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NCERT Exemplar · Class 10 Science Carbon and its Compounds

57 questions · 57 still being checked

Long Answer Questions 47–57 (part 6 of 6)

  1. Exercise 47

    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.

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    NCERT’s answer
    \(\displaystyle \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{CH}_3 \mathrm{COO} \mathrm{Na}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\) X is sodium ethanoate Gas evolved is carbon dioxide Hint— Activity Lime water will turn milky, a characteristic property of \(\displaystyle \mathrm{CO}_2\) gas NCERT_Solution_Class10_Science_Exemplar_Ch4_Q47_ncert
    Ethanoic acid displaces carbonic acid from the hydrogencarbonate, which breaks down to water and \(\displaystyle \mathrm{CO_2}\). \[\mathrm{CH_3COOH} + \mathrm{NaHCO_3} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2}\uparrow \] X is sodium ethanoate; the gas evolved is carbon dioxide. Activity: add ethanoic acid to solid \(\displaystyle \mathrm{NaHCO_3}\) in a test tube (brisk in the cold, no heating) and lead the gas, through an airtight cork and delivery tube, into lime water.The lime water turns milky, confirming \(\displaystyle \mathrm{CO_2}\). \[\mathrm{Ca(OH)_2} + \mathrm{CO_2} \rightarrow \mathrm{CaCO_3}\downarrow + \mathrm{H_2O} \] Stop bubbling as soon as it turns milky: excess \(\displaystyle \mathrm{CO_2}\) redissolves the precipitate and the milkiness clears. \[\mathrm{CaCO_3} + \mathrm{H_2O} + \mathrm{CO_2} \rightarrow \mathrm{Ca(HCO_3)_2} \] Answer: X = sodium ethanoate (\(\displaystyle \mathrm{CH_3COONa}\)); gas = \(\displaystyle \mathrm{CO_2}\), confirmed by the milky lime water.

  2. Exercise 48

    (a)
    What are hydrocarbons? Give examples.
    (b)
    Give the structural differences between saturated and unsaturated hydrocarbons with two examples each.
    (c)
    What is a functional group? Give examples of four different functional groups.

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    NCERT’s answer
    (a)
    Compounds of carbon and hydrogen are called hydrocarbons. Example, methane, ethane etc.
    (b)
    Saturated hydrocarbons contain carbon- carbon single bonds.
    Unsaturated hydrocarbons contain atleast one carbon - carbon double or triple bond.
    (c)
    Functional group - An atom/group of atoms joined in a specific manner which is responsible for the characteristic chemical properties of the organic compunds. Examples are hydroxyl group (- OH), aldehyde group (- CHO), carboxylic group (- COOH) etc.
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q48_ncert
    (a)
    Hydrocarbons contain only carbon and hydrogen, e.g. \(\displaystyle \mathrm{CH_4}\) (methane), \(\displaystyle \mathrm{C_2H_6}\) (ethane).
    (b)
    Saturated hydrocarbons have only single C–C bonds, e.g. \(\displaystyle \mathrm{CH_3-CH_3}\) (ethane), \(\displaystyle \mathrm{CH_3CH_2CH_3}\) (propane). Unsaturated hydrocarbons carry at least one double or triple bond, e.g. \(\displaystyle \mathrm{CH_2=CH_2}\) (ethene), \(\displaystyle \mathrm{CH\equiv CH}\) (ethyne).
    (c)
    A functional group is the atom or group of atoms that fixes a compound's chemical behaviour. Examples: \(\displaystyle -\mathrm{OH}\) (alcohol), \(\displaystyle -\mathrm{CHO}\) (aldehyde), \(\displaystyle -\mathrm{COOH}\) (carboxylic acid), \(\displaystyle -\mathrm{X}\) (haloalkane).
    Answer: (a) compounds of C and H only; (b) saturated = single bonds, unsaturated = double/triple bonds; (c) functional group – reactive site, e.g. \(\displaystyle -\mathrm{OH}, -\mathrm{CHO}, -\mathrm{COOH}, -\mathrm{X}\).

  3. Exercise 49

    Name the reaction which is commonly used in the conversion of vegetable oils to fats. Explain the reaction involved in detail.

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    NCERT’s answer
    Hint— Hydrogenation reaction NCERT_Solution_Class10_Science_Exemplar_Ch4_Q49_ncert
    The reaction is hydrogenation. \[\mathrm{R-CH=CH-R'} + \mathrm{H_2} \xrightarrow[\text{Ni}]{473\,\mathrm{K}} \mathrm{R-CH_2-CH_2-R'} \] Hydrogen adds across the C=C double bonds of the unsaturated fatty-acid chains in the vegetable oil, in the presence of a nickel catalyst at about $\displaystyle 473$ K, turning them into single C–C bonds. The product is a saturated fat (vanaspati ghee), solid unlike the starting liquid oil. Answer: Hydrogenation – catalytic addition of \(\displaystyle \mathrm{H_2}\) (Ni, \(\displaystyle 473\,\mathrm{K}\)) turns unsaturated oil into saturated fat.

  4. Exercise 50

    (a)
    Write the formula and draw electron dot structure of carbon tetrachloride.
    (b)
    What is saponification? Write the reaction involved in this process.

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    NCERT’s answer
    (a)
    \(\displaystyle \mathrm{CCl}_4\)
    (b)
    Saponification is the process of converting esters into salts of carboxylic acids and ethanol by treating them with a base.
    \[\mathrm{CH}_3 \mathrm{COO} \mathrm{C}_2 \mathrm{H}_5 \xrightarrow{\mathrm{NaOH}} \mathrm{CH}_3 \mathrm{COO} \mathrm{Na}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \]
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q50_ncert
    (a)
    Formula: \(\displaystyle \mathrm{CCl_4}\). Carbon uses all four of its valence electrons to form one shared pair with each chlorine atom; each Cl completes its octet with three lone pairs.
    (b)
    Saponification is the alkaline hydrolysis of an ester (fat/oil) by NaOH, giving soap and an alcohol.
    \[\mathrm{RCOOR'} + \mathrm{NaOH} \xrightarrow{\Delta} \mathrm{RCOONa} + \mathrm{R'OH} \]
    \(\displaystyle \mathrm{RCOONa}\) is the soap; for a fat, \(\displaystyle \mathrm{R'OH}\) is glycerol.
    Answer: (a) \(\displaystyle \mathrm{CCl_4}\) – C bonded to $\displaystyle 4$ Cl by single covalent bonds (electron-dot structure in figure). (b) Ester + NaOH \(\displaystyle \xrightarrow{\Delta}\) soap + alcohol (saponification).

  5. Exercise 51

    Esters are sweet-smelling substances and are used in making perfumes. Suggest some activity and the reaction involved for the preparation of an ester with well labeled diagram.

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    NCERT’s answer
    Activity
    Take $\displaystyle 1$ mL ethanol (absolute alcohol) and $\displaystyle 1$ mL glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
    Warm in a water-bath at about $\displaystyle 60$ C for at least $\displaystyle 15$ minutes as shown in the Figure (It should not be heated directly on flame as the vapours of ethanol catch fire)
    Pour into a beaker containing $\displaystyle 20$-$\displaystyle 50$ mL of water and smell the resulting mixture.
    \[\underset{\text{Ethanoic acid}}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{\text{Ethanol}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}} \xrightarrow{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} \underset{\text{Ester}}{\mathrm{CH}_3 \mathrm{COOCH}_2 \mathrm{CH}_3}+\mathrm{H}_2 \mathrm{O} \] NCERT_Solution_Class10_Science_Exemplar_Ch4_Q51_ncert
    Esterification: heat ethanoic acid with ethanol and a few drops of conc. \(\displaystyle \mathrm{H_2SO_4}\) (catalyst) in a test tube kept in a hot-water bath for a few minutes, then pour into a beaker of water. \[\mathrm{CH_3COOH} + \mathrm{C_2H_5OH} \xrightarrow[\Delta]{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_3COOC_2H_5} + \mathrm{H_2O} \]Heat only in a water bath, never on a direct flame – the mixture is flammable and can bump or splash. The fruity smell is the ester, ethyl ethanoate. Answer: Ester = ethyl ethanoate, \(\displaystyle \mathrm{CH_3COOC_2H_5}\), formed by heating ethanoic acid with ethanol and conc. \(\displaystyle \mathrm{H_2SO_4}\).

  6. Exercise 52

    A compound C (molecular formula, C2H4O2\displaystyle \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2 ) 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\displaystyle \mathrm{C}_3 \mathrm{H}_6 \mathrm{O}_2 ). 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.

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    NCERT’s answer
    C — Ethanoic acid
    R — Sodium salt of ethanoic acid (sodium acetate) and gas evolved is hydrogen
    A — Methanol
    S — Ester (Methyl acetete)
    (a)
    \(\displaystyle \underset{\text{(C)}}{2 \mathrm{CH}_3 \mathrm{COOH}}+2 \mathrm{Na} \rightarrow \underset{\text{(R)}}{2 \mathrm{CH}_3 \mathrm{COO} \mathrm{Na}}+\mathrm{H}_2\)
    (b)
    \(\displaystyle \underset{\text{(C)}}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{\text{(A)}}{\mathrm{CH}_3 \mathrm{OH}} \xrightarrow{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} \underset{\text{(S)}}{\mathrm{CH}_3 \mathrm{COOCH}_3}+\mathrm{H}_2 \mathrm{O}\)
    (c)
    \(\displaystyle \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH} \rightarrow \underset{\text{(R)}}{\mathrm{CH}_3 \mathrm{COO} \mathrm{Na}}+\mathrm{H}_2 \mathrm{O}\)
    (d)
    \(\displaystyle \mathrm{CH}_3 \mathrm{COOCH}_3+\mathrm{NaOH} \rightarrow \underset{\text{(R)}}{\mathrm{CH}_3 \mathrm{COO} \mathrm{Na}}+\underset{\text{(A)}}{\mathrm{CH}_3 \mathrm{OH}}\)
    C reacts with Na to evolve a gas that pops, and with NaOH to give R and water – so C is a carboxylic acid. \[2\,\mathrm{CH_3COOH} + 2\,\mathrm{Na} \rightarrow 2\,\mathrm{CH_3COONa} + \mathrm{H_2}\uparrow \] \[\mathrm{CH_3COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O} \] C (\(\displaystyle \mathrm{C_2H_4O_2}\)) is ethanoic acid; R is sodium ethanoate. With alcohol A and an acid catalyst, C forms the sweet-smelling ester S (\(\displaystyle \mathrm{C_3H_6O_2}\)), and base hydrolysis of S regenerates R and A – so A is methanol and S is methyl ethanoate. \[\mathrm{CH_3COOH} + \mathrm{CH_3OH} \xrightarrow[\Delta]{\mathrm{H^+}} \mathrm{CH_3COOCH_3} + \mathrm{H_2O} \] \[\mathrm{CH_3COOCH_3} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{CH_3OH} \] Answer: C = ethanoic acid; R = sodium ethanoate; A = methanol; S = methyl ethanoate.

  7. Exercise 53

    Look at Figure 4.1\displaystyle 4.1 and answer the following questions
    (a)
    What change would you observe in the calcium hydroxide solution taken in tube B?
    (b)
    Write the reaction involved in test tubes A and B respectively.
    (c)
    If ethanol is given instead of ethanoic acid, would you expect the same change?
    (d)
    How can a solution of lime water be prepared in the laboratory?
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q53

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    NCERT’s answer
    (a)
    It will turn milky
    (b)
    \(\displaystyle 2 \mathrm{CH}_3 \mathrm{COOH}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\) (Test tube A)
    \(\displaystyle \mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O}\) (Test tube B)
    With excess \(\displaystyle \mathrm{CO}_2\), milkiness disappears.
    \(\displaystyle \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \rightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2\)
    (c)
    As \(\displaystyle \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\) and \(\displaystyle \mathrm{Na}_2 \mathrm{CO}_3\) do not react, a similar change is not expected
    \(\displaystyle \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow\) No change
    (d)
    The lime water is prepared by dissolving calcium oxide in water and decanting the supernatent liquid.
    (a)
    Tube B turns milky (turbid).
    (b)
    Tube A:
    \[\mathrm{2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2\uparrow} \]
    Tube B:
    \[\mathrm{CO_2 + Ca(OH)_2 \rightarrow CaCO_3\downarrow + H_2O} \]
    (c)
    No — ethanol is not acidic enough to react with sodium carbonate, so no \(\displaystyle \mathrm{CO_2}\) evolves and B stays clear.
    (d)
    Add quicklime to water in small portions with constant stirring (highly exothermic); filter off the undissolved solid and keep the clear filtrate.
    \[\mathrm{CaO + H_2O \rightarrow Ca(OH)_2} \]
    Answer: (a) B turns milky; (c) no change; (d) dissolve quicklime in water and filter off the clear lime water.

  8. Exercise 54

    How would you bring about the following conversions? Name the process and write the reaction involved.
    (a)
    ethanol to ethene.
    (b)
    propanol to propanoic acid. Write the reactions.

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    NCERT’s answer
    Hint— (a) By the dehydration of ethanol in the presence of concentrated \(\displaystyle \mathrm{H}_2 \mathrm{SO}_4\).
    \[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[\text { conc. } \mathrm{H}_2 \mathrm{SO}_4]{\text { Hot }} \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O} \]
    (b)
    By the oxidation of propanol using oxidising agent such as alkaline \(\displaystyle \mathrm{KMnO}_4\).
    \[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[\text { Heat }]{\text { Alk. } \mathrm{KMnO}_4} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} \]
    (a)
    Ethanol \(\displaystyle \to\) Ethene — dehydration: conc. \(\displaystyle \mathrm{H_2SO_4}\) removes water from ethanol at $\displaystyle 443$ K.
    \[\mathrm{CH_3CH_2OH \xrightarrow[443\,K]{conc.\ H_2SO_4} CH_2=CH_2 + H_2O} \]
    (b)
    Propanol \(\displaystyle \to\) Propanoic acid — oxidation: alkaline \(\displaystyle \mathrm{KMnO_4}\) with heat (or acidified \(\displaystyle \mathrm{K_2Cr_2O_7}\)) supplies \(\displaystyle [O]\).
    \[\mathrm{CH_3CH_2CH_2OH + 2[O] \xrightarrow[heat]{alkaline\ KMnO_4} CH_3CH_2COOH + H_2O} \]
    Answer: (a) dehydration with hot conc. \(\displaystyle \mathrm{H_2SO_4}\); (b) oxidation with alkaline \(\displaystyle \mathrm{KMnO_4}\).

  9. Exercise 55

    Draw the possible isomers of the compound with molecular formula C3H6O\displaystyle \mathrm{C}_3 \mathrm{H}_6 \mathrm{O} and also give their electron dot structures.

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    NCERT’s answer
    Propanone \(\displaystyle \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CHO}\) Propanal NCERT_Solution_Class10_Science_Exemplar_Ch4_Q55_ncert NCERT_Solution_Class10_Science_Exemplar_Ch4_Q55_ncert_2 NCERT_Solution_Class10_Science_Exemplar_Ch4_Q55_ncert_3
    With C=O at the chain end the compound is an aldehyde; in the middle, a ketone: \[\mathrm{CH_3{-}CH_2{-}CHO} \quad \text{(propanal)} \] \[\mathrm{CH_3{-}CO{-}CH_3} \quad \text{(propanone)} \]Answer: Propanal \(\displaystyle \mathrm{CH_3CH_2CHO}\) and propanone \(\displaystyle \mathrm{CH_3COCH_3}\) — the two carbonyl (functional) isomers of \(\displaystyle \mathrm{C_3H_6O}\).

  10. Exercise 56

    Explain the given reactions with the examples
    (a)
    Hydrogenation reaction
    (b)
    Oxidation reaction
    (c)
    Substitution reaction
    (d)
    Saponification reaction
    (e)
    Combustion reaction

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    NCERT’s answer
    Hint— (a) Unsaturated hydrocarbons add hydrogen in the presence of nickel catalyst to give saturated hydrocarbons.
    (b)
    Ethanol is oxidised to ethanoic acid in the presence of alkaline \(\displaystyle \mathrm{KMnO}_4\) on heating.
    \[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[\text { Heat }]{\text { Alk. } \mathrm{KMnO}_4} \mathrm{CH}_3 \mathrm{COOH} \]
    (c)
    In the presence of sunlight, chlorine is added to hydrocarbons.
    \[\mathrm{CH}_4+\mathrm{Cl}_2 \xrightarrow{\mathrm{h} \upsilon} \mathrm{CH}_3 \mathrm{Cl}+\mathrm{HCl} \]
    (d)
    \(\displaystyle \underset{\text{Ester}}{\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5}+\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COO} \mathrm{Na}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
    Used in the preparation of soap
    (e)
    Most carbon compounds release a large amount of heat and light on burning
    \[\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}+\text { Heat and light } \]
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q56_ncert
    (a)
    Hydrogenation — addition of \(\displaystyle \mathrm{H_2}\) across a C=C bond over Ni; used to harden vegetable oils.
    \[\mathrm{CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3} \]
    (b)
    Oxidation — addition of oxygen/\(\displaystyle [O]\); an alcohol becomes an acid.
    \[\mathrm{CH_3CH_2OH + 2[O] \xrightarrow[heat]{alkaline\ KMnO_4} CH_3COOH + H_2O} \]
    (c)
    Substitution — in sunlight one H of a saturated hydrocarbon is replaced by a halogen, stepwise.
    \[\mathrm{CH_4 + Cl_2 \xrightarrow{sunlight} CH_3Cl + HCl} \]
    (d)
    Saponification — alkaline hydrolysis of an ester into the sodium salt of the acid and an alcohol (with a fat, that salt is soap).
    \[\mathrm{CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH} \]
    (e)
    Combustion — burning in \(\displaystyle \mathrm{O_2}\), releasing heat and light.
    \[\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + heat + light} \]
    Answer: (a) addition of \(\displaystyle \mathrm{H_2}\); (b) addition of oxygen; (c) H replaced by Cl; (d) alkaline hydrolysis of an ester; (e) burning in \(\displaystyle \mathrm{O_2}\).

  11. Exercise 57

    An organic compound A on heating with concentrated H2SO4\displaystyle \mathrm{H}_2 \mathrm{SO}_4 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\displaystyle \mathrm{CO}_2 and 3\displaystyle 3 moles of H2O\displaystyle \mathrm{H}_2 \mathrm{O}. Identify the compounds A, B and C and write the chemical equations of the reactions involved.

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    NCERT’s answer
    Since compound C gives $\displaystyle 2$ moles of \(\displaystyle \mathrm{CO}_2\) and $\displaystyle 3$ moles of \(\displaystyle \mathrm{H}_2 \mathrm{O}\), it shows that it has the molecular formula \(\displaystyle \mathrm{C}_2 \mathrm{H}_6\) (Ethane). C is obtained by the addition of one mole of hydrogen to compound B so the molecular formula of B should be \(\displaystyle \mathrm{C}_2 \mathrm{H}_4\) (Ethene). Compound B is obtained by heating compound A with concentrated \(\displaystyle \mathrm{H}_2 \mathrm{SO}_4\) which shows it to be an alcohol. So compound A could be \(\displaystyle \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\) (Ethanol) \[\underset{\text{A}}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \xrightarrow{\text { Hot conc. } \mathrm{H}_2 \mathrm{SO}_4} \underset{\text{B}}{\mathrm{C}_2 \mathrm{H}_4}+\mathrm{H}_2 \mathrm{O} \] \[\underset{\text{B}}{\mathrm{C}_2 \mathrm{H}_4}+\mathrm{H}_2 \xrightarrow{\mathrm{Ni}} \underset{\text{C}}{\mathrm{C}_2 \mathrm{H}_6} \] \(\displaystyle \underset{\text{C}}{2 \mathrm{C}_2 \mathrm{H}_6}+7 \mathrm{O}_2 \longrightarrow 4 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}+\) Heat and light
    One mole of \(\displaystyle C\) gives $\displaystyle 2$ mol \(\displaystyle \mathrm{CO_2}\) and $\displaystyle 3$ mol \(\displaystyle \mathrm{H_2O}\): \[n_C = 2\times1 = 2,\qquad n_H = 3\times2 = 6 \Rightarrow C=\mathrm{C_2H_6}\ \text{(ethane)} \] \[\mathrm{C_2H_6 + \tfrac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O} \]\(\displaystyle B\) is the alkene hydrogenated by Ni to give \(\displaystyle C\): \[\mathrm{CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3} \] so \(\displaystyle B=\mathrm{C_2H_4}\) (ethene).\(\displaystyle A\) is the alcohol whose dehydration by conc. \(\displaystyle \mathrm{H_2SO_4}\) gives \(\displaystyle B\): \[\mathrm{C_2H_5OH \xrightarrow[443\,K]{conc.\ H_2SO_4} CH_2=CH_2 + H_2O} \] so \(\displaystyle A=\) ethanol.Answer: \(\displaystyle A=\) ethanol \(\displaystyle \mathrm{(C_2H_5OH)}\), \(\displaystyle B=\) ethene \(\displaystyle \mathrm{(C_2H_4)}\), \(\displaystyle C=\) ethane \(\displaystyle \mathrm{(C_2H_6)}\).