
Write the main product formed when propanal reacts with the following reagent?
i)2 moles of \[C{{H}_{3}}OH\] present in dry $HCl$
ii)Dilute $NaOH$
iii)${{H}_{2}}N-N{{H}_{2}}$ Followed by heating with $KOH$ in ethylene glycol.
Answer
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Hint: in the given question we are asked about the reactions with various reagents. We will check the individual reaction and get the answer of the question.
Complete step by step solution:
The formula of propanal is $C{{H}_{3}}C{{H}_{2}}CHO$.
We are asked about the reactions of propanal.
In the first option the reactant is 2 moles of \[C{{H}_{3}}OH\] present in dry $HCl$
The reaction will be as follows.
\[C{{H}_{3}}C{{H}_{2}}CHO\,+\,2C{{H}_{3}}OH\,\xrightarrow{HCl}\,{{C}_{2}}{{H}_{5}}-CH(OC{{H}_{3}})\]
(1,1-diethoxy propane)
The major product formed in the above reaction is 1,1-diethoxy propane.
In the second option we will react with Dilute $NaOH$. The hydrogen atom present with the carbon atom just beside the aldehyde group is known as alpha hydrogen. Due to this presence of alpha hydrogen propanal will give this reaction. This reaction is known as the aldol condensation reaction.
The reaction is as follows:
${{C}_{2}}{{H}_{5}}-CHO\,+\,NaOH\,\to \,$
The product formed as a result of this reaction is shown above.
In the third option we are given ${{H}_{2}}N-N{{H}_{2}}$ Followed by heating with $KOH$ in ethylene glycol. This reaction is known as the Wolf-Kishner reaction. The carbonyl group of propanal is reduced to $C{{H}_{2}}$ group on treatment with hydrazine followed by heating with $KOH$ in ethylene glycol and the product formed will be propane.
The reaction is as follows: ${{C}_{2}}{{H}_{5}}-CHO\,\xrightarrow[-\,{{H}_{2}}O]{{{H}_{2}}N-N{{H}_{2}}}\,$ $\xrightarrow[heat]{KOH/ethylene\,glycol}\,{{C}_{2}}{{H}_{5}}-C{{H}_{3}}\,+\,{{N}_{2}}$
In the above reaction the major product is propane. This reaction is a type of reduction reaction.
Note: For solving this type of question the reaction should be known. Also it should be noted that the name reactions in organic chemistry should be learnt very nicely. The third reaction i.e. Wolf-Kishner reaction is a type of reduction reaction. In this the aldehyde group is changed to methyl group.
Complete step by step solution:
The formula of propanal is $C{{H}_{3}}C{{H}_{2}}CHO$.
We are asked about the reactions of propanal.
In the first option the reactant is 2 moles of \[C{{H}_{3}}OH\] present in dry $HCl$
The reaction will be as follows.
\[C{{H}_{3}}C{{H}_{2}}CHO\,+\,2C{{H}_{3}}OH\,\xrightarrow{HCl}\,{{C}_{2}}{{H}_{5}}-CH(OC{{H}_{3}})\]
(1,1-diethoxy propane)
The major product formed in the above reaction is 1,1-diethoxy propane.
In the second option we will react with Dilute $NaOH$. The hydrogen atom present with the carbon atom just beside the aldehyde group is known as alpha hydrogen. Due to this presence of alpha hydrogen propanal will give this reaction. This reaction is known as the aldol condensation reaction.
The reaction is as follows:
${{C}_{2}}{{H}_{5}}-CHO\,+\,NaOH\,\to \,$
The product formed as a result of this reaction is shown above.
In the third option we are given ${{H}_{2}}N-N{{H}_{2}}$ Followed by heating with $KOH$ in ethylene glycol. This reaction is known as the Wolf-Kishner reaction. The carbonyl group of propanal is reduced to $C{{H}_{2}}$ group on treatment with hydrazine followed by heating with $KOH$ in ethylene glycol and the product formed will be propane.
The reaction is as follows: ${{C}_{2}}{{H}_{5}}-CHO\,\xrightarrow[-\,{{H}_{2}}O]{{{H}_{2}}N-N{{H}_{2}}}\,$ $\xrightarrow[heat]{KOH/ethylene\,glycol}\,{{C}_{2}}{{H}_{5}}-C{{H}_{3}}\,+\,{{N}_{2}}$
In the above reaction the major product is propane. This reaction is a type of reduction reaction.
Note: For solving this type of question the reaction should be known. Also it should be noted that the name reactions in organic chemistry should be learnt very nicely. The third reaction i.e. Wolf-Kishner reaction is a type of reduction reaction. In this the aldehyde group is changed to methyl group.
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