Question

How will you convert ethanal into the following compounds?
A.$\left( 1 \right)$ Butane$ - 1,3 - $diol
B.$\left( 2 \right)$ But$ - 2 - $enal
C.$\left( 3 \right)$ But$ - 2 - $enoic acid

Answer 1

Hint: The two carbon-containing aldehyde functional groups are converted to four carbon molecules having different functional groups. Hence, to double the carbon numbers which have aldehyde or ketone as an initial functional there is a very famous common reaction taken place as a first step. Further, they are treated with different reagents used for reduction, oxidation, dehydration to get the desired product.

Complete step by step answer:
1) Ethanal into Butane$ - 1,3 - $diol: There will be an aldol condensation reaction in presence of base which is $dil.NaOH$ as a first step which gives $3 - $Hydroxybutanal and then further reacted by any reducing reagent which we used here is $NaB{H_4}$ that will convert the terminal aldehyde group into alcohol and get the desired Butane$ - 1,3 - $diol as the product.
Reaction scheme:
$C{H_3}CHO\xrightarrow[{Aldol}]{{dil.NaOH}}C{H_3} - CHOH - C{H_2} - CHO\xrightarrow[{\left( {reduction} \right)}]{{NaB{H_4}}}C{H_3} - CHOH - C{H_2} - C{H_2}OH$
2) Ethanal into But$ - 2 - $enal: The first step will be aldol condensation reaction in presence of a base which will give $3 - $Hydroxybutanal. Then it will further undergo a dehydration reaction by giving heat which will give But$ - 2 - $enal as the product.
Reaction scheme:
$C{H_3}CHO\xrightarrow[{Aldol}]{{dil.NaOH}}C{H_3} - CHOH - C{H_2} - CHO\xrightarrow[{ - {H_2}O\left( {Dehydration} \right)}]{\Delta }C{H_3} - CH = CH - CHO$
3) Ethanal into But$ - 2 - $enoic acid: The first step will be aldol condensation in presence of a base and form the product $3 - $Hydroxybutanal which will further undergo a dehydration reaction by giving heat which gives But$ - 2 - $enal as the product. Further, it undergoes an oxidation reaction in presence of Tollen’s reagent which converts the aldehyde group into a carboxylic group and gives the desired product But$ - 2 - $enoic acid.
Reaction scheme:
$C{H_3}CHO\xrightarrow[{Aldol}]{{dil.NaOH}}C{H_3} - CHOH - C{H_2} - CHO\xrightarrow[{ - {H_2}O\left( {Dehydration} \right)}]{\Delta }C{H_3} - CH = CH - CHO$
$C{H_3} - CH = CH - CHO\xrightarrow[{{\text{Tollen's reagent}}}]{{{{\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]}^ + }O{H^ - }}}C{H_3}CH = CH COOH$


Note:
Dehydration reaction is always undergone through heating at high temperature and removal of hydrogen from the carbon which is adjacent to $ - OH$ a group containing carbon which possesses the less number of hydrogen atoms is done which forms the highly substituted alkene product and ${H_2}O$ is eliminated.