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\mathrm{R-OH+PCl_5\rightarrow R-Cl+POCl_3+HCl}

The above reaction gives the best yield for

Option: 1

Primary alkyl halides


Option: 2

Tertiary alkyl halides


Option: 3

Secondary alkyl halides


Option: 4

All yields are similar irrespective of alcohol


Answers (1)

best_answer

As we learnt

Alkyl halides are produced by treating alcohol with \mathrm{PCl_{5}}

When alcohol treated with \mathrm{PCl_{5}}, OH group is replaced by Cl.

\mathrm{R-OH +PCl_{5}\rightarrow R-Cl+POCl_{3}+HCl}

 

Normally \mathrm{PCl_{5}} is use to produce primary alkyl halide but under extreme conditions, it can also produce secondary alkyl halides

Therefore, option (1) is correct.

Posted by

Devendra Khairwa

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