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How is chlorobenzene prepared from aniline? How is chlorobenzene converted into diphenyl?

Profile image of Aniket Singh
1 Year agoGrade
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1 Answer

Profile image of Askiitians Tutor Team
1 Year ago

Chlorobenzene can be prepared from aniline through a process called Sandmeyer reaction. The reaction involves the conversion of aniline to diazonium salt, followed by substitution of the diazonium salt with a chlorine source. Here are the steps involved in the preparation of chlorobenzene from aniline:

Diazotization: Aniline (C6H5NH2) is treated with nitrous acid (HNO2) to form the corresponding diazonium salt. This step is typically carried out by adding a solution of sodium nitrite (NaNO2) to aniline in the presence of hydrochloric acid (HCl) at low temperatures (around 0-5°C).

Sandmeyer Reaction: The diazonium salt formed in the previous step is then treated with a source of chlorine, such as cuprous chloride (CuCl) or sodium chloride (NaCl), in the presence of a catalyst like copper(I) chloride (CuCl). This leads to the replacement of the diazonium group (-N2+) with a chlorine atom (-Cl), resulting in the formation of chlorobenzene (C6H5Cl).

The reaction can be represented by the following chemical equation:

C6H5NH2 + HNO2 + HCl -> C6H5N2Cl + 2H2O
C6H5N2Cl + CuCl -> C6H5Cl + N2 + CuCl2

Now, regarding the conversion of chlorobenzene into diphenyl, this can be achieved through a reaction known as the Ullmann coupling reaction. The Ullmann reaction involves the reaction of two aryl halide molecules in the presence of a copper catalyst, resulting in the formation of a biaryl compound. Here are the steps involved in the conversion of chlorobenzene into diphenyl:

Activation of Chlorobenzene: Chlorobenzene is first activated by reacting it with a strong base, such as potassium tert-butoxide (KOtBu) or sodium tert-butoxide (NaOtBu), in a suitable solvent (e.g., dimethyl sulfoxide, DMSO). This step is necessary to generate the corresponding aryl anion.

Ullmann Coupling: The activated chlorobenzene is then reacted with another molecule of chlorobenzene under the influence of a copper catalyst, such as copper(I) iodide (CuI) or copper(I) oxide (Cu2O). This leads to the formation of diphenyl (C6H5C6H5) via a carbon-carbon bond formation between the two chlorobenzene molecules.

The reaction can be represented by the following chemical equation:

2C6H5Cl + 2KOtBu + 2CuI -> C6H5C6H5 + 2KCl + 2Cu + 2tBuOH

It's worth noting that the Ullmann coupling reaction is a general method for the synthesis of biaryl compounds and can be used to connect various aryl halides to form different biaryl structures.






question mark

is Pradeep Chemistry+Ncert sufficient for all of the theory of Jee Mains Chemistry?


Hi, I’m a self-study JEE 2027 aspirant with a large 11th backlog. Since I don’t think 6–8 hour one-shots are sufficient to cover the full theory + breadth of problem types of a JEE chapter, I’ve shifted from lecture-based preparation to book-based self-study.


I have a problem with using NCERT as my primary chemistry source. In my experience, it is concise and information-heavy, but often lacks detailed explanations, systematic organisation of concepts, and categorisation of problem types. This seems especially problematic for Physical and Organic Chemistry, where I feel that simply reading NCERT may not give enough depth or problem-solving preparation.


So I’m considering this approach:


Pradeep Chemistry → complete the chapter thoroughly from Pradeep → then read NCERT for that chapter → solve JEE Main PYQs/problems.


My main question is:


Is Pradeep + NCERT sufficient as the complete theory source for JEE Main-level Chemistry?


In other words, after thoroughly completing a chapter from Pradeep and then NCERT, can I consider the theory part of that chapter complete for JEE Main, with only PYQs/problem practice remaining?


I’m asking specifically about JEE Main level, not Advanced.


Also, if Pradeep is not sufficient, what specific gap does it leave, and what would you recommend as a better book-based alternative for self-study?

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