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Hydrocarbons are excellent fuels because they ignite easily at low temperature and liberate large amount of heat without producing harmful products.

(a) Ethyne is a highly reactive compound than ethene because of the presence of a triple bond between its two carbon atoms.

(b) Ethene is a highly reactive compound than ethane because of the presence of a double bond between its two carbon atoms.

(c) Hydrocarbons such as alkanes undergo combustion reactions with oxygen to produce carbon dioxide and water vapor. Alkanes are flammable which makes them excellent fuels.

Methane, for example, is the principal component of natural gas.

CH ₄ + 2O₂ → CO₂ + 2H₂O

Carbon (Z = 6) have electronic configuration of 1s² 2s² 2p² . It is not possible for the carbon to form either C⁴⁺ or C^4- ions to attain the nearest noble gas configuration as it requires large amount of energy. This implies that carbon cannot form ionic bond. So almost in all compounds of carbon, it form four covalent bonds.

The functional group present in acetic acid is Carboxyl – COOH.

1. Because of ability of carbon to catenate i.e forms straight chain, branched chains or ring like compounds. 

2. Because carbon monoxide is produced in an insufficient supply of air. This gas is extremely poisonous for human beings as it cuts off the oxygen supply by forming carboxy haemoglobin in the blood.

(D)  Dehydration of ethanol 

Correct option (D) 10, 3 

(B)   7 and 5

(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.

(d) Bronze

It is an alloy and a mixture of metals and all other are organic compounds.

(b) Acetone

Acetone shows functional group isomerism.

(c) Biochemical reactions

(c) Pyridine

Pyridine is an example of heterocylic aromatic compound.

(c) Cyclopropane

(a) Organic compounds : inflammable

(c) Pent-1-ene and pent-2-ene : Chain isomerism

(CH₃)₃C⁻ > (CH₃)₂CH⁻ > CH₃ − CH₂ − > CH₃⁻, D⁻,

Alcoholic – OH group present in CH₃OH. 

Carboxylic – COOH group present in CH₃COOH. 

Aldehydic – CHO group present in CH₃CHO.

The alkyl radical and the functional group are:

(i) 1 - Chloroethene

(ii) 1-methylcyclobutane

Pentatonic acid

(c) Kjeldahl’s method

Since the boiling point of liquid A is much higher than those of liquids B and C, therefore separate liquid A by simple distillation. Since boiling points of liquids B and C are quite close but much lower than liquid A therefore mixture of B and C will distil together leaving behind A. On further heating, A will distil over. Now place the mixture of liquids B and C in a flask fitted with fractionating column. Since the boiling point of liquid B is lower than that of C, on fractional distillation first liquid B will distil over and then liquid C.

BF₃, AlCl₃, SO₃ and ZnCl₂.

For testing sulphur, the sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test. If H₂SO₄ were used, lead acetate itself will react with H₂SO₄ to form white ppt. of lead sulphate which will interfere with the test.

Pb(OCOCH₃)₂ + H₂SO₄                  PbSO₄ + 2CH₃COOH

CO₂ is acidic in nature, therefore, it reacts with the strong base KOH to form K₂CO₃.

2KOH + CO₂                  K₂CO₃ + H₂O

The increase in the mass of U-tube containing KOH then gives the mass of CO₂ produced and from its mass, the percentage of carbon in the organic compound and can be estimated by using the equation:

% of Carbon = 12/44* Mass of CO₂ formed/ Mass of substance formed*100

% of nitrogen= 1.4*Molarity of the acid*Basicity of the acid* Volume of the acid used/Mass of the substance taken

= 1.4*1*2*10/0.5

= 56.0

It is because in a molecule, a reactive site has one or more unshared pairs of. electrons and a polar bond or an electron deficient atom. Alkanes have none of these.

 Name a gaseous hydrocarbon commonly Acetylene.

(b) 2 – Hydroxy Propanoic acid

(c) 3 – Bromo – 2 – methylbutanoic acid

(c) 3

The number of stereoisomers of 1, 2- dihydroxycyclopentane is 3.

(d) Glucose

Glucose is optically active.

Styrene (1-Phenyl ethene), sp²

1-sp³ and 2-sp²

chloro and bromo group . Ali cyclic . 5-bromo-6-chlorocyclohex-1-ene-3-yne 

a) is cyclohexanol and b) is 2-cyclohexylethanol . No they are not isomers. a) 6 b) 2

Alkyl Group : It is obtained by removing one hydrogen atom from a molecule of an alkane. 

Methane : Methyl (Alkyl group) 

Ethane : Ethyl (Alkyl group)

 C₇H₁₆ (Heptane). [∵2n + 2 = 16, 2n = 16 – 2, 2n = 14, n = 14/2 = 7]

(D)  Methyl propanoate

(A)  CH₃ CH₂ CH₂ CN

(i) Unsaturated open chain compound

(ii) Saturated open chain compound

(iii) Aromatic benzenoid compound

(iv) Alicylic compound