Aldehyde and Ketone

The complete reaction sequence is as follows

The benzaldehyde ( $X$ ) formed can form a 2,4-dinitrophenylhydrazone derivative and reduces ammoniacal silver nitrate solution.

Now,

Here, toluene reacts with $\mathrm{KMnO}_4-\mathrm{KOH} / \Delta, \mathrm{H}_3 \mathrm{O}^{+}$to form benzoic acid $(Y)$. The benzoic acid dissolves in $\mathrm{NaHCO}_3$, with the evolution of $\mathrm{CO}_2$.

Thus, $A$ and $B$ respectively are $\mathrm{CrO}_3+\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}, \mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{KMnO}_4-\mathrm{KOH} / \Delta, \mathrm{H}_3 \mathrm{O}^{+}$.

The complete reaction sequence is as follows

Benzene reacts with CO and HCl in presence of $\mathrm{AlCl}_3$ gives a compound $X$ (benzaldehyde).

$X$ can also be obtained by the following given process.

$ \text { Hence, option (c) is correct. } $

Example of Clemmensen reduction is given in option (d) i.e.

$ \text { The complete reaction sequence is as follows, } $

$ \text { The complete reaction sequence is as follows. } $

Thus, statement given in $B$ and $C$ are correct. i.e., $Y$ is the reduction product of $X$ and $Z$ on heating with sodalime gives benzene.

$ \text { The complete sequence is as follows. } $

The complete reaction is as follows,

Hence, A and C are position isomers.

The complete reaction sequence is as follows

The compound $ Z $ is toluene.

Benzaldehyde or its derivatives reacts with a primary amine to form a Schiff's base.

Thus Xand $ Y $ are $ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} $ and Schiff's base respectively.

The complete reaction is as follown

Reaction of conversion of $Y$ to $Z$ is a Cannizzaro reaction. It involves 2 moles of aldehydes. One mole undergo oxidation to carboxylic acid. The other molecules goes reduction to corresponding alcohol.

The complete reaction sequence is as follows.

So, the functional group present in ' $X^{\prime}$

A is correct, but R is not correct. The correct form of R is: Ketones are less electrophilic than aldehyde.

The positive inductive effect of second alkyl radical reinforces that of the firsy one decreasing still further the partial positive charge on carbonyl carbon atom. This reduce the attraction of aton for nucleophilic reagents. Hence, ketones are less electrophilic.

The complete reaction is as folloms

Benzene undergoes Friedel-Craft's acylation reaction to give acetophenone $(X) . X$ is reduced to ethyl benzene in presence of $\mathrm{Zn}-\mathrm{Hg}$ and conc. HCl .

The conversion of $X$ to $Y$ is known as Clemmensen reduction.

The main reactants involved in Ctard reaction are toluene.

Since, ethanal has $\alpha$-carbon, hence it does not undergo Cannizzaro reaction.

Aldehyde are more reactive towards nucleophilic addition reaction as compound to ketone, due to more electrophilic carbon.

$ \text { i.e. } \text { (III) }>\text { (I) }>\text { (IV) }>\text { (II) } $

$ \text { Thus, } B \text { in the given reaction is } $

The complete reaction is as follows:

Among the given reactions only (A) and (D) gives benzaldehyde as product. The complete reactions are as follows:

m-chlorobenzaldehyde on reaction with conc. NaOH at room temperature gives potassium $m$-chlorobenzoate and m-chlorobenzyl alcohol.

Cyanohydrin reaction occurs when an aldehyde gets treated with a HCN forming a cyanohydrin product.

Cyanohydrin is a compound which consist of hydroxy and cyano group on same carbon atom.

Hence, Statement I and Statement II are true.

1, 2-dibromoethane reacts with alcoholic KOH followed by $\mathrm{NaNH}_2$ to give ethyne as product $A$.

$A$ undergoes hydration in presence of $\mathrm{Hg}^{2+} / \mathrm{H}^{+}$to give unsaturated alcohol ${ }^{\prime} B .^{\prime} B$ ' undergoes tautomerism to form ' $C$ as major product.

In Gattermann-Koch reaction, benzene is treated with carbon monoxide ( CO ) in acidic medium in presence of anhy. $\mathrm{AlCl}_3$ to give benzaldehyde.

Sodium benzoate is heated with mixture of sodium hydroxide ( NaOH ) and CaO to give benzene (A). A undergoes Friedel-Crafts alkylation reaction to give toluene (B). B undergoes oxidation with chromyl chloride ( $\mathrm{CrO}_2 \mathrm{Cl}_2$ ) in $\mathrm{CS}_2$ to give benzaldehyde $(\mathrm{C})$. This conversion of (B) to $(\mathrm{C})$ is known as Etard's reaction.

The compound $(Z)$ formed undergoes Wolff-Kishner reduction and it also with yellow precipitate with $\mathrm{I}_2$ and NaOH solution.

Thus, statements given in I and III correct.

$\mathrm{NaBH}_4$ being a mild reducing agent reduces - CHO group to $-\mathrm{CH}_2 \mathrm{OH}$, but does not affect -COOMe

∴ Product formed will be

In this ammonia and its derivatives, $\mathrm{H}_2 \mathrm{~N}-Z$ undergo condensation reaction with carbonyl compound and this reaction is an example of reversible reaction.

(A) Benzaldehyde is formed by oxidation of toluene with $\mathrm{CrO}_2 \mathrm{Cl}_2$ in $\mathrm{CS}_2$ followed by hydrolysis this reaction is known as Etard reaction.

(B) Benzyl alcohol is converted to benzaldehyde by a catalytic action of copper (Cu).

Firstly, primary alcohol is converted to aldehyde $(P)$ using an oxidising agent PCC. The product $P$ then react with alkaline solution of ammonium silver nitrate to form carboxylate anion and silver mirror.

Benzene reacts with acetyl chloride in presence of anhydrous $\mathrm{AlCl}_3$ to give Friedel crafts acylation reaction. This reaction follows electrophilic substitution reaction. The product, $P$ formed is acetophenone.

$P$ on reacting with methyl magnesium bromide ( MeMgBr ) gives alkoxide product which upon hydrolysis gives alcohol product.

$ \text { Molecular formula of } Q \text { is } \mathrm{C}_9 \mathrm{H}_{12} \mathrm{O} \text {. } $

(A) Acid chlorides on treatment with $\mathrm{H}_2$, $\mathrm{Pd}-\mathrm{BaSO}_4$ form aldehyde. This reaction is called "Rosenmund reaction".

(B) Benzene can be converted to benzaldehyde by reacting it with $\mathrm{CO}, \mathrm{HCl}$, anhydrous $\mathrm{AlCl}_3$. This reaction is known as "Gatterman-Koch reaction".

(C) Alkene (acetylene) can be converted to aldehyde (acetaldehyde) by reacting it with $\mathrm{HgSO}_4$ and $\mathrm{H}_2 \mathrm{SO}_4$. This reaction is called "oxymercuration-demercuration reaction".

$ \text { (D) DIBAL-H reagent is used to convert esters into aldehydes. } $

Positive iodoform test is given by the compounds which contain the following unit

Out of given compounds, following compounds contain these units

Hence, compound II, IV and VI gives positive iodoform test.

$ \text { Molecular formula } \mathrm{C}_6 \mathrm{H}_{10} \mathrm{O} \text {. } $

Dialkyl cadmium reacts acid chloride to form ketone ( Cl is replaced by $R$ ).

Carbonyl compounds react with hydrazine to form hydrazone. When this hydrazone is heated with KOH and glycol, nitrogen is removed and corresponding alkane or cycloalkane is formed.

(i) Reaction of HBr with propene in the presence of peroxide gives $n$-propyl bromide. This addition reaction is an example of an anti-Markownikoff's addition reaction.

$ \begin{aligned} & R-\mathrm{O}-\mathrm{O}-\mathrm{R} \xrightarrow{\Delta} 2 \mathrm{R}-\mathrm{O}^* \\ & 2 \mathrm{R}-\mathrm{O}^*+\mathrm{H}-\mathrm{Br} \longrightarrow \mathrm{R}-\mathrm{OH}+\mathrm{Br}^* \end{aligned} $

(ii) Reaction of $n$-propyl bromide with Mg gives propyl magnesium bromide (Grignard reagent).

(iii) Reaction of propyl magnesium bromide with propanal gives nucleophilic additon reaction.

(iv) Reaction of hexan-3-ol with oxidising agent $\mathrm{CrO}_3$.

(A) Reaction of benzoyl chloride with benzene and aluminium chloride gives benzophenone.

(B) Reaction of benzoyl chloride with phenyl magnesium bromide (excess).

(C) Reaction of benzoyl chloride with diphenyl cadmium gives benzophenone.

Here, $(B)$ is $\mathrm{CH}_3 \mathrm{CHO}$ and $(C)$ is $\mathrm{CH}_3 \mathrm{CH}==\mathrm{CHCH}=\mathrm{NOH}$.

(i) On ozonolysis double bond get broken to give $\mathrm{CH}_3 \mathrm{CHO}$.

(ii) On reaction with NaOH followed by hydroxyl amine, we get $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CHCH}=\mathrm{NOH}$

Hence, option (d) is the correct answer.