Reaction:
$CH _3CH _2COONa + NaOH \rightarrow C _2H _6 + Na _2CO _3$

As ethanol has donated $H^{+}$ it suggests that it has acidic nature. Acid is defined as the substance which has replaceable $H^{+}$ions. In the products side, it has donated and became the reason for the formation of hydrogen molecule. Hence, option A is the correct answer. 

If a small piece of sodium is dropped into some ethanol, it reacts steadily to give off bubbles of hydrogen gas and leaves a colourless solution of sodium ethoxide, $CH _3CH _2ONa$ and hydrogen gas is evolved.

$2C _2H _5OH + 2Na \rightarrow 2C _2H _5ONa + H _2$ Option A is correct.

Vinegar is a liquid containing acetic acid and water. This acetic acid is produced by the fermentation of ethanol by acetic acid bacteria. It is mainly used as a cooking ingredient. It is a 5-8% solution of acetic acid in water and most easily available mild acid.

Acid and Base reacts to form Salt and Water. This reaction is called neutralisation reaction. In this, acid is acetic acid and base is sodium hydroxide. Salt ($CH _3COONa$) is formed.

$2CH _3CH _2OH + 2Na\longrightarrow 2CH _3CH _2ONa + H _2$
$2H _2O + 2Na\longrightarrow 2NaOH + H _2$

Vinegar is also known as acetic acid and water. Acetic acid is ethanoic acid produced by the fermentation of ethanol by acetic acid bacteria. Vinegar is a cooking ingredient and commercially produced by fast or slow fermentation processes. 

Acetic acid is Ethanoic acid $CH _3COOH$. It is also called vinegar.

Methyl acetate is obtained during reaction of ethanoic acid with diazomethane in presence of dil HCl.
$CH _3COOH + CH _2N _2 \xrightarrow {dil HCl} CH _3COOCH _3 + N _2$

When ethanoic acid reacts with chlorine in the presence of acetic anhydride as catalyst it forms chloroacetic acid.

The equation for the reaction between sodium and ethanol can be written as:
$2CH _{3}CH _{2}OH + 2Na \rightarrow 2CH _{3}CH _{2}ONa + H _{2}$
Therefore, the products formed are sodium ethoxide and hydrogen.

Reaction of ethanol with warm ethanoic acid to form ethyl acetate(an ester) and water can be written as:
$ C _{2}H _{5}OH +  CH _{3}COOH  \overset{H^{+}}{\rightarrow}   CH _{3}COOC _{2}H _{5}  +  H _{2}O$
(ethanol)      (ethanoic acid)             (ester)
Above reaction is an esterification reaction as an alcohol and an acid are combined to formed an ester. 

Saponification is the alkaline hydrolysis of esters to form soap and alcohol. Therefore, the correct answer is option D.

Ethanol is oxidised to ethanoic acid in the presence of alkaline
$KMnO _{4}$ on heating
$CH _{3}  CH _{2}  OH \xrightarrow[Heat]{Alk.  KMnO _{4}} CH _{3}COOH$

Ethanoic acid reacts with sodium bicarbonate, to form carbon dioxide which produces effervescence as: 

$C _2H _5OH$ has polar covalent $C-O$ bond due to significant electronegativity difference between $C$ and$ O.$ Similarly $O-H$ bond is also polar covalent due to significant electronegativity difference between $O$ and $H$ atoms.
A covalent bond is formed by sharing of electrons. When this electron sharing is to an equal extent, the bond is non polar covalent. Otherwise the bond is polar covalent.

Ethanol can be converted to ethene as:

Vinegar is a liquid having acetic acid and water as its mixture. So, ethanoic acid is the other name of acetic acid and this smell of vinegar is similar to ethanoic acid. 

Ethanol can be written as $CH _3-CH _2-OH$. The oxygen atom in the hydroxyl ($OH$) group is highly electronegative and hence attracts the electrons towards itself. Thus, ethanol is a polar compound with its electrons more towards the oxygen atom. Although ($C _2H _5$) group is non polar.

Na or Zn reacts with ethanoic acid to liberate hydrogen gas.

In industrial method, to obtain 100% pure alcohol, a small quantity of benzene is added to rectified spirit and the mixture is then distilled. Absolute alcohol is obtained in third fraction that distills over at 78.2°C (351.3 K).

None of the given method is correct. 

Option D is the answer.