SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Science Carbon and its Compounds

57 questions · 57 still being checked

Short Answer Questions 30–39 (part 4 of 6)

  1. Exercise 30

    Draw the electron dot structure of ethyne and also draw its structural formula

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Electron dot structure of ethyne (\(\displaystyle \mathrm{C}_2 \mathrm{H}_2\)) \(\displaystyle \mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \quad\) Structural formula of ethyne NCERT_Solution_Class10_Science_Exemplar_Ch4_Q30_ncert
    Valence electrons available: \[2(4)+2(1)=10\ e^- \;\Rightarrow\; 5\ \text{shared pairs} \]The two C share $\displaystyle 3$ pairs (triple bond) and each C shares $\displaystyle 1$ pair with an H, so every C has $\displaystyle 8$ electrons and every H has 2. \[\text{Structural formula: } \mathrm{H-C\equiv C-H} \] Answer: electron dot structure as drawn ($\displaystyle 3$ shared pairs between the two C, $\displaystyle 1$ shared pair on each C-H); structural formula \(\displaystyle \mathrm{H-C\equiv C-H}\).

  2. Exercise 31

    Write the names of the following compounds
    (a)
    (b)
    (c)
    (d)
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q31
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q31_2
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q31_3
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q31_4

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    Pentanoic acid
    (b)
    Butyne
    (c)
    Heptanal
    (d)
    Pentanol
    (a)
    $\displaystyle 5$-C chain ending \(\displaystyle \mathrm{-COOH}\): pentanoic acid (valeric acid). (b) $\displaystyle 4$-C chain with a terminal \(\displaystyle \mathrm{C\equiv C}\): but-$\displaystyle 1$-yne. (c) $\displaystyle 7$-C chain ending \(\displaystyle \mathrm{-CHO}\): heptanal. (d) $\displaystyle 5$-C chain ending \(\displaystyle \mathrm{-OH}\): pentan-$\displaystyle 1$-ol.
    Answer: (a) Pentanoic acid (b) But-$\displaystyle 1$-yne (c) Heptanal (d) Pentan-$\displaystyle 1$-ol

  3. Exercise 32

    Identify and name the functional groups present in the following
    compounds.
    (a)
    (b)
    (c)
    (d)
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q32
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q32_2
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q32_3
    NCERT_Question_Class10_Science_Exemplar_Ch4_Q32_4

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    — OH
    Hydroxyl/Alcohol
    (b)
    Carboxylic acid
    (c)
    Ketone
    (d)
    Alkene
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q32_ncert
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q32_ncert_2
    NCERT_Solution_Class10_Science_Exemplar_Ch4_Q32_ncert_3
    (a)
    chain ends \(\displaystyle \mathrm{-OH}\): alcoholic (hydroxyl) group. (b) chain ends \(\displaystyle \mathrm{-COOH}\): carboxylic group. (c) \(\displaystyle \mathrm{>C=O}\) inside the chain, both neighbours C: ketonic group. (d) \(\displaystyle \mathrm{C=C}\) double bond at the chain end: alkenic group.
    Answer: (a) Alcohol \(\displaystyle \mathrm{-OH}\) (b) Carboxylic acid \(\displaystyle \mathrm{-COOH}\) (c) Ketone \(\displaystyle \mathrm{>C=O}\) (d) Alkene \(\displaystyle \mathrm{C=C}\)

  4. Exercise 33

    A compound X is formed by the reaction of a carboxylic acid C2H4O2\displaystyle \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2 and an alcohol in presence of a few drops of H2SO4\displaystyle \mathrm{H}_2 \mathrm{SO}_4. The alcohol on oxidation with alkaline KMnO4\displaystyle \mathrm{KMnO}_4 followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of (a) carboxylic acid, (b) alcohol and (c) the compound X. Also write the reaction.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    Carboxylic acid is ethanoic acid
    (b)
    Alcohol is ethanol
    (c)
    X is ethyl ethanoate
    \(\displaystyle \underset{\text{Ethanoic acid}}{\mathrm{CH}_3-\mathrm{COOH}}+\underset{\text{Ethanol}}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \xrightarrow{\mathrm{H}_2 \mathrm{SO}_4} \underset{\text{Ethyl ethanoate}}{\mathrm{CH}_3-\mathrm{COOC}_2 \mathrm{H}_5}+\mathrm{H}_2 \mathrm{O}\)
    \(\displaystyle \mathrm{C_2H_4O_2}\) with a \(\displaystyle \mathrm{-COOH}\) group is \(\displaystyle \mathrm{CH_3COOH}\), ethanoic (acetic) acid. Oxidation of the alcohol must regenerate this same acid, so the alcohol is \(\displaystyle \mathrm{CH_3CH_2OH}\), ethanol: \[\mathrm{CH_3CH_2OH + 2[O]} \xrightarrow[\mathrm{then\ H^+}]{\mathrm{alk.\ KMnO_4}} \mathrm{CH_3COOH + H_2O} \] Esterification of the two, catalysed by \(\displaystyle \mathrm{H_2SO_4}\), gives X: \[\mathrm{CH_3COOH + C_2H_5OH} \xrightarrow{\mathrm{conc.\ H_2SO_4}} \underset{X}{\mathrm{CH_3COOC_2H_5}} + \mathrm{H_2O} \] Answer: (a) Ethanoic acid (acetic acid) \(\displaystyle \mathrm{CH_3COOH}\) (b) Ethanol \(\displaystyle \mathrm{C_2H_5OH}\) (c) X = ethyl ethanoate (ethyl acetate) \(\displaystyle \mathrm{CH_3COOC_2H_5}\)

  5. Exercise 34

    Why detergents are better cleansing agents than soaps? Explain.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Detergents work as cleansing agent both in hard and soft water. The charged ends of detergents do not form insoluble precipitates with calcium and magnesium ions in hard water.
    In hard water, soap's \(\displaystyle \mathrm{Ca^{2+}}\)/\(\displaystyle \mathrm{Mg^{2+}}\) salt is insoluble and separates as scum, so soap is used up before it lathers: \[2\,\mathrm{C_{17}H_{35}COONa} + \mathrm{Ca^{2+}} \rightarrow (\mathrm{C_{17}H_{35}COO})_2\mathrm{Ca}\!\downarrow + 2\,\mathrm{Na^+} \] Detergent (sodium alkyl sulphonate or sulphate): its \(\displaystyle \mathrm{Ca^{2+}}\)/\(\displaystyle \mathrm{Mg^{2+}}\) salts stay soluble, so no scum forms. Answer: Detergents lather in both hard and soft water because their \(\displaystyle \mathrm{Ca^{2+}}\)/\(\displaystyle \mathrm{Mg^{2+}}\) salts are soluble, so no scum forms.

  6. Exercise 35

    Name the functional groups present in the following compounds
    (a)
    CH3COCH2CH2CH2CH3\displaystyle \mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
    (b)
    CH3CH2CH2COOH\displaystyle \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}
    (c)
    CH3CH2CH2CH2CHO\displaystyle \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}
    (d)
    CH3CH2OH\displaystyle \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    Ketone
    (b)
    Carboxylic acid
    (c)
    Aldehyde
    (d)
    Alcohol
    (a)
    \(\displaystyle \mathrm{CH_3COCH_2CH_2CH_2CH_3}\): the \(\displaystyle \mathrm{-CO-}\) sits between two carbons – ketonic group. (b) \(\displaystyle \mathrm{CH_3CH_2CH_2COOH}\): terminal \(\displaystyle \mathrm{-COOH}\) – carboxylic acid group. (c) \(\displaystyle \mathrm{CH_3CH_2CH_2CH_2CHO}\): terminal \(\displaystyle \mathrm{-CHO}\) – aldehydic group. (d) \(\displaystyle \mathrm{CH_3CH_2OH}\): terminal \(\displaystyle \mathrm{-OH}\) – alcoholic group.
    Answer: (a) Ketone (b) Carboxylic acid (c) Aldehyde (d) Alcohol

  7. Exercise 36

    How is ethene prepared from ethanol? Give the reaction involved in it.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Ethanol on heating with excess concentrated sulphuric acid at $\displaystyle 443$ K results in the dehydration of ethanol to give ethene. \[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\text { Hot conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O} \]
    \[\mathrm{CH_3CH_2OH} \xrightarrow[443\,\mathrm{K}]{\text{conc. }\mathrm{H_2SO_4}\text{, excess}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O} \] Hold the temperature at $\displaystyle 443$ K; at $\displaystyle 413$ K with excess ethanol, ethoxyethane forms instead. Answer: heat ethanol with excess conc. \(\displaystyle \mathrm{H_2SO_4}\) at $\displaystyle 443$ K (dehydration) to get ethene.

  8. Exercise 37

    Intake of small quantity of methanol can be lethal. Comment.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Methanol is oxidised to methanal in the liver. Methanal reacts rapidly with the components of cells. It causes the protoplasm to coagulate. It also affects the optic nerve, causing blindness.
    \[\mathrm{CH_3OH + [O]} \xrightarrow{\text{liver}} \mathrm{HCHO + H_2O} \] Methanal coagulates the protoplasm of cells and affects the optic nerve, causing blindness. Answer: the liver oxidises methanol to methanal, so even a small intake can cause blindness or death.

  9. Exercise 38

    A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation of the reaction involved.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Gas evolved is hydrogen. \[2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+2 \mathrm{Na} \rightarrow 2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O}^{-} \mathrm{Na}^{+}+\mathrm{H}_2 \]
    \[2\,\mathrm{C_2H_5OH(l)} + 2\,\mathrm{Na(s)} \rightarrow 2\,\mathrm{C_2H_5ONa} + \mathrm{H_2(g)} \] The gas evolved is hydrogen; it burns with a pop near a flame. Answer: hydrogen gas, \(\displaystyle \mathrm{H_2}\) (sodium ethoxide, \(\displaystyle \mathrm{C_2H_5ONa}\), also forms).

  10. Exercise 39

    Ethene is formed when ethanol at 443\displaystyle 443 K is heated with excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation of this reaction.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Sulphuric acid acts as a dehydrating agent. \[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\text { Hot conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O} \]
    \[\mathrm{CH_3CH_2OH} \xrightarrow[443\,\mathrm{K}]{\text{conc. }\mathrm{H_2SO_4}\text{, excess}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O} \] Hold the temperature at $\displaystyle 443$ K; at $\displaystyle 413$ K with excess ethanol, ethoxyethane forms instead. Answer: conc. \(\displaystyle \mathrm{H_2SO_4}\) is a dehydrating agent: it removes \(\displaystyle \mathrm{H_2O}\) from ethanol to give ethene.