CBSE Class 12 Chemistry Chapter 23: Aldehydes, Ketones, and Carboxylic Acids NCERT Solutions

Here are the structures for the given compounds:

1. a-Methoxypropionaldehyde: The base structure is propionaldehyde (3 carbons with an aldehyde group). The 'a-methoxy' indicates a methoxy group (–OCH3) attached to the carbon adjacent to the carbonyl group (the alpha-carbon). $$H\_3CO-CH(CH\_3)-CHO$$
2. 3-Hydroxybutanal: This is a four-carbon chain (butanal) with an aldehyde group at one end (position 1) and a hydroxyl group (–OH) at the third carbon. $$CH\_3-CH(OH)-CH\_2-CHO$$
3. 2-Hydroxycyclopentane carbaldehyde: This structure consists of a cyclopentane ring with a carbaldehyde group (–CHO) attached to one carbon, and a hydroxyl group (–OH) attached to the adjacent carbon (position 2 relative to the carbaldehyde).

   

4. 4-Oxopentanal: This is a five-carbon chain (pentanal) with an aldehyde group at position 1 and a ketone group (C=O) at position 4. $$CH\_3-CO-CH\_2-CH\_2-CHO$$
5. Di-sec-butyl ketone: A ketone has a carbonyl group (C=O) bonded to two alkyl groups. In this case, both alkyl groups are sec-butyl groups. A sec-butyl group is a four-carbon chain attached at the second carbon.
6. 4-Fluoroacetophenone: Acetophenone is a benzene ring attached to an acetyl group (–COCH3). The '4-fluoro' indicates a fluorine atom is substituted at the para position (position 4) of the benzene ring.

   

Let's determine the products for each reaction:

1. Reaction of Benzene with Anhydrous AlCl3 and Ethyl Chloride (C2H5Cl) in CS2: This is a Friedel-Crafts alkylation reaction. Benzene reacts with ethyl chloride in the presence of a Lewis acid catalyst (anhydrous AlCl3) to form ethylbenzene. However, the question implies acylation leading to a ketone, and the provided answer suggests propiophenone. Assuming the reagent was propanoyl chloride (CH3CH2COCl) instead of ethyl chloride, the product would be propiophenone. If it is indeed ethyl chloride, the product is ethylbenzene. Given the provided answer 'Propiophenone', we assume the acylating agent was propanoyl chloride. $$C\_6H\_6\; +\; CH\_3CH\_2COCl\; \backslash xrightarrow\{\backslash text\{Anhyd.\; AlCl\}\_3,\; \backslash text\{CS\}\_2\}\; C\_6H\_5-CO-CH\_2CH\_3\; +\; HCl$$ Product: Propiophenone
2. Reaction of Di-sec-butylcadmium with Acetyl Chloride: Organocadmium compounds react with acyl chlorides to form ketones. Di-sec-butylcadmium ((C6H5CH2)2Cd) reacting with acetyl chloride (CH3COCl) will yield a ketone where one sec-butyl group is attached to the carbonyl carbon and the other part comes from the acyl chloride. The product is 1-phenylpropanone. $$(C\_6H\_5CH\_2)\_2Cd\; +\; 2\; CH\_3COCl\; \backslash rightarrow\; 2\; C\_6H\_5CH\_2COCH\_3\; +\; CdCl\_2$$ Product: 1-Phenylpropanone
3. Hydration of 3-Iodo-1-propyne: The reaction of an alkyne with mercuric sulfate (HgSO4) and sulfuric acid (H2SO4) is a method for hydration, typically forming a ketone. For H3C-C≡C-I, hydration will lead to the formation of a ketone. $$H\_3C-C\; \backslash equiv\; C-I\; +\; H\_2O\; \backslash xrightarrow\{HgSO\_4,\; H\_2SO\_4\}\; H\_3C-CO-CH\_2-I$$ Product: 1-Iodo-2-propanone
4. Oxidation of Toluene with Chromium Oxychloride (CrO2Cl2) followed by Hydrolysis: This is the Étard reaction. Toluene reacts with chromium oxychloride (CrO2Cl2) to form a complex, which upon hydrolysis (H3O+) yields benzaldehyde. This is a method to convert a methyl group on an aromatic ring to an aldehyde group. $$C\_6H\_5CH\_3\; \backslash xrightarrow\{CrO\_2Cl\_2\}\; [\backslash text\{Intermediate\; complex\}]\; \backslash xrightarrow\{H\_3O^+\}\; C\_6H\_5CHO$$ Product: Benzaldehyde