Aldehydes, Ketones and Carboxylic Acids

The lactone is a cyclic organic ester. The structure of lactone is given below.

The conversion of compound B to lactone is shown below:

The conversion of compound C to lactone is shown below:

Step 1:

The compound C is ortho-phthalaldehyde. The compound C when treated with conc. NaOH undergoes Cannizaro reaction as shown below:

Step 2:

On acidification, the COONa group is converted to COOH group. The compound now will have both $\mathrm{CH}_2 \mathrm{OH}$ and COOH groups.

Step 3:

The dehydration of the above compound gives lactone.

Note: Option B is not possible as it on reaction with conc. NaOH undergoes intermolecular cannizaro's reaction with possibility of forming pthalic acid as one of the products.

Pthalic acid undergoes dehydration in presence of acid to form an anhydride which is an unwanted product.

Step 1:

$ \text { Step 2: } $

$ \text { Step 3: } $

$ \text { Reaction between } \mathrm{CH}_3 \mathrm{COCH}_3 \text { and } \mathrm{NH}_2 \mathrm{OH} \text { : } $

The oxime formed is optically inactive as there is plane of symmetry present in it.

Reaction between $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COCH}_3$ and $\mathrm{NH}_2 \mathrm{OH}$

An oxime will be obtained which shows two enantiomers based on the position of bulky group, $\mathrm{CH}_3 \mathrm{CH}_2$ with respect to the OH group attached to N atom of the oxime.

The two enantiomers of the above compound are:

The two enantiomers are optically active as they lack a plane of symmetry and exist as $R$ and S isomers.

Hence, 3 different oximes are formed out of which 2 are optically active and one is optically inactive.