen atom and electro-negative element are present in different molecules, this type of H-bond formed is called inter-molecular hydrogen bond.

→ Resonance: To explain the characteristic properties of molecules, more than one structures are required and the actual structure is supposed as resonance hybrid of these structures.

→ Dipole moment is represented by an arrow pointing towards the more electro-negative element.

These MP Board Class 11th Chemistry Notes for Chapter 3 Classification of Elements and Periodicity in Properties help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 3 Classification of Elements and Periodicity in Properties

→ Periodic table is the arrangement of keeping similar elements together.

→ Dobereiners’s Triad : Dobereiner (1829) classified groups of three elements (triad) with similar physical and chemical properties. He also observed, that in each triad the atomic mass of the middle element was an average value of the other two elements.

→ Long form of Periodic Table : This form is based on Bohr Bury’s electronic concept. This periodic table consists of 7 periods and 18 groups. Each period starts with the filling of a new quantum number (n) and ends with the complete filling of outermost shell. In this, elements are arranged into four blocks (y-block,p-block, rf-block and^block) on the basis of electronic configuration.

→ Of all the known elements 78% of metals are placed on the left side of the periodic table.

→ Electron gain enthalpy (Electron-affinity) generally becomes more negative in a period and becomes less negative in a group.

→ Ionisation enthalpy (Ionisation energy) increases in a period, and decreases in a group.

→ Newland’s law of octaves: On arranging the elements in the increasing order of atomic mass the eighth and the first element shows periodicity like the eighth note of music.

→ Lother Meyer Curve : Atomic volume of elements are a periodic function of their atomic number.

→ Mendeleev’s Periodic Table : Mendeleev in 1869 stated that the physical and chemical properties of elements are a periodic fuhction of their atomic mass.

→ Period : Seven horizontal rows in which elements are arragned in the increasing order of atomic mass.

→ Group: Nine vertical columns in which 1 to 7 has subgroups A and B and 8lh group has three elements. Later Zero group was added.

→ Modern periodic-law : Physical and chemical properties of elements are periodic functions of their atomic numbers. Moseley proposed it in 1913.

→ Modern periodic table: Elements are placed according to their electronic structure. In long form 7 periods and 18 vertical columns (groups) are present.

→ Classification of elements :

1. s-block elements : Elements having last electron in s sub¬shell. Configuration of last shell is ns^{1 or 2}.
2. p-block elements : Electrons are filled in p-subshell of outermost orbit. Configuration is ns² np^{1 to 6}.
   Elements of 5-block and p-block are commonly known as typical elements.
3. Transition elements : Situated in between s- and p-block. There are 10 vertical rows in rf-block. Two outer shells are incomplete. General configuration (outer shell) is (n -1) ^{1 to 10}.
4. Inner transition elements : Also known as f-block elements. Three outer shells are incomplete. Last electrons are filled in (n -2) f subshells.

→ Ionization potential: Required energy to pull most loosely bounded electron out in isolated gaseous atom is called I.P. In period from left to right, its value increases while in group on moving below its value decreases.

→ Electron affinity: Energy released in accepting one electron by gaseous atom is called electron
affinity. Its value decreases in groups on moving below while in periods on moving left to right increases. .

→ Electronegativity: Tendency to accept electron is called electronegativity. Its value increases in periods from left to right and in groups decreases in below.

→ Atomic radii : Half of the length of single covalent bond exist between two atoms in a molecule. Atomic radii increases in groups on moving below while in periods from left to right, it decreases.

These MP Board Class 11th Chemistry Notes for CChapter 2 Structure of Atom help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 2 Structure of Atom

→  Atom: Smallest particle which has all properties of the element is called atom.

→ Atomic structure : Distribution of fundamental constituent particles of atom is called atomic structure.

→ Cathode rays : Negatively charged rays which move from cathode to anode in discharge tube. Anode rays: Rays of positively charged particles which move opposite the cathode in discharge tube are called anode rays.

→ Electron (⁰₁e): Particle with unit negative charge 1.60 × 10^-19 coulomb and mass 91 × 10^-31 kg.

→ Proton (¹₁e) : Fundamental particle of atom with unit positive charge 1.60 × 10^-19 coulomb and mass 1.67 × 10^-27 kg. Its mass is nearly equal to mass of hydrogen atom.

→ Neutron (¹₁n) : Fundamental particle of atom which has no charge. Mass of neutron is 1.6747 × 10^-27 kg. It is heavier than proton.

→ Nucleus: Central part of atom is called nucleus. Radius of nucleus is 10“23 cm. Total mass of atom and positive charge is in the nucleus.

→ Atomic number: Number of proton in the riucleus of atom or number of electrons in atom is . called atomic number.

→ Mass number : Sum of neutron and proton present in nucleus is called mass number.

→ Isotopes : Different atoms of elements which have same atomic number but different atomic mass are called isotopes.

→ Shell or Orbit : Electrons rotate in stable and definite circular orbits around the nucleus. These are called energy levels or shell or orbits.

→ Relation between frequency (υ) and wavelength (λ) υ = \(\frac{c}{\lambda}\) (Where c is velocity of light = 3 x 10⁸ms^-1)

→ Einstein equation E = mc²
Planck equation E = hυ = \(\frac{h c}{\lambda}\) (Where h = Planck’s constant = 6.626 x 10^-34 Js)

→ Photoelectric Effect hυ = hυ₀ + \(\frac { 1 }{ 2 }\)mv²
Where hυ = Energy of Striking photon
hυ₀ = w₀ = Work function
\(\frac { 1 }{ 2 }\)mv² = Kinetic energy of Ejected electron

(Where R = Rydberg constant = 1.09678 × 107 m^-1)

→ Rydberg formula \(\bar{v}=\frac{1}{\lambda}=\mathrm{RZ}^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\), n₂ > n₁
(Where R = Ryberg constant = 1.09678 × 10⁷ m^-1)
n₁ =1, n₂ = 2,3,4………………………….. UV region
n₁ = 2, n₂ = 3,4,5………………….. Visible region

→ Frequency of the absorbed or emitted radiation at two unstable states of transmission v = \(=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\) (Where E₁ and E₂ are the energies of lower and higher states)

→ Energy of stable state E_n = -R_H \(\left(\frac{1}{n^{2}}\right)\) where n = 1,2,3…………………..

→ Stable energy of H and species similar to H (Like : He⁺,Li²⁺,Be³⁺) i.e., one electron species ) E_n = -2.18 x 10^-18\(\left(\frac{z^{2}}{n^{2}}\right)\) J

→ Radius of n^th orbit r_n = \(\frac{n^{2} a_{0}}{Z}\) [where a₀ (Bohr’s radius of H) = \(\frac{h^{2}}{4 \mathrm{~A}^{2} m e^{2} k}\) = 0.529Å

→ de-Broglie equation \(\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m(\mathrm{KE})}}\)

→ Angular momentum mvr = \(\frac{n h}{2 \pi}\)

→ Heisenberg’s uncertainty principle Δx . Δp ≥ \(\frac{h}{4 \pi}\) or Δx . Δp ≥ \(\frac{h}{4 \pi}m\)
[Where Δx and Δv are uncertainty in position and principle]

→ Velocity in n shell v_n = 2.182 × 10⁶ × \(\frac{z}{n}\).

→ Ionisation Energy (IE)_H = ∆E ∝\(z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)

→ Position of an electron in an atom is determined by four quantum numbers (n, l, in. s
(i) n (Principal Quantum number) = 1, 2, 3, 4 ………………n
(ii) l (Azimuthal Quantum number) = 0, 1, 2 ………….. (n – 1)

(iii) m (Magnetic Quantum number) = – l to + l
(iv) s (Spin Quantum number) = \(+\frac{1}{2}\) or \(-\frac{1}{2}\)

→ Determination of subshell = nl

→ l also expresses the shape of orbital surrounded by electron : s – Spherical, p – dumbell, d – Double dumbell, f- Complex.

→ Total value of m – (2l + 1) = Number of orbitals in subshell = Number of spectrum lines in Magnetic or Electric field.

→ In an atom, l angular node, (n – l – 1) radial node i.e., Total nodes are (n – 1).

→ Electrons are filled in the various orbitals on the basis of the following rules :
(i) Aufbau’s Principle, (ii) Hund’s Rule, (iii) Pauli’s Exclusion Principle.

→ Electronic configuration of ₂₄Cr = [Ar]3r⁵4s¹

→ Electronic configuration of ₂₉CU = [Ar]3d¹⁰4s¹

→ Completely filled and half filled orbitals are more stable due to same symmetry and maxi¬mum energy exchange.

→ Energy of electron due to which it is bound to the nucleus = Work function (w₀) of metal.

→ Subshell: In a shell all electrons do not have same energy so shells are divided in different subshells. These subshells are s, p, d and f

→ Orbital: Space around the nucleus where probability to find a electron is maximum is called orbital. ‘

→ Spectrum: On jumping of electrons from higher energy level to lower energy level, obtained lines on photographic plate from the produced light is called spectrum.

→ Visible spectrum : It is a part of electromagnetic radiation which can be seen by our eyes. Wavelength range is 4000 Å to 7500 Å.

→ Invisible spectrum : The wavelength range of 7500 Å to 3 x 1.0⁶Å and before violet up to 4000A, which we cannot see is called invisible spectrum.

→ Hund’s rule : Pairing of electrons in orbitals of equivalent energy occurs when there is no vacant orbital.

→ Aufbau’s principle: Electrons are filled in subshells in increasing order of energy. Electrons are filled first in the shell whose (n + l) value is less. If (n + l) value is same, then electron will go to that shell whose in n value is lower.

→ Pauli’s exclusion principle: No two electrons in an atom have similar four quantum numbers

These MP Board Class 11th Chemistry Notes for Chapter 1 Some Basic Concepts of Chemistry help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

→ Laws of chemical combinations :

1. Law of conservation of matter : In a chemical change, total mass of reactants and total mass of products are the same.
2. Law of constant proportion : Ratio by weight of constituent elements in a chemical compound remains always the same.
3. Law of multiple proportion : When an element of definite weight combines with an other element of definite weight, two or more than two compounds are formed in a simple ratio.
4. Law of equivalence proportion : When two elements combine with a definite mass of an other element separately, the weight of both elements will be in the ratio in which they combine with each other or is simple multiple of that.
5. Gay Lussac’s law : When gases react, their volumes are in simple ratio and volume of products ifgaseous are also in simple ratio of reacting gases.

→ Atomic mass : Atomic mass of an element is the number which shows that one atom of element is how many times heavier than that of 1/12th part of C¹².

→ Molecular mass: Molecular mass of an element is the number which shows that one molecule of a compound is how many times heavier than 1/12 th part of C¹² atom.
Molecular mass = 2 × Vapour density.

→ Mole : It shows collection of particles of a substance. In one mole of substance (molecule, atom or ion), number of particles are 6023 × 10²³. It is also called as Avogadro’s number.

→ Volume of 1 mole of.a gas at N.T.R is 224 litre.

→ Percentage : Number of grams of solute present in 100 gm of a solution.

→ Molarity : No. of moles of solute in a litre of solution.

→ Percentage composition : Number of parts of each element present in 100 parts according to mass in a compound.

→ Chemical formula: Representation of molecule of a compound in terms of symbol of elements.
It is of three types :

1. Empirical formula: Formula which gives simple whole number ratio of different elements.
2. Molecular formula : Formula which gives exact number of different atoms present in a molecule.
3. Structural formula : Represents the arrangement of various atoms present in molecule of compound.

→ Chemical Equation : The short representation of a chemical reaction in terms of symbols and formula of reactants and products.

→ Element: Substance made up of only one kind of atoms.

→ Compound : Compound is formed by the combination of two or more elements in a definite ratio.

→ Mixture : Mixture is formed by mixing two or more substances in any ratio.

→ There are seven basic units of measurement. All other units are derived from these.

→ Limiting reactants are those which are present in less amount in chemical reaction.

→ Total number of digits in a number are called significant figures.

→ Some useful conversion fractions :

1 pm = 10^-12m, 1Å = 10^-10m, 1 nm= 10^-9 m
1L = 10^-3 m³ = 1 dm³, 760 torr = 101325 Pa = Nm^-2.
1 atm = 1.01325 bar
1 bar =10⁵ Nm^-2 = 10⁵ Pa, 1 atm = 760mm Hg = 760 torr 1 calorie =4184 joule, 1 Electron volt = 1-6022 x 10^-19 joule
Avogadro Number (N₀) = 6 022 x 10²³ (Atom, Molecule, ion or electron)
1 Atomic mass unit (amu or u) = \(\frac{\text { Atomic mass }}{\text { Avogadro Number }}=\frac{A}{N_{0}}\)

→ Normality (N) : Number of gram equivalent of solute per litre of solution.

→ Molality (m) : Number of moles of solute present in 1 kg solvent of the solution.

→ Mole Fraction : Ratio of number of moles of a component to the total number of moles of all the components of the solution.
In a mixture of solution number of moles of components 1 and 2 are n₁and n₂.
Mole fraction of component 1 will be
X₁ = X₁ + X₂ = 1
Thus X₁ = 1 – X₂

→ Average mole mass of an element (\(\bar{M}\))
= \(\frac{\text { Sum of mole mass of Isotopes } \times \% \text { amount (abundance) }}{100}\)
Sum of mole mass of Isotopes X % amount (abundance)
or \(\bar{M}\) = Σf_i × A_i
Where f_i = Fraction of amount of Isotope
A_i = Sum of masses of Isotopes.

These MP Board Class 11th Biology Notes for Chapter 22 Chemical Coordination and Integration help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 22 Chemical Coordination and Integration

→ Hormones are informational molecules, secreted by endocrine glands into venous blood to regulate the functions of other organs and tissues.

→ If young tadpoles are administered thyroxine, they metamorphosed prematurely and change into tiny frogs. On the other hand, if antithyroid substance like thiourea is given to tadpoles, it delays their metamorphosis.

→ Mexican axolotls which are the larval forms of Ambystoma tigrinum do not metamorphose due to the lack of thyroid hormone and ordinarily exist as axolotl larva and even reproduce in this form, but if thyroid extract is injected, they metamorphose into Ambystoma tigrinum, the adult terrestrial form. Breeding of Axolotl in the larval stage is called neoteny or pedogenesis.

→ Estrogens in human urine, increase twice during the menstrual cycle. Urine of a pregnant woman is a very rich source of estrogen, which appears to be produced by the placenta.

→ Medullary hormones are not essential for life.

→ Glucocorticoids reduce the number of circulating lymphocyte eosinophils (anti-leukaemic property) and have anti-inflammatory and anti-allergic properties. Glucocorticoids are also used in the treatment of rheumatoid arthritis.

→ Certain drugs like thiouracil, thiourea, para-amino benzoic acid, radioactive iodine, carbimazole and methimazole prevent the formation and secretion of thyroxine hormone.

→ Pars tuberalis a part of adenohypophysis of pituitary is rarely found in man.

→ Adrenogenital syndrome (Gallais, 1912) is due to the excess of adrenal cortex androgenic hormones. It is responsible for precocious growth of body “Infant Hercules”. This disease is found in females only.

→ Norepinephrine is known as hormonal secretion of adrenal medulla but it is seldom released as
norepinephrine from gland-what all is released is epinephrine only, not norepinephrine. The latter is secreted at nerve endings only.

→ Night urine usually contains more inactivated ADH than the day urine.

→ Alcohol inhibits ADH secretion.

→ Acromegaly is a kind of malformation of bones due to oversecretion of growth hormones in
adults.

→ Cachexia or Simroonds disease is caused due to undersecretion of growth hormone in adults.

→ Pituitary gland is also known to be as master gland because it controls the secretion of all the endocrine glands.

→ Pheromones or Ectohormones are chemical substances released into the external surroundings by some exocrine glands. The pheromones released by the members of a species are recognized by the other members of the same species. Hence, pheromones serve as means of communication among the members of same species, e.g. The path followed by a worker ant is recognized and followed by the other ants. This is possible because the rest of the ants recognize the pheromones released by the worker ant.

→ Thomas Addison is regarded as the father of Endocrinology.

→ Hyposecretion of thyroid gland results in the production of cretinism, myxoedema, goitre and Hashimoto disease.

→ Hyposecretion of adrenal cortex results in the production of Addisons, Hypoglycemia and Cons disease.

→ Pancreas is a mixed (exocrine as well endocrine) gland.

→ The humoral immune system defends against those pathogens that invade the body fluid (blood and lymph), i.e., extracellular pathogen.

→ Cell mediated immune system defends the body against those pathogens which invade the cells. i.e., intracellular pathogen.

These MP Board Class 11th Biology Notes for Chapter 21 Neural Control and Coordination help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 21 Neural Control and Coordination

→ If the blood flow to brain is interrupted for a few movements unconsciousness may result. Interruption of 1-2 minutes may weaken the brain cells by starving them of oxygen. Neurons are permanently injured if deprived of oxygen for 4 minutes.

→ If blood entering the brain has low glucose level, mental confusion, dizziness, convolutions and even unconsciousness may occur.

→ Myelin is white and made up of phospholipid. It is responsible for the colour of the white matter in brain, spinal cord and nerves.

→ Sir Charles Sherington (1861-1954) was the first person who applied the term synapse to the junctional points between two neurons. Actually synapse is a place of physiological continuity without anatomical continuity.

→ Parkinsonism or Parkinson’s disease is a disease of the basal ganglia deep in the substance of the brain that control movement. It occurs usually in the late middle age. Parkinson’s disease seems to be caused by the malfunction at the synapses.

→ Sciatica is the painful condition of the sciatic nerve or its branches. In this, pain passes from the back or thigh down its length into the leg, foot and toes.

→ Strabismus or Squint is an eye muscle disorder commonly known as crossed eyes. The eyeballs do not move in unison and the image does not fall upon corresponding points of the two retinas. It may be caused due to lack of coordination of the extrinsic eye muscles. As a result, two images are seen, a condition termed diplopia.

→ Taste of chillies, black pepper and hot sauces is not a true sensation. It is mainly a sensation of burning pain produced by the stimulation of pain receptors of the tongue by specific chemical substances in those goods.

→ Image formed on the retina is always inverted.

→ The hair cells in the human ear are variously estimated to be of 13,000 to 54,000 in number each one with about 40 cilia or hair at the receptive side and projecting into the endolymph.

→ The nervous system is derived from embryonic ectoderm.

→ The central nervous system consists of brain and spinal cord developing from neural ectoderm.

→ The brain and spinal cord are covered and protected by connective tissue membranes called meninges.

→ The wall of brain is made up of grey and white matter.

→ Grey matter differs from white matter in having cell bodies and non-myelinated proximal parts , of axons. The white matter contains myelinated parts of axons.

→ We lose our sense of smell when olfactory lobes are removed from brain.

→ The forebrain or prosencephalon is the largest part of the brain and cerebrum is the largest part of the forebrain.

→ The roof of the cerebrum is known as pallium.

→ Cranial nerves are arising from brain. There are 12 pairs of cranial nerves in mammals.

→ The cells of nervous system are called neurons.

→ In mammals, 31 pairs of spinal nerves are present.

→ The processes emerging out from cyton are known to be as dendrites.

→ Middle ear has three bones : (i) Malleus, (ii) Incus and (iii) Stapes.

→ Internal ear is made up of two tubules : (i) Vestibule and (ii) Semicircular tubules.

→ The path of nerve impulse in a reflex action is called reflex arc.

→ An automatic spontaneous involuntary and unconscious action brought about by nervous system is called reflex action.

→ Reflex actions are controlled by central nervous system.

→ Poliomyelitis is an acute viral infection that destroys the cell bodies of motor neurons in the anterior horn of the spinal cord.

→ Rods mainly enable the animal to see in the darkness, therefore rods are present in larger number in nocturnal animals.

→ Cones are chiefly concerned with the distinction in colour and light vision during day time.

→ The rods contain rhodopsin or visual purple pigment, whereas cones contain the iodopsin or visual violet pigment.

→ Acetylcholine functions like that of a synaptic stimulant.

These MP Board Class 11th Biology Notes for Chapter 20 Locomotion and Movement help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 20 Locomotion and Movement

→ Muscle fatigue is caused due to the accumulation of lactic acid.

→ Number of cervical vertebrae in mammals is always seven, even in case of camel and giraffe.

→ Sliding filament theory of muscle contraction was given by H. E. Huxley and A. F. Huxley in 1954.

→ Chemical changes which occur during the contraction of muscle fibres were studied by Albert Szent Gyorgyi and his associates for the first time.

→ Total number of bones in human is 206.

→ Locomotion of starfish (Asterias) takes place with the help of water vascular system.

→ Poliomyelitis is caused by the polio virus which affects the muscles of limbs.

→ In man caudal vertebra or coccyx is a small bony piece showing rudimentary caudal vertebra.

→ Haemopoiesis (formation of blood) takes place in the bone-marrow of long bones.

→ Actin is a contractile protein forming thin filaments of muscle fibres.

→ Myoglobin is a protein of muscle with haem.

→ Tube-feets are the locomotory organs of starfish.

→ The minimum intensity of stimulus that can stimulate the muscles to contract is called threshold stimulus.

→ Joints are structures where two bones are fixed with each other.

→ A single isolated contraction of the muscle fibre is called muscle twitch.

→ Sarcomere is a unit of myofibril which extends between two successive Z-lines. A sarcomere is made of A-band in the middle and one half of an I-band on its either side.

→ Oxygen debt is the extra oxygen required during relaxation over the resting state of muscle.

→ Osteology is a branch of biology that deals with structure, nature and development of bones.

→ Hydra locomotes with the help of tentacles and muscles.

→ Annelids locomote with the help of setae and parapodia, whereas in arthropods locomotion takes place with the help of jointed appendages.

→ In echinoderms locomotion takes place with the help of tube-feet.

→ Stapes is the smallest and femur is the largest bone of the human body.

→ Myalgia : Pain in the muscles.

→ Myology : Study of muscles. It is also called sarcology.

→ Arthrology : Study of joints.