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 C_nH_{2n -2}...............

Natural gas and Petroleum are sources for alkynes.

The general formula of alkynes are:

C_nH_2N-2

Kolbe

Acetic acid was synthesised by Kolbe.

aromatic benzenoid compound

Bromine solution,

 (i) (solid)

 (ii) (low

(iii) C_n H_2n+2

1. By treating ethyl chloride with aqueous KOH. 

2. By heating ethyl chloride with alcoholic KOH. 

3. By passing ethene into concentrated H₂SO₄ at 80°C and high pressure or by hydrating of ethene. 

4. By heating ethyl alcohol with concentrated H₂SO₄ at 170°C.

Berthiot

CH₃

The dipole moment of a molecule depends upon dipole-dipole interactions. Since cis-isomer has higher dipole moment, therefore it has higher boiling point. 

Pentane having a continuous chain of five carbon atoms has the highest boiling point (309.1K) whereas 2,2 – dimethylpropane boils at 282.5K. With an increase in a number of branched chains, the molecule attains the shape of a sphere. This results in a smaller area of contact and therefore weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperatures.

A mixture of camphor and CaSO₄ can be separated as:

(a) Camphor is sublimable but CaSO₄ is not, therefore, sublimation of the mixture gives camphor on the sides of the funnel while CaSO₄ is left in the china dish.

(b) Camphor is soluble in organic solvents like CHCl₃, CCl₄ etc. while CaSO₄ is not. Therefore, when the mixture is shaken with the solvent, camphor goes into the solution while CaSO₄ remains as residue. It is filtered and evaporation of the solvent gives camphor.

Because FeSO₄ gets hydrolysed if its solution is stored for long.

In Dumas method, Nitrogen present in organic compound on heating with CuO release free nitrogen gas, which collected by the downword displacement of KOH. From the volume of nitrogen gas collected at STP, Nitrogen could be estimated.

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/2 N₂ + (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.

Kjedahl 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 excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.

(a) When acetic acid is added to sodium bicarbonate, carbon dioxide is liberated.

CH ₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂

(b) When acetic acid is added to ethyl alcohol in presence of sulphuric acid ester (ethyl acetate) is formed.

CH₃COOH + C₂H₅OH + H₂SO₄ → CH₃COOC₂H₅ + H₂O

(c) When acetic acid is added to neutral FeCl₃, wine red color is produced

Sigma bonds are stronger than pi bond because the extent of overlap in sigma bond is greather than that in pi bond. Sigma bond is formed due to the axial overlap of two orbitals whereas pi bond is formed due to the lateral overlapping of two orbitals. Since the former is more effective, we can say that sigma bond is stronger than pi bond.

(a) If both assertion and reason are true, and reason is the true explanation of the assertion.

In trans-but-2-ene, the dipole moment of two C-CH₃ bonds are equal and opposite and hence they exactly cancel out each other. Thus trans-but-2-ene is non polar. 

In trans-pent-2-ene, the dipole moments of C-CH₃ and C-CH₂CH₃ bonds are unequal. Although these two dipoles oppose each other, yet they do not exactly cancel out each other and hence trans-pent2-ene has a small value of dipole moment and thus is polar.  

(i) There are following properties of resonance hybrid:

(a) Stability: A resonance hybrid is more stable than any of the contributing structures. Greater the number of equivalent contributing structures of a molecule, greater will be the stability of the resonance hybrid.

(b) Resonance energy: It is equal to the difference between the energy of the resonance hybrid and of the most stable resonating structure. More the resonance energy of a molecule, greater is the stability due to resonance, e.g., the resonance energy of benzene is 151 kJ mol^-1. It means benzene is more stable than any of its contributing structures by 151 kJ mol^-1.

(c) Bond lengths: The bond lengths of resonance hybrid are always different from that of any of the contributing structures e.g., in benzene.

(ii) Tert-Butyl carbocation is more stable than sec-propyl carbocation due to hyperconjugation. Similarly, hydrogen of other two methyl groups can undergo hyperconjugation resulting in 10-resonance structures. However, in case of sec-propyl carbocation, only 7-resonance structures are possible.

(i) Both A and R are correct and R is the correct explanation of A.

(iv) A is not correct but R is correct.

(i) Both A and R are correct and R is the correct explanation of A.

The isomers of Butene are:

(i) CH₃ − CH₂-CH = CH₂ ,But-1-ene

(ii) CH₃ − CH = CH − CH₃ , But-2-ene

(iii) CH₂ = C(CH₃) − CH₃ , 2-methyl propene

1. cyclo butane 

2. cyclo pentane 

3. cyclo hutene 

4. cyclo octane

C (carbon)..

2,4-dimethylpentanoic acid, carboxylic acid.

(a) IUPAC name: Propanal

Functional group: -CHO

(b) IUPAC name: Propanol

Functional group: -OH

(i), (ii) Hyperconjugation, also known as sigma-pi conjugation, is the delocalization of sigma electrons. Presence of αH with respect to double bond, triple bond or carbon containing positive charge (in carbonium ion) or unpaired electron (in free radical) is a condition required for hyperconjugation.

(i) C-Br because Br is more electronegative than H.

(ii) C-O because O is more electronegative than N.

(iii) C-O because O is more electronegative than S.

(i) Homologous, unsaturated

(ii) Double,  Addition.

Due to the presence of C = C (carbon – carbon double bond) alkenes are more reactive than alkanes. 

Reaction is given below :

CH₃C(CH₃)C=C(CH₃)CH₂CH₃   \(\overset{O_3}{\rightarrow}\) CH₃C(CH₃)CHO  + CH₃COCH₂CH₃

Paraffins is a Latin word which means (Parum = little and affinis = reactivity). Alkanes are called paraffins because they have a little affinity towards a general reagent.

Third homologous member of
Alkane ➝ Propane (C₃H₈)
Alkene ➝ Butene C₄H₈
Alkanone ➝ Pentanone (CH₃CH₂CH₂COCH₃)

-CH₂ = 12 + 2 = 14
The difference between the mass of two successive members of homologous series is 14.

(C)  geometrical

(a) 2,2- dimethylpropane

(b) 2-methyl butane

(c) Prop-1-ene

(d) 2,2- dimethyl pentane

(e) Pent-2-yne

(f) 3-methyl but-1-yne

(g) 2,3-dichloropentane

(h) 3-methylheptane

(i) 2-methyl butane

(j) Hept-2-yne

(k) 2,2- dimethyl hexanal

(l) Pentan-2-ol

(m) 4-methylpentanoic acid

(n) 2-bromo2-methyl butane

(o) 1- bromo3-methyl butane

(B)   Tautomerism

The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This phenomenon is known as stereoisomerism.

The structure of the following compounds are:

(a) Prop-1-ene

CH₃ – CH = CH₂

(b) 2,3-dimethylbutane

CH₃ − CH(CH₃) − CH(CH₃) − CH₃

(c) 2-methylpropane

CH₃ − CH(CH₃) − CH₃

(d) 3-hexene

CH₃ − CH₂ – CH = CH − CH₂ − CH₃

(e) 2-methylprop-1-ene

CH₃ − C(CH₃) = CH₂

(f) Alcohol with molecular formula C₄H₁₀O

CH₃ − CH₂ − CH₂ − CH₂ – OH

Optical isomerism:

1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.

2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.

3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign (+).

4. The optical isomer which rotates the plane of plane polarised light to the left or in anti- clockwise direction is said to be laveo rotatory and is denoted by the sign (-).

5. Dextrorotatory compounds are represented as ‘d’ (or) by (+) sign and leave rotatory compounds are represented as l (or) by (-) sign.

6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.

(I) Hex-4-ene-2-oI 

(ii) 3-ethyl-5-methyl heptane

(i) CH₂ − CH − CH₂ − CH − CH₂ 

(ii) CH₂ − CH − CH = CH − CH − CH₂ .

When methyl iodide is reduced by nascent hydrogen at ordinary room temperature then methane is formed.

CH₃l+2[H] → CH₄+Hl

When bromoethane is reduced by nascent hydrogen at ordinary room temperature then ethane is produced.

C₂H₅Br+2[H] →C₂H₆+HBr

(D)  diethyl ketone