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.

$ \text { The complete reaction, } $

Alcohol $/ \mathrm{KOH}, \Delta \mathrm{NaNH}_2 \downarrow$ Double elimination

The complete reaction sequence is as follows

$ \text { The complete reaction occurs as follows } $

Please note here, in question product of ' $X$ ' is mentioned, but it should be $X$.

The complete reaction is as follows

The electrophilicity of carbonyl carbon is increased by electron withdrawing group while it is decreased by electron donating group. Thus the correct decreasing order of electrophilicity of carbonyl carbon is

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

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

The mention reaction is Wolff-Kishner reduction reaction. The complete reaction is,

The correct match is A-IV, B-I, C-III, D-II

$ \text { The complete reaction sequence is as follow } $

$ \text { So, the number of } s p^3 \text { hybridised carbon in product } Y \text { is } 5 \text {. } $

Etard Reaction : is a method to oxidise toluene into benzaldehyde

$ \text { Complete reaction is as follows, } $

When $2^{\circ}$ alcohol is oxidised with $\mathrm{CrO}_{3^{\prime}}$ Ketone is obtained

$ \mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{C}_2 \mathrm{H}_5 \xrightarrow{\mathrm{CrO}_3} \mathrm{CH}_3 \mathrm{COC}_2 \mathrm{H}_5 $

The carbonyl compound $(X) \left(\mathrm{C}_8 \mathrm{H}_8 \mathrm{O}\right)$ giving yellow ppt. with NaOI indicates the presence of a methyl ketone. Also, since $X$ is a carbonyl compound not an alcohol thus $X$ must be a methyl ketone.

Now, the above two conditions specifies that $X$ is acetophenone.

Thus, hemiacetal of $X$ with methanol/dry HCl is

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

Among the given options only $\mathrm{KMnO}_4 / \mathrm{OH}^{-}, \mathrm{H}^{+}$and $\mathrm{CrO}_2 \mathrm{Cl}_2 / \mathrm{Cs}_2, \mathrm{H}_3 \mathrm{O}^{+}$ can convert toluene into benzaldehyde.

Let's analyze the reactions step by step:

For reagent A, $\mathrm{C_6H_5NHNH_2}$ (phenylhydrazine):

When a ketone reacts with phenylhydrazine, it forms a phenyl hydrazone.

From List II, "Phenyl hydrazone" is option iv.

For reagent B, $\mathrm{NH_2OH}$ (hydroxylamine):

Reaction with a ketone gives an oxime.

From List II, "Oxime" is option iii.

For reagent C, $\mathrm{C_6H_5NH_2}$ (aniline):

When a ketone reacts with a primary amine like aniline, a Schiff base (imine) is formed.

From List II, "Schiff base" is option i.

Thus, the correct matching is:

A → iv (Phenyl hydrazone)

B → iii (Oxime)

C → i (Schiff base)

So, the correct answer is: Option A, A-IV, B-III, C-I.

$ \text { The reactions involved are as follows } $

$ \text { Thus, the IUPAC name of the compound is but-3-en-2-one. } $

The incorrect statement about $B$ i.e. benzaldehyde is given in option (b) as it (an aromatic aldehyde) does not give test with Fehling's solution.

The complete reaction sequence is as follows

The reaction sequence according to question is,

So, the oxime of compound $X$ i.e. propanol is

The complete reaction is as follows.

$ X=\mathrm{NH}_2 \mathrm{OH} \text { and } Y=\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O} $

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.