CBSE Class 12 Chemistry Chapter 12: Aldehydes, Ketones and Carboxylic Acids - Intext Questions Solutions

a-Methoxypropionaldehyde: This compound has a propionaldehyde backbone (3 carbons with an aldehyde group) and a methoxy group (-OCH3) attached to the alpha-carbon (the carbon adjacent to the carbonyl group).

$$CH\_3O-CH(CH\_3)-CHO$$

3-Hydroxybutanal: This is a four-carbon chain (butanal) with an aldehyde group at one end and a hydroxyl group (-OH) on the third carbon atom.

$$HO-CH(CH\_3)-CH\_2-CHO$$

2-Hydroxycyclopentane carbaldehyde: This structure features a cyclopentane ring with a carbaldehyde group (-CHO) attached to one carbon, and a hydroxyl group (-OH) on the adjacent carbon (position 2).

$$\backslash begin\{array\}\{c\}\; \backslash text\{CHO\}\; \backslash \backslash \; |\; \backslash \backslash \; \backslash text\{C\}\_5H\_8-OH\; \backslash end\{array\}$$

4-Oxopentanal: This is a five-carbon chain (pentanal) with an aldehyde group at one end and a ketone group (C=O) on the fourth carbon atom.

$$CH\_3-CO-CH\_2-CH\_2-CHO$$

Di-sec-butyl ketone: This is a ketone where the carbonyl carbon is bonded to two sec-butyl groups. A sec-butyl group is a four-carbon chain attached at the second carbon.

4-Fluoroacetophenone: This is an acetophenone structure (a benzene ring attached to an acetyl group, -COCH3) with a fluorine atom substituted at the para (4th) position of the benzene ring.

$$\backslash begin\{array\}\{c\}\; \backslash operatorname\{F\}\; \backslash \backslash \; |\; \backslash \backslash \; \backslash text\{C\}\_6H\_4\; -\; CO\; -\; CH\_3\; \backslash end\{array\}$$

This reaction is a Friedel-Crafts acylation of benzene. Benzene reacts with ethyl chloride in the presence of anhydrous aluminum chloride (AlCl3) as a Lewis acid catalyst and carbon disulfide (CS2) as a solvent. The electrophile generated is the ethyl carbocation, which attacks the benzene ring, followed by deprotonation to form ethylbenzene.

Reactants: Benzene, Anhydrous AlCl3, C2H5Cl, CS2

Product: Ethylbenzene

$$C\_6H\_6\; +\; C\_2H\_5Cl\; \backslash xrightarrow\{\backslash text\{Anhyd.\; AlCl\}\_3,\; CS\_2\}\; C\_6H\_5-C\_2H\_5\; +\; HCl$$

The product is ethylbenzene.

This reaction involves the reaction of dibenzylcadmium with acetyl chloride. Organocadmium reagents react with acid chlorides to form ketones. In this case, two molecules of acetyl chloride react with one molecule of dibenzylcadmium to yield acetophenone and cadmium chloride.

Reactants: (C6H5CH2)2Cd, 2 CH3COCl

Product: Acetophenone

$$(C\_6H\_5CH\_2)\_2Cd\; +\; 2\; CH\_3COCl\; \backslash rightarrow\; 2\; C\_6H\_5CH\_2-COCH\_3\; +\; CdCl\_2$$

The product is 1-phenylpropanone (benzyl methyl ketone).

This reaction involves the hydration of an alkyne in the presence of mercuric sulfate and sulfuric acid. The alkyne is 1-iodo-1-propyne. Hydration of alkynes typically follows Markovnikov's rule, leading to the formation of a ketone or aldehyde. In this case, the addition of water across the triple bond, followed by tautomerization, will yield a ketone.

Reactants: $H\_3C-C\; \backslash equiv\; C-I$, Hg/H2SO4

Product: Acetyl iodide (CH3COI) or a related ketone structure after hydrolysis.

The reaction likely leads to the formation of a ketone after hydrolysis of an intermediate enol.

$$H\_3C-C\; \backslash equiv\; C-I\; \backslash xrightarrow\{Hg/H\_2SO\_4,\; H\_2O\}\; CH\_3-CO-CHI\_2\; \backslash rightarrow\; CH\_3-CO-CI\_3\; \backslash rightarrow\; CH\_3-CO-COOH\; \backslash text\{\; (after\; hydrolysis\; and\; further\; steps)\}$$

A more direct interpretation based on similar reactions might lead to a ketone like propanone if the iodine is displaced or involved in a complex manner. However, given the reagents, a complex reaction pathway is expected. Assuming a standard hydration followed by tautomerization, the product would be a ketone.

A likely product is 1-iodopropan-2-one or a related structure depending on the exact mechanism with the iodine substituent.

This reaction involves the oxidation of toluene. The reagent chromyl chloride (CrO2Cl2) in an inert solvent like carbon disulfide or carbon tetrachloride, followed by hydrolysis (H3O+), is used for the oxidation of a methyl group attached to a benzene ring to an aldehyde. This is known as the Étard reaction.

Reactants: Toluene, CrO2Cl2, then H3O+

Product: Benzaldehyde

$$C\_6H\_5-CH\_3\; \backslash xrightarrow\{CrO\_2Cl\_2\}\; [Intermediate]\; \backslash xrightarrow\{H\_3O^+\}\; C\_6H\_5-CHO$$

The product is benzaldehyde.