Question

Find the products A, B and C in the following reaction;
$C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}+B{{r}_{2}}\xrightarrow{hv}(A)\xrightarrow[(ii)CuI]{(i)Li}(B)\xrightarrow{C{{H}_{3}}I}(C)$

Answer 1

When alkane is made to react with halogen in the presence of light, it results in the formation of alkyl halide i.e. halogenation and when this is made to react with the lithium and copper iodide , it results in the formation of lithium alkyl cuprate which on further reaction with alkyl halide undergoes Corey-house synthesis reaction. Now, solve it.

Complete answer:
Alkanes are the simplest hydrocarbon and are saturated as it doesn’t consist of any double bond or single in it.
In the above given statement, is an alkane as there is no double and triple bond present in it. Since, the given alkane consists of three carbon atoms, so it is propane.
Propane when is made to react with the halogen such as F, Cl ,Br or I , it undergoes halogenation i.e. the one hydrogen atom of the alkane is replaced by the halogen. So, when propane is reacted with bromine, its hydrogen atom is replaced by bromine and forms bromopropane and HBr is removed. The reaction occurs as;
$C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}+B{{r}_{2}}\xrightarrow{hv}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br$
Now, when this alkyl halide i.e. 1-bromopropane is treated with lithium , it first results in the formation of alkyl lithium and when is made to react with copper iodide it results in the formation of lithium dialkylcuprate . The reaction occurs as;
$C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{Li}2C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Li\xrightarrow{CuI}{{(C{{H}_{3}}C{{H}_{2}}C{{H}_{2}})}_{2}}CuLi$
When this, lithium alkyl cuprate is made to react with further alkyl iodide, it results in the formation of alkane ( i.e. butane) and this reaction is known as Corey- house synthesis. The reaction occurs as;
${{(C{{H}_{3}}C{{H}_{2}}C{{H}_{2}})}_{2}}CuLi\xrightarrow{C{{H}_{3}}I}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}+C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cu+LiBr$
So, the overall reaction becomes as;
$C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}+B{{r}_{2}}\xrightarrow{hv}(A)\xrightarrow[(ii)CuI]{(i)Li}{{(C{{H}_{3}}C{{H}_{2}}C{{H}_{2}})}_{2}}CuLi\xrightarrow{C{{H}_{3}}I}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}$
So , the products A ,B and C are alkyl halide, lithium alkyl cuprate and alkane respectively.

Note:
Corey-house synthesis is the method of increasing the carbon chain and results in the formation of higher alkanes i.e. in this method, higher alkanes are formed from the alkyl lithium cuprates.