1st PUC Chemistry Question Bank Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

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Karnataka 1st PUC Chemistry Question Bank Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

General Introduction

Question 1.
What is organic chemistry?
Answer:
The study of carbon compounds is called organic chemistry.

Question 2.
Give the important applications of organic compounds.
Answer:

• Organic compounds are vital for sustaining life on earth.
• Bio complex molecules like DNA (carry genetic information), proteins, etc are organic compounds.
• Organic molecules appear in materials like cloth, fuels, polymers, dyes and medicines.

Question 3.
Name the type of forces believed to be responsible for the formation of organic compounds.
Answer:
Vital force.

Question 4.
Name the first organic compound prepared in the laboratory from inorganic compound by F Wohler.
Answer:
Urea (NH₂ – CO – NH₂).
By heating ammonium cyanate.

Tetra Valence of Carbon: Shapes of Organic Compounds

Question 1.
What are the unique properties of carbon?
Answer:
There are two unique properties of carbon.

• Carbon has a strong tendency to undergo catenation.
• Carbon is tetravalent, forming four bonds with neighbouring atoms.

Question 2.
What is catenation?
Answer:
Carbon atom has capacity to link with other carbon atoms to form chain and ring compounds. This unique property of carbon is called catenation.

Question 3.
Carbon forms large number of compounds. Why?
Answer:
Due to catonation.

Question 4.
Give the shapes, bond angle and percentage of s-character of carbon compounds based on hybridization.
Answer:

Question 5.
Give the relation between percentage s-character, bond length, bond enthalpy and bond angle.
Answer:
As the s-character decreases, bond length increases, bond enthalpy decreases and bond angle decreases.
Note:

• As the s-character increases, electronegativity of hybrid orbital increases.
• In organic compounds, single bond is always σ, double bond (1σ and 1π) and Triple bond (1σ and 2π)
• Carbon atom with single bond under goes sp³ hybridisation (tetrahedral shape), double bond undergoes sp² hybridisation (trigonal planar shape) and triple bond undergoes sp hybridisation (linear shape).

Question 6.
How many σ and π bonds are present in the following molecules?
(a) HC = C – CH = CHCH₃ (b) CH₂ = C = CHCH₃
Answer:

Question 7.
What is the type of hybridisation of each carbon in the following compounds?
(a) CH₃Cl,
(b) (CH₃)₂CO,
(C) CH₃CN,
(d) HCONH₂,
(e) CH₃CH = CHCN
Answer:
(a) sp³
(b) sp³, sp², sp³
(c) sp³, sp
(d) sp²
(e) sp³, sp², sp², sp

Question 8.
Write the state of hybridization of carbon in the following compounds and shapes of each of the molecules.
(a) H₂C = O
(b) CH₃F
(C) HC ≡ N
Answer:
(a) sp² hybridised carbon, trigonal planar.
(b) sp³ hybridised carbon, tetrahedral.
(c) sp hybridised carbon, linear.

Structural Representations of Some Organic Compounds

Question 1.
Discuss the different types of structural representations of organic compounds.
Answer:
Structures of organic compound are represented in several ways.
(a) Lewis dot structures: In this method, each covalent bond is represented by pair of dots [each dot represents the electron]. For example, Lewis dot structure of methane is

(b) Dash structure: In this method, each covalent bond is represented by Dash (-). A single dash (-) is used to represent single bond, two dashes (=) represent double bond and three dashes (≡) represent a triple bond. These type of structural representation are called complete structural formulae or graphic or displayed formulae. For example, dash line structure of methane, acetic acid and acetylene are as follows:

(c) Condensed structural formula: The structural formula represented by omitting some or all the covalent bonds and by indicating the identical groups attached to an atom by subscripts is called condensed formula. For example, the condensed structural formulae of propane, acetone and decane are
Propane: CH₃CH₂CH₃ Acetone: CH₃COCH₃ Decane:
CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₃

Condensed structural formulae can be further condensed by enclosing the repetitive structure unit within a bracket and placing an integer as a subscript indicating the number of times the structural unit is repeated.
For example, CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃ can be further condensed to CH₃(CH₂)₆CH₃.
Similarly, CH₃CH₂CH₂CH₂CH₂CH₂CHO can be shown as CH₃(CH₂)₅CHO.

Bond line structural formula: In this method, the carbon and hydrogen atoms are not shown, but only the bonds between the carbon atoms are shown as lines. The ends and vertices represent the carbon atoms. The number of hydrogens must be guessed according to the valence rules. In this method, lines have drawn in a zig-zag fashion. But all atoms other than carbon and hydrogen atoms such as oxygen, nitrogen, and halogen are shown on the zig-zag line. The terminal denote CH₃ group and unsubstituted intersection of two lines denotes a CH₂ group.
Double or triple bonds are represented by two or three parallel lines joining the same two atoms.
Cyclic compounds are represented by polygon without showing carbon and hydrogen atoms.

For example, bond line structure of butyraldehyde (CH₃CH₂CH₂CHO) can be represented as



In cyclic compounds, the bond – line formulas may be given as follows:

Question 2.
Expand each of the following condensed formulae into their complete structural formulae.
(a) CH₃CH₂COCH₂CH₃.
(b) CH₃CH = CH(CH₂)₃CH₃
Answer:

Question 3.
For each of the following compounds, write a condensed formula and also their bond-line formula.
(a) HOCH₂CH₂CH₂CH (CH₃)CH (CH₃)CH₃

Answer:
Condensed formula: (a) HO(CH₂)₃CH(CH₃)CH(CH₃)₂ (b) HOCH(CN)2
Bond-line formula:

Question 4.
Expand each of the following bond-line formulae to show all the atoms including carbon and hydrogen.

Answer:

Question 5.
What is wedge – dash formula? Write the wedge – dash representation of methane.
Answer:
A method of representing the three-dimensional structure of a molecule in which solid lines represent bonds in the plane of the image, solid wedge represents bond towards the viewer and dashed wedge represents bonds away from the viewer is called wedge dash formula. The wedge – dash representation of methane is

Question 6.
What are molecular models? How many types of molecular models are used to represent three dimensional structure of organic compounds. Give the importance of each model.
Answer:
Molecular models are physical devices that are used for a better visualization and perception of three dimensional shapes of organic molecules.
Commonly three types of molecular models are used.
(i) Frame work model – This model emphasizes the pattern of bonds of a molecule while ignoring the size of atoms only bonds are shown but not atoms.
For example frame work model of methane is

(ii) Ball and stick model: In this method, both the atoms and bonds are shown. Balls represent atoms and stick represents a bond. The ball and stick model of methane is

(iii) Space filling model: It imphasizes the relative size of each atom based on its van der Waals radius.

Classification of Organic Compounds

Question 1.
Give the classification of organic compounds.
Answer:
Organic compounds are broadly classified as follows:

Question 2.
What are acyclic or open chain compounds? Give examples.
Answer:
Organic compounds in which all the carbon atoms are linked to one another to form open chains (either straight or branched) are called acyclic or open chain or aliphatic compounds. Examples;

Question 3.
What are cyclic compounds? Give different types of cyclic compounds with examples.
Answer:
Compounds which have at least one ring or closed chain of atoms are called cyclic compounds. They are further classified into two types:
1. Homocyclic compounds or Carbocyclic compounds: The compounds containing rings which are made up of only carbon atoms are called homocyclic compounds. These are of two types:
i. Alicyclic compounds: These are carbocyclic compounds which resemble aliphatic compounds in their properties. Examples,

ii. Aromatic compounds: Compounds containing at least one benzene ring are called aromatic or benzenoid compounds, Examples,

Question 4.
What are heterocyclic compounds? Give different types of heterocyclic compounds with examples.
Answer:
Cyclic compounds having one or more heteroatoms (examples; O, N, S etc.) in the ring are called heterocyclic compounds. These are of two types:
a. Alicyclic heterocyclic compounds: Heterocyclic compounds resemble aliphatic compounds in their properties are called alicylic heterocyclic compounds. Examples;

b. Aromatic heterocyclic compounds: The heterocyclic compounds which resemble aromatic compounds in their properties are called aromatic heterocyclic compounds. Examples :

Question 5.
Give examples for benzenoid aromatic compound.
Answer:

Question 6.
Give an example for non benzenoid compound.
Answer:

Question 7.
What is a functional group? Give examples.
Answer:
An atom or a group of atoms which determines the properties of the organic compound is called a function group.
Examples

• Functional group of alcohols is – OH
• aldehydes is – CHO
• Carboxylic acids its – COOH, etc.

Question 8.
What is homologues series? Give examples.
Answer:
A series of same class of organic compounds which can be represented by the general formula and any two successive members differ by – CH₂– group is called a homologues series.
Each member of homologous series is called ‘homologue’. They have same chemical properties and regular gradation in physical properties with increase in molecular mass.
Examples: Alkanes, forms homologues series with the general formula C_nH_2n+2

Nomenclature of Organic Compounds

Question 1.
Discuss the common system of nomenclature of organic compounds.
Answer:
In the earlier days, the common name or trivial name was given to an organic compound

• Based on the source from which it is derived or
• Special property of the compound or
• Based on invention or
• Based on some other trivial reasons with out any reference to its structure.

For example, the organic compound HCOOH is called formic acid because it is obtained from dry distillation of red ants (in Latin formacum means red ants). Citric acid is named so because it is found in citrus fruits.

Question 2.
Give the common names of following organic compounds.
Answer:

Question 3.
Define different parts of IUPAC name of organic compound.
Answer:
The IUPAC nomenclature is a systematic method of naming organic compound recommended by the International Union of Pure and Applied Chemistry (IUPAC).
IUPAC name of an organic compound consists of three parts.
1. Root word
2. Suffix
3. Prefix
Root word: It indicates the number of carbon atoms in the longest which may or may not be a straight chain.

Suffix: It represents the functional group present in the organic compound.

Prefix: The prefix is used to indicate the side chains, substituents and low priority functional groups (which are considered as substituents). The prefix is added before the root word.

Question 4.
Give the general IUPAC rules in naming organic compounds.
Answer:
1. Longest chain rule: The longest possible continuous chain of carbon atoms in the molecule is chosen.

2. Position of the substituent: The carbon atoms of the parent chain are numbered as 1, 2, 3…. starting from that end which gives lowest numbers to the carbon atoms carrying substituents.

3. Lowest number for functional groups: In case, there is a functional group present in a molecule [even double bond or triple bonds], the parent chain is numbered from the end which in nearer to the functional group (even lowest set rule is violated).

4. Presence of identical groups: If two or more identical groups are present, then the locants are listed together, after the last locant hypen (-) is written & prefixes di, tri, tetra etc. are added before the name of the identical group.

5. Naming of different substituents: If two or more different substituents are attached to the parent chain, then they are named in the alphabetical order along with their appropriate positions.

6. Naming different substituents at equivalent position: If two different substituents are in equivalent position from the two ends of the chain (two sets of locants are equal), then the lowest number is given to the group which comes first in the alphabetical order.

7. Naming the branched side chains: If there is a branching in the side chain, the side chain is named as substituted alkyl groups by numbering the carbon atom of this group attached to the parent chain as 1. The name of such side chain is given in brackets to avoid confusion with the numbering of parent chain.

8. Naming of poly functional group: When a compound contains two or more different functional groups, one of the functional group is selected as principal functional group. The remaining functional groups are treated as substituents.
The following order for priority is given for functional groups.
-COOH, -SO₃H, -COOR, -COCl, -CONH₂, -CN, -CHO, > C = 0, – OH, – NH₂, > C = C <,-C ≡ C-

9. Naming nf cyclic compounds: The names of alicyclic compounds are obtained by adding the prefix “cyclo” to the name of the corresponding hydrocarbon.

Question 5.
Give the common and IUPAC names of following organic compounds.
Answer:

Question 6.
Give the IUPAC names following substituted benzene compounds.
Answer:

Question 7.
Write the structure of
(i) 2-ChIorohexane,
(ii) Pent-4-en-2-ol,
(iii) 3-Nitrocyclohexene,
(iv) Cydohex-2-en-l-ol,
(v) 6-Hydroxy-heptanal.
Answer:
(i) CH₃CH₂CH₂CH₂CH(Cl)CH₃.
(ii) CH₂ = CHCH₂CH (OH)CH₃.


(v) CH₃CH(OH)CH₂CH₂CH₂CH₂CHO.

Question 8.
Write the structural formula of:
(a) o-Ethylanisole,
(b) p-Nitroaniline,
(c) 2,3-Dibromo-l-phenylpentane,
(d) 4-Ethyl-l-fluoro-2-nitrobenzene.
Answer:

Isomerism

Question 1.
What is isomerism in organic compounds?
Answer:
Isomerism is the phenomenon in which two or more compounds having the same molecular formula but differ in their physical and chemical properties. Such compounds are called isomers.

Question 2.
Give the classification of isomerism.
Answer:
The flow chart of the classification of isomerism is as follows.

Question 3.
What is structural isomerism?- Give different types of structural isomerism.
Answer:
Two or more compounds having the same molecular formula but differ in their structural formula are called
structural isomers and the phenomenon is called structural isomerism. The types of structural isomerism are

1. Chain isomerism or Nuclear isomerism
2. Position isomerism
3. Functional isomerism
4. Metamerism
5. Tautomerism.

Question 4.
What is chain isomerism? Give examples.
Answer:
The isomers which differ in the length of the carbon chain present in their molecule are called chain isomers and the phenomenon is called chain isomerism.
Pentane (C₅H₁₂) exhibits the following three structures.

Question 5.
What is position isomerism? Give an example.
Answer:
Position isomerism is a structural isomerism which is arised due to the difference in the position of the same functional group on the carbon chain.
For example, Chloropropane (C₃H₇ – Cl) can have two position isomers given below.

Question 6.
What is functional isomerism? Give an example.
Answer:
Two compounds having the same molecular formula but different functional groups are called functional isomers and the phenomenon is called functional isomerism.
Examples for functional isomers are
(i) Alcohols and ethers

Question 7.
What is metamerism? Give an example.
Answer:
This type of isomerism arises due to unequal distribution of carbon atoms on either side of the functional group. For example, diethyl ether and methylpropyl ether are metamers.

Question 8.
What is stereoisomerism?
Answer:
Two or more compounds having same molecular formula and same structural formula but differ in the spacial arrangements of atoms or groups are called stereo isomers. This phenomenon is called stereo isomerism. There are two types of stereo-isomerism: geometrical isomerism and optical isomerism.

Fundamental Concepts in Organic Reaction Mechanism

Question 1.
What is meant by substrate in organic reaction?
Answer:
The major reactant containing carbon which is attacked by some other chemical species in a reaction is called a substrate.

Question 2.
What is meant by reagent in organic reactions?
Answer:
Species which attacks the substrate in order to get the product is called a reagent.

Question 3.
What is mechanism of a reaction?
Answer:
The stepwise study of chemical reaction which includes the movement of electrons, bond cleavage, the bond formation and rate of transformation is called the mechanism of a reaction.

Question 4.
Explain homolytic fusion.
Answer:
The symmetrical cleavage of a covalent bond resulting in the formation of free radicals is called hemolytic fusion or homolysis.

Question 5.
What are free radicals?
Answer:
Atoms or group of atoms containing an odd electron are called free radicals.
Examples:

Question 6.
Write a note on alkyl free radicals.
Answer:
The alkyl free radicals are planar and the carbon atom containing an odd electron in free radicals is sp² hybridised.

Alkyl free radicals may be classified as primary, secondary or tertiary depending upon whether one, two or three hydrocarbon groups are bonded to the carbon atom carrying an odd electron.
The stability of alkyl free radicals is in the order:

The mechanism of the reaction in which homolysis takes place is known as homolytic mechanism or free radical mechanism.

Question 7.
Explain heterolytic fission.
Answer:
The unsymmetrical cleavage of covalent bond is called heterolytic fission. This results in the formation of positive and negatively charged particles. Mechanism of the reaction in which heterolysis takes place is known as heterolytic mechanism or ionic mechanism.

Question 8.
What are carbocations? Give an example.
Answer:
The positively charged ions in which carbon carries a positive charge are called carbocations.

Question 9.
What are carbanions? Give an example.
Answer:
The negatively charged ions in which carbon carries a negative charge are called carbanions.
Example : \(: \overline{\mathrm{C}} \mathrm{H}_{3}\) (methyl carbanion).

Question 10.
What are electrophiles? Give example.
Answer:
Electron deficient species are called electrophiles.

Note: Electrophiles are Lewis acids because they are electron pair acceptors.

Question 11.
What are nucleophiles? Give examples.
Answer:
Electron rich species are called nucleophiles.
Examples-, H^–, OH^–, R – O^–, ^–CH₃, X^–, S H^–, etc.
Neutral nucleophiles: NH₃, R-NH₂, R₂NH, R₃N, NH₂-NH₂, H-O-H, R-O-H, R-O-R, etc.
Note: Nucleophiles are Lewis bases because they are electron pair donars.

Question 12.
Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
(a) CH₃ – SCH₃
(b) CH₃– CN
(C)CH₃ – CU
Answer:

Question 13.
Giving justification, categories the following molecules/ions as nucleophile or electrophile:

Answer:
Nucleophiles: HS^– C₂H₅O^–, (CH₃)₃N, H₂N:^–
These species have unshared pair of electrons, which can be donated and shared with an electrophile.

Reactive sites have only six valence electrons; can accept electron pair from a electroophile.

Question 14.
Identify electrophilic center in the following: CH₃CH = O, CH₃CN, CH₃I.
Answer:

will have partial positive charge due to polarity of the bond.

Question 15.
What type of notation is used to indicate the movement of electrons in organic reactions?
Answer:

Question 16.
What type of notation is used to indicate the movement of single electron?
Answer:

Question 17.
Write a note on inductive effect.
Answer:
The permanent polarity is developed in a molecule due to the displacement of bonded pair of sigma electrons towards more electronegative atom or group is called an Inductive effect.
In case of n-butyl chloride, inductive effect may be represented as below.

Types of inductive effects
There are two types of inductive effects. They are,
i) Negative inductive effect [-1 effect]
ii) Positive inductive effect [+1 effect]

i) Electron withdrawing inductive effect (-1 effect): It is the inductive effect in which the displacement of electrons takes place towards the substituent group.
The groups which show negative inductive effect are,
-NO₂> – CN > – F > – COOH > – Cl > – Br > – I > – OCH₃ > – OH > C₆H₅

ii) Electron releasing or donating inductive effect (+1 effect): It is an inductive effect in which the displacement of electrons takes place away from the substituent groups.
Groups which show ‘+I’ effect are, (CH₃)₃C – > – CH(CH₃)₂ > – CH₂ – CH₃ > – CH₃

Question 18.
Write the resonance structures of benzene and nitromethane.
Answer:
Resonance structures of benzene:

Resonance structure of nitromethane:

Question 19.
What is resonance stabilization energy?
Answer:
The difference in energy between energy of the resonance hybrid and the most stable form of the contributing structures is called resonance energy.

Question 20.
What is resonance or mesomeric effect?
Answer:
The polarity produced in a molecule due to interaction of two pi bonds or between the pi bond and lone pair of electrons in a conjugated system is called resonance effect or mesomeric effect.

Question 21.
What is negative resonance (-R) effect? Illustrate with an example.
Answer:
It is a resonance effect in which the displacement of electrons takes place towards the substituent group.
For example, in nitrobenzene the electron displacement can be represented as follows:

The groups showing -R effect, are.
-NO₂> -CHO, -CN, -COOH, -COR, -COOR etc.,

Question 22.
What is positive resonance (+R) effect? Illustrate with an example.
Answer:
It is a resonance effect in which the displacement of electrons takes place away from the substituent group attached to the conjugated system. For example, electron displacement in aniline is depicted as follows.

The group showing + R effect are
– NH₂ , -OH, – OR, -Cl, -Br, -1, etc.

Question 23.
What is electromeric effect?
Answer:
The complete transfer of pi electrons of a multiple bond in presence of an attacking agent is called electromeric effect.

Question 24.
What is positive electromeric (+E) effect? Illustrate with an example.
Answer:
It is an electromeric effect in which the shifting of pi electrons takes place towards the attacking reagent. Example:

Question 25.
What is negative electromeric (-E) effect? Illustrate with an example.
Answer:
It is an electromeric effect in which the shifting of pi electrons takes place away from the attacking agent.
Example:

Note:

• It is a temporary effect in which the complete transfer of pi electrons takes place.
• This effect takes place only in the presence of an attacking agent.

Question 26.
What is hyper conjugation? Illustrate with an example.
Answer:
Hyperconjugation is a special type of electromeric effect which involves the delocalization of a electrons of the C – H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with the unshared
p-orbital. This effect is also known as “no bond resonance” or “Baker-Nathan effect”.
Examples:
(1) In propene the a electrons of the C-H bond of methyl group can be delocalized into the p orbital of doubly bonded carbon as represented below.

(2) Hyper conjugated structures of ethyl cations are as follows:

(3) Hyper conjugated structures of ethyl radical

Question 27.
Give the difference between inductive effect and electromeric effect.
Answer:

	Inductive effect		Electronic effect
1.	It is a permanent effect.	1.	It is a temporary effect
2.	It is a weak effect.	2.	It is a strong effect
3.	This effect involves the shifting of sigma electrons.	3.	This effect includes the transfer of pi electrons
4.	This occurs in saturated system.	4.	This occurs in the unsaturated system.
5.	Presence of attacking agent is not required.	5.	This effects takes place only in the presence of attacking agent.
6.	Develops a partial charge in the molecule.	6.	Develops a complete charge separation in the molecule.

Question 28.
Give the difference between inductive effect and mesomeric effect.
Answer:

	Inductive effect		Mesomeric effect
1.	It takes place in a saturated system.	1.	It takes place in a conjugated system.
2.	It is a weak effect.	2.	It is a strong effect.
3.	It involves the displacement of sigma electrons.	3.	It involves the displacement of pi electrons
4.	It does not involves the delocalisation of electrons.	4.	It involves the delocalisation of electrons
5.	Partial charge is developed in a molecule	5.	Complete charge separation takes place
6.	This effect decreases with the increase of distance from the substituent group.	6.	The effect will be same throughout the conjugated system

Question 29.
Give the classification of organic reactions.
Answer:

• Substitution reactions
• Addition reactions
• Eliminations reactions
• Rearrangement reactions.

Methods of Purification of Organic Compounds

Question 1.
What are common techniques used for the purification of organic compounds?
Answer:
The following techniques are used for the purification of organic compound.

• Sublimation
• Crystallisation
• Distillation
• Differential extraction
• Chromatography

Question 2.
What is sublimation? What type of organic compounds are purified by this method.
Answer:
he process in which some solid substances change from solid to vapour state without passing through liquid state is called sublimation. This technique is used to purify sublimable organic compounds form non-sublimable impurities.

Question 3.
What is the basic principle involved in the purification of organic compound by crystallisaion?
Answer:
It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.

Question 4.
How do you purify organic compound by crystallization?
Answer:
The impure solid is dissolved in a solvent in which it is sparingly soluble at room temperature but more soluble at higher temperature. The hot solution of the compound is concentrated to get a saturated solution. The solution is filtered to remove insoluble impurities. On cooling the solution, pure compound crystallises out and is separated by filtration. The solid is dried. Any coloured impurities present in a compound is removed by adding activated charcoal to the solution during crystallization. Repeated crystallisation is required for the purification of compounds containing impurities of comparable solubilities.

Question 5.
What type of organic compounds are purified by distillation?
Answer:
This method is used separate

• Volatile liquids from non volatile impurities
• Liquids having sufficient difference in the boiling pints.

Question 6.
Explain the purification of organic liquids by distillation with neat labelled diagram.
Answer:
This is an important method used to separate the liquids having sufficient difference in their boiling points. The liquid mixture is taken in a distillation flask and the apparatus is arranged as shown in the figure.

The liquid mixture is heated slowly. The liquid with a lower boiling point begins to boil and vapours are formed. The vapours of liquid are condensed in a condenser and collected in the receiver.

Question 7.
Describe the separation of organic liquids by fractional distillation with neat labelled diagram.
Answer:

This method is employed to separate the liquids with smaller difference in their boiling points. The liquid mixture is taken in a distillation flask and the distillation flask is fitted with a fractionating column. The fractionating column is carefully filled with glass beads. These beads improve the separation between the liquids being distilled. The mixture is heated. The liquids are vapourised and vapours are enter into the fractionating column. The vapours of the liquid with higher boiling point condensed in the fractionating column before the vapours of the liquid with lower boiling point. The vapours of the low boiling component reaches the top of the fractionating column and enters into the condenser. The vapours are condensed and collected in the receiver in the pure form.

Question 8.
Describe the separation of organic compounds by distillation under reduced pressure with neat labelled diagram.
Answer:
This method is used to purify liquids having very high boiling points and those which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of water pump or vaccum pump. Glycerol can be separated from spent-lye in soap industry by using this technique.

Question 9.
Describe the separation of organic compounds by steam distillation with neat labelled diagram.
Answer:

This technique is applied to separate substances which are steam volatile and are immiscible with water. In steam distillation, steam from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the vapours of organic compound is condensed in the condenser and collected in the receiver. The compound is later separated from water using a separating funnel.

Question 10.
Describe the separation of organic compounds by differential extraction.
Answer:

When an organic compound is present in aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be ‘immiscible with each other so that they form two distinct layers which can be separated by separating funnel. The organic solvent is later removed by distillation or by evaporation to get back the compound. Differential extraction is carried out in a separating funnel as shown in figure.

Question 11.
What is chromatography?
Answer:
Chromatography is a technique used to separate mixture in to their components, to purify compounds and also to test the purity of compounds.

Question 12.
What is adsorption chromatography?
Answer:
Adsorption chromatography is a separation technique in which different compounds are adsorbed on the adsorbent to different degrees (differential adsorption).

Question 13.
What is column chromatography? How do you separate the organic compounds from the mixture by this technique?
Answer:

Fig. Separation of organic compounds by column chromatography It is a separation technique which involves the separation of components of the mixture over the column of adsorbent. A column like burette having a stopcock at the bottom is used with a glass wool placed at the bottom. The tube uniformly packed with adsorbent using a slurry of adsorbent like silica gel or alumina. The mixture of substances (if liquids) or the solution of substance (if solids) is placed on the top of the column. The eluent (mobile Student’s illuminator IPU Chemistry phase) which is a liquid or a mixture of liquids is allowed to flow down the column slowly. The most readily adsorbed substance are retained near the top and other come down at various distances in chromatographic column. Different components of mixtures are collected in separate conical flaks. On distillation of the eluent, components are obtained in the pure form.

Question 14.
Write a note on thin layer chromatography.
Answer:

It involves the separation of components of a mixture over a thin layer of an adsorbent coated on a glass plate. A thin layer of 0.2 mm thickness of adsorbent is spread over a glass plate. It is called chromo plate. The solution of the mixture is placed at a small spot about 2cm above the end of the TLC plate. The glass plate is carefully placed in a closed jar containing the eluent. As the eluent rise up, the components are separated. When the solvent has moved 3/4^th of the plate, the plate is removed and the solvent front is marked. If the spots are coloured, they will be visible on TLC plate. If the spots are colorless, the plate is placed in a covered jar containing iodine crystals or in UV-light. A ninhydrin reagent can be used as a spray reagent to identify amino acids. By measuring the distance of solvent front and solute front from the base line, relative absorption of each component is measured using Rf value (Retention factor)

Question 15.
What is partition chromatography?
Answer:
Partition chromatography is based on continuous differential partitioning of components of mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography.

Question 16.
Describe the separation of organic compound by paper chromatography.
Answer:

In paper chromatography, a special quality paper known as chromatography paper is used. Acts as the stationary phase. Solution of the mixture is placed at small spot about 2cm above the end of the chromatographic paper. It is placed in closed jar containing the eluent. As the eluent rises up, the components get separated. When the solvent has moved 3/4th of the chromatographic paper, the paper is removed. If the spots are coloured, they will be visible at different heights from the base line. If the spots are colorless they are observed under UV-light or by using appropriate spray reagent.

Qualitative Analysis of Organic Compounds

Question 1.
How do you detect carbon and hydrogen by copper oxide method?
Answer:
The given organic compound is mixed with dry cupric oxide. The mixture is taken in a hard glass test tube and heated. The carbon and hydrogen of the organic compound are oxidised to CO₂ and H₂O respectively. The CO₂ turns lime water milky confirms the presence of carbon and H₂O vapours turns white coloured anhydrous CuSO₄ into blue coloured hydrated copper sulphate. Confirms the presence of hydrogen.
c + 2CuO → CO₂ + 2Cu
2H + CuO → H₂O + Cu
CO₂ + Ca (OH)₂ → CaCO₃ + H₂O

Question 2.
How do you prepare sodium fusion extract? (Lassaigne’s reagent).
Answer:
A small piece of sodium metal is taken in a fusion tube and heated till it melts. The small amount of organic compound is added to the tube. The tube is heated strongly till it becomes red hot. The red hot tube is dropped into 10 mL of distilled water taken in china dish. The contents are heated and then filtered. The filtrate is called sodium fusion extract or Lassaigne’s reagent.

Question 3.
Give the Lassaigne’s reagent for the detection of nitrogen.
Answer:

Question 4.
Give the reactions involved in the detection of nitrogen by Lasaigne’s reagent test.
Answer:
Reactions: Na + C + N → NaCN
FeSO₄ + 2NaCN → Fe(CN)₂ + Na₂SO₄

Question 5.
How do you detect sulphur by Lassaigne’s reagent?
Answer:

Question 6.
Explain the detection of halogens by sodium fusion extract.
Answer:

Question 7.
Give a test to detect phosphorus in organic compound.
Answer:
The organic compound is heated with an oxidising agent say sodium peroxide. The phosphorus in the organic compound is oxidised to PO^3-₄. The solution is boiled with HNO₃ and treated with ammonium molybdate. The formation of canary yellow precipitate indicates the presence of phosphorous in the compound.
Na₃PO₄ + 3HNO₃ → H₃PO₄ + 3NaNO₃

Quantitative Analysis

Question 1.
Describe the estimation of carbon and hydrogen by Liebig’s method
Answer:
A known mass of the organic compound is completely oxidised with cupric oxide in Leibig’s apparatus as shown in figure. The carbon and hydrogen of the organic compound are oxidised to carbon dioxide and water respectively. The mass of CO₂ and H₂O formed are determined by passing through anhydrous CaCl₂ tube and KOH solution tube. From these masses of CO₂ and H₂O, the % of carbon and hydrogen are calculated as follows:

Calculations
Mass of organic compound taken = w g
Mass of CO₂ formed = x g
Mass of H₂O formed = y g

Percentage of carbon:
44 g of CO₂ contains 12 g of carbon

Percentage of hydrogen:
18 g of water contains 2 g of Hydrogen

Question 2.
0.2346 g of an organic compound containing carbon, hydrogen and oxygen only was analysed by the combustion method. The increase in weight of the U-tube and the potash bulbs at the end of the operation was found to be 0.2754 g and 0.4488 g respectively. Determine the percentage composition of the compound.
Answer:
Mass of CO₂ (increase in mass of potash bulb) = 0.4488 g
Mass of H₂O (increase in mass of U-tube) = 0.2754 g

∴ Percentage of oxygen = 100 – (52.18 + 13.04) = 34.78

Question 3.
On complete combustion, 0.246 g of an organic compound gave 0.198 g of carbon dioxide and 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in the compound.
Answer:

Question 4.
An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Answer:

Question 5.
Describe estimation of nitrogen by Kjeldahls method.
Answer:
A known mass of an organic compound containing nitrogen is heated with concentrated sulphuric acid, nitrogen is converted into ammonium sulphate. The ammonium sulphate is distilled with excess of sodium hydroxide to give ammonia. The ammonia evolved is passed into a known volume of standard hydrochloric acid. The unreacted acid is estimated by back titration with standard sodium hydroxide solution. By knowing the volume of standard HCl reacts with ammonia, percentage of nitrogen in an organic compound is calculated.

Calculation:
Mass of the organic compound taken = ‘w’ g
Let ‘V’ cm³ of ‘M’ molar HCl is used for the neutralization of ammonia.
V cm³ of ‘M’ molar HCl = V cm³ of ‘M’ molar ammonia.
1000 cm³ of 1M NH₃ contain 17 g of NH₃ or 14 g nitrogen.
∴ Mass of Nitrogen present in V cm³ ‘M’ molar NH₃ = \(\frac{14 \times V \times M}{1000 \times 1} g\)
= ‘x’ g
Percentage of Nitrogen = \(\frac{x}{w} \times 100\)

Note:
1. The percentage of nitrogen in an organic compound is directly calculated by using the formula

2. If sulphuric acid is used to neutralise ammonia, then

Question 6.
A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralization. Find the percentage composition of nitrogen in the compound.
Answer:
Step 1: To determine the volume of H₂SO₄ used.
Volume of acid taken = 50 mL of 0.5 MH₂SO₄
Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH
Volume of 0.5 M sulphuric acid required to neutralise 60 ml of 0.5 M NaOH = \(\frac { 60 }{ 2 }\) = 30 mL
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)
∴ Volume of acid used by ammonia = 50 – 30 = 20 mL

Step 2: The percentage of nitrogen

Question 7.
During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄. Find out the percentage of nitrogen in the compound.
Answer:
Percentage of nitrogen

Question 8.
In Kjeldahl’s method the gas evolved from 1.325 g sample of a fertiliser is passed into 50 cm³ of 0.2030 M sulphuric acid. 25.32 cm³ of 0.2030 M NaOH are required for the titration of the unused acid. Calculate the percentage of nitrogen in the fertilizers.
Answer:
Mass of organic compound taken = 1.325 g
Volume of stilphuric acid taken = V₁ = 50 cm³
Volume of 0.203 M sulphuric acid reacted with 25.32 cm³ 0.203 M NaOH = V₂ = \(\frac{25 \cdot 32}{2}\) = 12 . 66 mL

Volume of sulphuric neutralised with ammonia = V₁ – V₂
= 50 – 12.66 = 37.34 mL.
Percentage of nitrogen =

Question 9.
Describe estimation of nitrogen by Dumas method.
Answer:
The known mass of an organic compound containing nitrogen is heated with copperoxide in an atmosphere of carbondioxide. Nitrogen gas, carbondioxide and water are formed. The mixture of gases produced is collected over an aqueous potassium hydroxide which absorbs carbondioxide. Nitrogen is collected in the upper part of the graduated tube as shown in the figure.

Calculation:
Let the mass of organic compound = wg
Volume of nitrogen at STP = \(\frac{\mathrm{p}_{1} \mathrm{V}_{1} \times 273}{760 \times \mathrm{T}_{1}}\)
Where p₁ and V₁ are the pressure and volume of nitrogen. p₁ is different from the atmospheric pressure at which nitrogen gas is collected. The value of P₁ is Obtained by the relation.
P₁ = Atmospheric pressure – Aqueous tension
22400 mL N₂ at STP weights 28 g.
Percentage of nitrogen = \(\frac{28 \times V \times 100}{22400 \times w}\)

Question 10.
In Dumas method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)
Answer:
Volume of nitrogen collected at 300 K and 715 mm pressure is 50 mL
Actual pressure = 715 – 15 = 700 mm
Volume of nitrogen at STP = \(\frac{273 \times 700 \times 50}{300 \times 760}=41 \cdot 9 \mathrm{mL}\)
Percentage of Nitrogen = \(\frac{28}{22400} \times \frac{V_{0} \text { of } \mathrm{N}_{2}}{\mathrm{w}} \times 100\)
\(=\frac{28 \times 41 \cdot 9 \times 100}{22400 \times 0 \cdot 3}=17 \cdot 46\)

Question 11.
In the estimation of nitrogen present in an organic compound by Dumas method, 0.200 g yielded 20.7 mL of dry nitrogen at 15° C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
Answer:
We known that
Volume of nitrogen at S.T.P.

[Note: For dry gas f = 0]
Percentage of nitrogen = \(\frac{28}{22400} \times V_{0} \times \frac{100}{w}\)
\(=\frac{28}{22400} \times 19.486 \times \frac{100}{0.2}=12.178\)

Question 12.
Describe estimations of halogens (Cl, Br and I) by Carius method
Answer:

A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube in a furnace. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.
Let the mass of organic compound taken = ‘w’ g
Mass of AgX formed = ‘m’ g
1 mol of AgX contains = 1 mol of ‘X’

Question 13.
0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Answer:
Mass of substance taken = 0.3780 g
Mass of AgCl formed = 0.5740 g
Molar mass of AgCl = 143.5 g mol^-1
Percentage of chlorine

Question 14.
In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.
Answer:
Molar mass of AgBr = 108 + 80 = 188 g mol^-1
Percentage of bromine

Question 15.
0.50 g of a substance containing bromine gave 0.86 g of silver bromide. Find the percentage of bromine in the compound.
Answer:

Question 16.
0.197 g of an organic substance when heated with excess of strong nitric acid and silver nitrate gave 0.3525 g of silver iodide. Find the percentage of iodine in the compound.
Answer:

Question 17.
How do you estimate the sulphur present in given organic compound?
Answer:
A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.
Let the mass of organic compound taken = ‘w’ g
Mass of barium sulphate formed = ‘m’ g
1 mol of BaSO₄ = 233 g BaSO₄ = 32 g sulphur

Question 18.
In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Answer:
Mass of the substance taken = 0.468 g
Mass of BaSO₄ formed = 0.668 g
Molar mass of AgCl = 22.3 g mol^-1
Percentage of sulpur

Question 19.
In an estimation of sulphur by the Carius method 0.2175 g of the substance gave 0.5825 g of barium sulphate. Calculate the percentage of sulphur in the substance.
Answer:
We known that

Question 20.
In an experiment 0.36 g of an organic compound gave 0.35 g of BaS04. Calculate the percentage of sulphur in the compound.
Answer:

Question 21.
How do you estimate phosphorous present in organic compound?
Answer:
A known mass of an organic compound is heated with fuming nitric acid. The phosphours present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄.12MoO₃, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH₄PO₄ by adding magnesia mixture which on ignition yields Mg₂P₂O₇.
Let the mass of organic compound taken = ‘w’ g
Mass of ammonium phosphomolydate = m g
Molar mass of (NH₄)₃PO₄ ,12MoO₃ = 1877g
Percentage of phosphorus = \(\frac{31 \times m \times 100}{1877 \times w}\)

If phosphorus is estimated as Mg₂P₂O₇,

Question 22.
0.30 g of an organic compound gave 0.550 g of magnesium pyrophosphate by usual analysis. Calculate the percentage of phosphorus in the compound.
Answer:

Question 23.
Describe the estimation of oxygen present organic compound.
Answer:
The percentage of oxygen in an organic compound is found by taking the difference between total percentage composition (100) and the sum of the percentage of all other elements.
Percentage of oxygen 100 (%C + %H + ………. +)