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Carbon and its Compounds · CBSE STD 10 (CBSE - English Medium). 13 questions.

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Question 15 Marks
Why detergents are better cleansing agents than soaps? Explain.
Answer

A substance capable of removing grease and dirt from any fabric or body is called detergent. The detergents are of two types Le. soapy and non-soapy detergents. The soapy detergents are soaps whereas the non-soapy detergents are synthetic detergents or simply detergents. Although both are cleansing agents, they differ in chemical composition. In the present chapter, we shall briefly discuss the composition and cleansing action of soaps and synthetic detergents.

Soaps are the sodium and potassium salts of long chain fatty acids with general formula RCOONa or RCOOK. The acids present have the formula RCOOH where R may have following values.

C15H31COOH
Palmitic acid
(R = C15H31)
C17H33COOH
Oleic acid
(R = C17H33)
C17H35COOH
Stearic acid
(R = C17H35)

 These fatty acids exist as triesters of glycerol which is a trihydric alcohol. The triesters are also called triglycerides or simply glycerides and are the constituents of edible oils and fats. These are of animal and vegetable origin e.g. castor oil, linseed oil or soyabean oil. Chemically the triglycerides are formed as a result of esterification reaction.

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Question 25 Marks
What is a functional group? Give examples of four different functional groups.
Answer
For the definition of functional group and example,

Functional Group

Family

Representation

$-\text{OH}$

 Alcohols

$\text{R}-\text{O}-\text{H}$

$\ \ \ \ \text{O}\\\ \ \ \ ||\\-\text{C}-\text{H}$

 Aldehydes

$\ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\text{H}$

$\ \ \ \ \text{O}\\\ \ \ \ ||\\-\text{C}-$

 Ketones

$\ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\text{R}$

$\ \ \ \ \text{O}\\\ \ \ \ ||\\-\text{C}-\text{OH}$

 Carboxylic acids

$\ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\text{OH}$

$-\text{N}-\text{H}\\\ \ \ \ |\\\ \ \ \text{H}$

 Amines

$\text{R}-\text{NH}_2$

$-\text{O}-$

 Ethers

$\text{R}-\text{O}-\text{R}$

$-\text{X}\ (\text{halogen})$

 Haloalkanes

$\text{R}-\text{X}$

$-\text{NO}_2$

 Nitroalkanes

$\text{R}-\text{NO}_2$
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Question 35 Marks
What are hydrocarbons? Give examples.
Answer
Hydrocarbons are the organic compounds containing only carbon and hydrogen atoms as their constituents. These may be alkanes, alkenes and alkynes.
Alkane: $\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \text{Methane}$  $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \text{Ethane}$  $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ \ \text{Propane}$
Alkyl group: $\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\\ \text{Methyl group}$  $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\\\ \ \ \text{Ethyl group}$  $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{C}-\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\ \ \ \ \text{H}\\\ \ \ \ \ \ \text{Propyl group}$
The structures of compounds belonging to a few families are:
$\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{Cl}\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}$  $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{OH}\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}$  $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{C}-\text{NH}_2\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\ \ \ \ \text{H}$
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Question 45 Marks
Name the reaction which is commonly used in the conversion of vegetable oils to fats. Explain the reaction involved in detail.
Answer

The reaction that is used to convert vegetable oil to fat is called as hydrogenation reaction. Vegetable oils contain unsaturated hydrocarbons which can exhibit addition reaction with hydrogen to form saturated hydrocarbons or fats. The reaction occurs in the presence of metal catalyst like finely divided nickel or palladium at 200°C and forms saturated vegetable fats.

The hydrogenation reaction is an industrial method for the manufacturing of vanaspati ghee from vegetable oil.

$\text{Vegetable oil}+\text{H}_2\xrightarrow{\ \ \text{Ni}, 473\text{K}\ \ }\text{Vanaspati ghee}$

Here, R is any alkyl group which usually contains a large number of carbon atoms in vegetable oils.

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Question 55 Marks
How would you bring about the following conversions? Name the process and write the reaction involved.
  1. Ethanol to ethene.
  2. Propanol to propanoic acid. Write the reactions.
Answer
  1. From ethanol: This method involves the slow oxidation of a dilute solution of ethanol (10-15 per cent) by oxygen present in air in the presence of an enzyme acetobactor.

$\text{CH}_3\text{CH}_2\text{OH}+\text{O}_2\xrightarrow{\ \ \text{Acetobactor}\ \ }\text{CH}_3\text{COOH}+\text{H}_2\text{O}\\\ \ \ \ \text{Ethanol}\ \ \ \ \ (\text{Air})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanoic acid}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{as vinegar})$

The acid obtained is in the form of dilute solution called vinegar.

We have also studied under ethanol that it gets oxidised to ethanoic acid in the presence of dilute solution of alkaline KMnO4 or acidified K2Cr2O7.

  1. From methanol: These days ethanoic acid is manufactured by the reaction between methanol and carbon monoxide in the presence of iodine-rhodium (I2—Rh) catalyst mixture.

$\text{CH}_3\text{OH}+\text{CO}\xrightarrow{\ \ \text{I}_2-\text{Rh Catalyst}\ \ }\text{CH}_3\text{COOH}\\\ \ \text{Methanol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanoic acid}$​​​​​​​​​​​​​​

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Question 65 Marks
Give the structural differences between saturated and unsaturated hydrocarbons with two examples, each.
Answer
Saturated hydrocarbons or alkanes contain either C—C or C—H bonds in their molecules. These are represented by the general formula CnH2n + 2 For example,

$\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \text{Ethane}$  $\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ \ \text{Propane}$

Unsaturated hydrocarbons contain either atleast one >C = C< bond or triple —C ≡ C— bond in their molecules. These may be either alkenes or alkynes in nature. The general formula of alkenes is CnH2n while that of alkynes is CnH2n - 2 for example,

$\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\text{H}-\text{C}=\text{C}-\text{H}\\\ \ \ \ \text{Ethene}$  $\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}=\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \text{Propene}$  $\text{H}-\text{C}\equiv\text{C}-\text{H}\\\ \ \ \ \ \ \text{Ethyne}$  $\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}\equiv\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \text{Propyne}$

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Question 75 Marks
Esters are When ester treated with an alkali, the reaction gives ethanol and sodium ethanoate. This reaction is called saponification reaction because this reaction forms the basis of preparation of soap. This can be given by following equation.
Answer
Esters as pointed, are pleasant smelling compounds. These are therefore, commonly used as flavouring agents and also in perfumes. When an ester is reacted with water in the presence of a dilute acid like dilute HCl, acid and alcohol are formed as the product. The reaction is called ester hydrolysis.

$\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}(\text{l})\xrightarrow{\ \ \text{H}^+\ \ }\text{CH}_3\text{COOH}(\text{aq})+\text{C}_2\text{H}_5\text{OH(aq)}\\\ \ \ \text{Ethylethanoate}$

Ester hydrolysis is the reverse of esterification reaction.

When an ester is reacted with an aqueous solution of base like NaOH or KOH, the product is an alcohol and salt of the acid. For example,

$\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{NaOH}(\text{aq})\xrightarrow{\ \ \ \ }\text{CH}_3\text{COONa (aq)}+\text{C}_2\text{H}_5\text{OH}(\text{aq})\\ \ \ \text{Ethyl ethanoate}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sod. ethanoate (salt.)}$

$\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{KOH}(\text{aq})\xrightarrow{\ \ \ \ \ \ }\text{CH}_3\text{COOK}(\text{aq})+\text{C}_2\text{H}_5\text{OH}(\text{aq})\\\ \ \ \ \ \ \ \text{Ethylethanoate}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Pot. ethanoate (salt)}$

The reaction is known as saponification reaction because it is the basis for the formation of soap.

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Question 95 Marks
Answer
  1. Calcium hydroxide solution in the test tube B becomes milky.
  2. Reaction in test tube A:

CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O

Reaction in test tube B:

Ca(OH)2 + CO2 → CaCO3 + H2O

  1. Ethanol does not react with sodium hydrogen carbonate. Hence, same change cannot be observed.
  2. For this, take distilled water in a beaker and mix calcium carbonate powder in it. After stirring thoroughly, wait till the mixture settles. Decant the clear liquid from the beaker. This liquid is lime water.
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Question 105 Marks
A salt X is formed and a gas is evolved when ethanoic acid reacts with sodium hydrogencarbonate. Name the salt X and the gas evolved. Describe an activity and draw the diagram of the apparatus to prove that the evolved gas is the one which you have named. Also, write chemical equation of the reaction involved.
Answer

Ethanoic (CH3COOH) acid reacts with sodium hydrogen carbonate (NaHCO3) to form sodium ethanoate (CH3COONa), water and carbon dioxide gas. Hence salt X is sodium ethanoate and the gas evolved is CO2. The chemical equation of the reaction involved is:

CH3COOH + NaHCO3 → CH3COONa + H2O + CO2

Activity for observing the gas evolved in above reaction:

Set up the apparatus and take sodium hydrogen carbonate in test tube. Add ethanoic acid to it and observe brisk effervescence of carbon dioxide (CO2) gas. Now pass this gas through freshly prepared lime water (Ca(OH)2). It will turn milky due to the formation of insoluble calcium carbonate (CaCO3).

$\text{Ca(OH)}_2+\text{CO}_2\xrightarrow{\ \ \ }\text{CaCO}_3+\text{H}_2\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Milky})$

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Question 115 Marks
An organic compound A on heating with concentrated H2SO4 forms a compound B which on addition of one mole of hydrogen in presence of Ni forms a compound C. One mole of compound C on combustion forms two moles of CO2 and 3 moles of H2O. Identify the compounds A, B and C and write the chemical equations of the reactions involved.
Answer
Since one mole of compound C on combustion forms two moles of CO2 and three moles of H2O the compound C is a hydrocarbon with formula C2H6. It is ethane. The compound B which forms C2H6 upon addition of hydrogen is ethene (C2H4). The organic compound A which forms ethene upon acidic dehydration is ethanol. The chemical equations for the reactions involved are:

$\text{C}_2\text{H}_5\text{OH}\xrightarrow[\ \ \ \text{heat}\ \ \ ]{\ \text{H}_2\text{SO}_4\ }\text{CH}_2=\text{CH}_2\xrightarrow[\ \ \text{heat}\ \ \ ]{\ \ \text{H}_2\text{Ni}\ \ }\text{CH}_3-\text{CH}_3\\\text{Ethanol (A)}\ \ \ \ \ \ \ \ \ \text{Ethene (B)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethane (C)}$

$\text{C}_2\text{H}_6+3\text{O}_2\xrightarrow{\ \ \ \ (\text{Combustion})\ \ \ }2\text{CO}_2+3\text{H}_2\text{O}\\\ \ (\text{C)}$

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Question 125 Marks
A compound C (molecular formula, C2H4O2) reacts with Na-metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, C3H6O2). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A.

Identify C, R, A, S and write down the reactions involved.

Answer
Compound C with molecular formula C2H4O2 contains two oxygen atoms so it can be either ester or carboxylic acid.

Since it reacts with sodium metal to form compound R and evolves a gas which burns with pop sound, therefore it should be a carboxylic acid which forms sodium alkanoate and hydrogen gas with sodium metal.

$2\text{CH}_3\text{COOH}+2\text{Na}\xrightarrow{\ \ \ \ \ }2\text{CH}_3\text{COONa}+\text{H}_2\uparrow$

The gas which burns with pop sound is hydrogen gas.

Reaction of ethanoic acid with alcohol in the presence of an acid (Conc. H2SO4) forms sweet smelling ester. So compound S that is formed due to reaction of ethanoic acid and methanol (A) is methyl ethanoate with molecular formula C3H6O2 and structural formula CH3COOCH3.

$\text{CH}_3\text{COOH}+\text{CH}_3\text{OH}\xrightarrow{\ \ \text{Conc. H}_2\text{So}_4\ \ }\text{CH}_3\text{COOCH}_3+\text{H}_2\text{O}$

Hence compound C = Ethanoic acid (CH3COOH), R = Sodium ethanoate (CH3COONa), A = Methanol (CH3OH) and S = Methyl merhanoate (CH3COOCH3)

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Question 135 Marks
A compound X is formed by the reaction of a carboxylic acid C2H4O2 and an alcohol in presence of a few drops of H2SO4. The alcohol on oxidation with alkaline KMnO4 followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of:
  1. Carboxylic acid.
  2. Alcohol.
  3. The compound X. Also, write the reaction.
Answer
  1. Carboxylic acid with molecular formula C2H4O2 is acetic acid or ethanoic acid having the structure as:

  1. Given alcohol forms acetic acid on oxidation with alkaline KMnO4 followed by acidification. Therefore, it must be ethanol with structure CH3 ─ CH2 ─ OH.

$\text{CH}_3\text{CH}_2\text{OH}\xrightarrow[\ \ \ \text{Acidification}\ \ \ ]{\ \ \text{Alkaline KMnO}_4\ \ }\text{CH}_3\text{COOH}$

  1. Reaction of ethanoic acid with ethanol in presence of a few drops of conc. H2SO4 is an esterification reaction that forms an ester, ethyl ethanoate (CH3COOC2H5).

$\text{CH}_3\text{COOH}+\text{C}_2\text{H}_5\text{OH}\xrightarrow{\ \ \text{Conc. H}_2\text{SO}_4\ \ }\text{CH}_3\text{COOC}_2\text{H}_5+\text{H}_2\text{O}$

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