Aldehydes, Ketones and Carboxylic Acids

The reactions involved are

In reaction I, in basic medium, with each $\alpha$-halogenation the rate of reaction increases and the complete haloform reaction takes place.

In reaction II, in acidic medium only one halogenated product is obtained with 1 mol of halogen.

The reaction is

The $\beta$-keto acid given in option (B) undergoes decarboxylation most easily because it results in formation of stable product on rearrangement. The reaction taking place is

In the given reaction, an aromatic aldehyde is condensed with an acid anhydride (Perkin condensation) in the presence of a base to form $\alpha$,$\beta$-unsaturated acid (cinnamic acid). This gives effervescence on reaction with NaHCO₃ and positive test with Bayer's reagent (1% alkaline KMnO₄).

The reaction sequence is

The structures of the given compounds are as follows. Only aspirin has $-$COOH group.

The reactions involved in the transformation are:

The reaction is

The reactions involved are as follows:

Both phenol and benzoic acid are soluble in NaOH, so they cannot be separated.

C₆H₅OH + NaOH $\to$ C₆H₅ONa + H₂O

C₆H₅COOH + NaOH $\to$ C₆H₅COONa + H₂O

Benzel alcohol C₆H₅CH₂OH is not soluble in NaOH and NaHCO₃, while benzoic acid is soluble in both, so these can be separated. Both C₆H₅OH and C₆H₅CH₂OH are insoluble in NaHCO₃, so they cannot be separated. C₆H₅CH₂COOH is soluble in NaOH and NaHCO₃, whereas benzyl alcohol is not soluble in NaOH and NaHCO₃. So, they can be separated.

(i) When 3,3-dimethyl but-1-yne reacts with dil. $\mathrm{H}_2 \mathrm{SO}_4$ in presence of $\mathrm{HgSO}_4$, ketone is formed.

(ii) The ketone is reduced to secondary alcohol in presence of sodium borohydride $\left(\mathrm{NaBH}_4\right)$ ethanol).

(iii) Reaction of acid with alcohol results in dehydration forming an alkene. During the process, methyl shift results in rearrangement of carbocation which further leads to the formation of alkene.

The product 2,3-dimethyl but-2-ene undergoes ozonolysis to form two moles acetone.

Two moles of acetone is the final product of the reaction.

Compound P is 3,3-dimethyl butyne.

(i) When 3,3-dimethyl but-1-yne reacts with dil. $\mathrm{H}_2 \mathrm{SO}_4$ in presence of $\mathrm{HgSO}_4$, ketone is formed.

(ii) The ketone is reduced to secondary alcohol in presence of sodium borohydride $\left(\mathrm{NaBH}_4\right)$ ethanol).

The given product is an ester, obtained by condensation of a hydroxy acid obtained through hydrolysis of a cyanohydrin :

Acid above is obtained by acid hydrolysis of cyanohydrin S as

S is obtained by nucleophile addition of HCN on R, hence R is

$R$ is obtained by treatment of $P$ and $Q$ with aqueous $\mathrm{K}_2 \mathrm{CO}_3$ through aldol condensation reaction as

The given product is an ester, obtained by condensation of a hydroxy acid obtained through hydrolysis of a cyanohydrin :

Acid above is obtained by acid hydrolysis of cyanohydrin S as

S is obtained by nucleophile addition of HCN on R, hence R is

$R$ is obtained by treatment of $P$ and $Q$ with aqueous $\mathrm{K}_2 \mathrm{CO}_3$ through aldol condensation reaction as

The given product is an ester, obtained by condensation of a hydroxy acid obtained through hydrolysis of a cyanohydrin :

Acid above is obtained by acid hydrolysis of cyanohydrin S as

S is obtained by nucleophile addition of HCN on R, hence R is

$R$ is obtained by treatment of $P$ and $Q$ with aqueous $\mathrm{K}_2 \mathrm{CO}_3$ through aldol condensation reaction as

(A) The compound undergoes nucleophilic substitution due to formation of stable carbocation and elimination reaction in which the alkene formed is stabilized due to conjugation with aromatic nucleus.

(D) The compound undergoes nucleophilic substitution reaction

The structure of carbonyl compound P is

The structures for compounds Q and R are

The structure of compound S is

Hex-3-ynal on reduction with sodium borohydride produces an alcohol.

The $-$OH group of alcohol reacts with phosphorus tribromide to form alkyl bromide (compound I).

The alkyl bromide reacts with Mg to form alkyl magnesium bromide (Grignard reagent) which attacks carbon dioxide followed by hydrolysis to form a carboxylic acid (compound J).

It reacts with thionyl chloride (compound K) to form acid chloride.

Acid chloride which on hydrogenation with Pd supported on barium sulphate partially poisoned with quinoline to form compound L. Here, a triple bond is selectively reduced to a double bond which has $cis$ configuration.

Its open structure is

(A)

(B)

(C) CN$^-$ is negatively charged, so it is a nucleophile

$\mathrm{CN^-+AgNO_3\rightarrow AgCN\downarrow + ~NO^-_3}$

Thus involves in cyanohydrin fomration.

(D) $\mathrm{I^-+AgNO_3\rightarrow AgI\downarrow +~NO^-_3}$

We know that hydrogen are of two types :

1. Intermolecular hydrogen bonding

2. Intramolecular hydrogen bonding

In any organic compound, if there is ortho substituent and hydrogen bonding is possible in that compound then always the bonding will be intramolecular hydrogen bonding. If the substituent is at para position, then intermolecular hydrogen bonding exists. Therefore, we can say that o-Hydroxybenzoic acid will show intramolecular benzoic acid and p-hydroxybenzoic acid will show intermolecular hydrogen bonding.

The strength of intermolecular hydrogen bonding is more than intramolecular hydrogen bonding. Therefore we can say that p-hydroxybenzoic acid because of intermolecular hydrogen bonding has a higher boiling point than o-hydroxybenzoic acid. Therefore, statement 1 is false but statement 2 is correct.

Now, if we look at the reason it states that o-hydroxybenzoic acid has intramolecular hydrogen bonding. From the above explanation we can say that this statement is true.

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.