Aldehyde and Ketone

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.

Friedel-Crafts acylation of toluene gives p-acylated product.

Product A in presence of $\mathrm{KMnO}_4 / \mathrm{KOH}$ gives benzoic acid derivative. Here, methyl group is oxidised to $-$COOH group.

Product B in presence of $\mathrm{H}_2 \mathrm{SO}_4$ get hydrolysed to form $-$COOH group at para-position as well.

In nucleophilic addition reaction, nucleophile attacks at the electrophilic carbonyl centre. The order of electrophilicity is

So, the increasing order of their reactivity for nucleophilic addition reaction is

i.e. A < B < C < D

LiAlH$_4$ (Lithium aluminium hydride) reduces acetic acid to ethyl alcohol which further oxidises in presence of Cu to ethanal and undergoes aldol condensation.

When CH$_3$COCl react with (CH$_3$)$_2$Cd, it will form acetone as a product (P). On further step, acetone (P) react with acetaldehyde to give aldol condensation reaction and formed $\alpha,\beta$-unsaturated alkene product.

Hence, total products in final step are 4.

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.