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Why does ethene undergo addition reactions?

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प्रश्न

Why does ethene undergo addition reactions?

अति संक्षिप्त उत्तर
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उत्तर

Ethene is a reactive compound because it contains a double bond. Hence, it undergoes addition reactions.

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अध्याय 14: Unsaturated Hydrocarbons: Alkenes - Exercises [पृष्ठ २७३]

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फ्रैंक Chemistry Part 2 [English] Class 10 ICSE
अध्याय 14 Unsaturated Hydrocarbons: Alkenes
Exercises | Q 12. (ii) | पृष्ठ २७३

संबंधित प्रश्न

What is vital force theory?


Arrange the following according to the instruction given in the bracket:

Ethane, methane, ethene, ethyne (In the increasing order of the molecular weight) [H = 1. C = 12].


Give the names of any three alkyl groups. How are they formed?


Write the IUPAC name of the foloowing:

CH3CH2CH2CH2CH3


Write the composition of the following:

Spurious alchohol


What is the special feature of the structure of Ethyne?


Write the equation of the following laboratory preparation:

Ethane from sodium propionate.


Give a chemical test to distinguish between Saturated and Unsaturated?


An organic compound ‘A’ is widely used as a preservative and has the molecular formula C2H4O2. This compound reacts with ethanol to form a sweet-smelling compound ‘B’.

Identify the compound ‘A’.


The structures of six organic compounds are shown:

A

\[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H - C - C = C - C - H}\\
|\phantom{.............}|\\
\ce{H}\phantom{............}\ce{H}
\end{array}\]

B

\[\begin{array}{cc}
\phantom{.......}\ce{H}\phantom{.....}\ce{O}\\
\phantom{.......}|\phantom{.....}//\\
\ce{H - C - C}\\
\phantom{......}|\phantom{.....}\backslash\\
\phantom{...........}\ce{H}\phantom{.....}\ce{O - H}
\end{array}\]

C

\[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - C - H}\\
|\phantom{....}|\phantom{....}|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}
\end{array}\]

D

\[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\\
\ce{H - C - C - H}\\
|\phantom{....}|\\
\ce{H}\phantom{...}\ce{O}\\
\phantom{.......}\backslash\\
\phantom{.........}\ce{H}
\end{array}\]

E

\[\begin{array}{cc}
\ce{H}\\
|\\
\ce{H - C - H}\\
\ce{H}\phantom{....}\phantom{....}\ce{H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H - C - C - C - H}\\
|\phantom{....}|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}\phantom{...}\ce{H}
\end{array}\]

F

\[\begin{array}{cc}
\ce{H}\phantom{...}\ce{H}\\
|\phantom{....}|\\
\ce{C = C}\\
|\phantom{....}|\\
\ce{H}\phantom{...}\ce{H}
\end{array}\]

  1. Identify two of the compounds that are members of the same homologous series but are not isomers.
  2. Which two compounds are isomers of each other?
  3. F can be prepared from D. Give a chemical equation for the reaction.

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