Question

How do you convert ethanol to \[but - 2 - yne\] ?

Answer 1

Hint: In order to answer this question, we have to follow a certain set of steps. These steps involve the ethanol to ethyl halide followed by the addition of alcoholic KOH which converts the halide to alkene. We then treat it with bromine molecules.

Complete step-by-step answer:
We have to convert ethanol which is an alcohol to \[but - 2 - yne\] which is an unsaturated hydrocarbon.
The structure of ethanol is written as $C{H_3} - C{H_2} - OH$. The structure of \[but - 2 - yne\] is $C{H_3} - C \equiv C - C{H_3}$

Let us now perform the steps of conversion of ethanol to \[but - 2 - yne\]
i) Firstly, we convert the given alcohol to a halide. In order to do this, we react ethanol with $P/B{r_2}$ to get the product as ethyl bromide.
We can now write this reaction as $C{H_3} - C{H_2} - OH\xrightarrow{{P/B{r_2}}}C{H_3} - C{H_2} - Br$
ii) Conversion of ethyl bromide to ethene: In order to do this, we have to treat the produced ethyl bromide with alcoholic KOH. The product would be ethene.
$C{H_3} - C{H_2} - Br\xrightarrow{{alc.KOH}}C{H_2} = C{H_2}$
iii) Addition of Bromine molecule: We now treat this ethene with Bromine molecule to get a dibromide.
$C{H_2} = C{H_2}\xrightarrow{{B{r_2}}}Br - C{H_2} - C{H_2} - Br$
iv) Substitution of sodium ions: Let us now react this bromide with sodium in presence of ammonia.
$Br - C{H_2} - C{H_2} - Br\xrightarrow{{Na/N{H_3}}}N{a^ + }{C^ - } \equiv {C^ - }N{a^ + }$
v) Reaction with methyl bromide: We can write this reaction as follows.
$N{a^ + }{C^ - } \equiv {C^ - }N{a^ + }\xrightarrow{{C{H_3}Br}}C{H_3} - C \equiv C - C{H_3}$
Therefore, \[but - 2 - yne\] is formed from ethanol.

Note: It is to be noted that in order to solve problems like these where we convert organic compounds, we should mainly learn and understand the reactions that increase or decrease the number of carbons in the reactant molecule.