
The addition of HCl to 1-phenylpropene gives:-
A) ${{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}$
B) ${{C}_{6}}{{H}_{5}}CHCHClC{{H}_{3}}$
C) ${{C}_{6}}{{H}_{5}}CHCHC{{H}_{2}}Cl$
D) ${{C}_{6}}{{H}_{5}}CH(C{{H}_{3}})C{{H}_{2}}Cl$
Answer
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Hint: When the 1-phenylpropene is made to undergo reaction with the HCl, it results in the formation of the compound which consists of the -C-C single bonds i.e. unsaturation changes to saturation. Now you can easily answer the given statement accordingly.
Complete answer:
The 1-phenylpropene is itself formed by the reaction between the benzene ring and the propene. The chemical reaction is supposed to take place as;-
${{C}_{6}}{{H}_{6}}+{{H}_{2}}C=CHC{{H}_{3}}\to {{C}_{6}}{{H}_{5}}CH=CHC{{H}_{3}}$
Now considering the statement as;-
When 1-phenylpropene is treated with the hydrochloric acid i.e. HCl , it results in the formation of a saturated compound i.e. double bond between the carbon atoms breaks into single bonds leading to the formation of alkane. In this reaction, two compounds are formed. Out of which , one is a major compound and the other is a minor compound.
The chemical reaction is supposed to take place as;-
$\begin{align}
& {{C}_{6}}{{H}_{5}}CH=CHC{{H}_{3}}+HCl\to {{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}+{{C}_{6}}{{H}_{5}}CHCHClC{{H}_{3}} \\
& 1-phenylpropene\text{ major minor} \\
\end{align}$
In this reaction, ${{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}$ is major product because the chlorine is attached to the carbon next to the benzene ring and is thus, more stable than the minor product in which the chlorine is attached to the second carbon of the propene.
So, the addition of HCl to 1-phenylpropene gives ${{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}$.
Thus, option (a) is correct.
Note:
Don’t get confused in the saturated and unsaturated compounds. Saturated compounds are those compounds which do not consist of any double or triple bonds i.e. in them the carbon atoms are held together by the single bonds. Whereas on the other hand unsaturated compounds are those compounds which consist of double or triple bonds in between the carbon atoms.
Complete answer:
The 1-phenylpropene is itself formed by the reaction between the benzene ring and the propene. The chemical reaction is supposed to take place as;-
${{C}_{6}}{{H}_{6}}+{{H}_{2}}C=CHC{{H}_{3}}\to {{C}_{6}}{{H}_{5}}CH=CHC{{H}_{3}}$
Now considering the statement as;-
When 1-phenylpropene is treated with the hydrochloric acid i.e. HCl , it results in the formation of a saturated compound i.e. double bond between the carbon atoms breaks into single bonds leading to the formation of alkane. In this reaction, two compounds are formed. Out of which , one is a major compound and the other is a minor compound.
The chemical reaction is supposed to take place as;-
$\begin{align}
& {{C}_{6}}{{H}_{5}}CH=CHC{{H}_{3}}+HCl\to {{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}+{{C}_{6}}{{H}_{5}}CHCHClC{{H}_{3}} \\
& 1-phenylpropene\text{ major minor} \\
\end{align}$
In this reaction, ${{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}$ is major product because the chlorine is attached to the carbon next to the benzene ring and is thus, more stable than the minor product in which the chlorine is attached to the second carbon of the propene.
So, the addition of HCl to 1-phenylpropene gives ${{C}_{6}}{{H}_{5}}CHClCHC{{H}_{3}}$.
Thus, option (a) is correct.
Note:
Don’t get confused in the saturated and unsaturated compounds. Saturated compounds are those compounds which do not consist of any double or triple bonds i.e. in them the carbon atoms are held together by the single bonds. Whereas on the other hand unsaturated compounds are those compounds which consist of double or triple bonds in between the carbon atoms.
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