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Carbon And Its Compounds

Class 10th Science Tamilnadu Board Solution
Part-a
  1. Assertion: Chemical bonds in organic compounds are covalent in nature.Reason: Covalent…
  2. Assertion: Diamond is the hardest crystalline form of carbon.Reason: Carbon atoms in…
  3. Assertion: Due to catenation a large number of carbon compounds are formed.Reason: Carbon…
  4. Buckminster fullerene is the allotropic form of _______ .(Nitrogen / Carbon / Sulphur)…
  5. Even though it is a non-metal, graphite conducts electricity. It is due to the presence of…
  6. The formula of methane is CH4 and its succeeding member ethane is expressed asC2H6. The…
  7. IUPAC name of the first member of alkyne is ___________ . (ethene / ethyne)…
  8. Out of ketonic and aldehydic group, which is the terminal functional group?…
  9. Acetic acid is heated with Na2CO3 in a test tube. A colourless and odourless gas (X) is…
  10. Assertion: Denaturation of ethyl alcohol makes it unfit for drinking purpose.Reason:…
Part-b
  1. Write down the possible isomers and give their IUPAC names using the formula C4H10.…
  2. Diamond is the hardest allotrope of Carbon. Give reason for its hardness.…
  3. An organic compound (A) is widely used as a preservative in pickle and has a molecular…
  4. An organic compound (A) of molecular formula C2H6O on oxidation with alkaline KMnO4…
  5. C2H6O is the molecular formula for two compounds A and B. They have different structural…
  6. Rewrite the following choosing the correct word from each pair given in brackets:The…
  7. Identify the compounds using the clues given below:i) This is a dark coloured syrupy…
  8. Read each description given below and say whether it fits for ethanol or ethanoic acid.i)…
  9. Match these words /sentences with appropriate statements given below:(methanol,…
Part-c
  1. Fill the blanks in the given table using suitable formulae.No.AlkaneAlkeneAlkyne1.C2H6…
  2. Homologous series predict the properties of the members of the series. Justify this…
  3. Write the common name and IUPAC name of the following:i) CH3CH2CHO ii) CH3COCH3iii) iv)…
  4. Look at the diagram and answer the following questions:i) What type of structure do…
  5. CnH2n+2 is the general formula of a homologous series of hydrocarbons.i) Is this series…
  6. Ethanol is heated with excess concentrated H2SO4 at 443K.i) Name the reaction that occurs…
  7. Complete the following table:Molecular FormulaCommon NameIUPAC NameCH3CH2CH2CH2OHDimethyl…
  8. Ethanoic acid is a member of Homologous series with general formulaCnH2n+1 COOH.i) Name…
  9. i) Identify A B.ii) Convert ethanol into power alcohol. Mention one of its uses.iii)…
  10. Write a balanced equation using the correct symbols for these chemical reactions:i) Action…
  11. Look at the picture and identify what happens. Support your answer with equations.i) How…
  12. Organic compounds ‘A’ and ‘B’ are the isomers with the molecular formula C2H6O.Compound…
  13. Organic compound ‘A’ of molecular formula C2H6O liberates hydrogen gas with sodium metal.…
  14. Organic compound ‘A’ of molecular formula C2H4O2 gives brisk effervescence with sodium…

Part-a
Question 1.

Assertion: Chemical bonds in organic compounds are covalent in nature.

Reason: Covalent bond is formed by the sharing of electrons in the bonding atoms.

Does the reason satisfy the given assertion?


Answer:

Yes, the reason satisfies the given assertion.

Explanation: The carbon atom has:


i. Atomic number = 6


ii. Electronic configuration = 2,4



iii. Four electrons in the outermost shell


To achieve noble gas configuration (2,8) carbon atom shares its valence electrons (electrons in the outermost shell) with other atoms of carbon or with atoms of other elements.


In the covalent bond, sharing of electrons takes place. For


example: CH4




Question 2.

Assertion: Diamond is the hardest crystalline form of carbon.

Reason: Carbon atoms in diamond are tetrahedral in nature (Verify the suitability of reason to the given Assertion mentioned above)


Answer:

i. In diamond, every carbon atom is covalently bonded to four


neighbouring atoms in a tetrahedral way.


ii. This leads to a tetrahedral three dimensional structure.


iii. Structure of diamond is given below:



iv. This fact is responsible for its hardness and rigidity.



Question 3.

Assertion: Due to catenation a large number of carbon compounds are formed.

Reason: Carbon compounds show the property of allotropy.

Does the reason hold good for the given Assertion?


Answer:

Yes, the reason holds good for the given assertion.

Explanation: Carbon compounds show the property of allotropy.


Allotropy is defined as the property by which an element can exist in more than one form.


i. The forms are physically different.


ii. But chemically similar.


Due to catenation, carbons atoms form a large number of compounds (allotropes)



Question 4.

Buckminster fullerene is the allotropic form of _______ .(Nitrogen / Carbon / Sulphur)


Answer:

Carbon

Explanation: Fullerene is an another types of carbon allotropes.


i. The first discovered fullerene contains 60 carbon atoms in the shape of a football. (C-60).


ii. It is named as Buck Minster Fullerene.




Question 5.

Even though it is a non-metal, graphite conducts electricity. It is due to the presence of ___________ . (free electrons / bonded electrons)


Answer:

free electrons

Explanation: Being a non-metal, graphite is a good conductor of electricity due to the following reasons:


i. In graphite, carbon atoms are arranged in different layers.


ii. In each layer, every carbon atom is linked to three


neighbouring carbon atoms.


iii. Thus, the fourth electron of each carbon atom is free to move


continuously within the entire layer.


iv. Because of the presence of these free electrons in different


layers, graphite becomes a god conductor of electricity.



Question 6.

The formula of methane is CH4 and its succeeding member ethane is expressed as

C2H6. The common difference of succession between them is _______ .(CH2 / C2 H2)


Answer:

CH2

Explanation: The common difference between CH4 and C2H6 :


C2H6 – CH4 = CH2


A homologous series is a group or a class of organic compounds having same general molecular formula and similar chemical properties in which the successive members differ by a CH2 group.



Question 7.

IUPAC name of the first member of alkyne is ___________ . (ethene / ethyne)


Answer:

ethyne

Explanation: Alkyne is an unsaturated hydrocarbon.


i. The unsaturated hydrocarbons containing carbon to carbon triple bond are called alkynes.


ii. Alkynes are named by replacing suffix of alkane with “-yne”.


iii. The first member of alkane is ethane.


iv. Thus, the first member of alkyne is named as ethyne.


Note: The unsaturated hydrocarbons are those which contain carbon to carbon double bonds C=C or carbon to carbon triple bonds -CΞC- in their molecules.



Question 8.

Out of ketonic and aldehydic group, which is the terminal functional group?


Answer:

Aldehydic group is the terminal functional group.

i. Terminal functional group is the one that occur at the end of carbon chain.


ii. Aldehyde is a terminal functional group because it occurs at the end of the group.



iii. Ketone is not a terminal functional group because it does not occur at the end of the group. It occurs at the centre.




Question 9.

Acetic acid is heated with Na2CO3 in a test tube. A colourless and odourless gas (X) is evolved. The gas turns lime water milky. Identify X.


Answer:

When acetic acid (CH3COOH) is heated with Na2CO3 , the following reaction takes place:


i. In this reaction, when sodium carbonate comes in contact


with the gas released in the form of an effervescence, it turns


milky. This is chemical test for carbon dioxide gas.


ii. Hence, the gas (X) is carbon dioxide.



Question 10.

Assertion: Denaturation of ethyl alcohol makes it unfit for drinking purpose.

Reason: Denaturation of ethyl alcohol is carried out by pyridine.

Check whether the reason is correct for assertion.


Answer:

Yes, the reason is correct for assertion.

Explanation: Pyridine is a harmful substance. If it is added to ethyl alcohol. It makes ethyl alcohol unfit for drinking purpose




Part-b
Question 1.

Write down the possible isomers and give their IUPAC names using the formula C4H10.


Answer:

Possible isomers of C4H10 are:


i. The IUPAC name of isomer A is butane.


ii. The IUPAC name of isomer B is iso-butane.


Note: Isomers are the two or more compounds having same molecular formula but different chemical structures.



Question 2.

Diamond is the hardest allotrope of Carbon. Give reason for its hardness.


Answer:

Diamond is the hardest allotrope of carbon because:

a) In diamond, each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral way.


b) This leads to a tetrahedral three dimensional structure.


c) Three dimensional structure of diamond is:



d) The above reason is responsible for its hardness.



Question 3.

An organic compound (A) is widely used as a preservative in pickle and has a molecular formula C2H4O2. This compound reacts with ethanol to form a sweet smelling compound

(B).

(i) Identify the compounds A and B.

(ii) Name the process and write the corresponding chemical equation.


Answer:

Given:

a) Organic compound has a molecular formula: C2H4O2


b) Used as a preservative.


c) Reacts with ethanol to form a sweet smelling substance.


(i) As the organic compound is a preservative, hence it is an acetic acid (ethanoic acid).


Ethanoic acid reacts with ethanol to form a sweet smelling substance called ester(B).


Thus, A is ethanoic acid and B is ester.


(ii) The reaction taking place:



The above process is called esterification.



Question 4.

An organic compound (A) of molecular formula C2H6O on oxidation with alkaline KMnO4 solution gives an acid (B) with the same number of carbon atoms. Compound A is used as an antiseptic to sterilize wounds, in hospitals. Identify A and B. Write the chemical equation involved in the formation of B from A.


Answer:

Given:

i. Organic compound has a molecular formula = C2H6O


ii. Used as an antiseptic to sterilize wounds.


iii. Oxidation with KMnO4 gives an acid.


As the organic compound is used as an antiseptic to sterilize wounds, it must be an ethanol.


Ethanol (A) undergoes oxidation with KMnO4 to form a carboxylic acid, i.e., ethanoic acid (B)


The reaction takes place is given below:



Thus, A is ethanol and B is ethanoic acid.



Question 5.

C2H6O is the molecular formula for two compounds A and B. They have different structural formula.

i) What is this phenomenon known as?

ii) Give the structural formula of A and B.

iii) Write down their common and IUPAC names.

iv) Mention the functional groups of A and B.


Answer:

i) C2H6O is the molecular formula for two compounds A and B but have different structural formula. This phenomenon is called isomerism.


ii) The structural formula of A (CH3CH2OH) is:



The structural formula of B (CH3—O—CH3) is:



iii) Structure A


Common name: Ethyl alcohol


IUPAC name: Ethanol


Structure B


Common name: Dimethyl ether


IUPAC name: Methoxymethane


iv) The functional group of A is alcohol


The functional group of B is ether



Question 6.

Rewrite the following choosing the correct word from each pair given in brackets:

The hydrocarbons containing at least one carbon to carbon _______________ (double/triple) bond are called __________(alkenes/alkynes).They have the general formula CnH2n.These were previously called ______________(olefins/paraffins). When this compound is treated with ____________(bromine/lime) water, decolourisation occurs because it is _____________(saturated/unsaturated).


Answer:

The hydrocarbons containing at least one carbon to carbon double bond are called alkenes. They have the general formula CnH2n. These were previously called olefins. When this compound is treated with bromine water, decolourisation occurs because it is unsaturated.


Explanation:


The hydrocarbons having general formula CnH2n are alkenes which contain at least one carbon to carbon double bond (C=C)


In earlier times, alkenes are known as olefins (meaning-oil) because the lower gaseous members of the family form oily products when treated with chlorine.


When alkene is treated with bromine water (Br2), no change of colour takes place (decolourization) because it is an unsaturated compound (have at least one bond)



Question 7.

Identify the compounds using the clues given below:

i) This is a dark coloured syrupy liquid containing 30% of sucrose.

ii) During manufacture of ethanol this is added as food for yeast.

iii)This enzyme converts sucrose into glucose and fructose.

iv) This compound contains 95.5% ethanol and 4.5% water.

v) This compound contains 100% pure alcohol.


Answer:

i) Molasses


Explanation: Molasses is a dark coloured syrupy liquid which is used in the manufacture of ethanol.


It contains 30% of sucrose which cannot be separated by the crystallization process.


ii) Nitrogenous matter


Explanation: Molasses usually contain nitrogenous matter which act as food for yeast during fermentation.


iii) Invertase


Explanation: The enzyme invertase present in yeast, bring about the conversion of sucrose into glucose and fructose.



iv) Rectified spirit


Explanation: Rectified spirit is an aqueous solution of ethanol which contains 95.5% of ethanol and 4.5% of water.


v) Absolute alcohol


Explanation: The mixture of “rectified spirit” is heated under reflux over quicklime for about 5 to 6 hours. Then it is allowed to stand for 12 hours.


On distillation of the above mixture, pure alcohol (100%) is obtained. This is called absolute alcohol.



Question 8.

Read each description given below and say whether it fits for ethanol or ethanoic acid.

i) It is a clear liquid with a burning taste.

ii) It is used to preserve biological specimens in laboratories.

iii) It is used to preserve food and fruit juices.

iv) On cooling, it is frozen to form ice flakes which look like a glacier.


Answer:

i) Ethanol is a clear liquid and has a burning taste.


ii) Ethanol is used as a preservative for biological specimens in labs.



iii) Ethanoic acid (acetic acid) is used for making vinegar which is used as a preservative in food and fruit juices.


iv) On cooling, pure ethanoic acid is frozen to form ice like flakes. They look like glaciers, so it is called glacial acetic acid.



Question 9.

Match these words /sentences with appropriate statements given below:

(methanol, fermentation, catenation, homologous series, hydrogen gas)

i) The ability of carbon to form large number of compounds through self-linking property.

ii) Alcohols react with sodium to give this element.

iii) This series helps in giving knowledge and enables systematic study of members.

iv) Formation of simple molecules from complex organic compounds using enzymes.

v) Unlike ethanol, the intake of this compound in very small quantities can cause death.


Answer:

i) Catenation


Explanation: Carbon has:


a) Ability to form covalent bonds with other atoms of carbon.


b) This gives rise to a larger number of molecules through self-linking property.


c) This property is called catenation.


ii) Hydrogen gas


Explanation: Carbon compounds such as alcohols react with sodium to liberate hydrogen gas.


Example: 2CH3CH2OH + 2Na → 2CH3CH2ONa + H2


iii) Homologues series


Explanation: Homologues series:


a) All the members are differ by a CH2 group.



b) It helps to predict the properties of the members of the series that are yet to be prepared.


c) Knowledge of homologous series gives a systematic study of the members.


iv) Fermentation


Explanation: Fermentation is the process of formation of simple molecules from complex organic compounds using enzymes.



v) Methanol


Explanation: Methanol:


a) Intake of methanol is very dangerous and can even cause death.


b) If it is consumed, it gets oxidized into methanal in the liver.


c) It affects the liver and causes death.




Part-c
Question 1.

Fill the blanks in the given table using suitable formulae.



Answer:



Explanation: All the members of homologues series have general formula given below:


Alkane: CnH2n + 2


Alkene: CnH2n


Alkyne: CnH2n-2




Question 2.

Homologous series predict the properties of the members of the series. Justify this statement through its characteristics.


Answer:

The characteristics of homologues series are:

i. Each member of the series differs from the preceding or succeeding member by a common difference of CH2


ii. All members of each homologous series contain same elements and same functional groups.


iii. All members of each homologous series have same general molecular formula.


Alkane: CnH2n + 2


Alkene: CnH2n


Alkyne: CnH2n-2


iv. The members in each homologous series show a regular gradation in their physical properties with respect to increase in molecular mass.


v. The chemical properties of the members of each homologous series are similar.


vi. All members of each homologous series can be prepared by using same general method.



Question 3.

Write the common name and IUPAC name of the following:

i) CH3CH2CHO ii) CH3COCH3

iii)

iv) CH3COOH v) HCHO


Answer:

i) CH3CH2CHO


Common name: Propionaldehyde


IUPAC name: Propanal


ii) CH3COCH3


Common name: Dimethyl ketone (acetone)


IUPAC name: Propanone


iii)


Common name: Isopropyl alcohol(or) secondary propyl alcohol


IUPAC name: 2-propanol


iv) HCHO


Common name: Formaldehyde


IUPAC name: Methanal



Question 4.



Look at the diagram and answer the following questions:

i) What type of structure do diamond and graphite have?

ii) Why are diamonds used in cutting tools?

iii) Why is graphite used in electrical circuits?

iv) Name the force that accounts for the softness of graphite.

v) Name the precious diamond you know and give its weight in grams.


Answer:

i) In diamond:


Diamond exists in tetrahedral three dimensional structure.


In Graphite:


Every carbon atom is bonded to three other carbon atoms by covalent bonds. Graphite exists in hexagonal layer structure.


ii) Diamond is used to cut metals due to the following reasons:


a. Brilliant, pure diamond is an extremely hard substance on the earth.


b. It is really very harder than metals and rocks present in the earth.


iii) Free electrons move continuously within the entire layer of graphite, thus it is a good conductor of electricity and mstly used in electrical circuits.


iv) Graphite is a soft metal due to the following reasons:


a. A graphite crystal is hexagonal.


b. The regular arrangement of atoms in graphite is in hexagonal layers.


c. The layers are loosely bounded with each other.


d. As there is weak Vander Walls forces present between them, hence the layers slide with each other.


e. This makes the graphite soft.



v) Blue moon diamond si one of three precious diamond.


In carat, its weight is 12.03 carat.


In grams


1 carat = 0.2 grams


12.03 carats =


⇒ 12.03 × 0.2 grams


⇒ 2.406grams


In grams, blue diamond weight is 2.406 grams



Question 5.

CnH2n+2 is the general formula of a homologous series of hydrocarbons.

i) Is this series saturated or unsaturated?

ii) Name the series described above. Give the formula and name of the member with two carbon atoms.

iii) Draw the structural formula of the first member of this series.

iv) Define the homologous series and find the common difference between the successive members of this family.

v) Write the formula of n-butane and n-pentane.


Answer:

i) The series is saturated.


Explanation: CnH2n+2 is the general formula of a homologous series of alkane which is a saturated hydrocarbon.


Note: Saturated hydrocarbons are those organic compounds which contain carbon–carbon single bond. These were earlier named as paraffins.


ii) The given is homologues series of alkane hydrocarbons.


The formula and name of the member with two carbon atoms:


n=2


C2H2×2+2 = C2H6 (ethane)


Thus, formula is C2H6 and name is ethane


iii) The first member of the series is:


n=1


C1H2×1+2 = CH4 (methane)


The structural formula of methane is:



iv) Homologues series: A homologous series is a group or a class of organic compounds having same general molecular formula and similar chemical properties in which the successive members differ by a CH2 group.


v) In butane, there are four carbon atoms and in pentane there are five carbon atoms.




Question 6.

Ethanol is heated with excess concentrated H2SO4 at 443K.

i) Name the reaction that occurs and explain it.

ii) Write the equation for the above reaction.

iii) What is the product formed? What happens when this gas is passed through bromine water?

iv) When ethanol vapour is passed through bromine water, why does no change occur?


Answer:

i) When ethanol is heated with excess conc. H2SO4 at 443K, it undergoes intra molecular dehydration (i.e. removal of water within a molecule of ethanol) to give ethene.


The reaction is intra molecular dehydration reaction.


ii) The equation for the above reaction is:



iii) The product formed is ethane (alkene), an unsaturated hydrocarbon which contains a double bond.


When this is passed through bromine water, it decolourises the water. It means decolourisation occurs.


iv) When ethanol vapour is passed through bromine water, no changes occur because it is neither a saturated nor an unsaturated hydrocarbon.


Ethanol is a functional group, hence no changes occur.



Question 7.

Complete the following table:



Answer:




Question 8.

Ethanoic acid is a member of Homologous series with general formula

CnH2n+1 COOH.

i) Name the series and give its functional group.

ii) Give the molecular formula and the common name of ethanoic acid.

iii) If this compound is mixed with ethanol in the presence of Conc.H2SO4, a sweet smelling compound is formed. Give the equation and name the compound.

iv) Ethanoic acid reacts with carbonates. Which gas is liberated during this reaction?

v) Write the balanced equation for the reaction of ethanoic acid with carbonate.

vi) Your grandmother has prepared mango pickle. What has she added to preserve it for a long time?


Answer:

i) The given is a member of homologues series of carboxylic acid.


Its functional group is carboxylic.


ii) Ethanoic acid:


Molecular formula: CH3COOH


Common name: Acetic acid


iii) Ethanoic acid reacts with ethanol in the presence of concentrated H2SO4, a sweet smelling compound “ester” is formed. The reaction taking place:



The above process is called esterification.


iv) Ethanoic acid reacts with carbonates and bicarbonates and produces brisk effervescence due to the evolution of carbon dioxide.


v) The reaction of ethanoic acid with a carbonate (Na2CO3) is:


2CH3COOH + Na2CO3→ 2CH3COONa + CO2↑ + H2O


vi) She has added vinegar to preserve it for long time. Acetic acid is used for making vinegar.



Question 9.

i) Identify A & B.



ii) Convert ethanol into power alcohol. Mention one of its uses.

iii) What should be added to obtain denatured spirit?

iv) Give one use of denatured spirit.


Answer:

i) A is methylated spirit which is a mixture of 95% of ethanol and 5% of methanol.


B is rectified spirit which is a mixture of 95.5% of ethanol and 4.5% of water.


ii) By adding petrol to ethanol, we can get power alcohol.


Power alcohol is used in automobiles. It acts as best fuel for automobiles.


iii) Pyridine should be added to obtained denatured spirit. This makes alcohol unfit for drinking.


iv) Denatured spirit is used as a solvent and a fuel.



Question 10.

Write a balanced equation using the correct symbols for these chemical reactions:

i) Action of hydrogen on ethene in the presence of nickel catalyst.

ii) Combustion of methane evolving carbondioxide and water.

iii) Dehydrogenation of ethanol.

iv) Decarboxylation of Sodium salt of ethanoic acid.


Answer:

i) When hydrogen reacts with ethene (unsaturated hydrocarbon) in the presence of nickel catalyst, it gives ethane (saturated hydrocarbon). The reaction takes place given below:



ii) Combustion of methane evolving carbon dioxide and water:


CH4 + 2O2→ CO2 + 2H2O


iii) When the vapour of ethanol is passed over heated copper catalyst at 573 K, it is dehydrogenated to acetaldehyde. This reaction is called dehydrogenation reaction.



iv) Decarboxylation (Removal of CO2) : When sodium salt of ethanoic acid is heated with soda lime (solid mixure of 3 parts of NaOH and 1 part of CaO), methane gas is formed.




Question 11.

Look at the picture and identify what happens. Support your answer with equations.

i) How is B formed from A?



ii) What happens when acetic acid is treated with carbonate salt. Name the gas produced. What happens when this gas is treated with lime water?



iii) What happens when acetic acid is treated with ethanol in the presence of concentrated H2SO4? Give the equation.




Answer:

i. Preparation of ethanoic acid from ethanol:


Ethanol undergoes oxidation with KMnO4 to form a carboxylic acid, i.e., ethanoic acid.


The reaction takes place is given below:



ii. Ethanoic acid reacts with sodium bicarbonate, it gives brisk effervescence of carbon dioxide.


The reaction of ethanoic acid with a carbonate (Na2CO3) is:



When the released gas is treated with limewater, it turns


milky.


iii. When acetic acid is treated with ethanol in the presence of concentrated H2SO4, it forms a sweet smelling substance called ester. This reaction is called esterification.




Question 12.

Organic compounds ‘A’ and ‘B’ are the isomers with the molecular formula C2H6O.

Compound ‘A’ produces hydrogen gas with sodium metal, whereas compound ‘B’ does not. Compound ‘A’ reacts with acetic acid in the presence of concentrated H2SO4 bto form compound ‘C’ with a fruity flavour. What are the isomers ‘A’, ‘B’ and the compound ‘C’?


Answer:

It is given that C2H6O has two isomers A and B.

Compound A produces hydrogen gas with sodium metal,


i. Compound A must be an ethanol. Ethanol reacts with sodium metal to produce hydrogen gas.



Whereas compound B does not react with sodium metal.


ii. Ethanol reacts with acetic acid in the presence of concentrated H2SO4, a sweet smelling compound “ester” is formed. The reaction taking place:


C2H5OH + CH3COOH → CH3COOC2H5 + H2O


EthanolEthanoic acidEster


(A) (C)


The above process is called esterification.


Thus, A is CH3CH2OH (ethanol)


B is CH3—O—CH3(diethyl ether)


C is ester



Question 13.

Organic compound ‘A’ of molecular formula C2H6O liberates hydrogen gas with sodium metal. ‘A’ gives ‘B’ of formula C4H10O, when it reacts with concentrated H2SO4 at 413K. At 443K with concentrated H2SO4 ‘A’ gives compound ‘C’ of formula C2H4. This compound ‘C’ decolourises bromine water. What are ‘A’, ‘B’ and ‘C’?


Answer:

It is given that organic compound liberates hydrogen gas with sodium metal.


i. It must be an ethanol. Ethanol reacts with sodium metal to form sodium ethoxide and hydrogen gas.



ii. When ethanol (A) reacts with concentrated H2SO4 at 413K, it gives diethylether (B).



iii. When ethanol (A) reacts with concentrated H2SO4 at 413K, it gives ethene (C)



Ethene (C) when reacts with bromine water, it delcolourises it.


Thus, A is ethanol.


B is diethyl ether.


C is ethene.



Question 14.

Organic compound ‘A’ of molecular formula C2H4O2 gives brisk effervescence with sodium bicarbonate solution. Sodium salt of A on treatment with soda lime gives a hydrocarbon ‘B’ of molecular mass 16. It belongs to the first member of the alkane family. What are ‘A’ and ‘B’ and how will you prepare ‘A’ from ethanol?


Answer:

It is given that organic compound gives brisk effervescence with sodium bicarbonate solution.


i. As we know that when ethanoic acid reacts with sodium bicarbonate, it gives brisk effervescence of carbon dioxide.


Hence, the organic compound “A” must be ethanoic acid


(CH3COOH)


The reaction of ethanoic acid with a carbonate (Na2CO3) is:



ii. When sodium salt of ethanoic acid is heated with soda lime, methane (alkane) gas is formed.



Methane (B) formed belongs to the first member of the alkane family


iii. Preparation of ethanoic acid (A) from ethanol:


Ethanol undergoes oxidation with KMnO4 to form a carboxylic acid, i.e., ethanoic acid (A)


The reaction takes place is given below:



Thus, A is ethanoic acid and B is methane.