Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as a nucleophile and which acts as an electrophile.

seo-qna
Last updated date: 20th Jun 2024
Total views: 373.8k
Views today: 6.73k
Answer
VerifiedVerified
373.8k+ views
Hint: Aldol has two functional groups together i.e., aldehyde and alcohol. Similarly, ketols have ketonic and alcoholic groups. Ketol and aldol both lose water ready to give $\alpha ,\beta $-unsaturated carbonyl compounds, hence this reaction is called Aldol condensation.

Complete step by step answer:
In order to answer the question, we need to learn about the aldol condensation reaction. In the aldehydes and ketones, the carbonyl group is present. The carbon is $s{{p}^{2}}$ hybridized and has a geometry of triangular planar. There is an electronegativity difference between the oxygen and carbon atoms and this makes the bond polarised. As oxygen is more electronegative, it pulls the shared pair of electrons towards itself and the carbon becomes positively charged. This makes the aldehydes and ketones prone to nucleophilic reactions. An increase in the size of alkyl or aryl groups around 'C' atom produces steric hindrance to the attacking nucleophile. It is due to the crowding of space around carbonyl carbon atoms. Hence greater the number of alkyl/aryl groups or the bigger the size of groups, the more will be the steric hindrance and the lesser will be the tendency of nucleophiles to attack the carbonyl carbon.

Aldehydes and ketones having at least one a-Hydrogen atom take part in Aldol condensation in the presence of dilute alkali like NaOH. Two molecules of carbonyl compound condense to form a $\beta $ hydroxy aldehyde or ketone which gets dehydrated on heating into $\alpha ,\beta $ unsaturated ketone.
The acidity of $\alpha $H atoms of carbonyl compounds is due to the strong e withdrawing effect of a carbonyl group and resonance stabilization of the conjugate base. Now, let us see the four products of the aldol condensation:
(1)
\[C{{H}_{3}}C{{H}_{2}}CHO+C{{H}_{3}}C{{H}_{2}}CHO\xrightarrow{dil\,NaOH}C{{H}_{3}}C{{H}_{2}}CHOHCHC{{H}_{3}}CHO\]
Here, one propanal acts as nucleophlile and other propanal acts as electrophile
(2) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHO+C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHO\xrightarrow{dil\,NaOH}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHOHCHC{{H}_{2}}C{{H}_{3}}CHO\]
Here, one butanal acts as nucleophile and other butanal acts as electrophile
(3)
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHO+C{{H}_{3}}C{{H}_{2}}CHO\to C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHOHCHC{{H}_{3}}CHO\]
Here, butanal acts as electrophile and propanal acts as electrophile
(4)
\[C{{H}_{3}}C{{H}_{2}}CHO+C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHO\to C{{H}_{3}}C{{H}_{2}}CHOHCHC{{H}_{2}}C{{H}_{3}}CHO\]
Here, propanal is the electrophile and butanal is the nucleophile.
So, we get the structures of all four products.

Note: Aldehydes which do not have any alpha H atom such as HCHO undergo self oxidation and self-reduction (i.e., disproportionation) on treatment with conc. alkali. In this one molecule is reduced to alcohol and the other is oxidized to a salt of carboxylic acid.