Properties Questions in English

(D) $1$. $CH_3COOH + SOCl_2 \rightarrow CH_3COCl$ (Compound $A$ is acetyl chloride).
$2$. $CH_3COCl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5COCH_3$ (Friedel-Crafts acylation,Compound $B$ is acetophenone).
$3$. $C_6H_5COCH_3 + HCN \rightarrow C_6H_5C(OH)(CH_3)CN$ (Nucleophilic addition,Compound $C$ is acetophenone cyanohydrin).
$4$. $C_6H_5C(OH)(CH_3)CN + H_2O \rightarrow C_6H_5C(OH)(CH_3)COOH$ (Hydrolysis of nitrile to carboxylic acid,Compound $D$ is $2-$hydroxy$-2-$phenylpropanoic acid).

(C) The starting material is $3$-bromobenzoic acid.
$1$. Formation of $X$:
Reaction with $B_2H_6$ followed by $H_3O^+$ is a selective reduction of the carboxylic acid group $(-COOH)$ to a primary alcohol $(-CH_2OH)$ without affecting the bromine $(-Br)$ substituent on the benzene ring. Thus,$X$ is $3$-bromobenzyl alcohol.
$2$. Formation of $Y$:
$(i)$ Esterification: $3$-bromobenzoic acid reacts with $C_2H_5OH/H^+$ to form ethyl $3$-bromobenzoate.
(ii) Reduction: $DIBAL-H$ is a selective reducing agent that reduces esters $(-COOC_2H_5)$ to aldehydes $(-CHO)$.
(iii) Hydrolysis: The final step yields $3$-bromobenzaldehyde as $Y$.
Therefore,$X$ is $3$-bromobenzyl alcohol and $Y$ is $3$-bromobenzaldehyde. The correct option is $C$.

(B) Step $1$: $C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl (A) + SO_2 + HCl$. This is the conversion of benzoic acid to benzoyl chloride.
Step $2$: $C_6H_5COCl + C_6H_6 \xrightarrow{\text{anhy. } AlCl_3} C_6H_5COC_6H_5 (B) + HCl$. This is a Friedel-Crafts acylation reaction producing benzophenone.
Step $3$: $C_6H_5COC_6H_5 \xrightarrow[\text{(ii) } KOH / (CH_2OH)_2]{\text{(i) } NH_2-NH_2} C_6H_5CH_2C_6H_5 (C)$. This is the Wolff-Kishner reduction of benzophenone to diphenylmethane.

(C) Step $1$: Ozonolysis of $2-$methylpropene $(CH_3)_2C=CH_2$ gives acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$.
$(CH_3)_2C=CH_2 \xrightarrow[\text{(2) } Zn/H_2O]{\text{(1) } O_3} CH_3COCH_3 (X) + HCHO (Y)$
Step $2$: The reaction of acetone with formaldehyde in the presence of dilute $NaOH$ and heat is an aldol condensation reaction.
Acetone has $\alpha$-hydrogens,so it acts as the nucleophile,and formaldehyde acts as the electrophile.
The product formed is $4-$hydroxybutan$-2-$one $(CH_3COCH_2CH_2OH)$.
Upon heating $(\Delta)$,it undergoes dehydration to form but$-3-$en$-2-$one $(CH_3COCH=CH_2)$.
However,the question asks for the product '$Z$' formed after the reaction with dilute $NaOH$ and $\Delta$. The aldol product is $4-$hydroxybutan$-2-$one,and the dehydrated product is but$-3-$en$-2-$one.
Given the options,the $IUPAC$ name of the final dehydrated product '$Z$' is but$-3-$en$-2-$one.

(C) The general structure for these derivatives is $C=N-Z$.
For oximes,the derivative is formed by the reaction of a carbonyl compound with hydroxylamine $(NH_2OH)$. Thus,the group $Z$ attached to the nitrogen is $-OH$.
For semicarbazones,the derivative is formed by the reaction of a carbonyl compound with semicarbazide $(NH_2NHCONH_2)$. Thus,the group $Z$ attached to the nitrogen is $-NHCONH_2$.
Therefore,for oxime $(I)$,$Z = -OH$ and for semicarbazone $(II)$,$Z = -NHCONH_2$.

(B) $(I)$ Aldehydes and ketones that are less sterically hindered at the carbonyl carbon show a higher rate of nucleophilic addition reaction.
$(II)$ The presence of electron-donating groups at the carbonyl carbon decreases the electrophilicity of the carbon atom,thereby decreasing the rate of reaction.
$(III)$ Comparing the given compounds: $CH_3 CHO$ is an aldehyde with the least steric hindrance and the least number of electron-donating groups attached to the carbonyl carbon.
$\therefore$ Nucleophilic addition reaction will be most favoured in $CH_3 CHO$.

(C) Alkenes $(X)$ are electron-rich due to the presence of a $\pi$-electron cloud,making them susceptible to attack by electrophiles. Therefore,they undergo electrophilic addition reactions.
Carbonyl compounds $(Y)$ contain a polar $C=O$ bond where the carbon atom is electron-deficient (electrophilic). Consequently,they undergo nucleophilic addition reactions.

(B) Carbonyl compounds (aldehydes and ketones) form solid crystalline addition products with sodium bisulphite $(NaHSO_3)$.
These solid bisulphite addition products can be separated from the reaction mixture by filtration,leaving impurities behind.
The pure carbonyl compound is then regenerated from the solid addition product by treatment with dilute acid or base.
The reaction is as follows:
$CH_3CHO + NaHSO_3 \rightarrow CH_3CH(OH)SO_3Na$ (Solid addition product)
$CH_3CH(OH)SO_3Na + H_2O \rightarrow CH_3CHO + NaHSO_3$ (Pure aldehyde)

(C) The conversion of $3-$hexene to propane involves two main steps:
Step-$I$: Ozonolysis of $3-$hexene $(CH_3CH_2CH=CHCH_2CH_3)$ using $(i) O_3$ followed by $(ii) Zn, H_2O$ yields two molecules of propanal $(CH_3CH_2CHO)$.
Step-$II$: Reduction of propanal $(CH_3CH_2CHO)$ to propane $(CH_3CH_2CH_3)$ is achieved using Clemmensen reduction,which employs $Zn(Hg)$ and $HCl$.

(C) The Clemmensen reduction uses zinc-amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$ to reduce carbonyl groups $(C=O)$ of aldehydes and ketones to methylene groups $(CH_2)$.
Additionally,under these acidic conditions,secondary and tertiary alcohols can undergo substitution reactions with $HCl$ to form alkyl chlorides.
In the given molecule,the ketone group is reduced to an ethyl group $(-CH_2CH_3)$,and the secondary alcohol group is converted into a chloro group $(-Cl)$.
Therefore,the major product is $1$-chloro-$3$-ethyl-$5$-methylcyclohexane.

(A) The reaction conditions $(i) \ N_2H_4, (ii) \ NaOH, \text{Ethylene glycol}, \Delta$ represent the Wolff-Kishner reduction. This reaction specifically reduces a carbonyl group $(C=O)$ to a methylene group $(CH_2)$. In the given substrate,the acetyl group $(-COCH_3)$ is reduced to an ethyl group $(-CH_2CH_3)$,while the phenolic $-OH$ and nitro $-NO_2$ groups remain unaffected.

(B) Let us analyze each reaction:
$(A)$ $CH_3CHO$ is reduced by $LiAlH_4$ to ethanol $(CH_3CH_2OH)$. This is correct.
$(B)$ $CH_3COCH_3$ undergoes Clemmensen reduction with $Zn(Hg)/HCl$ to form propane $(CH_3CH_2CH_3)$. The product given in the option is incorrect as it shows a diol.
$(C)$ $CH_3CHO$ undergoes the iodoform reaction with $I_2/NaOH$ to form iodoform $(CHI_3)$ and sodium formate $(HCOONa)$. This is correct.
$(D)$ $CH_3CH_2CHO$ is oxidized by $KMnO_4$ to propanoic acid $(CH_3CH_2COOH)$. This is correct.
Therefore,the incorrect product is in option $(B)$.

(A) The reaction sequence is as follows:
$1$. Styrene $(C_6H_5CH=CH_2)$ reacts with $HBr$ in the presence of peroxide via anti-Markovnikov addition to form $1-$bromo$-2-$phenylethane ($X$,$C_6H_5CH_2CH_2Br$).
$2$. $X$ reacts with $Mg$ in dry ether to form the Grignard reagent $C_6H_5CH_2CH_2MgBr$. This reacts with $CO_2$ followed by hydrolysis to yield $3-$phenylpropanoic acid ($Y$,$C_6H_5CH_2CH_2COOH$).
$3$. $Y$ reacts with $PCl_5$ to form $3-$phenylpropanoyl chloride $(C_6H_5CH_2CH_2COCl)$.
$4$. $3-$phenylpropanoyl chloride undergoes Rosenmund reduction with $H_2, Pd-BaSO_4$ to form $3-$phenylpropan$-1-$al ($Z$,$C_6H_5CH_2CH_2CHO$).
The structure of $Z$ corresponds to the image provided in option $A$.

(A) The reaction sequence is as follows:
$1$. $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$ (Markovnikov addition).
$2$. $CH_3-CH(Br)-CH_3 + NaOH(aq) \rightarrow CH_3-CH(OH)-CH_3$ (Nucleophilic substitution to form propan$-2-$ol).
$3$. $CH_3-CH(OH)-CH_3 \xrightarrow{Cu/573 \ K} CH_3-CO-CH_3$ (Dehydrogenation of secondary alcohol to acetone).
$4$. $2 CH_3-CO-CH_3 \xrightarrow{Ba(OH)_2, \Delta} (CH_3)_2C=CH-CO-CH_3$ (Aldol condensation of acetone to form mesityl oxide,which is $4-$methylpent$-3-$en$-2-$one).

(C) $1$. $Propene$ $(CH_3-CH=CH_2)$ reacts with $HBr$ to form $2-bromopropane$ $(A)$ $(CH_3-CH(Br)-CH_3)$.
$2$. $2-bromopropane$ reacts with $Mg$ in $Dry \ ether$ to form $isopropylmagnesium \ bromide$ $(B)$ $(CH_3-CH(MgBr)-CH_3)$.
$3$. $Isopropylmagnesium \ bromide$ $(B)$ reacts with $Pentan-3-one$ $(X)$ $(CH_3CH_2-CO-CH_2CH_3)$ followed by acid hydrolysis $(H_3O^+)$ to form $3-ethyl-2-methylpentan-3-ol$ $(Y)$.

(C) $1$. $Propyne$ $(CH_3-C \equiv CH)$ reacts with $Hg^{2 } / H_2O, H $ to undergo hydration to form $Acetone$ $(CH_3-CO-CH_3)$ as $P$.
$2$. $Acetone$ $(P)$ undergoes $Aldol$ condensation in the presence of $Ba(OH)_2$ to form $Diacetone$ $alcohol$ $(Q)$,which is $CH_3-CO-CH_2-C(OH)(CH_3)_2$.
$3$. Upon heating $(\Delta)$,$Diacetone$ $alcohol$ $(Q)$ undergoes dehydration to form $Mesityl$ $oxide$ $(R)$,which is $CH_3-CO-CH=C(CH_3)_2$.

(C) $1$. Decarboxylation of $C_6H_5COONa$ with soda lime $(NaOH/CaO)$ gives $X = C_6H_6$ (Benzene).
$2$. Gattermann-Koch reaction of benzene with $CO/HCl$ in the presence of anhydrous $AlCl_3$ gives $Y = C_6H_5CHO$ (Benzaldehyde).
$3$. The reaction of benzaldehyde with concentrated $NaOH$ is a Cannizzaro reaction.
$4$. In the Cannizzaro reaction,$C_6H_5CHO$ undergoes disproportionation to give $A = C_6H_5CH_2OH$ (Benzyl alcohol,the reduction product) and $B = C_6H_5COONa$ (Sodium benzoate,the oxidation product).

(A) Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic factors. Sterically,the presence of two relatively large substituents in ketones hinders the approach of the nucleophile to the carbonyl carbon more than in aldehydes,which have only one such substituent. Electronically,aldehydes are more reactive because the two alkyl groups in ketones reduce the electrophilicity of the carbonyl carbon more effectively than in aldehydes. Benzophenone $(a)$ is a ketone and is the least reactive. Among the aldehydes,$p$-tolualdehyde $(b)$ is less reactive than benzaldehyde $(c)$ because the $CH_3$ group at the para position increases electron density on the carbonyl carbon through the hyperconjugation effect,making it less electrophilic. In $p$-nitrobenzaldehyde $(d)$,the $NO_2$ group is strongly electron-withdrawing due to both $-I$ and $-R$ effects,which decreases electron density on the carbonyl carbon and significantly increases its reactivity towards nucleophilic attack. Therefore,the increasing order of reactivity is $a < b < c < d$.

(A) The reaction proceeds in two steps:
$1$. Friedel-Crafts acylation of benzene with $CH_3CH_2COCl$ in the presence of anhydrous $AlCl_3$ gives propiophenone $(C_6H_5COCH_2CH_3)$.
$2$. Clemmensen reduction of the ketone using $Zn-Hg$ and concentrated $HCl$ reduces the carbonyl group $(C=O)$ to a methylene group $(-CH_2-)$,yielding propylbenzene $(C_6H_5CH_2CH_2CH_3)$.

(D) The reaction sequence is as follows:
$1$. Chlorobenzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form $p$-chloroacetophenone as the major product $(X)$.
$2$. $X$ ($p$-chloroacetophenone) undergoes Clemmensen reduction using $Zn-Hg$ and $conc. HCl$,which reduces the carbonyl group $(-COCH_3)$ to an ethyl group $(-CH_2CH_3)$.
$3$. Therefore,the final product is $1-chloro-4-ethylbenzene$,and $\underline{Z}$ is the ethyl group,$-CH_2CH_3$.

(C) Step $1$: Benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form acetophenone $(P)$,which is $C_6H_5COCH_3$.
Step $2$: Acetophenone $(P)$ reacts with methyl magnesium bromide $(CH_3MgBr)$ followed by hydrolysis $(H_2O)$ to form $2$-phenylpropan-$2$-ol $(Q)$.
The reaction is: $C_6H_5COCH_3 + CH_3MgBr$ $\rightarrow C_6H_5C(OMgBr)(CH_3)_2$ $\xrightarrow{H_2O} C_6H_5C(OH)(CH_3)_2$.
The structure of $Q$ is $C_6H_5C(OH)(CH_3)_2$,which contains $9$ carbon atoms,$12$ hydrogen atoms,and $1$ oxygen atom.
Thus,the molecular formula of $Q$ is $C_9H_{12}O$.

(D) $NaBH_4$ is a mild reducing agent. It selectively reduces the aldehyde group to a primary alcohol while leaving the ester group $(-COOCH_3)$ and the alkene $(-C=C-)$ intact.

(C) $1$. Cyclohexanone reacts with $CH_3MgI$ followed by $H_2O$ to form $1$-methylcyclohexan-$1$-ol.
$2$. Dehydration with $H_3PO_4$ gives $1$-methylcyclohexene.
$3$. Hydroboration-oxidation $(B_2H_6, H_2O_2/NaOH)$ adds $-OH$ group anti-Markovnikov to the double bond,forming $2$-methylcyclohexan-$1$-ol.
$4$. Oxidation with $PCC$ converts the secondary alcohol into a ketone,resulting in $2$-methylcyclohexanone.

(A) The correct matching is as follows:
$A$. $\text{Ketone} + NaBH_4 \rightarrow 2^\circ$-alcohol $(III)$.
$B$. $\text{Ester} + LiAlH_4 \rightarrow 1^\circ$-alcohol $(I)$.
$C$. $RMgX + \text{Ketone} \rightarrow 3^\circ$-alcohol $(II)$.
Thus,the correct sequence is $A-III, B-I, C-II$.

(C) $1$. Dehydration of ethanol: $CH_3CH_2OH \xrightarrow[443 \ K]{\text{Conc. } H_2SO_4} CH_2=CH_2$ ($X$ is ethene).
$2$. Ozonolysis of ethene: $CH_2=CH_2 \xrightarrow[(2) \ Zn/H_2O]{(1) \ O_3} 2HCHO$ ($Y$ is formaldehyde).
$3$. Cannizzaro reaction: Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen,so it undergoes the Cannizzaro reaction in the presence of concentrated $NaOH$ to form methanol $(CH_3OH)$ and sodium formate $(HCOONa)$.
$4$. Thus,the conversion of $Y$ to $Z$ is the Cannizzaro reaction.

(D) $1$. Primary alcohols are oxidized to aldehydes by pyridinium chlorochromate $(PCC)$. Thus,the product $P$ is an aldehyde.
$2$. Aldehydes react with ammoniacal silver nitrate solution (Tollens' reagent) to undergo oxidation,forming a carboxylate anion along with the deposition of metallic silver (silver mirror test).

(B) The iodoform test is given by compounds containing a $CH_3CO-$ group (methyl ketones) or compounds that can be oxidized to such a group (like acetaldehyde or secondary alcohols with a methyl group at the alpha position).
$(A)$ $CH_3CH_2CH_2COC_6H_5$ (Butyrophenone): Does not have a $CH_3CO-$ group.
$(B)$ $C_6H_5COCH_3$ (Acetophenone): This is a methyl ketone $(CH_3CO-C_6H_5)$,so it gives a positive iodoform test.
$(C)$ $CH_2=CHCH_2COC_6H_5$: Does not have a $CH_3CO-$ group.
$(D)$ $C_6H_5CH_2COC_2H_5$: Does not have a $CH_3CO-$ group.
Thus,option $B$ is the correct answer.

(C) The reaction involves the nucleophilic addition of the Grignard reagent,$PhMgBr$,to the carbonyl group of $5$-bromo-$2$-pentanone.
$1$. The phenyl group $(Ph^-)$ attacks the electrophilic carbonyl carbon of the ketone to form an intermediate alkoxide,$CH_3-C(OMgBr)(Ph)-CH_2-CH_2-CH_2Br$ (labeled as $P$).
$2$. The alkoxide oxygen then performs an intramolecular nucleophilic substitution $(S_N2)$ on the carbon bearing the bromine atom,displacing the bromide ion and forming a five-membered cyclic ether ring.
$3$. This results in the formation of $2$-methyl-$2$-phenyltetrahydrofuran as the major product $(Z)$.

(D) The reaction of $HC \equiv C-CH_2-CH_2-COCH_3$ with $HgSO_4$ and dilute $H_2SO_4$ is a hydration reaction of an alkyne. The terminal alkyne undergoes hydration to form a ketone. The product $A$ is $CH_3COCH_2CH_2COCH_3$ (hexane-$2,5$-dione).
Next,the reaction of $CH_3COCH_2CH_2COCH_3$ with $EtONa$ and $EtOH$ is an intramolecular aldol condensation. The base $EtO^-$ abstracts an $\alpha$-hydrogen from one of the methyl groups,forming an enolate,which then attacks the other carbonyl group to form a five-membered ring. Subsequent dehydration yields $3$-methylcyclopent-$2$-en-$1$-one as product $B$.

(A) The reaction sequence is as follows:
$C_2H_5NH_2$ $\xrightarrow{NaNO_2/HCl} [C_2H_5N_2^+Cl^-] (X)$ $\xrightarrow{H_2O} C_2H_5OH (Y)$ $\xrightarrow{Cu, 573 \ K} CH_3CHO (Z)$
$Z$ is acetaldehyde $(CH_3CHO)$.
$I$. $Z$ is an aldehyde. (Correct)
$II$. $Z$ has $\alpha$-hydrogen atoms,so it does not undergo Cannizzaro reaction. (Incorrect)
$III$. $Z$ contains a $CH_3CO-$ group,so it gives a positive iodoform test. (Correct)
$IV$. $Z$ is an aldehyde,so it gives a positive test with Tollens' reagent. (Incorrect)
Thus,statements $I$ and $III$ are correct.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$(I)$ $CH_3CH_2COCH_2CH_3$: Does not contain $CH_3CO-$ group.
$(II)$ $CH_3CH(OH)CH_3$: Contains $CH_3CH(OH)-$ group,so it gives a positive test.
$(III)$ $1$-indanone: Does not contain $CH_3CO-$ group.
$(IV)$ $CH_3CH_2COCH_3$: Contains $CH_3CO-$ group,so it gives a positive test.
$(V)$ $PhCOPh$: Does not contain $CH_3CO-$ group.
$(VI)$ $PhCOCH_3$: Contains $CH_3CO-$ group,so it gives a positive test.
Therefore,compounds $(II), (IV),$ and $(VI)$ give a positive iodoform test.

(A) Alcohols containing the $CH_3CH(OH)-$ group and carbonyl compounds containing the $CH_3CO-$ group give a yellow precipitate with iodine and $NaOH$ solution. This reaction is known as the iodoform test.
Thus,$CH_3CHO$ gives a yellow precipitate with $I_2$ and $NaOH$ due to the presence of the $CH_3CO-$ group.
The reaction is: $CH_3CHO + 3I_2 + 4NaOH \longrightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$.

(B) Salicylaldehyde ($2$-hydroxybenzaldehyde) contains a strongly activating $-OH$ group,which directs electrophilic substitution to the ortho and para positions.
Since the ortho position is already occupied by the $-CHO$ group,the bromine atoms substitute at the $3$ and $5$ positions (which are ortho and para to the $-OH$ group respectively).
When treated with $2$ equivalents of $Br_2$ in glacial acetic acid,the reaction yields $3,5-$dibromosalicylaldehyde as the major product.

(D) The reagent $DIBAL-H$ (diisobutylaluminium hydride,$AlH(i-Bu)_2$) is a selective reducing agent. It reduces nitriles $(-CN)$ to aldehydes $(-CHO)$ after hydrolysis,while leaving other functional groups like carbon-carbon double bonds $(C=C)$ unaffected. Therefore,the reaction of $Ph-CH=CH-CH_2-CH_2-CN$ with $DIBAL-H$ followed by $H_2O$ yields $Ph-CH=CH-CH_2-CH_2-CHO$.

(B) The reaction sequence is as follows:
$1$. Benzaldehyde $(C_6H_5CHO)$ is oxidized by $KMnO_4/H^+$ to form Benzoic acid $(C_6H_5COOH)$,which is '$X$'.
$2$. Benzoic acid reacts with $SOCl_2$ to form Benzoyl chloride $(C_6H_5COCl)$.
$3$. Benzoyl chloride undergoes Rosenmund reduction $(H_2, Pd/BaSO_4)$ to form Benzaldehyde $(C_6H_5CHO)$,which is '$Y$'.
$4$. Benzaldehyde undergoes Clemmensen reduction $(Zn-Hg/Conc. HCl)$ to form Toluene $(C_6H_5CH_3)$,which is '$Z$'.
Therefore,'$Z$' is Toluene.

(C) The reaction sequence is as follows:
$1$. Sodium benzoate reacts with $NaOH + CaO$ (sodalime) to form benzene $(A)$.
$2$. Benzene $(A)$ undergoes Friedel-Crafts alkylation with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene $(B)$.
$3$. Toluene $(B)$ reacts with chromyl chloride $(CrO_2Cl_2)$ in $CS_2$ followed by hydrolysis $(H_3O^+)$ to form benzaldehyde $(C)$.
This specific oxidation of the methyl group of toluene to a formyl group using chromyl chloride is known as the $Etard$ reaction.

(D) The reaction proceeds in two main steps:
$1$. Stephen reduction: $CH_3-CN$ $\xrightarrow{SnCl_2, HCl} CH_3-CH=NH$ $\xrightarrow{H_3O^+} CH_3CHO$.
$2$. Reduction of aldehyde: $CH_3CHO \xrightarrow{HI, \text{Red } P} CH_3-CH_3$ (Ethane).
In the starting material,the carbon of the $-CN$ group is $sp$ hybridized.
In the final product,ethane $(CH_3-CH_3)$,the carbons are $sp^3$ hybridized.
The functional group changes from $-CN$ (nitrile) to $-CH_3$ (alkyl group).
Thus,the correct change is $A: -CN \rightarrow -CH_3$ and $B: sp \rightarrow sp^3$.

(C) The reactions are as follows:
$A$. Benzaldehyde $(C_6H_5CHO)$ undergoes Clemmensen reduction with $Zn-Hg/HCl$ to form Toluene $(C_6H_5CH_3)$. Thus,$A \rightarrow III$.
$B$. Benzoyl chloride $(C_6H_5COCl)$ undergoes Rosenmund reduction with $H_2/Pd-BaSO_4$ to form Benzaldehyde $(C_6H_5CHO)$. Thus,$B \rightarrow IV$.
$C$. Cyclohexanone undergoes Wolff-Kishner reduction with $NH_2NH_2/KOH$ to form Cyclohexane $(C_6H_{12})$. Thus,$C \rightarrow V$.
Therefore,the correct match is $A-III, B-IV, C-V$.

(A) The reaction sequence is as follows:
$1$. The conversion of $Toluene$ to $Benzaldehyde$ is an oxidation reaction known as $Etard's$ reaction,which uses $CrO_2Cl_2$ in $CS_2$ followed by hydrolysis $(H_3O^+)$. Thus,$X = (i) \ CrO_2Cl_2/CS_2, (ii) \ H_3O^+$.
$2$. $Benzaldehyde$ contains a $-CHO$ group,which is a deactivating and $meta$-directing group for electrophilic substitution reactions.
$3$. Nitration of $Benzaldehyde$ using a nitrating mixture $(HNO_3 + H_2SO_4)$ at $273-283 \ K$ yields $m-nitrobenzaldehyde$ as the major product. Thus,$Y = m-nitrobenzaldehyde$.

(A) The reaction sequence is as follows:
$1$. $C_2H_2$ (acetylene) when passed through a red hot iron tube undergoes cyclic polymerization to form benzene. Benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ (Gattermann-Koch reaction) to form $X$ (Benzaldehyde).
$2$. Benzaldehyde $(X)$ reacts with aniline $(Y)$ to form a Schiff's base.
$3$. Benzaldehyde $(X)$ undergoes nitration with $HNO_3/H_2SO_4$ at $273 \ K$ to form $Z$ ($m$-nitrobenzaldehyde) because the $-CHO$ group is meta-directing.
Therefore,$X$ is benzaldehyde,$Y$ is aniline,and $Z$ is $m$-nitrobenzaldehyde. The correct option is $(a)$.