Properties Questions in English

(A) The reaction of acetaldehyde $(CH_{3}CHO)$ with ammonia derivatives in the presence of $H^{+}$ follows the general reaction: $CH_{3}CHO + H_{2}N-Z \rightarrow CH_{3}CH=N-Z + H_{2}O$.
$A$. Ammonia $(NH_{3})$ reacts to form $CH_{3}CH=NH$ (Acetaldehyde imine) $\rightarrow A-iii-o$.
$B$. Methylamine $(CH_{3}NH_{2})$ reacts to form $CH_{3}CH=NCH_{3}$ (Acetaldehyde methyl imine) $\rightarrow B-i-q$.
$C$. Hydroxylamine $(NH_{2}OH)$ reacts to form $CH_{3}CH=NOH$ (Acetaldehyde oxime) $\rightarrow C-iv-n$.
$D$. Hydrazine $(NH_{2}NH_{2})$ reacts to form $CH_{3}CH=NNH_{2}$ (Acetaldehyde hydrazone) $\rightarrow D-ii-m$.
Correct matching: $A-iii-o, B-i-q, C-iv-n, D-ii-m$.

(A-II, B-III, C-IV, D-I) $HCHO$ reacts with $\text{Cyclohexyl-MgBr}$ to form a primary alcohol, $\text{Cyclohexylmethanol}$ $(ii)$.
$(B)$ $NaBH_4$ selectively reduces the ketone group to an alcohol, yielding $\text{2-(2-hydroxycyclohexyl)acetate}$ $(iii)$.
$(C)$ $HCHO$ reacts with $CH_3-CH_2-CH(CH_3)-CH_2-MgBr$ to form a primary alcohol, $\text{3-methylpentan-1-ol}$ $(iv)$.
$(D)$ $\text{Benzene}$ reacts with $\text{Oleum}$ $(H_2SO_4 + SO_3)$ to undergo sulfonation, forming $\text{Benzenesulfonic acid}$ $(i)$.
Thus, the correct matching is: $(A-ii, B-iii, C-iv, D-i)$.

(A) The reaction of Grignard reagents $(RMgX)$ with acetaldehyde $(CH_3CHO)$ followed by acidic hydrolysis $(H_3O^+)$ yields secondary alcohols.
$(i) CH_3CHO + C_6H_5MgBr$ $\rightarrow CH_3CH(OMgBr)C_6H_5$ $\xrightarrow{H_3O^+} CH_3CH(OH)C_6H_5$ $(1-\text{phenylethanol})$
$(ii) CH_3CHO + C_6H_{11}MgBr$ $\rightarrow CH_3CH(OMgBr)C_6H_{11}$ $\xrightarrow{H_3O^+} CH_3CH(OH)C_6H_{11}$ $(1-\text{cyclohexylethanol})$
$(iii) CH_3CHO + (CH_3)_2CHMgBr$ $\rightarrow CH_3CH(OMgBr)CH(CH_3)_2$ $\xrightarrow{H_3O^+} CH_3CH(OH)CH(CH_3)_2$ $(3-\text{methylbutan-2-ol})$
Thus,the correct products are $1-\text{phenylethanol}$,$1-\text{cyclohexylethanol}$,and $3-\text{methylbutan-2-ol}$.

(C) is propanal $(CH_{3}CH_{2}CHO)$. Reaction with $CH_{3}MgBr$ gives butan$-2-$ol $(CH_{3}CH_{2}CH(OH)CH_{3})$ as $C$.
Butan$-2-$ol is a $2^{\circ}$ alcohol,gives positive Iodoform test,and Lucas test turbidity in $5$ minutes.
$B$ is acetone $(CH_{3}COCH_{3})$. Reaction with $CH_{3}MgBr$ gives $2$-methylpropan$-2-$ol $(CH_{3}C(CH_{3})(OH)CH_{3})$ as $D$.
$2$-methylpropan$-2-$ol is a $3^{\circ}$ alcohol,gives immediate turbidity with Lucas reagent,and negative Iodoform test.
Thus,$C$ is $CH_{3}CH_{2}CH(OH)CH_{3}$ and $D$ is $CH_{3}C(CH_{3})(OH)CH_{3}$.

(B) $1$. The reaction of $2$-methyl-$2H$-chromene with excess $HBr$ under heating leads to the cleavage of the ether ring. The $C-O$ bond breaks,and $HBr$ adds across the double bond and the ring-opened position,resulting in a phenolic compound with a dibromoalkyl side chain: $2-(3,5-dibromopentyl)phenol$ (or similar depending on regioselectivity,but the provided solution image shows the specific transformation).
$2$. Treatment with alcoholic $KOH$ followed by $H^+$ causes dehydrohalogenation,forming a conjugated diene side chain attached to the phenol ring: $2-(penta-1,3-dienyl)phenol$.
$3$. Finally,reductive ozonolysis $(O_3, Zn/H_2O)$ cleaves the double bonds in the side chain. The terminal double bond yields formaldehyde $(HCHO)$,the internal double bond yields glyoxal $(CHO-CHO)$,and the side chain attached to the benzene ring yields salicylaldehyde $(2-hydroxybenzaldehyde)$.
$4$. Among the options,the major aromatic product $C$ is salicylaldehyde.

(A) The reactivity of carbonyl compounds towards nucleophilic addition of $HCN$ depends on the electrophilicity of the carbonyl carbon and steric hindrance. Electron-withdrawing groups ($-I$ or $-R$ effect) increase the electrophilicity of the carbonyl carbon,thereby increasing reactivity. Electron-donating groups ($+R$ or $+I$ effect) decrease the electrophilicity,thereby decreasing reactivity.
$(i)$ $m$-Methoxybenzaldehyde: $-OCH_3$ group shows $-I$ effect (electron-withdrawing) from the meta position.
(ii) $o$-Nitrobenzaldehyde: $-NO_2$ group shows both $-I$ and $-R$ effects (strong electron-withdrawing).
(iii) $o$-Methoxybenzaldehyde: $-OCH_3$ group shows $+R$ effect (strong electron-donating) from the ortho position.
(iv) $m$-Nitrobenzaldehyde: $-NO_2$ group shows only $-I$ effect from the meta position.
Comparing the effects:
- (iii) has a strong electron-donating group $(+R)$,making it the least reactive.
- $(i)$ has a weak electron-withdrawing group $(-I)$.
- (iv) has a stronger electron-withdrawing group ($-I$ only).
- (ii) has the strongest electron-withdrawing effect ($-I$ and $-R$),making it the most reactive.
Thus,the increasing order of reactivity is: $(iii) < (i) < (iv) < (ii)$.

(A) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: Smaller groups around the carbonyl carbon increase reactivity. Aldehydes are more reactive than ketones.
$2$. Electronic effects: Electron-donating groups decrease the electrophilicity of the carbonyl carbon,thus decreasing reactivity.
Comparing the compounds:
- $Propanal$ $(CH_3CH_2CHO)$: Most reactive (Aldehyde,least steric hindrance).
- $Benzaldehyde$ $(C_6H_5CHO)$: Less reactive than $Propanal$ due to resonance stabilization of the carbonyl group by the phenyl ring.
- $Propanone$ $(CH_3COCH_3)$: Ketone,less reactive than aldehydes due to steric hindrance and $+I$ effect of two methyl groups.
- $Butanone$ $(CH_3COCH_2CH_3)$: Least reactive due to greater steric hindrance compared to $Propanone$.
Thus,the increasing order of reactivity is: $Butanone < Propanone < Benzaldehyde < Propanal$.

(A) The product formed from $D$ by the action of $Ba(OH)_2$ and $\Delta$ is mesityl oxide $(CH_3-C(CH_3)=CH-CO-CH_3)$,which is a self-aldol condensation product of acetone. Therefore,$D$ is acetone $(CH_3-CO-CH_3)$.
Since ozonolysis of $B$ gives $C$ and $D$ (acetone),$B$ must contain the $(CH_3)_2C=$ group.
When $A$ $(C_6H_5-CH_2-CO-CH_3)$ reacts with $CH_3MgBr$ followed by dehydration with $Conc. H_2SO_4$,it forms $B$ $(C_6H_5-CH=C(CH_3)_2)$.
$C_6H_5-CH_2-CO-CH_3$ $\xrightarrow{CH_3MgBr} C_6H_5-CH_2-C(OH)(CH_3)_2$ $\xrightarrow{H^{+}, \Delta} C_6H_5-CH=C(CH_3)_2 (B)$.
Ozonolysis of $B$: $C_6H_5-CH=C(CH_3)_2 \xrightarrow{O_3/Zn, H_2O} C_6H_5-CHO (C) + CH_3-CO-CH_3 (D)$.

(B) The reaction involves the acid-catalyzed dehydration of a tertiary alcohol.
$1$. Protonation of the hydroxyl group by $H_2SO_4$ forms a good leaving group $(-OH_2^+)$.
$2$. Loss of water molecule generates a stable carbocation at the tertiary carbon.
$3$. Elimination of a proton from the adjacent carbon (alpha to the carbonyl group) occurs to form the most stable conjugated alkene.
$4$. The double bond forms between the carbon bearing the ethyl group and the adjacent carbon to form a conjugated system with the carbonyl group,resulting in $3-$ethylcyclohex$-2-$en$-1-$one.

(D) The reaction of acetone $(CH_3COCH_3)$ with methylmagnesium chloride $(CH_3MgCl)$,which is a Grignard reagent,involves the nucleophilic attack of the methyl group on the carbonyl carbon of acetone.
This forms an intermediate magnesium alkoxide: $(CH_3)_3C-OMgCl$.
Subsequent hydrolysis of this intermediate with water $(H_2O)$ yields $2$-methylpropan-$2$-ol,commonly known as tert-butyl alcohol $((CH_3)_3COH)$.

(A) The reaction between two different carbonyl compounds (benzaldehyde and acetophenone),where at least one of them has an $\alpha$-hydrogen atom (acetophenone has three $\alpha$-hydrogens),in the presence of a dilute base like $NaOH$ is called $Cross$ $Aldol$ $condensation$.
Benzaldehyde $(C_6H_5CHO)$ does not have an $\alpha$-hydrogen,while acetophenone $(C_6H_5COCH_3)$ has $\alpha$-hydrogens.
This specific reaction leads to the formation of a $\beta$-hydroxy ketone,which upon dehydration gives an $\alpha,\beta$-unsaturated ketone.

(A) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone.
In $CH_3-C(CH_3)_2-CHO$ ($2,2$-dimethylpropanal),the $\alpha$-carbon atom is bonded to three methyl groups and has no hydrogen atoms attached to it.
Therefore,it does not undergo aldol condensation.

(D) The given reaction is an intramolecular Aldol condensation reaction.
$1$. The base $(OH^-)$ abstracts an $\alpha$-hydrogen from the methyl group of the ketone,forming an enolate ion.
$2$. This enolate ion performs a nucleophilic attack on the carbonyl carbon of the aldehyde group.
$3$. This leads to the formation of a cyclic $\beta$-hydroxy ketone intermediate.
$4$. Upon heating $(\Delta)$,the intermediate undergoes dehydration (loss of $H_2O$) to form an $\alpha, \beta$-unsaturated carbonyl compound.
Comparing the final structure with the given options,the product corresponds to option $D$.

(B) Step $1$: Cyclohexanol is oxidized by $PCC$ (Pyridinium chlorochromate) to form cyclohexanone,which is product $(A)$.
Step $2$: Cyclohexanone undergoes self-aldol condensation in the presence of dilute $NaOH$ and heat $(\Delta)$ to form the $\alpha,\beta$-unsaturated ketone,which is cyclohexylidenecyclohexanone,product $(B)$.

(B) The first compound is a cyclic acetal ($2$-methoxytetrahydrofuran),which is stable in basic conditions and does not react with Tollen's reagent.
The second compound is a cyclic hemiacetal ($2$-hydroxytetrahydrofuran),which exists in equilibrium with its open-chain hydroxy-aldehyde form in aqueous solution.
The open-chain hydroxy-aldehyde contains a free aldehyde group $(-CHO)$,which gives a positive Tollen's test (formation of silver mirror).
Therefore,Tollen's reagent can be used to differentiate between the two.

(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in $5-$methylenecyclohex$-2-$enone.
First,$HBr$ adds to the exocyclic double bond. Protonation occurs to form a stable $3^{\circ}$ carbocation at the $5-$position,which is then attacked by the bromide ion to give a $5-$bromo$-5-$methylcyclohex$-2-$enone intermediate.
Second,$HBr$ adds to the remaining endocyclic double bond. According to Markovnikov's rule,the proton adds to the carbon with more hydrogens,and the bromide adds to the carbon that forms a more stable carbocation (stabilized by the inductive effect of the carbonyl group).
The final product is $3-$bromo$-5-$bromo$-5-$methylcyclohexanone.

(A) In a Cross-Cannizzaro reaction between formaldehyde $(HCHO)$ and acetaldehyde $(CH_3CHO)$,formaldehyde is more reactive towards nucleophilic attack by $OH^-$ because it lacks electron-donating alkyl groups.
Consequently,formaldehyde is oxidized to sodium formate $(HCO_2Na)$,and acetaldehyde is reduced to ethanol $(CH_3CH_2OH)$.
Reaction: $HCHO + CH_3CHO \xrightarrow{Conc. NaOH} HCO_2Na + CH_3CH_2OH$

(A) Nucleophilic addition reactivity depends on the electrophilicity of the carbonyl carbon.
$1.$ Aldehydes are more reactive than ketones due to less steric hindrance and lower electronic repulsion.
$2.$ Electron-donating groups $(EDG)$ decrease the electrophilicity of the carbonyl carbon,thereby decreasing reactivity.
$3.$ The order of electron-donating strength is: $CH_{3}O$ (strong $+M$) $> CH_{3}$ (weak $+I$) $> H$ (no effect).
$4.$ Comparing the compounds:
$(1)$ $C_{6}H_{5}COC_{6}H_{5}$ (Ketone,least reactive)
$(2)$ $C_{6}H_{5}CHO$ (Benzaldehyde,most reactive among these)
$(3)$ $p-CH_{3}C_{6}H_{4}CHO$ (Aldehyde with $+I$ group)
$(4)$ $p-CH_{3}OC_{6}H_{4}CHO$ (Aldehyde with $+M$ group)
Thus,the reactivity order is: $2 > 3 > 4 > 1$.

(C) The reaction of an aldehyde or ketone with $CH_{3}MgBr$ (Grignard reagent) followed by hydrolysis yields an alcohol.
For the resulting alcohol to give a positive iodoform test,it must contain a $CH_{3}CH(OH)-$ group.
Starting with $C_{7}H_{10}O$,if $A$ is cyclohex$-3-$ene$-1-$carbaldehyde,the reaction is:
$C_{6}H_{9}CHO + CH_{3}MgBr \rightarrow C_{6}H_{9}CH(OMgBr)CH_{3}$
Upon hydrolysis,this gives $C_{6}H_{9}CH(OH)CH_{3}$,which is a secondary alcohol with a $CH_{3}CH(OH)-$ group.
This structure gives a positive iodoform test with $I_{2}/NaOH$.
Thus,the structure of $A$ is cyclohex$-3-$ene$-1-$carbaldehyde.

(B) $DIBAL-H$ (Diisobutylaluminium hydride) acts as a selective reducing agent. It is capable of reducing both esters $(-COOR)$ and nitriles $(-CN)$ to aldehydes $(-CHO)$ at low temperatures. In the given substrate,both the ester group $(-COOCH_3)$ and the nitrile group $(-CN)$ are reduced to aldehyde groups $(-CHO)$. Therefore,the product is a cyclohexane$-1,4-$dicarbaldehyde. Thus,option $(B)$ is the correct answer.

(D) $NaBH_4$ (Sodium borohydride) is a selective reducing agent that specifically reduces aldehydes and ketones to their corresponding alcohols.
It does not reduce esters or carbon-carbon double bonds $(C=C)$ under standard conditions.
In the given substrate,there is a ketone group on the cyclohexene ring and an ester group $(-COOCH_3)$ on the side chain.
Therefore,$NaBH_4$ will selectively reduce the ketone to a secondary alcohol $(-OH)$ while leaving the ester and the double bond intact.
The product $X$ is $2\text{-hydroxycyclohex-3-en-1-yl} \text{ acetate derivative}$ (specifically,the structure shown in option $D$).

(C) $DIBAL-H$ (Diisobutylaluminum hydride) is a selective reducing agent that reduces esters to aldehydes at low temperatures $(-78^{\circ}C)$.
In the given reaction,$DIBAL-H$ reduces the cyclic ester (lactone) to a lactol,which then opens to form a hydroxy-aldehyde.
$DIBAL-H$ does not reduce the carbon-carbon double bond $(C=C)$ present in the molecule.
Therefore,the reaction converts the lactone group into a hydroxy-aldehyde group while keeping the rest of the structure intact,which matches option $C$.

(B) The reaction of a ketone with ethylene glycol in the presence of an acid catalyst $(H^+)$ is a standard method for the protection of the carbonyl group as a cyclic acetal.
In the given molecule,ethyl acetoacetate,there are two carbonyl groups: a ketone and an ester.
Ketones are more reactive towards nucleophilic addition than esters.
Therefore,the ethylene glycol selectively reacts with the ketone group to form a cyclic acetal,while the ester group remains unaffected.
The product $A$ is the cyclic acetal of ethyl acetoacetate.

(D) The reaction is an Aldol condensation of cyclopentanone.
$1$. In the presence of $dil. NaOH$,two molecules of cyclopentanone undergo self-aldol condensation to form a $\beta$-hydroxy ketone,which is $X$ ($2$-($1$-hydroxycyclopentyl)cyclopentanone).
$2$. Upon heating with $H^+$,the $\beta$-hydroxy ketone undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone,which is $Y$ ($2$-cyclopentylidenecyclopentanone).

(B) The oxidation of an aldehyde to a carboxylate ion by Tollen's reagent is represented by the following balanced chemical equation:
$RCHO + 2[Ag(NH_{3})_{2}]^{+} + 3OH^{-} \rightarrow RCOO^{-} + 2Ag + 4NH_{3} + 2H_{2}O$
In this reaction,the aldehyde group $(-CHO)$ is oxidized to a carboxylate group $(-COO^{-})$,which involves the loss of $2$ electrons.
These $2$ electrons are transferred to the two silver ions $(Ag^{+})$ present in the Tollen's reagent,reducing them to metallic silver $(Ag^{0})$.
Therefore,the total number of electrons transferred to the Tollen's reagent per aldehyde group is $2$.

(D) The reaction proceeds in two steps:
$1$. $3 CH \equiv CH \xrightarrow{\text{Red hot Fe tube, } 873 \text{ K}} C_6H_6$ (Benzene).
$2$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$ (Benzaldehyde) via Gattermann-Koch reaction.
In benzaldehyde $(C_6H_5CHO)$,there are $6$ carbon atoms in the benzene ring,all of which are $sp^{2}$ hybridized.
The carbonyl carbon atom $(C=O)$ is also $sp^{2}$ hybridized.
Therefore,the total number of $sp^{2}$ hybridized carbon atoms is $6 + 1 = 7$.

(B) The reaction involves the reduction of a carbonyl group $(C=O)$ to a methylene group $(-CH_{2}-)$ in the presence of a carbon-carbon double bond $(C=C)$.
Wolff-Kishner reduction,which uses hydrazine $(NH_{2}NH_{2})$ followed by a strong base like sodium ethoxide $(C_{2}H_{5}O^{-}Na^{+})$ and heat,is specifically used to reduce carbonyl groups to alkanes without affecting the $C=C$ double bond.
$NaBH_{4}$ reduces carbonyls but is typically used for alcohols,and $Ni/H_{2}$ would reduce both the carbonyl and the $C=C$ double bond.

(B) $2,4-DNP$ stands for $2,4-dinitrophenylhydrazine$.
This reagent is used to identify carbonyl compounds,which include both aldehydes and ketones.
Among the given options,aldehyde is a carbonyl compound.

(A) The given reaction is an intramolecular aldol condensation of $o$-phenylenediacetaldehyde.
$1$. The base $(OH^-)$ abstracts an $\alpha$-hydrogen from one of the aldehyde groups to form an enolate ion.
$2$. This enolate ion performs a nucleophilic attack on the carbonyl carbon of the other aldehyde group,leading to the formation of a cyclic $\beta$-hydroxy aldehyde.
$3$. Upon heating in the presence of an acid or base (dehydration),the $\beta$-hydroxy aldehyde undergoes elimination of a water molecule to form an $\alpha,\beta$-unsaturated aldehyde.
$4$. The final product is $2$-formyl$-1,2,3,4-$tetrahydronaphthalene.

(A) The given reaction is the haloform reaction. Acetophenone $(C_6H_5COCH_3)$ reacts with bromine $(Br_2)$ in the presence of a base $(KOH)$.
This reaction converts the methyl ketone group $(-COCH_3)$ into a carboxylate salt $(-COOK)$ and a haloform $(CHBr_3)$.
The reaction is: $C_6H_5COCH_3 + 3Br_2 + 4KOH \rightarrow C_6H_5COOK + CHBr_3 + 3KBr + 3H_2O$.
Thus,the major products are $A = C_6H_5COOK$ and $B = CHBr_3$.

(C) The reaction involves the nucleophilic addition of cyanide to an aldehyde,catalyzed by a base like $NaOH$.
$1$. The first step is the formation of a cyanohydrin: $R-CHO + HCN \xrightarrow{NaOH} R-CH(OH)CN$.
$2$. The second step is the reduction of the nitrile group $(-CN)$ to a primary amine $(-CH_2NH_2)$ using a strong reducing agent like $LiAlH_4$.
$3$. Thus,$X$ is $NaOH$ and the final product $Y$ is $CH_3CH_2CH(CH_3)CH(OH)CH_2NH_2$.

(D) The reaction involves the bromination of valerophenone ($1$-phenylpentan$-1-$one) using $Br_2$ in the presence of $AlBr_3$ and diethyl ether $(Et_2O)$.
$AlBr_3$ is a Lewis acid catalyst used for electrophilic aromatic substitution.
The carbonyl group $(-CO-R)$ attached to the benzene ring is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
This makes the benzene ring deactivated and meta-directing for electrophilic substitution reactions.
Therefore,the electrophile $Br^+$ attacks the meta-position of the benzene ring to form $1-$($3$-bromophenyl)pentan$-1-$one as the major product.

(D) Step $1$: $CH_3-C \equiv CH + NaNH_2 \rightarrow CH_3-C \equiv C^- Na^+ (A) + NH_3$.
Step $2$: The nucleophile $(A)$ reacts with $4$-bromopentan-$2$-ol to form $B$ $(CH_3-C \equiv C-CH_2-CH_2-CH(OH)-CH_3)$.
Step $3$: Hydrogenation with $H_2/Pd-C$ reduces the alkyne to an alkane,yielding $C$ $(CH_3-CH_2-CH_2-CH_2-CH_2-CH(OH)-CH_3)$.
Step $4$: Oxidation with $CrO_3$ converts the secondary alcohol into a ketone,yielding $D$ $(CH_3-CH_2-CH_2-CH_2-CH_2-CO-CH_3)$.

(A) The given reaction is a haloform reaction,which is characteristic of methyl ketones $(R-CO-CH_3)$.
In the presence of a hypohalite reagent like $NaOCl$ followed by acidic workup $(H_3O^+)$,a methyl ketone is oxidized to a carboxylic acid and chloroform $(CHCl_3)$ is produced as a byproduct.
The product shown in the reaction is crotonic acid $(CH_3-CH=CH-COOH)$.
Therefore,the starting compound $P$ must be the corresponding methyl ketone,which is $CH_3-CH=CH-COCH_3$ (pent$-3-$en$-2-$one).

(D) The reaction involves the treatment of a compound containing both a nitrile $(-CN)$ group and a ketone $(C=O)$ group with $2$ equivalents of $CH_3MgBr$.
$1$. The $CH_3MgBr$ acts as a nucleophile and attacks both the nitrile carbon and the ketone carbonyl carbon.
$2$. Upon acidic workup $(H_3O^+)$,the intermediate imine is hydrolyzed to a ketone,and the alkoxide is protonated to an alcohol.
$3$. The final step involves dehydration using $H_2SO_4$ and heat,which removes the hydroxyl group and a hydrogen atom from the adjacent carbon to form a double bond.
$4$. The final product is a ketone with a substituted double bond,as shown in the solution image.

(A) Step $1$: $CH_3COOH + SOCl_2 \longrightarrow CH_3COCl (A) + SO_2 + HCl$.
Step $2$: $CH_3COCl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5COCH_3 (B) + HCl$. This is a Friedel-Crafts acylation reaction.
Step $3$: $C_6H_5COCH_3 + KCN \xrightarrow{H^+} C_6H_5C(OH)(CH_3)CN (C)$. This is a nucleophilic addition reaction of $CN^-$ to the carbonyl group of acetophenone,followed by protonation to form a cyanohydrin.

(C) The given reaction is an intramolecular Cannizzaro reaction because the substrate $4-formylbenzyl alcohol$ contains both an aldehyde group $(-CHO)$ and a hydroxymethyl group $(-CH_2OH)$,but no $\alpha$-hydrogen atoms.
In the presence of concentrated $NaOH$ and heat,the aldehyde group undergoes disproportionation (oxidation and reduction).
The oxidation of the $-CHO$ group leads to the formation of a carboxylic acid group $(-COOH)$,while the reduction of the $-CHO$ group leads to the formation of a primary alcohol group $(-CH_2OH)$.
The final products are $1,4-benzenedimethanol$ (a diol) and $4-(hydroxymethyl)benzoic acid$ (a compound with both alcohol and acid functional groups).
Comparing these products with the options:
$A$. Compound with both alcohol and acid functional groups: Formed $(4-(hydroxymethyl)benzoic acid)$.
$B$. Monocarboxylic acid: Formed ($4-(hydroxymethyl)benzoic acid$ is a monocarboxylic acid).
$C$. Dicarboxylic acid: Not formed.
$D$. Diol: Formed ($1,4-benzenedimethanol$ is a diol).
Therefore,the compound that is not formed is a dicarboxylic acid.

(B) Statement $I$ is true. The nucleophilic addition of sodium hydrogen sulphite $(NaHSO_3)$ to an aldehyde or a ketone involves the attack of the sulphite nucleophile on the carbonyl carbon,followed by a proton transfer to the oxygen atom to form a stable bisulphite addition product (a white crystalline solid).
Statement $II$ is false. The nucleophilic addition of hydrogen cyanide $(HCN)$ to an aldehyde or a ketone yields a cyanohydrin,not an amine. The reaction involves the addition of the cyanide ion $(CN^-)$ to the carbonyl carbon followed by protonation of the oxygen.
Therefore,Statement $I$ is true but Statement $II$ is false.

(A) Tollen's reagent is used to detect the presence of aldehydes.
$I$ is an aldehyde (isobutyraldehyde),which gives a positive Tollen's test.
$II$ is an enol that tautomerizes to acetone (a ketone),which does not give a positive Tollen's test.
$III$ is an enol that tautomerizes to an aldehyde (cyclohexanecarbaldehyde),which gives a positive Tollen's test.
$IV$ is a hemiacetal,which exists in equilibrium with its open-chain aldehyde form in the presence of base,thus giving a positive Tollen's test.
Therefore,only compound $II$ does not form a silver mirror.

(C) The reaction of acetone with semicarbazide produces acetone semicarbazone and water.
The chemical reaction is: $(CH_3)_2C=O + H_2N-NH-CONH_2 \rightarrow (CH_3)_2C=N-NH-CONH_2 + H_2O$.
In the structure of acetone semicarbazone,$(CH_3)_2C=N-NH-CONH_2$,we can count the nitrogen atoms:
$1$. One nitrogen atom is in the $C=N$ group.
$2$. One nitrogen atom is in the $-NH-$ group.
$3$. One nitrogen atom is in the $-NH_2$ group.
Therefore,the total number of nitrogen atoms in a molecule of acetone semicarbazone is $3$.

(D) The given reaction is an intramolecular aldol condensation.
$1$. The base $(KOH)$ abstracts an $\alpha$-hydrogen from the ketone group to form an enolate.
$2$. This enolate attacks the carbonyl carbon of the aldehyde group,forming a cyclic $\beta$-hydroxy ketone.
$3$. Subsequent dehydration (catalyzed by $H^+, \Delta$) leads to the formation of an $\alpha,\beta$-unsaturated ketone.
$4$. The resulting product is a spiro-compound where a cyclohexenone ring is fused to a cyclohexane ring.

(D) $1$. The reaction of $5,5$-dimethylcyclopentanone with $CH_3CHO$ in the presence of $NaOH$ is an aldol condensation reaction,which produces a $\beta$-hydroxy ketone intermediate.
$2$. This intermediate contains a $CH_3CH(OH)-$ group,which is susceptible to the iodoform reaction.
$3$. Treatment with $I_2/NaOH$ performs the iodoform reaction,converting the $CH_3CH(OH)-$ group into a carboxylate group $(-COO^-Na^+)$ and producing $CHI_3$ (yellow precipitate).
$4$. The filtrate containing the carboxylate is then acidified with $HCl$ to yield the final product $'X'$,which is $5,5$-dimethyl$-2-$oxocyclopentanecarboxylic acid.

(A) The reaction involves the addition of a Grignard reagent $(C_2H_5MgBr)$ to an $\alpha,\beta$-unsaturated ketone (but$-3-$en$-2-$one).
Grignard reagents typically undergo $1,2$-addition to the carbonyl group of $\alpha,\beta$-unsaturated ketones to form an allylic alcohol.
Step $1$: The ethyl group $(C_2H_5^-)$ from $C_2H_5MgBr$ attacks the carbonyl carbon,forming an alkoxide intermediate.
Step $2$: Subsequent protonation with $H_2O$ yields the allylic alcohol,$3$-methylpent-$1$-en-$3$-ol.
Although $HCl$ is mentioned in the reaction conditions,under standard laboratory conditions for Grignard reactions,the primary product isolated is the alcohol. Among the given options,the structure corresponding to $3$-methylpent-$1$-en-$3$-ol is the correct major product.

(D) The reaction of benzonitrile $(C_{6}H_{5}CN)$ with one equivalent of methylmagnesium bromide $(CH_{3}MgBr)$ followed by acid hydrolysis $(H_{3}O^{+})$ proceeds as follows:
$1$. Nucleophilic attack of $CH_{3}^{-}$ on the carbon of the nitrile group forms an imine magnesium salt intermediate: $C_{6}H_{5}C(CH_{3})=NMgBr$.
$2$. Hydrolysis of this intermediate yields acetophenone $(C_{6}H_{5}COCH_{3})$ and ammonia $(NH_{3})$.
$3$. Acetophenone contains the acetyl group $(CH_{3}CO-)$ attached to a phenyl ring. Compounds containing the $CH_{3}CO-$ group attached to a hydrogen or carbon atom give a positive iodoform test with $I_{2}/NaOH$.

(D) The $2,4-$dinitrophenylhydrazine $(2,4-DNP)$ test is a characteristic reaction used to identify carbonyl compounds,specifically aldehydes and ketones.
$2,4-DNP$ reacts with the carbonyl group $(>C=O)$ of aldehydes and ketones to form a crystalline orange or yellow precipitate of $2,4-$dinitrophenylhydrazone.
Among the given options:
$A$. Ethyl benzoate is an ester.
$B$. Salicylic acid is a carboxylic acid.
$C$. Ethyl salicylate is an ester.
$D$. Acetophenone is a ketone.
Since only acetophenone contains a carbonyl group,it will give a positive $2,4-DNP$ test.

(B) $1$. The first step is the Gattermann-Koch reaction where benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ to form $A$,which is benzaldehyde $(C_6H_5CHO)$.
$2$. The second step is the Claisen-Schmidt condensation reaction between benzaldehyde $(A)$ and a ketone $(B)$ in the presence of a base $(OH^-)$ to form the product,which is chalcone $(C_6H_5-CH=CH-CO-C_6H_5)$.
$3$. For the formation of chalcone,the ketone $B$ must be acetophenone $(C_6H_5COCH_3)$.
$4$. Therefore,the substance $B$ is acetophenone.

(C) The reactions of carbonyl compounds are as follows:
$(a).$ Cyanohydrin is formed by the reaction of carbonyl compounds with $HCN$ $(iv)$.
$(b).$ Acetal is formed by the reaction of aldehydes with alcohol in the presence of dry $HCl$ $(iii)$.
$(c).$ Schiff's base is formed by the reaction of carbonyl compounds with primary amines $(RNH_2)$ $(ii)$.
$(d).$ Oxime is formed by the reaction of carbonyl compounds with hydroxylamine $(NH_2OH)$ $(i)$.
Therefore,the correct matching is $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(D) The boiling point order for compounds of comparable molecular masses is: Alcohols > Aldehydes/Ketones > Hydrocarbons.
Statement $I$ is correct: Aldehydes and ketones possess a polar carbonyl group,leading to dipole-dipole interactions,which are stronger than the weak London dispersion forces found in non-polar hydrocarbons.
Statement $II$ is correct: Alcohols exhibit intermolecular $H$-bonding,which is significantly stronger than the dipole-dipole interactions present in aldehydes and ketones,resulting in higher boiling points for alcohols.
Therefore,both statements are correct.

(A) The reaction between acetone $(CH_3COCH_3)$ and $2-$pentanone $(CH_3COCH_2CH_2CH_3)$ in the presence of dilute $NaOH$ and heating is a cross-aldol condensation reaction.
Acetone has two equivalent $\alpha-$hydrogens,while $2-$pentanone has two types of $\alpha-$hydrogens (at $C-1$ and $C-3$).
Self-aldol products of acetone and $2-$pentanone can also form.
The product shown in option $A$ requires a structure that cannot be derived from the condensation of these two specific ketones under the given conditions,as it would require a different carbon skeleton or starting materials.
Therefore,the structure in option $A$ is not formed.