Question

Write the equations for the preparation of \[1 - iodobutane\] from:
\[(i){\text{ 1}} - butanol\]
\[(ii)1 - chlorobutane\]
\[(iii)but - 1 - ene\]

Answer 1

Hint: In \[1 - iodobutane\], we can understand that there is presence of an aliphatic four carbon chain and an Iodo group. From the given options we need to find out the main functional group and understand what can convert them.

Complete answer:
In the first point, \[(i){\text{ 1}} - butanol\] we observe that it has an alcoholic group.
If any alkyl alcohol is reacted with red phosphorus and in the presence of \[{I_2}\], it gives out alkyl iodide.
In this first question we react \[(i){\text{ 1}} - butanol\] reacted with red phosphorus and in presence of \[{I_2}\]
, it gives \[1 - iodobutane\].
Reaction of the following:
\[ROH\xrightarrow{{\operatorname{Re} d{\text{ P, }}{{\text{I}}_2}}}RI\]
\[C{H_3}C{H_2}C{H_2}OH\xrightarrow{{\operatorname{Re} d{\text{ P, }}{{\text{I}}_2}}}C{H_3}C{H_2}C{H_2}I\]
(ii) In second, \[1 - chlorobutane\] we observe that it has a halogen group.
If any alkyl chloride is reacted in the presence of NaI, there is exchange of weaker nucleophile in presence of stronger nucleophile.
In this part we react, \[1 - chlorobutane\]reacts in the presence of NaI, it gives \[1 - iodobutane\].
The following reaction:
\[{R_2}CH = CH\xrightarrow{{HI}}{R_2}C{H_2}I\]
\[RCl\xrightarrow{{NaI}}RI\]
\[C{H_3}C{H_2}C{H_2}Cl\xrightarrow{{NaI}}C{H_3}C{H_2}C{H_2}I\]
(iii) In third, \[but - 1 - ene\] we observe that it has one unsaturated group. If any alkene is reacted with HI it undergoes markovnikov’s addition, and forms alkyl iodide.
In this part we react \[but - 1 - ene\] we react with HI and then it undergoes Markovnikov's addition reaction. It gives out \[1 - iodobutane\].
The reaction is as follows: \[{R_2}CH = CH\xrightarrow{{HI}}{R_2}C{H_2}I\]
\[C{H_3}C{H_2}CH = CH\xrightarrow{{HI}}C{H_3}C{H_2}C{H_2}I\]

Note:
Do not forget to add reagents when and where required. They act as catalytic agents or sometimes help in getting the end product with ease. They are very important because they can accelerate the reaction. Always find the correct functional group attached, otherwise the whole procedure of finding the product can lead to formation of something else.