# Write the equations for the preparation of 1-iodobutane from 1-butanol. Name the reaction.

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Hint: According to the question, we need to perform a reaction using 1-butanol and another reagent to produce 1-iodobutane. To achieve this, we need to substitute the alcohol group, i.e. the –OH group from 1-butanol with iodine.

Complete Step by Step Solution:
We can easily substitute the –OH group with the –I group by using HI as a reagent with the presence of anhydrous ZnCl2.
$C{H_3} - C{H_2} - C{H_2} - C{H_2} - OH + HI\xrightarrow{{ZnC{l_2}}}C{H_3} - C{H_2} - C{H_2} - C{H_2} - I + {H_2}O$

Anhydrous ZnCl2 is added with hydrohalic acid for a specific purpose. ZnCl2 is a strong Lewis acid added to the reaction as Cl is considered as a better nucleophile than I. It happens due to its bulky size and Cl makes the alcohol group (-OH) a better leaving ability group than other halides. When alcohols are reacted with a strong Lewis acid, it forms a complex, [Zn(OH)Cl2].

Then in the following rearrangement reaction, the R-Cl product is changed to R-I and thus we get our desired haloalkanes. The reagent used, [HI + ZnCl2] is also known as the Lucas reagent and the reaction is commonly referred to as the Lucas reaction or the substitution of the –OH group.

Additional Information: Lucas reagent is a solution of anhydrous zinc chloride mixed with concentrated hydrochloric acid. This solution is used to identify and classify low molecular weight alcohols. The reaction taking place is a substitution in which the chloride ion replaces a hydroxyl group. If the colour of the solution changes from colourless to turbid with an oily layer on top of the solution, the test for alcohol is positive and it signals the formation of a chloroalkane.

Note: ZnCl2 reacts with the alcohol group in the following mechanism.