a) When carbon dioxide is passed through the calcium hydroxide solution, it turns milky due to the formation of insoluble calcium carbonate.
b) Reaction in Test tube A:
\(\underset{Acetic\ or\ ethnoic\ acid}{2CH_3COOH(1)}+\underset{Sodium\ carbonate}{Na_2CO_3(s)}\rightarrow\underset{Sodium\ acetate}{2CH_3COONa}+\underset{Water}{H_2O}+\underset{Carbon\ ioxide}{CO_2(g)}\)
Reaction in Test tube B:
\(\underset{Calciumhydroxide\\ (Lime\ water)}{Ca(OH)_2(aq)}+\underset{Carbondioxide}{CO_2(g)}\rightarrow\underset{Calcium\ carbonate\\ (White\ ppt)}{CaCO_3(s)\downarrow}\underset{Water}{H_2O}\)
c) No with ethanol no reaction takes place.
d) When a quick lime i.e., calcium oxide is added to water in a test tube, it forms calcium hydroxide which is lime water. This solution is filtered and the clear solution is obtained
\(CaO(s)+H_2O(l)\rightarrow Ca(OH)_2\)

i) Since compound C on combustion forms two moles of CO₂ and 3 moles of H₂ O, therefore compound C must contain two carbon atoms and six hydrogen atmos. Thus compound C must be C₂H₆ (ethane)
\(\underset{Ethene(C)}{2C_2H_6}+\underset{Oxygen}{7O_2}\xrightarrow{Heat}\underset{Carbon\ dioxide}{4CO_2}+\underset{Water}{6H_2O}\)
ii) Since compound (C) is obtained by addition of 1 mole of H₂ in presence of Ni to compound (B), therefore (B) must be ethane.
\(\underset{Ethene(B)}{CH2}=CH_2+H_2\xrightarrow{Ni/Heat}\underset{Ethene(C)}{CH_3-CH_3}\)
iii) Since compound (B) is formed by heating compound (A) with cone. H₂SO₄ therefore compound (A) must be ethanol.
\(\underset{Ethanol(A)}{CH_3CH_2OH}\xrightarrow{Conc. H_2SO_4,443K}CH_2=CH_2+\underset{Ethene(B)}{H_2O}\)

a)\(\underset{Ethanol}{CH_3CH_2OH}\xrightarrow{Conc.H_2SO_4.170^0C}CH_2=\underset{Ethene}{CH_2}+H_2O\)
b)\(\underset{Propanol}{CH_3CH_2CH_2OH+2[O]}\xrightarrow[or\ Aci.K_2Cr_2O_7,\Delta]{AlK.KMnO_4, \Delta}\underset{Propanoic\ acid}{CH_3CH_2COOH}+H_2O\)

Compound with molecular formula C₂H₄O₂ should be Ethanoic acid reacts with Na metal to give sodium ethanoate.
\(\underset{\underset{Ethanoic\ acid}{(C)}}{2CH_3COOH}+2Na\rightarrow\underset{\underset{Sodium\ ethanoate}{(R)}}{2CH_3COONa}+H_2\)
\(\underset{\underset{Ethanoic\ acid}{(C)}}{CH_3COOH}+\underset{\underset{methanol}{(A)}}{CH_3OH}\xrightarrow{conc. H_2SO_4}\underset{\underset{Methyl\ acetate}{(S)}}{CH_3COOCH_3+H_2O}\)
\(CH_3COOH+\underset{Sodium\\ hydroxide}{NAOH}\rightarrow\underset{\underset{Sodium\ ethanoate}{(R)}}{CH_3COONa}+H_2O\)
\(\underset{\underset{Methyl\ acetate}{(s)}}{CH_3COOCH_3}+NaOH\rightarrow\underset{\underset{sodium\ ethanoate}{(R)}}{CH_3COONa}+\underset{\underset{Method}{(A)}}{CH_3OH}\)

Carboxylic acid reacts with alcohols to form ester. For example, when ethanoic acid is warmed with ethanol in presence of a few drops of concentrated sulphuric acid as catalyst, an ester (ethyl ethanoate, commonly called ethyl acetate) and water are formed.
\(\underset{Acetic\ or\ ethanoic\ acid}{CH_COOH}+\underset{Ethanol}{C_2H_5OH}\xrightarrow{Conc. H_2SO_4}\underset{Ethyl\ ethanoate}{CH_3COOC_2H_5}+\underset{Water}{H_2O)}\)
Ester are sweet -smelling substances. So they are widely used in making perfumes and flavoring agents. Further, the reaction between a carboxylic acid and an alcohol to form a sweet ester is used as a test for alcohols as well as carboxylic acids.

\(\underset{Carbon}{C}+\underset{Chlorine}{2Cl_2}\xrightarrow{Heat}\underset{Carbon\ tetrachloridse}{CCl_4}\)
Its electron dot structure and structural formulae are:

b) Saponification - Alkaline hydrolysis of an ester to give the salt of the corresponding
acid and the alcohol is called saponification. It is reverse of esterification reaction.
For example.
\(\underset{Ester}{CH_3COOC_2H_5}+\underset{sodium\ hydroxide}{NaOH}\xrightarrow{Heat}\underset{Sodium\ acetate}{CH_3COONa}+\underset{Ethanol}{CH_3CH_2OH}\)
Dehydration means removal of a molecule of water. When ethanol is heated with
conc. H2SO₄ at 443 K, it undergoes dehydration to form ethene.
\(\underset{Ethanol}{CH_3CH_2OH}\xrightarrow{Conc. H_2SO_4,443K}\underset{Ethene}{CH_2=CH_2+H_2O}\)

Addition of hydrogen to understand hydrocarbons in presence of a catalyst such as nickel or palladium to form saturated hydrocarbons is called hydrogenation. For example,
\(\underset { Ethyne }{ CH } \equiv CH+{ H }_{ 2 }\xrightarrow [ ]{ Ni/473K} \underset { Ethene }{ { CH }_{ 2 } } ={ CH }_{ 2 }+{ H }_{ 2 }\xrightarrow [ ]{Ni/473K } { CH }_{ 3 }-\underset { Ethane }{ { CH }_{ 3 } } \)

a) Compounds of carbon and hydrogen only are called hydrocarbons. For example
methane (CH₄), ethane (C₂H₆), ethene (C₂H₄), ethyne (C₂H₂), cydohexane
(C₆H₁₂), benzene (C₆H₆) etc.

b) Saturated hydrocarbons contains only C - C and C - H single covalent bonds while
unsaturated hydrocarbons contain at least one C = C double bond (besides C - C
and C - H single covalent bonds) or one triple bond.

c) That portion of the organic molecule which largely determine its chemical properties is called the functional group. For example, -OH (alcoholic group), (keto group) - CHO (aldehyde group), -COOH (carboxyl group).

\(\underset{Ethanoic\ acid}{CH_3COOH} + \underset {Sodium\ hydrogen\\ Carbonate}{NaHCO_3} \rightarrow \underset{sodium\ ethanoate}{CH_3COONa}+ \underset{Cabon\ dioxide}{CO_2} + \underset{Water}{H_2O}\)
Activity to show that ethanoic acid reacts with metal hydrogen carbonate to liberate carbon dioxide.
Procedure:
i) Take two test tubes and label them as A and B.
ii) Now, take few grams of sodium carbonate in test tube A.
iii) Take few grams of sodium hydrogen carbonate in test tube B.
iv) Now, add ethanoic acid to both the test tubes.
v) Note your observation.
vi) Pass the gas liberated in test tube A and B through lime water or calcium hydroxide solution.
vii) Record your observation.
Observation
a) A brisk effervescence is formed.
b) The lime water turns milky in both the test tubes.

When carboxylic acids react with sodium carbonate or sodium bicarbonates, it liberates carbon dioxide gas which turns lime water milky.
\(\underset{Calcium\ hydroxide\\ (Lime\ wate)}{Ca(OH)_2(aq)}+\underset{Carbon\ dioxide}{CO_2(g)}\rightarrow \underset{Calcium\ carbonate\\ (White \ ppt)}{CaCO_3(s)\downarrow}+\underset{Water}{H_2O(l)}\)