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Carbon undergoes three types of hybridisations:
(i) sp3 in saturated compounds,
(ii) sp2 in all unsaturated compounds containing a double bond,
(iii) sp in all unsaturated compounds containing a triple bond.
What are aromatic compounds?
What are heterocyclic compounds?
(i) Planar molecules —sp2
(ii) Linear molecules—sp.
What are isomers?
Compounds having some molecular formula but different chemical and physical properties are called isomers.
Which characteristic is common to different isomers of a compound?
Name the types of structural isomerism shown by alkanes.
Chain isomerism and position isomerism.
Name the four main types of structural isomerism.
(i) Chain isomerism
(ii) Position isomerism
(iii) Functional isomerism
(iv) Tautomerism.
What types of structural isomerism is shown by the following pairs of organic compounds ?
i) Position Isomerism
ii) Functional Isomerism
Draw the structure of the tautomer of phenol and write its IUPAC name.
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Define ring - chain isomerism. Given one example.?
Compounds having the same molecular formula but possessing open chain and cyclic structures are called ring chain isomers and the phenomenon is called ring-chain isomerism. For example propene and cyclopropane are ring chain isomers.
Write the metamer of diethyl ether ?
What is a functional group?
The atom or group of atoms present in a molecule which largely determines its chemical properties is called a functional group.
What is the functional group of:
(i) an aldehyde
(ii) carboxylic acid
(iii) a nitro compound?
Name the alkyl groups obtained from isobutane.
Give the IUPAC name of the compound
CH2 = CH – CH(CH3)2.
The IUPAC compound of the given formula is,
3-methyl but-l-ene.
What are reaction intermediates and why are they highly reactive?
The highly reactive intermediate species formed from the reactants during the reaction are called reaction intermediates. The moment these are formed in the reaction, these get consumed. The high reactivity of reaction intermediates is due to the fact that these are charged species and moreover have an incomplete octet.
These are:
(i) Free radicals (ii) Carbocations
(iii) Carbanions (iv) Carbenes
(v) Nitrenes (vi) Arynes.
Classify the following into electrophilic and nucleophilic reagents:
Which of the two electrophilic and nucleophilic reagents would attack carbonium ion and why?
What is mesomeric effect?
The electron-pair displacement caused by an atom or group along a chain by a conjugative mechanism is called the mesomeric effect of that atom or group.
Which groups have +M and -I effects ?
Atoms having lone pairs and greater electronegativity have +M and -I effects. For example, halogens have greater electronegativity(-I) but have lone pairs of electrons (+M).
What is resonance energy?
The difference between the actual energy of a molecule and that of the most stable canonical form is known as resonance energy.
Write the various canonical forms of urea.
What is the state of hybridisation of positively charged carbon of carbocation?
How is triphenyl methyl anion generated?
It is generated by treating triphenyl methane with base.
Why are free radicals extremely reactive?
Their greater reactivity is due to the tendency of the odd electron in them to get paired up.
Tell the structure of:
(i) Free radical
(ii) Carbocation and
(iii) Carbanion.
(i) Trigonal (ii) Trigonal (iii) Pyramidal.
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Which of the following carbocation is most stable?
B.
The order of stability of carbocation is 3° > 2° > 1°
The reaction
is classified as
electrophilic substitution
nucelophilic substitution
elimination
addition
B.
nucelophilic substitution
Option (b) is correct. This reaction is a nucleophilic substitution reaction since the nucleophile I- is replaced by the nucleophile OH– ion.Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
The components present can be separated by sublimation. Camphor is a volatile substance and will sublime upon heating whereas calcium sulphate does not change into vapours.
Suggest a suitable method for the purification of acetone (b.pt 329 K) mixed with methyl alcohol (b.Pt 338 K).
A compound 'A' has boiling point of 480 K but it starts decomposing at 410 K. Which type of distillation
How will you affect the separation of two organic solids varying in solubility in the same solvent?
What is the principle of paper chromatography?
Name a nitrogenous compound which does not respond to Lassaigne's test.
Name two methods of purification of organic compounds.
How would you extract oil from sunflower seed?
By different extraction using Soxhlet apparatus.
Name a compound which contains nitrogen but does not give ammonia on heating with soda lime.
Aniline (C6H5NH2).
What is the purpose of fusing the organic compound with sodium metal before testing for N, S or halogen?
Sodium metal reacts with some of the elements present in the organic compound and forms corresponding water soluble salts.
Blood red, because compound contains both N and S.
Fe3[Fe(CN)6]4
B.
Fe4[Fe(CN)6]3 The prussian blue colour is due to the formation of K4[Fe(CN)6]3.Why Lassaigne's extract is boiled with concentrated HCl while detecting the presence of nitrogen in the compound ?
It is boiled with concentrated HCl to dissolve the greenish precipitate of ferrous hydroxide formed by an excess of NaOH with FeSO4, which would otherwise mark the Prussian blue colour.
Why is it necessary to boil the sodium extract with conc. HNO3 before testing the halogen?
It is boiled with concentrated nitric acid to decompose any sodium cyanide or sodium sulphide present in the compound containing nitrogen or sulphur also. 
Name the method, if any, for the estimation of oxygen in an orangic compound.
There is no direct method for the estimation of oxygen in an organic compound.
Why a solution of potassium hydroxide is used to absorb carbon dioxide evolved during the estimation of carbon present in an orangic compound?
In Kjeldahl's method for estimation of nitrogen, potassium sulphate is also added along with conc. H2SO4 for digestion of a compound. Why?
What type of substances do not respond positively to the Kjeldahl’s method for the estimation of nitrogen?
Nitro and diazo compounds.
What is collected by the downward displacement of KOH solution in Schiff's nitrometer?
Moist nitrogen gas
What is taken in nitrometer where N2 gas is collected in Duma's method and why?
Potassium hydrogen sulphate is used in the Schiff's nitrometer and its function is to absorb CO2 and H2O.
When does Carius method fail while estimating halogens?
The atomic number of carbon is 6 and its electronic configuration is 2, 4. In other words, carbon has four electrons in the valence shell and thus needs four more electrons to complete its octet. Therefore carbon is tetravalent. Since carbon atom has high ionisation energy and moderate electron affinity, therefore it is very difficult for carbon to either lose or gain four electrons to achieve the nearest inert gas configuration. Consequently, carbon always combines with other atoms by mutual sharing of electrons and thus forms covalent bonds. Thus, carbon is always tetravalent i.e. it forms four covalent bonds with other atoms.
According to Le Bel and van’t Hoff, the four bonds of a carbon atom are directed towards the four corners of a regular tetrahedron. The carbon atom lies at the centre of a regular tetrahedron and the angle between any two adjacent bonds is 109° – 28' or 109.5°. This arrangement of carbon atom is regarded as a tetrahedral model or carbon or space model.
State and explain hybridisation.
The phenomenon of intermixing of atomic orbitals of slightly different energies of an atom so as to redistribute their energies to form the same number of new orbitals of equivalent energies and identical shapes is called hybridization. The new orbitals, thus formed are called hybrid orbitals or hybridised orbitals.
Explanation: In order to understand hybridisation, let us take an example of carbon (Z = 6). Its ground state electronic configuration is
Since it has two filled orbitals, therefore, the valency of the carbon atom should be 2. But actually, carbon atom always exhibits a valency of four (tetravalent). To achieve this, an electron is promoted from 2s filled orbital to the vacant higher energy 2pzorbital. This is called excited state of a carbon atom.
In the excited state of carbon and p orbitals have different energies. Consequently, four bonds of carbon must be of two types. Three of the bonds should be of one type (s-p bonds) while the fourth bond should be of a different type (s-s bond). However, experimental evidence indicates that all the four bonds in case of CH4 (methane) are equivalent. To explain the equivalence of all the four bonds in case of methane, the concept of hybridisation is used i.e. all the four orbitals in the valence shell of carbon may get mixed, redistribute energies and give orbitals of new energy and shape. These equivalent orbitals are called hybrid orbitals.
What are the necessary conditions for hybridization?
Conditions for hybridization:
(i) The orbitals taking part in hybridization must have only a small difference of energies.
(ii) The orbitals undergoing hybridization generally belong to the valence of the atom.
(iii) It can take place between completely filled, half-filled or empty orbitals.
(iv) All the orbitals of the valence shell may or may not take part in hybridization.
Discuss the different types of hybridisation in s and p atomic orbitals
These are of three types:
(i) sp3-type (Tetrahedral hybridisation)
(ii) sp2-type (Trigonal hybridisation)
(iii) sp-type (Diagonal hybridisation)
(i) sp3 hybridisation : Electronic configuration of carbon (Z=6) in the excited state is 1s2
. This type of hybridisation involves the mixing of all the four half filled orbitals. i.e. one s and three p orbitals to form four new orbitals called sp3 hybrid orbitals of equivalent energies and identical shapes.
Fig. Representation of sp3 hybridisation
The four sp3 hybrid orbitals are directed towards the four corners of a tetrahedral. The angle between two adjacent sp3 hybrid orbitals is of 109° – 28'. Each sp3 hybrid orbital has 1/4 s-character and p-character. sp3 hybridisation is also known as tetrahedral hybridisation.
Examples of compounds in which sp3-hybridisation takes place are:
(a) All saturated compounds of carbon such as alkanes, cycloalkanes etc.
(b) Water, alcohols, ethers, ammonia and amines.
(ii) sp2 hybridisation: The electronic configuration of carbon (Z = 6) in the excited state is 
. In this type of hybridisation one s and two p-orbtials of the valence shell of carbon atom take part in hybridisation to give three new sp2 hybrid orbitals. These sp2 hybrid orbitals lie in a plane and are directed towards the corners of an equilateral triangle with carbon atom in the centre. The unhybridised 
orbital lies perpendicular to the three hybridised orbitals, three hybridised orbitals.
sp2 hybridisation is also known as trigonal hybridisation. Each sp2 hybrid orbital has 1/3 s-character and 2/3 p-character.
Examples of compounds in which sp2- hybridisation takes place are:
(a) All compounds of boron i.e. BF3, BH3 etc.
(b) All compounds of carbon containing C=C double bonds such as alkenes, arenes etc.
(iii) sp-hybridisation. The electronic configuration of carbon atom (Z = 6) in the excited state is 
In this type of hybridisation, one s and one p orbital of valence shell of carbon atom take part in hybridisation to give two sp hybrid orbitals of equivalent energies and shape. These sp hybrid orbitals are directed in space at an angle of 180° with respect to each other. Therefore, sp hybridisation is also called linear hybridisation or diagonal hybridisation. Each sp-hybrid orbital has 1/2 s-character and 1/2 p-character. The two unhybridised orbitals (2p and 2pz) which are left lie in different planes at right angles to each other and also to the plane of hybridised orbitals.
Examples of compounds in which sp-hybridisation takes place are:
(i) All compounds of carbon containing C ≡ C and C ≡ N triple bonds.
(ii) All compounds of beryllium such as BeF2, BeH2 etc.
Discuss the shape of methane molecule.
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Discuss the shape of ethane molecule.
Discuss the shape of ethylene (first member of alkenes).
bond, The 
bond consists of two 
electron clouds which lie above and below the plane of carbon and hydrogen atoms. 
; the bond between carbon-carbon atom. The bond length of carbon-carbon double bond of ethylene is 134 pm (1.34 Å).
bond. Similarly, the unhybridised 2pz orbitals from the two carbon atoms overlap sidewise to form one more 
; bond.
; bonds lie in a plane at right angles to each other and each 
; bond consists of two 
; electron clouds. The electron cloud of one 
; bond is considered to lie above and below the line joining the carbon atoms. Thus, the four 
; electron clouds overlap and merge to form a single cylindrical cloud about the internuclear axis. 
The – C ≡ C – bond in acetylene consists of one strong a bond (sp -sp) and two weak σ; bonds (p – p). The bond length of carbon-carbon triple bond is 120 pm (1.20 Å).
In acetylene, carbon atoms as well as hydrogen atoms lie along a line and each C – C – H bond angle is 180°. Thus, acetylene has a linear structure.
Indicate the 
bonds in the following molecules:
Indicate the 
bonds in the following molecules:
CH2 = C = CH2
These are organic compounds which contain only two elements viz carbon and hydrogen.
Classification. These have been classified into two main classes:
(a) Open chain hydrocarbons or acyclic hydrocarbons
(b) Closed chain or cyclic hydrocarbons.
(a) Open chain hydrocarbons: These hydrocarbons contain an open chain of carbon atoms in their molecules which may be either straight chains or branched chains in nature.
These are also called aliphatic hydrocarbons and burn with a non-smoky flame. These have been further divided into two classes:
(a) Saturated hydrocarbons: These are open chain hydrocarbons which contain single covalent bonds between either carbon and carbon atoms (C – C) or carbon and hydrogen atoms (C – H). These are represented by the general formula CnH2n+2 where n is the number of carbon atoms. e.g.
(b) Unsaturated hydrocarbons: These are open chain hydrocarbons which contain carbon-carbon multiple bonds in their molecules. These are further classified as:
(i) Alkenes: These are unsaturated hydrocarbons which contain a double bond between two carbon atoms (>C= C<) in their molecules. These are represented by the general formula 
where n is the number of carbon atoms. e.g.
(ii) Alkynes: These are unsaturated hydrocarbons which contain a triple bond between two carbon atoms 
in their molecules.
These are represented by the general formula 
where n is the number of carbon atoms. e.g. 
II. Closed chain or cyclic hydrocarbons. These hydrocarbons contain closed chains or ring of carbon atoms in their molecules. They are subdivided into two classes:
(i) Alicyclic hydrocarbons: These are hydrocarbons which contain rings of three or more carbon atoms in their molecules. They resemble aliphatic hydrocarbons in most of their characteristics. For example
All hydrocarbons containing a benzene type ring in their molecules are known as aromatic hydrocarbons or arenes.
Thus organic compounds containing one or more fused or isolated benzene rings and their functionalized derivatives are called aromatic compounds. For example,
Cyclic compounds containing one or more heteroatoms (atoms other than C and H) in their rings are called heterocyclic compounds. They are further classified into two types:
(i) Alicyclic heterocyclic compounds: Aliphatic cyclic compounds containing one or more heteroatoms in their rings are called alicyclic heterocyclic compounds. For example,
(ii) Aromatic heterocyclic compounds: Aromatic cyclic compounds containing one or more heteroatoms in their molecules are called aromatic heterocyclic compounds. For example,
Write a short note on homologous series.
Write a short note on:
(a) Generic formula
(b) Primary, secondary, tertiary and quaternary carbon atoms.
: It is that which is linked to one or no carbon atom. e.g. in the compound, 
carbon atoms, 1 and 4 are primary carbon atoms.
: It is that which is linked to two carbon atoms. e.g. in the compound 
carbon atoms, 2 or 3 are secondary carbon atoms.
: A tertiary carbon atom is linked to three carbon atoms. eg. in the compound carbon 2 is a tertiary carbon atom.
: A quaternary carbon atom is linked to four carbon atoms e.g. in the compound
Expand each of the following condensed formulae into their complete structural formula ?
Structural formula:
For each of the following compounds, write a more condensed formula and also their bond-line formula?
Condensed formula:
Make the primary, secondary, tertiary and quaternary carbon atoms in the following molecule:
Also find the number of primary, secondary and tertiary carbon atoms in it.
The given molecule is
Name of a carbon atom. Primary (5), secondary (1), tertiary (1).
Types of isomerism . Isomerism may be divided into following types:
1. Structural isomerism
2. Space or Stereoisomerism.
1. Structural isomerism: When the same molecular formula represents two or more compounds which differ in structures, then such compounds are called structural isomers and the phenomenon is called structural isomerism. Structural isomerism may be further classified as:
(i) Chain isomerism
(ii) Position isomerism
(iii) Functional group isomerism
(iv) Metamerism
(v) Tautomerism.
2. Stereoisomerism: In this case, isomers have the same structural formulae but they differ in the distribution of atoms or groups in space. It is of three types:
(i) Conformational isomerism
(ii) Optical isomerism
(iii) Geometrical isomerism.
What is chain isomerism?
Chain isomerism: When two or more compounds have a similar molecular formula but different carbon skeletons, these are referred to as chain isomers and the phenomenon is termed as chain isomerism. For example, C5H12 represents three compounds:
Write the different chain isomers of butane. Give their IUPAC as well as common names.
Write the different chain isomers of pentane. Give their IUPAC as well as common names.
It has five structural isomers. 
In the isomers of C6H14, indicate primary, secondary, tertiary and quaternary carbon atoms.
Define Position isomerism and Functional isomerism.
Define metamerism and tautomerism.
What do you mean by nomenclature of aliphatic hydrocarbons?
The commonly adopted systems for naming organic compounds are:
(i) Common or Trivial system: In this system, the name of the organic compound is based on its origin/ history or some property. Though common names are short and easy to remember, yet a particular compound may be given a number of names. e.g. methyl alcohol is named wood-spirit because it is obtained by destructive distillation of wood.
2. IUPAC system: This is the best system for naming the aliphatic organic compounds. To evolve a systematic method of nomenclature, the international group of chemists have several conferences and recommended the most definite set of organic nomenclature rules called International Union of Pure and Applied Chemistry rules (IUPAC rules). In the IUPAC system, name of organic compound consists of three parts:
(i) Word root (ii) Prefix (iii) Suffix.
Explain the following with suitable examples:
(i) Word root (ii) Prefix (iii) Suffix
|
Chain length |
Word root |
Chain length |
Word root |
|
C1 |
Meth |
C6 |
Hex |
|
C2 |
Eth |
C7 |
Hept |
|
C3 |
Prop |
C8 |
Oct |
|
C4 |
But |
C9 |
Non |
|
C5 |
Pent |
C10 |
Dec |
|
Nature of bond |
Primary suffix |
General name |
|
C–C single bond |
–ane |
Alkane |
|
C = C double bond |
–ene |
Alkene |
|
C≡ C triple bond |
–yne |
Alkyne |
(b) Secondary suffix: A secondary suffix is added after the primary suffix to indicate the nature of the functional group. These are given below:
How do you write the IUPAC name of a given organic compound?
(i) Prefix: Bromo
(ii) Word root: Pent
(iii) Primary suffix: yne
(iv) Secondary suffix: oic acid
∴ Complete name of the compound is 4- Bromopent-2 yne-1-oic acid.
Here 4, 2 and 1 represent the positions of prefix, primary suffix and of the secondary suffix in the carbon atom chain.
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Alkanes or paraffin are saturated hydrocarbons having C–C and C–H bonds in their molecules. These are represented by the general formula CnH2n+2 where n may have value 1, 2, 3, 4......etc.
There is hardly any difference between IUPAC and common names of the members except that prefix n-(normal or straight chain) is used for common names of alkanes with four or more carbon atoms present in straight chains. The common name of the compound is written in the bracket.
When
n = 1 CH4 Methane (Methane)
n = 2 C2H6 Ethane (Ethane)
n = 3 C3Hg Propane (Propane)
n = 4 C4H10 Butane (n-Butane)
n = 5 C5H12 Pentane (n-Pentane)
n = 6 C6H14 Hexane (n-Hexane)
n = 7 C7H16 Heptane (n-Heptane)
n = 8 C8H18 Octane (n-Octane)
n = 9 C9H20 Nonane (n-Nonane)
n=10 C10H22 Decane (n-Decane)
Name three hydrocarbons with no difference in the IUPAC and common name.
What is an alkyl group? Name the different alkyl groups which originate from:
(i) Ethane (ii) Propane
(iii) n-Butane (iv) Iso-butane.
What are alkenes? Give the IUPAC and common names of few members of alkenes.
Alkenes are unsaturated hydrocrbons containing >C = C< bond in their molecules. These are represented by the general formula, 
Here n may have the value 2, 3, 4....
The IUPAC and common names of few important are:
What are alkynes or acetylenes? Give the IUPAC and common names of few members of alkynes.
Alkynes are unsaturated hydrocarbons containing –C ≡ C– bond in their molecules. These are represented by the general formula Cn H2n-2. Here n may have the value 2,3,4....
The IUPAC and common names of few important members are:
(i) Mono halogen derivatives: These are halogen derivatives of alkanes.
General formula. CnH2n+1 – X or R–X where X is Cl, Br or 1.
Examples:
(ii) Polyhalogen derivatives:
2. Alcohols or alkanols (suffix-ol)
(i) Monohydric alcohols or alkanols.
General formula CnH2n+1 – OH
Functional group –OH
(ii) Dihydric alcohols or Alkanediols (suffix-diol)
General formula: CnH2n(OH)2
Two–OH groups are linked to two different carbon atoms.
iii) Trihydric alcohols or alkane triols (suffix – triol)
General formula: CnH2n–1 (OH)3
Three –OH groups are linked to three different carbon atoms.
3. Ethers or alkoxy alkanes (prefi-alkoxy) General formula:
4. Aldehydes or Alkanals (suffix-al).
General formula CnH2n+1 – CHO or R–CHO
Functional group –CHO
Structure | IUPAC name | Common name |
HCHO | Methanal | Formaldehyde |
ch3cho | Ethanal | Acetaldehyde |
CH3CH2CHO | Propanal | Propionaldehyde |
CH3CH2CH2CHO | Bulanal n-Butyraldehyde |
(ii) Dicarboxylic acid or Alkanedioic acid.
General formula C2H2n (COOH)2
7. Acyl halides or Alkanoyl halide (suffix – oyl chloride)
General formula RCOCl;
Functional group –COCl
8. Acid amides or Alkanamides (suffix – amide).
General formula RCONH2;
Functional group –CONH2
Structure | IUPAC name | Common name |
HCONH2 | Methanamide | Formamide |
CH3CONH2 | Ethanamide | Acetamide |
CH3CH2CONH2 | Proponamide | Propionamide |
CH3CH2CH2CONH2 | Butanamide | n-Butyramide |
9. Esters or AIkyI alkanoate (suffix-oate).
General formula RCOOR or RCOOR'
Structure | IUPAC name | Common name |
HCOOCH3 | Methyl methanoate | Methyl formate |
HCOOC2H5 | Ethyl methanoate | Ethyl formate |
CH3COOCH3 | Methyl ethanoate | Methyl acetate |
CH3COOC2H5 | Ethyl ethanoate | Ethyl acetate |
11. Primary amines (1°) or Alkanamines (suffix amine).
General formula R – NH2
Functional group – NH2 (Amino group)
The common name of an amine is always written as one word.
12. Nitroalkanes or nitro paraffines.
General formula R – NO2
Functional group –NO2 (called nitro group)
Nitro compounds are named under IUPAC system only.
Describe briefly the rules for writing the IUPAC names of long chain hydrocarbons.
(i) 4-Ethyl-3, 4-dimethyl heptane.
(ii) 2,3 dimethyl 3 -propyl hexane.
(iii) 4-(1, 1-dimethyl-ethyl) heptane.
(iv) 2-ethyl 3,3 dimethyl heptane
In naming the compounds containing double or triple bonds, the following rules are followed:
Rule 1. Select the longest continuous chain containing the carbon atoms carrying multiple bonds (double or triple). This gives the parent name of alkene or alkyne.
Rule 2. Once the longest chain is selected, the carbon atoms of the chain are numbered as 1,2,3......from one end to the other keeping in mind that the carbon atoms involved in multiple bonds should get a lowest possible number. For example
Rule 3. If the organic compound contains only one double bond or the triple bond, its locant on the positional number is always placed before its suffix, e.g.
Rule 4. All the rules for naming side chains or substituents are then followed as in alkanes. e.g.
Rule 5. If there are more than one double bond in the molecule, their positions are indicated separately and the prefix (word root) is followed by a diene, a triene etc. e.g.
Rule 6. If there are more than one triple bond in the molecule, their positions are indicated separately and the prefix (word root) is followed by a diene, triene, etc. e.g.
Rule 7. If both double and triple bonds are present, the numbering of the parent chain should always be done from that end which is nearer to the double or the triple bond (lowest sum for the multiple bonds must be followed).
Rule 8. If the double bond and the triple bond are present at equal distances from the end of the chain, then double bond is always given the first preference over the triple bond in numbering, e.g.
Write the IUPAC names of the following compounds:
(i) 2-Methylbuta-1, 3-diene.
(ii) Hexa-1, 3-dien -5-yne.
(iii) 4-methyl pent-2-yne.
Structures and IUPAC names of some hydrocarbons are given below: Explain why the names given in the parentheses are incorrect:
(a) Sum of the locant = 2 + 5 + 6=13 Sum of the locant = 3 + 4 + 7= 14 Lowest locant number 2, 5, 6 is lower than 3, 4, 7
(b) Substituents are in equivalent position, therefore lower number is given to the one that comes first in the name according to alphabetic order.
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-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne.
The correct IUPAC names from each pair is given below:
(a) 2, 2-Dimethylpentane
(b) 2, 4, 7-Trimethyloctane
(c) 2-Chloro-4-methylpentane
(d) But-3-yn-1-ol
Discuss the rules for IUPAC nomenclature of compounds containing one functional group, multiple bonds and substituents.
(A) Lowest number for the functional group.
Rule. 1. Select the longest continuous chain containing the carbon atom having the functional group. e.g.
Rule 2. The numbering of atoms in the parent chain is done in such a way that the lowest number must be always given to the functional group e.g.
Rule 3. All the rules for naming side chains as substituents are then followed as in the case of alkanes.
Rule 4. When a chain containing functional groups such as –CHO, – COOH, –COOR, – CONH2, –COCl, –C ≡ N etc. is present it is always given number 1 and number l is usually omitted from the final name of the compound.
Rule 5. If the organic compound contains a functional group, multiple bond, side chain or substituent, the following order of preference should be followed:
Functional group > Double bond or Triple bond > Substituent/side chain, e.g.
Rule 6. If the compound contains two or more like groups, the numerical prefixes di, tri, tetra etc. are used and the terminal ‘e’ from the primary suffix is retained.
Discuss the rules for IUPAC nomenclature of compounds containing two or more functional groups.
(i) 6-Methyloctan-3-ol
(ii) Hexane-2, 4-dione
(iii) 5-Oxohexanoic acid
(iv) Hexa-1, 3-dien-5-yne
Give the IUPAC names of the following compounds:
(a) Propylbenzene.
(b) 3-Methylpentanenitrile.
(c) 2, 5-Dimethylheptane.
(d) 3-Brome-3-chloroheptane.
(e) 3-chloropropanal.
(f) 2, 2-Dichloroethanol.
Given condensed and bondline structural formulae and identify the functional group (s) present, if any for:
(i) 2, 2, 4 - Trimethylpentane
(ii) 2- Hydroxy - 1, 2, 3 - propanetricarboxylic acid
(iii) Hexanedia
Functional groups:
(i) No functional group present,
(ii) –OH, –COOH
(iii) –CHO
(i) Methyl propanoate
(ii) Aminomethanamide
(iii) 2-Hydroxyethanoic acid.
Give both the IUPAC and common names and find the number of primary, secondary and tertiary carbon atoms in
Which of the following represents the correct IUPAC name for the compounds concerned:
(a) 2, 2-Dimethylpentane or 2- Dimethyl-pentane.
(b) 2, 3-Dimethylpentane or 3, 4-Dimethyl-pentane.
(c) 2, 4, 7-Trimethyloctane or 2,5,7-Trimethyl- octane.
(d) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane.
(e) But -3- yn-l-ol or But-4-ol-l-yne.
The correct IUPAC name for given compound are :
(a) 2, 2-Dimethylpentane
(b) 2, 3-Dimethylpentane
(c) 2, 4, 7-Trimethyloctane
(d) 2-Chloro-4-methylpentane
(e) But-3-yn-l-ol
Discuss the rules for naming alicyclic compounds.
Rule 1. The name of alicyclic compounds is obtained by prefixing cyclo to the name of the corresponding straight-chain hydrocarbon (alkane, alkene or alkyne).
Rule 2. If one alkyl group is present, the same rules are applied as discussed earlier. For example.
Rule 3. If two or more alkyl groups or other substituent groups are present in the ring, their positions are indicated as 1,2, 3...... etc. While numbering the carbon atoms of the ring, the substituent which comes first in the alphabetical order is given the lowest number provided it does not violate the lowest sum rule. For example,
the order is given the lowest number provided it does not violate the lowest sum rule. For example,
Rule 4. If the ring contains more equal number of carbon atom as the alkyl group attached to it, the compound is named as a derivative of cycloalkane and the alkyl group is treated as the substituent eroup. For example.
Rule 5. If the ring contains less number of carbon atoms as the alkyl group attached to it, the compound is named as a derivative of alkane and the cycloalky! group is taken as the substituent group.
Rule 6. If some functional group is present in the ring, it is named as
Rule 7. If the side chain contains a multiple bond or a functional group, the alicycli ring is treated as the substituent irrespective of the size of the ring. For example, 
Rule 8. If a multiple (double or tiple) bond and some other substituents are present in the ring, the numbering is done in such a way that the multiple bond gets the lowest number. For example. 
Rule 9. If some functional group along with other substituent groups is present in the ring, it is indicated by some appropriate prefix or suffix and its position is indicated by numbering the carbon atom of the ring in such a way that the functional group gets the lowest number. 
Discuss short and convenient method (bond line notations) of representing organic molecules ?
In these notations, the bonds are represented by lines and carbon atoms by line ends and intersections. For example.
(i) n-Hexane may be represented as:
(ii) 2-Methylbuta-1, 3-diene may be represented as:
(iii) Hexa -1, 3, 5-triene may be represented as:
Give the IUPAC names of the following compounds:
i)Propene
ii) But-1-ene
iii) But-2-ene
iv) Buta-1,3-diene
v) 2-methylprop-1-ene
vi) 3-methylprop1-ene
vii) 3-methylbut-1-ene
Discuss the nomenclature of simple aromatic compounds.
Aromatic compounds contain one or more isolated or fused benzene rings. Aromatic compound consists of two parts:
(i) nucleus (ii) side chain
(i) Nuclear substituted compounds are those in which the functional group is directly attached to the benzene ring. In the IUPAC system, they are named as derivatives of benzene. The position of the substituents in disubstituted benzene are indicated as 1,2; 1, 3 and 1, 4. These are respectively called ortho (or o-), meta (or m–) and para (or p-).
(ii) Side chain substituted compounds are those in which the functional group is present in the side chain of the benzene ring. Both in the common and IUPAC system, these are usually named as phenyl derivatives of the corresponding aliphatic compounds (except arenes which are named as derivatives of benzene in the IUPAC system). The position of the substituents on the side chain including the benzene ring are indicated as α, β, γ....in the common system and I, 2, 3 etc. in the IUPAC system.
1. Aromatic hydrocarbons (Arenes). Hydrocarbons which contain both aliphatic and aromatic units are called arenes.
Identify the functional groups in the following compounds:
Identify the functional groups in the following compound:
Identify the functional groups in the following compound:
Write the IUPAC names of the following compounds:
(a) 1, 1 1-Trichloro-2, 2-diphenylethane.
(b) Phenyl 2-methyl propanoate.
(c) 2, 4-Dinitrobenzenamine.
What do you understand by:
(i) Homolytic fission
(ii) Heterolytic fission?
Explain the terms:
(i) Electrophiles
(ii) Nucleophiles
Or
What are electrophiles and nucleophiles? Explain with examples.
(Nitronium ion),
(water), 
(ammonia) 
(ethers) etc.Using the curved-arrow notation, show the formation of reactive intermediate when the following covalent bonds undergo heterolytic cleavage.
(a) CH3 – SCH3 (b) CH3 – CN (C) CH3 – Cu
Nucleophile:
These have a unshared pair of electrons which can be donated and shared with an electrophile.
Electrophiles: 
Positive sites have only six valence electrons, can accept pair from a nucleophile.
Identify the electrophilic centre in the following:
CH3CH = O, CH3CN, CH3I
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.
Here, HO- acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus seeking species.
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
Here CH3-C+=O act as electrophiles.
Explain the term inductive effect.
Whenever an electron - withdrawing atom such as halogen i.e. - X(or a group such as nitro) is attached to the end of a carbon atom, the σ; electrons of the C – X bond are displaced towards the more electronegative halogen atom. As a result, the atom X acquires a partial negative charge (δ–). C1 on the other hand acquires a partial positive charge (δ+). The C1 –X bond thus becomes a polar bond.
The partial positive charge on C1, in turn, attracts, the σ;-electrons, of the C1 – C2 bond towards it. As a result, C2 acquires a partial positive charge (i.e. δ'+ ), of course, smaller than that of C1. Similarly, C3 will acquire a small positive charge (δ"+ ) that will still be smaller than that on C2. Thus, positive charge (δ"+ ) that will still be smaller than that on C2. Thus, positive charge (δ"+ ) that will still be smaller than that on C2. 
This permanent displacement of electrons along the chain of carbon atoms due to the presence of an atom or group of different electronegativity at the end of the carbon atom is called inductive effect or I-effect. It is represented by an arrow as shown below:
The atom X exerts an inductive effect only up to a certain length of the chain depending upon its electronegativity. Beyond that, the effect disappears. For practical purposes, it is ignored after the third atom.
The inductive effect is also called transmission effect (T-effect) because the polarity is transmitted along the carbon atom chain.
Types of Inductive effect. It is of two types:
(i) - I effect (Electron withdrawing inductive effect). If the substituent attached to the end of the carbon chain is electron withdrawing, the effect is called - I effect. For example,
The decreasing order of -I effect of some common groups is as follows:
(ii) + I effect (Electron releasing inductive effect). If the substituent attached to the end of the carbon chain is electron donating, the effect is called +I effect. For example. 
The decreasing order of some groups having +I effect is as follows:
Between formic acid and acetic acid which is stronger acid and why?
The Formic acid is more acidic than acetic acid. 
This is because the methyl group is an electron releasing group (+I effect). It displaces the electrons towards the carbon chain. As a result, electron density on the oxygen atom of O-H bond increases. Therefore, the release of H+ ion become difficult, so acidic character decreases.
With the help of inductive effect, show that monochloroacetic acid is stronger acid than acetic acid.
Which bond is more polar in the following pairs of molecules:
(i) H3C – H, H3C – Br
(ii) H3C – NH2, H3C – OH
(iii) H3C – OH, H3C – SH
(i) C – Br bond is more polar because Br is more electronegative than H.
(ii) C – O bond is more polar because O is more electronegative than N.
(iii) C – O bond is more polar because O is more electronegative than S.
In which C-C bond of CH3CH2CH2Br, the inductive effect is expected to be the least?
Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
Or
How will you explain the following correct orders of acidity of the carboxylic acids?
(i) The above order can be explained by - I effect of chlorine atoms. 
As the number of halogen atoms decreases, the overall-I effect decreases and the acid strength decreases accordingly.
(ii) The above order can be explained by + I effect of the methyl group.
As the number of halogen atoms decreases, the overall-I effect decreases and the acid strength decreases accordingly.
(iii) The above order can be explained by +I effect of the methyl group.
As the number of alkyl groups increases, the +I effect increases and the acid strength decreases accordingly.
Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why ?
Explain the term electromeric effect.
which shows the direction of transference of electrons. The atom which receives the electron pair gets negatively charged while the other atoms get positively charged.
electron pair comes back to its original position forming multiple bonds again. 
Give a brief account of resonance effect or mesomeric effect.
It is a process of permanent displacement of electrons from one part of the conjugated 
to the other reacting centres of low and high electrons density due to the phenomenon of resonance and is called effect (R-effect) or mesomeric effect (M-effect). The resonance effect is transmitted through 
-electrons and operates in unsaturated system particularly those containing conjugated systems. For this reason, it is also called conjugative effect. Mesomeric effect operates:
(i) Through conjugative mechanism of electron displacement such as
(ii) When any atom contains a lone pair of electrons and the system is conjugated.
Types of resonance or mesomeric effect. The resonance effect (R) or mesomeric effect (M) are two types:
(i) Groups having electron withdrawing resonance effect. Groups such as 
etc tend to withdraw the electrons from the multiple bonds through resonance and are said to have -R or -M effect. For example,
(ii) Groups having electron-releasing resonance effect. Groups such as
etc. which can release electrons through resonance are said to have +R or +M effect. For example. 
What is resonance? On its basis, explain the stability of benzyl carbocation.
Resonance may be defined as a phenomenon in which a single compound is supposed to be existing as a hybrid of two or more structures differing in the distribution of electrons and not of atoms. These different structures of a molecule are known as contributing structures or resonating structures or canonical forms. No one of the contributing structure truly represents the molecule, but each one of them contributes to the final i.e. actual structure is intermediate between all the contributing structures. Resonance effect is a process of permanent displacement of electrons in a molecule brought about by resonance. Resonance effect is transmitted through 
;-electrons and operates in unsaturated systems particularly those containing a suitable atom or group in conjugation with the unsaturated system. The stability of benzyl carbocation can be explained by a resonance effect. Benzyl carbocation is a resonance hybrid of the following contributing structures:
Describe the main characteristic features of resonance.
The main characteristic features of resonance are:
(i) Resonance involves only the displacement of electrons over the same atomic nuclei.
(ii) Resonance occurs only when all the atoms lie in the same plane.
(iii) The resonating structures must have the same number of paired and unpaired electrons.
(iv) The energies of various resonating structures should be nearly the same.
(v) A more stable contributing form has a greater percentage representation towards the actual molecule.
Write resonance structures of CH3COO– and show the movement of electrons by curved arrows.
Write the structure and put unshared pairs of valence electrons on appropriate atoms. Then draw the arrows to indicate the movement of electrons. 
Write resonance structure of C6H5NH2 and show the movement of electrons by curved arrows.
The following pairs of structures do not constitute resonance structure: (a), (b) and (d).
Write the structures and put unshared pairs of valence electrons on appropriate atoms. Then draw the arrows to indicate the movement of electrons.
C6H5OH (Phenol):
Electron curved arrow notation:
Resonance structure of the given compound :
Resonance structures of the given compound:
Write resonance structures of CH2 = CH–CHO, indicate relative stability of the contributing structures.
Structure 1 is the most stable because each carbon and oxygen atom has an octet and no carbon or oxygen atom has a charge.
In structure II, the negative charge on the more electronegative atom and a positive charge on the more electropositive atom.
Structure III is least stable because more electronegative oxygen has a positive charge and electropositive carbon has a negative charge.
Give reasons why the following two structures I and II cannot be the major contributors to the real structure of CH3COOCH3:
The two structures are less important contributors as they contain charge separation. Also, the structure I contain a carbon atom with an incomplete octet.
Explain hyperconjugation effect or no-bond resonance.
This effect is also called anchimeric effect or Balker Nathan effect.
This effect is explained on the basis of:
(i) molecular orbital concept and
(ii) resonance effect.
(iii) Explanation of the molecular orbital concept. Hyperconjugation is the derealization of the electrons brought about by the sideway overlapping of the p-orbitals of the double bond and σ; orbital of the C – H bond of the alkyl group pz electrons of the double bond and (sp3 – s) σ;-orbital of C – H bond. For example, let us consider the delocalisation of electrons in the alkene 
where 
represents an alkyl group. The delocalisation may be shown as:
Thus electron pair forming C – H bond of the 
-carbon not only binds these atoms (C and H) together but also binds the two carbon atoms of the double bond to some extent. Similarly, the electron pair forming the 
;-bond not only binds these doubly bonded carbon atoms together but also binds the carbon and hydrogen atoms to some extent. In other words, delocalisation helps in bonding together all the four atoms to some extent i.e. there are four electrons delocalised over four nuclei (C1 C2, C3 and H).
(ii) Explanation of resonance effect. According to this concept, if an alkyl group carrying at least one hydrogen is attached to an unsaturated carbon atom, it releases the electrons of C–H bond towards the multiple bonds. For example, propylene can be considered to be resonance hybrid of the following four structures:
Structures I, II and III are hyperconjugation structures. Since there is no bond between carbon and hydrogen atom in these structures, hyperconjugation is also called no-bond resonance. Therefore, more the number of such 
-hydrogen atoms, more are the number of hyperconjugation structures and hence greater is the inductive effect. The number of hydrogen atoms is three with the methyl group, two with the ethyl group, one with the isopropyl group and none with the tertbutyl group. Thus, the order of hyperconjugation effect decreases in the order:
CH3 – > CH3CH2 – > (CH3)2CH – > (CH3)3C –
Explain why alkyl groups act as electron donors when attached to a 
Alkyl group has no lone pair of electrons but it acts as an electron donor when attached to a 
- electron system because of hyperconjugation. Carbon is slightly more electronegative than hydrogen. Thus, the carbon atom in an alkyl group has higher electron density around it as compared with an H atom. As a result, alkyl group are able to donate electrons inductively when attached to a pi system. Let us illustrate this by taking an example of propylene. The various resonating structures are as follows:
What are carbocations? Discuss the relative stabilities of primary, secondary and tertiary carbocations.
are two simple carbocations known as methyl cation and ethyl cations respectively? Carbocations are very short-lived and highly reactive species. Among primary (1°), secondary (2°) and tertiary (3°) carbocations, 3° is most stable.
(i) Inductive effect: Alkyl groups have +I effect. In the carbocation, the alkyl group releases electrons to the positive carbon and thus reduces its charge and in turn itself becomes somewhat positive. Greater the dispersal of charge, greater will be the stability of carbocation. Thus, tertiary carbocations with three alkyl groups are more stable than secondary (with two alkyl groups) which in turn is more stable than primary (with one alkyl group). The methyl carbonium ion is least stable as it has no alkyl group.
(ii) Hyperconjugative effect. Greater the hyper conjugative structures, greater will be the stability of the ion.
Since tertiary butyl carbocation has maximum (10) number of canonical forms, so it is most stable.
The order of stability is 
Give two methods for the generation of carbocation. Describe its structure.
(i) Direct ionisation: The carbon-halogen bond (C-X) in many organic halides generates carbocations in the presence of a highly polar medium.
(ii) Protonation of alkene:
The structure of carbocation: The positively charged carbon of the carbocation is sp2 state of hybridisation. The three sp2 hybridised orbitals which lie in the same plane are involved in the formation of three bonds with other atoms. The unhybridised p-orbital remains vacant. Thus, the carbocation has a flat structure. 
The bond angle around the positively charged carbon is nearly 120° each.
What are carbanions? How are these generated? Discuss the relative stabilities of primary, secondary and tertiary carbanions.
Carbanions: Carbanions may be defined as negatively charged ions, in which carbon is having a negative charge and it has eight electrons in the valence shell. For example,
Carbanions are very short-lived and highly reactive species.
Generation of carbanion: These are mostly generated in the presence of a base by heterolytic cleavage. 
Stability of carbanions. Amongst primary (1°) secondary (2°) and tertiary (3°) carbanions, 1° is the most stable.
The above stability order can be explained by inductive effect. Alkyl group has +I effect. Thus electron releasing group intensifies the negative charge on the carbon atom and destabilises the carbanion. In 3° carbanion due to the presence of three alkyl groups with +I effect, a negative charge is intensified on the carbon atoms and the carbanion gets destabilised. So this is the least stable carbanion. Hence primary carbanion with one alkyl group is, therefore, more stable than secondary (with two alkyl groups) which in turn is more stable than tertiary (with three alkyl groups). In methyl carbanion, H has not any appreciable inductive effect, so it is most stable.
Why carbanions are very negative? Discuss the structure of carbanion.
What are free radicals? How are these formed? Discuss the structure of free radicals.
The order of the relative stabilities of different alkyl free radicals are:
There is no scope of any hyperconjugation in methyl free radical. But ethyl free radical is regarded as resonance hybrid of the following four structures:
Therefore, ethyl free radical is more stable than the methyl free radical. In the case of isopropyl free radical, there are six contributing structures in addition to the normal structure.
Similarly, nine contributing structures are possible for the tertiary butyl free radical in addition to its normal structure. Therefore, it is still more stable and thus, the order of the relative stabilities of the different alkyl free radicals can be justified.
Give the hybridisation state of each carbon in the following species:
The hybridisation of the given compound :
For the following bond cleavages, used curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
For the following bond cleavages, used curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
For the following bond cleavages, used curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
For the following bond cleavages, used curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Structural isomers (actually position isomers as well as metamers.
Resonance contributors because they differ in the position of electrons but not atoms.
What do you understand by:
(i) Substitution reactions
(ii) Addition reactions?
What do you mean by:
(i) Elimination reactions
(ii) Rearrangement reactions?
(i) Elimination reactions: An elimination reaction is one which involves the loss of two atoms or groups either from the same or adjacent atoms of a substance leading to the formation of multiple bond i.e. double or triple bond. In such reactions at least two σ; bonds are lost and one 
; bond is created. These are two types:
(a) β-elimination reaction. Here loss of two atoms or groups takes place from the adjacent carbon atoms in the molecule. For example,
(i) Dehydrohalogenation of alkyl halides with an alcoholic solution of KOH.
(ii) Dehydration of alcohols in the presence of concentrated H2SO4 upon heating:
(iii) 
-Elimination reactions. In such reactions, there is loss or elimination of two atoms or groups from the same carbon atom in the molecule. For example.
(b) Dehydrohalogenation of chloroform with NaOH
(ii) Dehydrogenation of primary or secondary alcohols with reduced copper at 573K.
(iii) Rearrangement reactions Reactions involving the shift or migration of an atom or a group from one atom to another within the same molecule are called rearrangement reactions for example.
(i) n-Butane rearranges to form 2-methyl propane when heated in a sealed tube in the presence of AlCl3/HCl.
(ii) Ammonium cyanate upon heating rearranges to form urea.
What do you understand by:
(i) Isomerisation reaction
(ii) Condensation reaction?
(i) Isomerisation reaction. All those reactions in which interconversion of isomers take place without affecting the molecular formulae and carbon skeletons of reactants and products are called Isomerisation reactions. For example,
(ii) Condensation reaction. All those reactions in which two same or different organic reactants unite to give a product with or without the elimination of another, simple molecule are called condensation reaction. For example condensation of aldehyde or ketone.
Classify the following reactions in one of the reaction type studied in this unit. 
Classify the following reactions in one of the reaction type studied in this unit. 
It is a Electrophilic addition reaction.
Classify the following reactions in one of the reaction type studied in this unit. 
Classify the following reactions in one of the reaction type studied in this unit. 
In this reaction, substitution takes place, followed by a rearrangment of atoms and groups of atoms.
It is rearrangment reaction.
Classify the following transformations according to the reaction type.
In this reaction, substitution takes place, followed by a rearrangment of atoms and groups of atoms.
Classify the following transformations according to the reaction type.
It is a Isomerisation reaction.
Classify the following transformations according to the reaction type.
It is a Addition reaction.
Classify the following transformations according to the reaction type.
Substitution reaction.
Classify the following transformations according to the reaction type.
It is a Elimination reaction.
Classify the following transformations according to the reaction type.
It is a Condensation reaction.
List various methods used for the purification of organic solids.
(i) Crystallisation (ii) Fractional cyrstallistaion (iii) Sublimation (iv) Differential extraction (v) Chromatography
How will you effect the separation of:
(i) Two immiscible liquids
(ii) Two miscible liquids
(iii) Two organic solids differing in solubility in same solvent ?
(i) By extraction with a separating funnel.
(ii) By fractional distillation.
(iii) By fractional crystallisation.
It is based on the difference in the solubilities of the compound and impurities in a suitable solvent.
The process of repeating crystallisation for the separation of two or more solid substances (which are soluble in the same solvent but to a different extent) or cooling the hot saturated solution containing these is called fractional crystallisation. This technique is applied:
(i) to purify a solid compound containing solid impurities having different solubilities in the same solvent.
What is sublimation?
How will you effect the separation of a mixture of acetone and methyl alcohol which are miscible with each other?
Or
Give a brief description of fractional distillation.
Fractional distillation: This process is generally used for those mixtures, where the difference in the boiling points of the two liquids is less than 15 K. Acetone boils at 330 K while methyl alcohol at 338 K, therefore, these two liquids are separated by the fractional distillation process. This is similar to the ordinary distillation method with the only exception that a fractionating column is introduced in between the distillation flask and the condenser.
The process of separation of the components of a liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of these vapours is called fractional distillation.
The fractionating columns used for the purpose are of different shapes as shown in the figure.
When is the process of steam distillation employed? What is its principle?
Steam distillation. This technique is applicable for the purification of solid or liquid organic compounds which are immiscible with water and are volatile in steam from non-volatile impurities which are left behind in the distillation flask. The apparatus used for this purpose is shown.
Principle: The principle underlying steam distillation is that the vapour pressure of a mixture of two immiscible liquids is equal to the sum of the vapour pressures of the two liquids when taken separately. Thus the mixture of two liquids boils at a lower temperature than the boiling point of either liquid.
Steam distillation is employed for the:
(i) recovery of various essential oils from plants and flowers.
(ii) separation of a mixture of o-nitrophenol (steam volatile) from p-nitro phenol.
Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
The essential condition for a liquid to boil is that its vapour pressure should be equal to the external pressure. In steam distillation, the liquid boils when the sum of vapour pressure due to the organic liquid (p1) and water (p2) become equal to the atmospheric pressure (p) i .e. P = P1 + P2.
Since p1 is lower than p, the organic liquid vaporises at low temperature than its boiling point.
What types of liquids are purified by distillation under reduced pressure? What is its principle?
Distillation under reduced pressure: This technique is applied for the purification and separation of high boiling liquids or liquids which decompose partially or completely at or below their normal boiling points. The apparatus for distillation is shown.
Principle: A liquid boils at a temperature at which its vapour pressure becomes equal to the atmospheric pressure. Thus, by decreasing the outer pressure, the boiling point also decreases and it gets distilled without decomposition.
This technique is very helpful in the recovery of glycerol from spent lye in the soap industry. Glycerol boils with decomposition at 563 K at normal pressure; when distilled at 12 mm pressure, glycerol boils at 453K. without decomposition.
Extraction with solvent (differential extraction): The method is based on the fact that organic substances are more soluble in organic solvents than in water. The organic substance is extracted from its aqueous solution adopting the following procedure:
1. The aqueous solution containing organic substance is shaken with a suitable organic solvent which dissolves the substance but is immiscible with water. Two layers are formed—organic layer and an aqueous layer.
2. The solvent layer containing the organic substance (organic layer) is separated using a separating funnel. The impurities remain in the aqueous layer.
3. The organic solvent is removed by distillation to obtain the organic substance.
What is the principle of chromatography? Name the important types of chromatography.
Or
Give a brief description of the principle of chromatography?
The important types of chromatography are:
(i) Column chromatography
(ii) TLC or thin layer chromatography
(iii) Gas chromatography
(iv) Paper chromatography.
Explain the principle of paper chromatography.
The analysis of unknown substances by the flow of solvent on a filter paper is known as paper chromatography. This technique proceeds by a mechanism which is partly partition (distribution) and partly adsorption. The constituents of a mixture are distributed between the water held in the filter paper (water thus acts as a stationary phase) and an organic solvent (mobile phase).
In this method, a drop of the test solution is applied as a small spot near one edge of the filter paper and spot is dried. When the end of the paper strip is dipped into a developing solvent, the solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of the initial spot on the chromatogram. The spots of the separated colourless compounds may be made visible either by ultraviolet light or by the use of a suitable spray reagent.
How will you detect the presence of carbon and hydrogen in an organic compound ?
The detection of carbon and hydrogen in an organic compound is done by a single experiment. A small quantity of dry and powdered organic compound is mixed with 4-5 times its weight of dry cupric oxide. The mixture is taken in a hard glass test tube fitted with a delivery tube having a small bulb in it. The other end of the delivery tube is immersed in freshly prepared lime water taken in another test tube. In the bulb of the delivery tube, a small amount of anhydrous copper sulphate (white) is placed. The mixture is heated strongly when carbon and hydrogen present are oxidised to carbon dioxide and water respectively.
Carbon dioxide turns lime water milky whereas water vapours turn colourless copper sulphate blue. 
Turning of lime water milky and of colourless copper sulphate blue shows the presence of carbon and hydrogen respectively.
Discuss soda lime test for the detection of nitrogen.
Soda lime test: A small amount of an organic compound is heated strongly with soda lime (CaO + NaOH). The liberation of ammonia indicates the presence of nitrogen in the organic compound. 
Limitation: This test is not given by all nitrogenous compounds. Nitro and azo compounds do not respond to this test.
Discuss the chemistry of Lassaigne's test.
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by 'Lassaigne's test'. The elements present in the compound are converted from the covalent form into the ionic form by fusing the compound with sodium metal.
Reactions involved during fusion.
Cyanide, sulphide and halide of sodium so formed in sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
Discuss the Lassaigne's test for the detection of nitrogen.
Lassaigne’s test: This test consists of two steps:
(i) Preparation of Lassaigne’s or sodium extract. A pea-size sodium metal is heated gently in a fusion tube. When it melts to a shining globule, a small amount of the organic compound is added. The fusion tube is heated first gently and then strongly till it becomes red hot. The red hot tube is plunged into distilled water taken in a china dish. The contents in the dish are boiled for a few minutes, cooled and then filtered. The filtrate is known as sodium extract or Lassaigne’s filtrate.
(ii) Test for nitrogen The Lassaigne’s filtrate is usually alkaline. If not, it may be made alkaline by adding a few drops of sodium hydroxide solution. To a part of the alkaline solution, a small amount of a freshly prepared ferrous sulphate solution is added. The contents are then boiled and cooled. A few drops of ferric chloride are then added and the solution is acidified with dilute hydrochloric acid. The appearance of a green or Prussian blue colour confirms the presence of nitrogen in the organic compound.
Chemistry of the test. During fusion, carbon and nitrogen present in the organic compound combines with sodium to form sodium cyanide.
If the organic compound contains both nitrogen and sulphur, it forms sodium sulphocyanide on fusion with sodium. This gives blood red colouration with ferric chloride due to the formation of ferric sulphocyanide.
How will you detect the presence of sulphur in an organic compound by oxidation test?
Oxidation test (for volatile compounds): The organic compound is fused with a mixture of potassium nitrate and sodium carbonate. Sulphur present gets oxidised to sodium sulphate. 
The fused mass is extracted with water and filtered. The filtrate is acidified with hydrochloric acid and then a solution of barium chloride is added. Formation of white precipitate indicates the presence of sulphur in the organic compound. 
How will you detect the presence of sulphur in an organic compound by Lassaigne's test?
Lassaigne’s test: This test involves following two steps:
(i) Preparation of Lassaigne’s or sodium extract (As described under detection of nitrogen in :
Sulphur present in the organic compound combines with sodium on fusion to form sodium sulphide. 
Test for sulphur: The extract is divided into two parts and tested for sulphur as follows:
(i) Lead acetate test: A portion of the Lassaigne's filtrate is acidified with acetic acid and then lead acetate solution is added. Formation of black precipitate confirms the presence of sulphur in the organic compound. 
(ii) Sodium nitroprusside test:To another portion of the Lassaigne's filtrate, few drops of sodium nitroprusside are added. The appearance of purple colour confirms the presence of sulphur in the organic compound.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
When sodium extract is acidified with acetic acid and lead acetate is added to it, a black precipitate of lead sulphide is formed.
PbS is insoluble in acetic acid and soluble in sulphuric acid.Also,because SO42- ions from H2SO4 reacts with Pb++ from Lead Acetate, giving a White ppt of PbSO4, though sulphur is not present in the organic compound.
How will you test the presence of halogen in an organic compound by Lassaigne's test?
Lassaigne's test:
This test involves following two steps;
1. Preparation of Lassaigne's or sodium extract:
where X may be Cl, Br or I.
2. Test for halogen: The Lassaigne’s filtrate is boiled with concentrated nitric acid, cooled and then treated with silver nitrate solution. Formation of precipitate indicates the presence of halogen.
(i) White precipitate soluble in ammonium hydroxide indicates the presence of chlorine in the compound then treated with silver nitrate solution. Formation of precipitate indicates the presence of halogen.
(ii) Light yellow precipitate partially soluble in ammonium hydroxide indicates the presence of bromine in the organic compound.
(iii) Bright yellow precipitate insoluble in ammonium hydroxide indicates the presence of iodine in the compound.
Why is it necessary to boil the Lassaigne's extract with concentrated nitric acid before testing for halogens?
Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
How will you detect the presence of bromine and iodine present in an organic compound by carbon disulphide test?
(i) Test for bromine and iodine. A small portion of Lassaigne's filtrate is acidified with dilute hydrochloric acid and few drops of carbon disulphide (or carbon tetrachloride) are added to it which forms a separate layer at the bottom. An excess of chlorine water is then added to it and shaken vigorously.
(a) An orange colour in carbon disulphide layer confirms bromine.
(b) Violet colour in the carbon disulphide layer confirms iodine.
Write a short note on Beilstein test.
Beilstein test: This test is employed to detect the presence of halogen in an organic compound. A clean copper wire flattened at one end is heated in the oxidising flame till it imparts no colour to the flame. The appearance of blue or green colour due to the formation of volatile copper halide indicates the presence of halogens in the organic compound.
Limitations. This test is not very reliable because:
(i) Compounds like pyridine, urea and thiourea, which do not contain halogens, also respond to this test.
(ii) It does not tell which member of the halogen family is present in the organic compound.
How will you detect the presence of phosphorus in an organic compound?
Ammonium molybdate test: The organic compound is fused with a mixture of sodium carbonate and potassium nitrate. Phosphorus present is oxidised to sodium phosphate.
The residue is extracted with water, boiled with some concentrated nitric acid, and then a hot solution of freshly prepared ammonium molybdate is added to it in excess. Formation of yellow precipitate or colouration indicates the presence of phosphorus in the given organic compound.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
What do you understand by the estimation of elements? Discuss breifly a method for the estimation of carbon and hydrogen in an organic compound.
Estimation of elements. Estimation of elements means to determine the percentage composition of each element present in the organic compound.
Estimation of carbon and hydrogen (Liebig’s method): A known weight of the organic compound is strongly heated with an excess of dry copper oxide in an atmosphere of dry and pure oxygen or air. Carbon of the compound is oxidised to carbon dioxide and hydrogen to water.
Water vapours and carbon dioxide are bubbled through weighed U-tube containing anhydrous calcium chloride and weighed potash bulbs containing a strong solution of potassium hydroxide. The increase in mass of U-tube gives the mass of water formed and increase in mass of potash bulbs gives the mass of carbon dioxide formed. Knowing the masses of water and carbon dioxide formed, the percentage of hydrogen and carbon can be calculated.
Calculations:
Let the mass of organic compound taken = Wg
Let the mass of water formed = a g
(increase in mass of anhydrous CaCl2 U- tube)
Let the mass of carbon dioxide formed = bg
(increase in mass of potash bulbs)
(i) Percentage of hydrogen.
18 g of water contains hydrogen = 2g
(ii) Percentage of carbon.
44 g of carbon dioxide contains carbon = 12 g
Thus percentage of carbon
On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in the compound.
We have given,
Mass of Organic compound= 0.246 g
Mass of Carbon dioxide = 0.198 g
Mass of Water = 0.1014 g
Percentage of carbon:
Mass of carbon dioxide = 0.198 g
Percentage of carbon
Percentage of hydrogen:
Mass of water = 0.1014 g
0.25 g of an organic compound containing carbon, hydrogen and oxygen was analysed by the combustion method. The increase in mass of calcium chloride tube and the potash bulbs at the end of the operation was found to be 0.15 mg and 0.1837 g respectively. Calculate the percentage composition of the compound.
(i) Percentage of carbon.
Increase in mass of potash bulbs = 0.1837 g
i.e. mass of CO2 formed = 0.1837 g
(ii) Percentage of hydrogen.
Increase in mass of calcium chloride = 0.15 g
i.e. mass of water formed = 0.15 g
(iii) Percentage of oxygen.
Percentage of oxygen
= 100 - (Percentage of C + Percentage of H)
= 100 - (20.04 + 6.66) = 73.30
How is nitrogen estimated in a given organic compound by Duma's method?
= (273 + t)K
= p mm
V' mL of nitrogen gas at N.T.P. weighs
In a Duma's nitrogen estimation method, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300K and 715 mm pressure. Calculate the percentage of nitrogen in the compound. (Aqueous tension of water at 300 K is 15 mm).
Mass of organic compound = 0.3 g
(i) To calculate the volume of nitrogen at N.T.P.
(Given) (At N.T.P.)
Applying the general gas equation
(ii) To calculate the percentage of nitrogen:
Now 22400 mL of nitrogen at N.T.P. weighs = 28 g.
How is nitrogen estimated in a given organic compound by Kjeldahl’s method?
taken
Mass of the organic compound taken = 0.5 g
Volume of 
required for neutralising ammonia = 10 mL
Mass of organic compound taken = 0.50 g
Volume of standard H2SO4 taken
= 50 mL of 0.5 M solution
After the adsorption of ammonia,
Volume of alkali required for the excess acid
= 60 mL of 0.5 M solution
60 mL of 0.5 M NaOH = 30 mL of 0.5 M
H2SO4
ie.. the acid left unused = 30 mL of 0.5 M
H2SO4
Discuss a suitable method for the estimation of halogens.
Carius method for the estimation of halogens: A known mass of the organic compound containing halogens is heated with an excess of fuming nitric acid and silver nitrate in a sealed tube called Carius tube. Cabron, hydrogen and sulphur (if present) are oxidised to CO2, H2O and H2SO4 respectively whereas halogen forms a precipitate of silver halide. The precipitate is separated, washed with distilled water, dried and weighed. The percentage of halogens is then calculated. Calculations:
Let the mass of the compound taken = Wg Let the mass of halide (AgX) formed = ag
Atomic mass of halogen = x
Molecular mass of AgCl = 108 + 35.5 = 143.5 g mol-1.
Now 143.5 g AgCl contains 35.5 g chlorine
0.5740 g AgCl would contain
Mass of organic compound is =0.3780
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.
Mass of organic compound 0.15g
Mass of AgBr = 0.12g
Molecular mass of AgBr = 108+80 = 188 g mol-1
Now 188g AgBr contains 80 g bromine
0.15 g of iodoform gave 0.2682 g of silver iodide. Calculate the percentage of iodine.
Mass of the compound = 0.15 g, Mass of silver iodide = 0.2682 g
The molecular mass of silver iodide.
(Ag I) = 108 + 127 = 235
235g of AgI contains = 127 g of iodine
Describe briefly Carius method for the estimation of sulphur.
A known mass of an organic compound containing sulphur is heated with an excess of fuming nitric acid in a sealed Carius tube. The whole of sulphur present in the compound is converted to sulphuric acid which is treated with a slight excess of barium chloride solution. Thus, barium sulphate gets precipitated. It is filtered, washed, dried, and weighed. From this, the percentage of sulphur can be calculated.
Calculations.
Mass of organic compound taken = W
Mass of barium sulphate formed BaSO4 = ag
Molecular mass of barium sulphate
= 
Now 233 g of barium sulphate contain sulphur = 32 g.
In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?
Mass of Organic compound =0.157 g
Mass of barium sulphate = 0.4813 g
Molecular mass of BaSO4
= 137 + 32 + 64 = 233 g mol-1
233 g BaSO4 contains 32 g sulphur
Mass of organic compound = 0.468g
Mass of barium sulphate= 0.668 g
Molecular mass of BaSO4
= 137 + 32 + 64
= 233 g mol-1
Now 233 g BaSO4 contain 32g sulphur
H3PO4
Determine the percentage of phosphorus in an organic compound weighing 0.155 of which gave 0.111 g of Mg2P2O7 in a Carius tube.
Mass of organic compound taken = 0.155 g
Mass of 
formed = 0.111 g
Now 222 g of 
contains phosphorus = 62 g
Draw formulas for the first five members of each homologous series beginning with the following compounds.
(a) H–COOH
(b) CH3COCH3
(c) H–CH=CH2
First five-member of given compounds:
a) H-COOH
i) H-COOH, = methanoic acid
ii)CH3COOH, = Ethanoic acid
iii)CH3CH2COOH=proponic acid
iv)CH3CH2CH2COOH= butanoic acid
v) CH3CH2CH2CH2COOH= pentanoic acid
(b) CH3COCH3
i) CH3COCH3 =2-propanone
ii)CH3CH2COCH3=2-butanone .
iii)CH3CH2CH2COCH3=2-pentanone
iv)CH3CH2CH2CH2COCH3=2-hexanone
v)CH3CH2CH2CH2CH2COCH3= 2-heptanone
(c) H–CH=CH2
i) CH2=CH2 = ethene
ii) CH3CH2=CH2= Propene
iii) CH3CH2CH2=CH2 =Butene
iv) CH3CH2CH2CH2=CH2 =pentene
v) CH3CH2CH2CH2CH2=CH2 =Hexene
What are electrophiles and nucleophiles? Explain with examples.
(b) Neutral electrophiles. These are electrophilic reagents in which the electron deficient atom does not carry any charge. For example, AlCl3, FeCl3, BF3, SnCl4. Carbenes also act as electrophiles because the carbon in them has only six electrons.
Since both positively charged and neutral electrophiles are short by a pair of electrons, they have strong tendency to attract electrons from other sources and hence behave as Lewis acids.
Electrophiles always attack the substrate molecule at the point of high electron density.
(ii) Nucleophiles (or Nucleophilic reagents): Nucleophiles are nucleus loving chemical species containing an atom having an unshared or lone pair of electrons. Nucleophiles are two types:
(a) Negatively charged ions or Negative nucleophiles: These have excess electron pairs I and carry a negative charge. For example OH– J (hydroxyl ion), X– (halide ion), RO– (alkoxide ion), CN– (cyanide ion) and carbanions.
(b) Neutral nucleophiles: These are I nucleophilic reagents which contain atoms with lone pairs of electrons but do not carry any charge. For example, 
(water); 
(ethers) etc.
Since both negatively charged and neutral nucleophiles contain at least one unshared pair of electrons, they have strong tendency to donate this pair of electrons to electron deficient species and hence behave as Lewis bases.
Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
The two compounds with different solubilities in a solvent S can be separated by the method of crystallisation. Crystallisation is used to separate two compounds with different solubilities in a solvent 'S'. the less soluble component will crystallise first then and the more soluble component will crystallise on heating again and then cool it. This process is known as fractional crystallisation.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Simple Distillation: A large number of a liquid boil under ordinary pressure without decomposition. If they contain non-volatile impurities, they are purified by simple distillation. In this method, the impure liquid is vaporised by heating. The vapours formed are passed through a condenser whereas non-volatile impurities are left behind. Condensation of the vapours gives back the liquid in pure form which is collected in a separate vessel.
Distillation under Reduced pressure:Some liquids decompose on boiling under atmospheric pressure. such liquids cannot be purified by simple distillation. they are generally purified by distillation under a reduced pressure.
In this method, the pressure on the liquid surface is reduced with the help of a suitable pump. The liquid, therefore, boils and distills at a temperature which is much lower than its normal boiling point. Thus, decomposition of the liquid is prevented.
Steam Distillation: 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 volatile organic compound is condensed and
collected. The compound is later separated from water using a separating funnel. In steam distillation, the liquid boils when the sum of vapour pressures due to the
organic liquid (p1) and that due to water (p2) becomes equal to the atmospheric pressure (p), i.e. p =p1+ p2. Since p1 is lower than p, the organic liquid vaporises
at lower temperature than its boiling point.
Differentiate between the principle of estimation of nitrogen in an organic compound by: (i) Dumas method and (ii) Kjeldahl’s method.
There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.
(i) Dumas method: The nitrogen-containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
CxHyNz + (2x + y/2) CuO ⎯→
x CO2 + y/2 H2O + z/2N2 + (2x + y/2) Cu
Traces of nitrogen oxides formed if any are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube.
ii) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate. The resulting acid mixture is then heated with an excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of a standard solution of sulphuric acid.
The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction.
Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation (b) Distillation
(c) Chromatography
Crystallisation:
This is a most common method for the purification of solid organic compounds. It is based on the fact that certain organic compounds are partly soluble in a solvent at room temperature and solubility increases with increases in temperature. Example: separation of sugar from a mixture of sugar and common salt by using ethanol.
Distillation:
This method is based on the principle that at constant pressure every pure liquid boils at a definite temperature called its boiling point. The method is used for the purification of those liquids which boil without decomposition provided the impurities are non-volatile. The method is applied for the purification when the two liquid differs in the boiling points by 30-50K.
Chromatography
The technique dependes on the distribution of the mixture between two phases, one fixed and the other mobile. The fixed phase may bean absorbent column (a solid chemical compound) or a paper strip. The moving phase may be a liquid or gas. the mixture to be separted is dissolved in the moving phase and passed over the fixed phase. There are three types of chromatography.
i) adsorption chromatography
ii) partition chromatography
iii) gas chromatography
Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Estimation of phosphorus:
A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3 PO4.12MoO3, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7.
Estimation of halogens:
Carius method: 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. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.
Estimation of sulphur:
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 an excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
In the test of halogens from sodium extract, a small amount of dilute HNO3 is added. This is because in testing the halogen from the Lassigne's extract, it contains halogens. IN case nitrogen and sulphur are also present along with halogens in the organic compound, then along with sodium halide compound, we also get sodium sulphide and sodium cyanide.
Nitric acid decomposes sodium cyanide and sodium halide. else, they precipitate in test and misguide the result.
NaCN +HNO3 -> NaNO3 +HCl
Na2S +2HNO3 ->2NaNO3 + H2S
Therefore, dilute nitric acid is added before testing halogens to expel all the gases if evolved.
The hottest region of Bunsen flame shown in the figure below is:
region 2
region 3
region 4
region 1
A.
region 2
region 1 (pre-heating zone)
region 2 ( Primary combustion zone, hottest zone)
region 3 (Internal zone)
region 4 (secondary reaction zone)
The distillation technique most suited for separating glycerol from spent-lye in the soap industry is:
fractional distillation
steam distillation
distillation under reduced pressure
simple distillation
C.
distillation under reduced pressure
Glycerol with high boiling point (290oC) can be separated from spent lye by distillation under reduced pressure. This process is used to purify liquids having very high boiling points. By this process, the liquid is made to boil at a lower temperature than its boiling point by lowering the pressure on its surface.
Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono substituted alkyl halide?
Tertiary butyl chloride
Neopentane
Isohexane
Neohexane
B.
Neopentane
CnH2n+2 = 12n+ (2n+2)
n=5
Thus, hydrocarbon is C5H12
Identify the compound that exhibits tautomerism
2-Butene
Lactic acid
2-Pentanone
Phenol
C.
2-Pentanone
Which of the following molecules is least resonance stabilized?
D.
is non-aromatic and hence least reasonance stabilized.
The IUPAC name of neopentane is
2-methylbutane
2, 2-dimethylpropane
2-methylpropane
2,2-dimethylbutane
B.
2, 2-dimethylpropane
D.
2,2-dimethylbutane
neopentane is 2, 2 dimethyl propane
The electrophile, E⊕ attacks the benzene ring to generate the intermediate σ-complex. Of the following, which σ-complex is of lowest energy?
B.
-NO2 is electron withdrawing which will destabilize σ - complex.
Which one of the following is the correct statement?
Boric acid is a protonic acid
Beryllium exhibits coordination number of six
Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
B2H6.2NH3 is known as ‘inorganic benzene
C.
Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl
gets oxidised by oxalic acid to chlorine
furnishes H+ions in addition to those from oxalic acid
reduces permanganate to Mn2+
oxidises oxalic acid to carbon dioxide and water
C.
reduces permanganate to Mn2+
HCl being stronger reducing agent reduces MnO4− to Mn2+ and result of the titration becomes unsatisfactory.
Due to the presence of an unpaired electron, free radicals are:
Chemically reactive
Chemically inactive
Anions
Cations
A.
Chemically reactive
The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution hydroxide solutio for complete neutralization. The organic compound is
acetamide
thiourea
urea
benzamide
C.
urea
Using Kjeldhal Process to calculate the percentage of nitrogen inorganic compound.
Let unreacted 0.1 M (= 0.2 N) H2SO4 = V'mL
Therefore, 20 mL of 0.5 M NaOH = V' mL ofg 0.2 NH2SO4
∴ 20 x 0.5 =V' x 02
∴ V' = 50 mL
used H2SO4 = 100-50 = 50 mL
therefore %Nitrogen = 14NV/w
Where N = normality of H2SO4
V = volume of H2SO4 used
% Nitrogen = 1.4 x 0.5 x 50/0.30 = 46.67%
Urea = NH2CONH2 = 28 x 100/60 = 46.67%
When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is
Fe
Zn
Ca
Al
D.
Al
Al(OH)3 = white gelatinous ppt.soluble in excess of NaOH and form Na[Al(OH)4]
Al2O3 used as an adsorbent in chromatography So metal is Al.
The order of stability of the following tautomeric compound is 
I>Ii>III
III > II>I
Ii> I> III
II> III> I
B.
III > II>I
The rolls of beta-dicarbonyl compounds are more stable because of conjugation and intramolecular H-bonding. Thus, the order of stability is,
Less stable as (=) bond is not in conjugation with carbonyl group.
The radical 
because it has
6 p -orbitals and 6 unpaired electrons
7 p- orbitals and 6 unpaired electrons
7 p- orbitals and 7 unpaired electrons
6 p- orbitals and 7 unpaired electrons
A.
6 p -orbitals and 6 unpaired electrons
Due to the presence of 6p -orbitals and 6 unpaired electrons, it is aromatic in nature as these unpaired electrons delocalised in p-orbitals. 
The structure of isobutyl group in an organic compound is
A.
'Iso' mean one Me group is present six chain. Hence, the structure of isobutyl group in an organic compound is
(Yl' suffix is used to represent one -H less than the parent hydrocarbon.)
Which of the statements is not true?
On passing H2S through acidified K2Cr2O7 solution, milky colour is observed.
Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis
Na2Cr2O7 solution in acidic medium is orange
K2Cr2O7 solution becomes yellow on increasing the pH beyond 7
B.
Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis
Being hygroscopic, sodium dichromate Na2Cr2O7 cannot be used in volumetric analysis. All other given statements are true.
Which nomenclature is not according to IUPAC system?
A.
In IUPAC system of nomenclature preference is given to multiple bonds than halogen substituent, so the correct name of
Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear.
CH3 - C ≡ C - CH3
CH2 = CH - CH2 - C ≡ CH
CH3 - CH2 - CH2 - CH3
CH3 - CH = CH - CH3
A.
CH3 - C ≡ C - CH3
CH3 - C ≡ C - CH3 linear, because C2 and C3 are sp hybridised.
The Lassaigne's extract is boiled with conc HNO3 while testing for halogens. By doing so it
helps in the precipitation of AgCl
increases the solubility product of AgCl
increases the concentration of NO3- ions
decomposes Na2S and NaCN, if formed
D.
decomposes Na2S and NaCN, if formed
Na2S and NaCN, if present in the extract, will be decomposed to H2S and HCN by HNO3.
NaCN + HNO3 → NaNO3 + HCN
Na2S +2HNO3 → 2 NaNO3 + H2S
These will escape from the solution and will not interfere with the test for halogens.
The correct IUPAC name of the compound of 
3 - ethyl - 4 ethenyl heptane
3 - ethyl - 4 - propylhex- 5-ene
3-(1-ethyl propyl) hex - 1 ene
4 - ethyl - 3 - propyl hex - 1- ene
D.
4 - ethyl - 3 - propyl hex - 1- ene
Which one of the following is most reactive towards electrophilic reagents?
A.
Which of the following statement is correct for a nucleophile?
Nucleophile is s Lewis acid
Ammonia is a nucleophile
Nucleophiles attack low electrons density sites
Nucleophiles are not electron seeking
A.
Nucleophile is s Lewis acid
Nucleophiles are electron rich species. Hence, act as a Lewis base but not Lewis acid.
The number of structural isomers possible from the molecular formula C3H9N is
4
5
2
3
A.
4
Structural isomers of C3H9N are 
Two possible stereo-structures of CH3CHOH.COOH, which are optically active, are called
diastereomers
atropisomers
enantiomers
mesomers
C.
enantiomers
Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by
oxidation
cracking
distillation under reduced pressure
hydrolysis
B.
cracking
Lower hydrocarbon exists in a gaseous state while higher ones are in liquid state or solid state.
On cracking or pyrolysis, the hydrocarbon with higher molecular mass gives a mixture of hydrocarbons having lower molecular mass. Hence, we can say that by cracking a liquid hydrocarbon can be converted into a mixture of gaseous hydrocarbons.
In the following the most stable conformation of n-butane is
B.
The conformation in which the heavier groups are present at maximum possible distances so that the forces of repulsion get weak is more stable.
Among the given conformations of n- butane, the conformation is shown in option (b) ie, anti conformation is most stable as in it the bulkier groups I(ie, CH3 group) are present at the maximum possible distance.
Which of the following species is not electrophilic in nature?
Cl+
BH3
H3O+
N+O2
C.
H3O+
Electrophiles are electron deficient species. Among the given H3O+ has lone pair of electron for donation, thus, it is not electron deficient and hence, does not behave like an electrophile.
The IUPAC name of the compound having the formula 
3-butene-1-yne
1-butyn-3-ene
but-1-yne-3-ene
1-butene-3 -yne
D.
1-butene-3 -yne
Double bond have preference over triple bond while naming 
The stability of carbanions in the following
is in the order of
(1) > (2) > (3) > (4)
(2) > (3) > (4) > (1)
(4) > (2) > (3) > (1)
(1) > (3) > (2) ) (4)
A.
(1) > (2) > (3) > (4)
When s character increases, then electrons becomes more closer to the nucleus and structure is of lower energy and is more stable.
The stability order of carbanions is as:
The general molecular formula, which represents the homologous series of alkanols is:
CnH2nO2
CnH2nO
CnH2n+1O
CnH2n+2O
D.
CnH2n+2O
Alkanols are the derivatives of alkanes which are derived from the replacement of -H of alkanes with -OH (hydroxyl groups)
The IUPAC name of is:
3,4-dimethylpentanoyl chloride
1-chloro-1-oxo-2, 3-dimethyl pentane
2-ethyl -3-ethylbutanonyl chloride
2,3-dimethylpentanoyl chloride
D.
2,3-dimethylpentanoyl chloride
The correct statement regarding electrophile is
Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile
Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
D.
Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
The most suitable method of separation of 1 : 1 mixture of ortho and para-nitrophenols is
Sublimation
Chromatography
Crystallisation
Steam distillation
D.
Steam distillation
Steam distillation is the most suitable method of separation of 1: 1 mixture of ortho and para nitrophenols as there are intramolecular H-bonds in ortho nitrophenol.
Which one of the following pairs of species has the same bond order?
CO, NO
O2, NO+
CN–, CO
N2, O2
D.
N2, O2
CN(–) and CO have bond order 3 each.
Which of the following is correct with respect to – I effect of the substituents? (R = alkyl)
-NH2 <- OR < - F
-NR2 <- OR <-F
-NR2 >-OR>-F
-NH2>-OR>-F
A.
-NH2 <- OR < - F
-I effect increases on increasing electronegativity of the atom. So, the correct order of -I effect is
-NH2 <-OR < -F
Also,
-NR2<-OR<-F
Sponsor Area
Sponsor Area
; bonds are present in each of the following molecules:
bonds in the following molecules:
bonds in the following molecules:
bonds in the following molecules:
bonds in the following molecules:
