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Question

Question: Write the equations for the preparation of \[1 - iodobutane\] from: \[(i){\text{ 1}} - butanol\] ...

Write the equations for the preparation of 1iodobutane1 - iodobutane from:
(i) 1butanol(i){\text{ 1}} - butanol
(ii)1chlorobutane(ii)1 - chlorobutane
(iii)but1ene(iii)but - 1 - ene

Explanation

Solution

In 1iodobutane1 - 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) 1butanol(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 I2{I_2}, it gives out alkyl iodide.
In this first question we react (i) 1butanol(i){\text{ 1}} - butanol reacted with red phosphorus and in presence of I2{I_2}
, it gives 1iodobutane1 - iodobutane.
Reaction of the following:
ROHRed P, I2RIROH\xrightarrow{{\operatorname{Re} d{\text{ P, }}{{\text{I}}_2}}}RI
CH3CH2CH2OHRed P, I2CH3CH2CH2IC{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, 1chlorobutane1 - 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, 1chlorobutane1 - chlorobutanereacts in the presence of NaI, it gives 1iodobutane1 - iodobutane.
The following reaction:
R2CH=CHHIR2CH2I{R_2}CH = CH\xrightarrow{{HI}}{R_2}C{H_2}I
RClNaIRIRCl\xrightarrow{{NaI}}RI
CH3CH2CH2ClNaICH3CH2CH2IC{H_3}C{H_2}C{H_2}Cl\xrightarrow{{NaI}}C{H_3}C{H_2}C{H_2}I
(iii) In third, but1enebut - 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 but1enebut - 1 - ene we react with HI and then it undergoes Markovnikov's addition reaction. It gives out 1iodobutane1 - iodobutane.
The reaction is as follows: R2CH=CHHIR2CH2I{R_2}CH = CH\xrightarrow{{HI}}{R_2}C{H_2}I
CH3CH2CH=CHHICH3CH2CH2IC{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.