Question:

Ramya performs an experiment in the following two steps:
Step 1: Heats ethanol with conc. \(\mathrm{H_2SO_4}\) at 170\(^\circ\)C to get product A.
Step 2: Passes hydrogen gas over product A, using Ni as a catalyst and gets product B.
(i) Name the products A and B.
(ii) Write the reactions involved in both the steps.

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Conc. sulphuric acid at 170 C removes water from ethanol. Nickel with hydrogen adds hydrogen to a double bond.
Updated On: Nov 3, 2026
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Solution and Explanation

Step 1: Understand step 1:
Heating ethanol with excess conc. \(\mathrm{H_2SO_4}\) at 170\(^\circ\)C (443 K) removes one molecule of water from ethanol. This is a dehydration reaction.
The product has a double bond between the two carbon atoms. So A is ethene, \(\mathrm{CH_2=CH_2}\).

Step 2: Understand step 2:
Ethene is an unsaturated hydrocarbon. When hydrogen is passed over it with nickel as a catalyst, hydrogen adds across the double bond. This is called hydrogenation or an addition reaction.
The double bond becomes a single bond and the product is the saturated hydrocarbon ethane, \(\mathrm{CH_3CH_3}\). So B is ethane.

Step 3: Part (i): Names:
Product A = ethene.
Product B = ethane.

Step 4: Part (ii): Reactions:
Step 1 (dehydration):
\[ \mathrm{CH_3CH_2OH \xrightarrow{\text{conc. }H_2SO_4,\ 170^\circ C} CH_2=CH_2 + H_2O} \]
Step 2 (hydrogenation):
\[ \mathrm{CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3CH_3} \]

Final Answer:
A is ethene and B is ethane. Step 1 is dehydration of ethanol and step 2 is hydrogenation of ethene. \[ \boxed{A=\text{ethene},\ B=\text{ethane}} \]
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