Properties Questions in English

(B) The reaction between chloroform $(CHCl_3)$ and acetone $(CH_3COCH_3)$ in the presence of a base (like $KOH$) is a nucleophilic addition reaction. This reaction produces $1,1,1$-trichloro$-2-$methylpropan$-2-$ol,which is commonly known as Chloretone. Chloretone is used as a hypnotic agent in medicine.

(B) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. $2-$Hydroxypropane $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test.
$2$. Methyl acetate $(CH_3COOCH_3)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test.
$3$. Benzophenone $(C_6H_5COC_6H_5)$ does not contain a $CH_3CO-$ group attached to a hydrogen or another carbon,so it does not give the iodoform test.
$4$. Acetamide $(CH_3CONH_2)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test.
Therefore,Benzophenone does not give a yellow precipitate with iodine and alkali.

(D) The reaction involves the condensation of acetaldehyde $(CH_3CHO)$ and benzaldehyde $(C_6H_5CHO)$ in the presence of a dilute base.
$1$. $X$ is formed by the self-aldol condensation of two molecules of acetaldehyde: $2CH_3CHO$ $\rightarrow CH_3-CH(OH)-CH_2-CHO$ $\xrightarrow{-\Delta, -H_2O} CH_3-CH=CH-CHO$ (But$-2-$enal,$C_4H_6O$). This is a simple aldol condensation product.
$2$. $Y$ is formed by the cross-aldol condensation between acetaldehyde and benzaldehyde: $CH_3CHO + C_6H_5CHO \rightarrow C_6H_5-CH=CH-CHO$ (Cinnamaldehyde,$C_9H_8O$). This is a cross-aldol condensation product.
$3$. The specific cross-aldol condensation between an aromatic aldehyde and an aliphatic aldehyde/ketone is known as the Claisen-Schmidt reaction.
Therefore,all statements are correct.

(D) The given reactant is $2,3,5-\text{trimethylhepta-2,4-diene}$.
Upon reductive ozonolysis,the double bonds break to form carbonyl compounds:
$1$. The first double bond at $C_2-C_3$ yields acetone ($CH_3COCH_3$,$C_3H_6O$) as $X$.
$2$. The central part yields methylglyoxal ($CH_3COCHO$,$C_3H_4O_2$) as $Y$.
$3$. The terminal double bond yields acetaldehyde ($CH_3CHO$,$C_2H_4O$) as $Z$.
Compounds containing the $CH_3CO-$ group or those that can be oxidized to it (like ethanol or acetaldehyde) give a positive iodoform test.
- $X$ (Acetone,$CH_3COCH_3$) contains the $CH_3CO-$ group.
- $Y$ (Methylglyoxal,$CH_3COCHO$) contains the $CH_3CO-$ group.
- $Z$ (Acetaldehyde,$CH_3CHO$) contains the $CH_3CO-$ group.
Therefore,all three compounds $X, Y,$ and $Z$ give a positive iodoform test.

(A) Carbon monoxide $(CO)$ reacts with chlorine $(Cl_2)$ in the presence of sunlight (or activated charcoal) to form carbonyl chloride,which is commonly known as phosgene $(COCl_2)$.
The chemical equation is: $CO + Cl_2 \xrightarrow{h\nu} COCl_2$.

(D) Benzaldehyde $(C_6H_5CHO)$ reacts with a Grignard reagent $(CH_3MgI)$ to form an addition product,which upon hydrolysis yields $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$.
Step $1$: $C_6H_5CHO + CH_3MgI \rightarrow C_6H_5CH(OMgI)CH_3$
Step $2$: $C_6H_5CH(OMgI)CH_3 + H_2O \rightarrow C_6H_5CH(OH)CH_3 + Mg(OH)I$
Therefore,the correct reagent is methylmagnesium iodide $(CH_3MgI)$,which is formed from methyl iodide and magnesium.

(B) In the reactant $CH_3-CH_2OH$,the oxidation state of the $C$ atom in the $CH_3$ group is $-3$,and the oxidation state of the $C$ atom in the $CH_2$ group is $-1$.
In the product $CH_3-CHO$,the oxidation state of the $C$ atom in the $CH_3$ group remains $-3$,while the oxidation state of the $C$ atom in the $CHO$ group becomes $+1$.
Since the oxidation state of the $C$ atom in the $CH_2$ group increases from $-1$ to $+1$,it undergoes oxidation.

(C) Step $1$: Hydration of ethyne $(HC \equiv CH)$ in the presence of $1\% \ HgSO_4$ and $20\% \ H_2SO_4$ (Kucherov reaction) yields acetaldehyde $(CH_3CHO)$ as product $A$.
Step $2$: Reaction of acetaldehyde $(CH_3CHO)$ with methylmagnesium halide $(CH_3MgX)$ followed by hydrolysis $(H_2O)$ gives propan$-2-$ol $(CH_3CHOHCH_3)$ as product $B$.
Step $3$: Oxidation $([O])$ of propan$-2-$ol $(CH_3CHOHCH_3)$ yields acetone $(CH_3COCH_3)$ as the final product $C$.

(D) $3-\text{Hydroxybutanal}$ is formed by the $\text{Aldol condensation}$ of acetaldehyde $(CH_3-CHO)$ in the presence of dilute $NaOH$.
Therefore,$X = CH_3-CHO$,$Y = CH_3-CHO$,and $Z = NaOH$.
$2CH_3-CHO \xrightarrow{\text{dilute } NaOH} CH_3-CH(OH)-CH_2-CHO$

(C) In the Cannizzaro reaction,the rate-determining step involves the transfer of a hydride ion $(H^-)$ from the gem-diol dianion intermediate to another molecule of the aldehyde.
The ability of the intermediate to act as a hydride donor depends on the electron density at the carbon atom attached to the hydrogen. An electron-donating group (like $-OCH_3$) increases the electron density on the carbon,making the $C-H$ bond more polarized and facilitating the release of the hydride ion.
Conversely,an electron-withdrawing group (like $-NO_2$) decreases the electron density,making it a poorer hydride donor.
Comparing the options:
- Option $A$ is a mono-anion,which is not the primary hydride donor.
- Option $B$ is the unsubstituted benzaldehyde gem-diol dianion.
- Option $C$ has a methoxy group $(-OCH_3)$,which is an electron-donating group ($+M$ effect),making it the best hydride donor.
- Option $D$ has a nitro group $(-NO_2)$,which is an electron-withdrawing group ($-M$ effect),making it the poorest hydride donor.
Therefore,the intermediate with the electron-donating group is the best hydride donor.

(B) Acetone reacts with ethylene glycol in the presence of dry $HCl$ gas to form a cyclic ketal,specifically $2,2-dimethyl-1,3-dioxolane$.
The reaction involves the elimination of a water molecule between the carbonyl oxygen of the ketone and the hydroxyl hydrogens of the diol.
The reaction is: $CH_3-CO-CH_3 + HO-CH_2-CH_2-OH \xrightarrow{HCl} \text{Cyclic Ketal} + H_2O$.

(C) Aldol condensation is a reaction given by aldehydes or ketones that contain at least one $\alpha-$ hydrogen atom.
In the molecule $(CH_3)_3CCHO$ ($2$,$2$-dimethylpropanal),the carbonyl carbon is attached to a tertiary butyl group $(CH_3)_3C-$.
There is no hydrogen atom attached to the $\alpha-$ carbon (the carbon adjacent to the carbonyl group).
Due to the absence of $\alpha-$ hydrogen atoms,it cannot form an enolate ion,and therefore,it does not undergo aldol condensation.

(D) The haloform reaction requires the presence of a methyl ketone group $(-COCH_3)$ or a group that can be converted into one,such as a secondary alcohol $(-CH(OH)CH_3)$.
In the presence of a base and halogen,the $\alpha$-hydrogens are replaced by halogen atoms to form a trihalo derivative $(-COCX_3)$.
This intermediate then undergoes nucleophilic attack by $OH^-$,followed by the cleavage of the $C-C$ bond to release the haloform $(CHX_3)$.
Compound $ (i) \ CH_3CH_2COCH_2Cl $ has $\alpha$-hydrogens on the $CH_2Cl$ group,which can be further halogenated to $COCCl_3$.
Compound $ (ii) \ C_6H_5COCH_3 $ is a methyl ketone.
Compound $ (iii) \ C_6H_5COCHCl_2 $ already has a dihalo group and can be further halogenated to $COCCl_3$.
Compound $ (iv) \ CH_3CH_2COCCl_3 $ already contains the $COCCl_3$ group,which is the necessary intermediate for the haloform cleavage.
Therefore,all four compounds can undergo the haloform reaction.

(C) The Cannizzaro reaction of formaldehyde $(HCHO)$ involves the nucleophilic attack of the hydroxide ion $(OH^-)$ on the carbonyl carbon to form the hydroxyalkoxide monoanion $(HO-CH_2-O^-)$.
In the presence of a high concentration of base,a second $OH^-$ can deprotonate the hydroxyl group to form the dianion $(O^--CH_2-O^-)$.
Both of these species can act as intermediates that transfer a hydride ion to another formaldehyde molecule to complete the disproportionation process.

(A) The carbonyl group $(C=O)$ is polar due to the difference in electronegativity between carbon $(2.5)$ and oxygen $(3.5)$.
Oxygen is more electronegative,which pulls the electron density of the $\pi$-bond towards itself,creating a partial positive charge $(\delta+)$ on the carbon atom and a partial negative charge $(\delta-)$ on the oxygen atom.
This electrophilic carbon center makes the carbonyl compound susceptible to attack by nucleophiles.
Furthermore,the resulting intermediate is an alkoxide ion where the negative charge is stabilized on the highly electronegative oxygen atom,making the addition reaction favorable.

(B) Acetophenone $(C_6H_5COCH_3)$ is a ketone.
Ketones do not reduce Tollen's reagent,so it does not form a silver mirror.
It reacts with $2,4-DNP$ as it contains a carbonyl group.
It gives a positive iodoform test due to the presence of the $CH_3CO-$ group.
Oxidation of acetophenone with alkaline $KMnO_4$ followed by hydrolysis yields benzoic acid.

(A) The reaction of benzaldehyde $(C_6H_5CHO)$ with a Grignard reagent $(CH_3MgBr)$ follows a nucleophilic addition mechanism.
First,the nucleophilic $CH_3^-$ group attacks the carbonyl carbon of benzaldehyde to form an intermediate alkoxide: $C_6H_5CHO + CH_3MgBr \rightarrow C_6H_5CH(OMgBr)CH_3$.
Second,acid hydrolysis of this intermediate yields the final product: $C_6H_5CH(OMgBr)CH_3 + H_2O/H^+ \rightarrow C_6H_5CH(OH)CH_3 + Mg(OH)Br$.
The product $C_6H_5CH(OH)CH_3$ is $1$-phenylethanol,which is a secondary $(2^o)$ alcohol.

(B) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$ in the presence of dilute alkali is a Claisen-Schmidt condensation.
Acetaldehyde acts as the nucleophile by forming an enolate ion,which attacks the carbonyl carbon of benzaldehyde.
The initial aldol product formed is $C_6H_5-CH(OH)-CH_2-CHO$ ($3$-hydroxy$-3-$phenylpropanal).
This product undergoes spontaneous dehydration in the presence of base and heat to form the more stable conjugated product,cinnamaldehyde $(C_6H_5-CH=CH-CHO)$.

(A) The ester $(A)$ is $C_6H_5COOC_2H_5$ (ethyl benzoate).
Reaction with excess $CH_3MgBr$ proceeds as follows:
$C_6H_5COOC_2H_5 + 2CH_3MgBr \rightarrow C_6H_5-C(OH)(CH_3)_2 + Mg(OC_2H_5)Br + Mg(OH)Br$.
Dehydration with $H_2SO_4$ gives the olefin $(B)$: $C_6H_5-C(CH_3)=CH_2$ ($2$-phenylpropene).
Ozonolysis of $(B)$ yields $C_6H_5COCH_3$ (acetophenone) and $HCHO$.
Acetophenone $(C_6H_5COCH_3)$ contains a $CH_3CO-$ group and thus gives a positive iodoform test.

(D) Aldol condensation occurs in aldehydes and ketones that possess at least one $\alpha$-hydrogen atom.
$1$. Propanal $(CH_3CH_2CHO)$ has $2$ $\alpha$-hydrogen atoms.
$2$. Trichloroethanal $(Cl_3CCHO)$ has no $\alpha$-hydrogen atoms.
$3$. Methanal $(HCHO)$ has no $\alpha$-hydrogen atoms.
$4$. Ethanal $(CH_3CHO)$ has $3$ $\alpha$-hydrogen atoms.
$5$. Benzaldehyde $(C_6H_5CHO)$ has no $\alpha$-hydrogen atoms.
Therefore,trichloroethanal $(2)$,methanal $(3)$,and benzaldehyde $(5)$ do not undergo aldol condensation.

(A) Aldol condensation is observed only in carbonyl compounds (aldehydes and ketones) that possess at least one $\alpha$-hydrogen atom.
$(a)$ Chloral $(CCl_3CHO)$ has no $\alpha$-hydrogen atom because the $\alpha$-carbon is bonded to three chlorine atoms.
$(b)$ Phenylacetaldehyde $(C_6H_5CH_2CHO)$ has two $\alpha$-hydrogen atoms.
$(c)$ Hexanal $(CH_3(CH_2)_4CHO)$ has two $\alpha$-hydrogen atoms.
$(d)$ Nitromethane $(CH_3NO_2)$ is not an aldehyde or ketone,but it does possess $\alpha$-hydrogen atoms and can undergo a condensation reaction similar to aldol,but strictly speaking,the question refers to carbonyl compounds. Among the options,chloral is the classic example that does not undergo aldol condensation.

(A) Acetaldehyde $(CH_3CHO)$ contains $\alpha$-hydrogen atoms attached to the $\alpha$-carbon.
In the presence of a dilute alkali,it undergoes an aldol condensation reaction to form $3-$hydroxybutanal (aldol).
Therefore,the presence of $\alpha$-hydrogen is the reason for the aldol condensation reaction.

(A) Grignard reagents $(RMgX)$ are strong bases and highly reactive nucleophiles.
They react rapidly with acidic hydrogen atoms,such as those present in the hydroxyl $(-OH)$ group of hydroxyketones,to form an alkoxide and an alkane $(R-H)$.
Therefore,to perform a Grignard reaction on a hydroxyketone,the hydroxyl group must first be protected (e.g.,by silylation) to prevent this side reaction.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or compounds that can be oxidized to this group (like ethanol or secondary alcohols with a methyl group).
Isobutanal is $(CH_3)_2CHCHO$. It does not contain the $CH_3CO-$ group,so it does not give the iodoform test.
The reason provided is incorrect because isobutanal does have an $\alpha$-hydrogen atom (the hydrogen attached to the carbon adjacent to the carbonyl group).
Therefore,the Assertion is correct,but the Reason is incorrect.

(B) The Assertion is correct because protonation of the carbonyl oxygen atom increases the positive charge on the carbon atom,thereby enhancing its electrophilic character.
The Reason is also correct because the oxygen atom of the carbonyl group acts as a nucleophile (due to lone pairs) and accepts a proton $(H^+)$,which is an electrophile.
However,the Reason is not the correct explanation for the Assertion,as the increased electrophilicity is a consequence of the resonance effect and inductive effect induced by the positive charge on oxygen,not simply the fact that an electrophile was added.

(B) The Assertion is correct because aldol condensation involves the formation of a nucleophile (either an enolate ion in base or an enol in acid) which attacks the carbonyl carbon.
Both acids and bases can catalyze this process.
The Reason is also correct because $\beta$-hydroxyaldehydes or ketones (aldols) are unstable and undergo dehydration to form $\alpha,\beta$-unsaturated carbonyl compounds in the presence of an acid.
However,the Reason does not explain why the condensation itself is catalyzed by both acids and bases; it describes a subsequent step (dehydration).
Therefore,both are correct,but the Reason is not the correct explanation of the Assertion.

(A) The Cannizzaro reaction is a disproportionation reaction (self-oxidation and reduction) that occurs in aldehydes lacking $\alpha-hydrogen$ atoms.
$2, 2-$ Dimethylpropanal (pivalaldehyde) is $(CH_3)_3CCHO$. It has no $\alpha-hydrogen$ atom attached to the carbonyl carbon.
Therefore,it undergoes the Cannizzaro reaction in the presence of concentrated $NaOH$.
Both the Assertion and the Reason are correct,and the Reason correctly explains why the compound undergoes the reaction.

(D) Both Assertion and Reason are incorrect.
Acetoacetic ester $(CH_3COCH_2COOC_2H_5)$ does not give a positive iodoform test under standard conditions because the active methylene group $(CH_2)$ between the two carbonyl groups is highly acidic and reactive,leading to cleavage rather than the formation of iodoform.
The Reason is also incorrect because the structure of acetoacetic ester clearly contains the $CH_3CO-$ (acetyl) group.

(D) The Assertion is incorrect because benzaldehyde is less reactive than ethanol towards nucleophilic attack. This is due to the steric hindrance of the bulky phenyl group and the resonance stabilization of the carbonyl carbon by the phenyl ring.
The Reason is also incorrect because the $+R$ effect of the phenyl group is electron-donating,which increases the electron density on the carbonyl carbon,thereby reducing its electrophilicity.

(A) Aldehydes and ketones possess a polar carbonyl group $(C=O)$,which creates a permanent dipole moment.
Due to this polarity,there exist dipole-dipole interactions between the molecules of aldehydes and ketones.
These intermolecular dipole-dipole interactions are stronger than the weak van der Waals forces present in hydrocarbons and ethers of comparable molecular masses.
Consequently,more energy is required to overcome these forces,leading to higher boiling points for aldehydes and ketones compared to hydrocarbons and ethers.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) The oxidation of primary alcohols with strong oxidizing agents like $K_2Cr_2O_7$ or $KMnO_4/H^+$ yields carboxylic acids. Thus,phenylmethanol $(C_6H_5CH_2OH)$ gives benzoic acid $(C_6H_5COOH)$ in options $A$ and $D$.
The haloform reaction of acetophenone $(C_6H_5COCH_3)$ with $NaOCl$ followed by acid workup $(H_3O^+)$ is a standard method to produce benzoic acid,as shown in option $B$.
$PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes primary alcohols only up to the aldehyde stage. Therefore,the reaction of phenylmethanol with $PCC$ yields benzaldehyde $(C_6H_5CHO)$ as the major product,not benzoic acid. Hence,option $C$ is the correct answer.

(D) is formed by the reaction of $A$ with $CH_3MgBr$.
Since $B$ gives $2\text{-methyl-2-butene}$ $(CH_3-C(CH_3)=CH-CH_3)$ upon heating with $Cu$ at $573 \text{ K}$,$B$ must be a tertiary alcohol,specifically $2\text{-methyl-2-butanol}$ $(CH_3-C(OH)(CH_3)-CH_2-CH_3)$.
Therefore,$A$ must be butanone $(CH_3-CO-CH_2-CH_3)$.
The molecular formula of $A$ is $C_4H_8O$.
The molar mass of $A = (4 \times 12) + (8 \times 1) + 16 = 72 \text{ g/mol}$.
The mass of carbon in $A = 4 \times 12 = 48 \text{ g}$.
The mass percentage of carbon $= \frac{48}{72} \times 100 = 66.67\%$.

(A) The reaction sequence starts with $(A)$,which gives a positive iodoform test,indicating the presence of a $CH_3CO-$ group.
$1$. $(A)$ reacts with $CH_3MgBr$ followed by $H_3O^+$ to form a tertiary alcohol.
$2$. Dehydration with conc. $H_2SO_4/\Delta$ yields a cyclic compound $(B)$ (an indene derivative).
$3$. Ozonolysis $(O_3/Zn, H_2O)$ of $(B)$ breaks the double bond to form a dicarbonyl compound.
Based on the provided solution image,$(A)$ is $3$-methyl-$4$-phenylbut-$3$-en-$2$-one. This structure contains the $CH_3CO-$ group (positive iodoform test) and matches the reaction pathway shown.

(C) Step $1$: Hydration of alkyne $(X)$
$CH_3-CH(CH_3)-C\equiv CH + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_3-CH(CH_3)-CO-CH_3$ ($X$ is $3-methylbutan-2-one$)
Step $2$: Grignard reaction
$CH_3-CH(CH_3)-CO-CH_3 + C_2H_5MgBr \xrightarrow{H_2O} CH_3-CH(CH_3)-C(OH)(C_2H_5)-CH_3$
Step $3$: Dehydration
$CH_3-CH(CH_3)-C(OH)(C_2H_5)-CH_3 \xrightarrow{conc. H_2SO_4, \Delta} CH_3-C(CH_3)=C(C_2H_5)-CH_3$ (Major product $Y$ follows Saytzeff rule)

(N/A) $(i)$ $CH_3-CH=CH_2$ (Propene) $\xrightarrow{H_2O/H^+}$ $CH_3-CH(OH)-CH_3$ (Propan$-2-$ol)
$(ii)$ $NaBH_4$ selectively reduces the ketone group to a secondary alcohol while leaving the ester group intact. The product is Methyl ($2$-hydroxycyclohexyl) ethanoate.
$(iii)$ $CH_3-CH_2-CH(CH_3)-CHO$ ($2$-Methylbutanal) $\xrightarrow{NaBH_4}$ $CH_3-CH_2-CH(CH_3)-CH_2OH$ ($2$-Methylbutan$-1-$ol)

(N/A) The carbon atom of the carbonyl group of $benzaldehyde$ is less electrophilic than the carbon atom of the carbonyl group present in $propanal$.
In $benzaldehyde$,the phenyl group is attached to the carbonyl carbon,which provides resonance stabilization to the carbonyl group. This resonance effect delocalizes the positive charge on the carbonyl carbon,thereby reducing its electrophilicity.
Additionally,the phenyl group exerts a steric hindrance that makes the approach of a nucleophile more difficult compared to the smaller alkyl group in $propanal$.
Therefore,$benzaldehyde$ is less reactive than $propanal$ towards nucleophilic addition reactions.

(A) forms a $2,4-DNP$ derivative,which indicates it is an aldehyde or a ketone. Since it does not reduce Tollens' or Fehling's reagent,$(A)$ must be a ketone.
$(A)$ responds to the iodoform test,which confirms it is a methyl ketone.
The molecular formula of $(A)$ indicates a high degree of unsaturation,yet it does not decolourise bromine water or Baeyer's reagent,suggesting the unsaturation is due to an aromatic ring.
Compound $(B)$,being an oxidation product of a ketone,is a carboxylic acid. The molecular formula of $(B)$ corresponds to benzoic acid $(C_6H_5COOH)$.
Therefore,$(A)$ is phenyl methyl ketone (acetophenone,$C_6H_5COCH_3$).
The reactions are as follows:
$1$. Formation of $2,4-DNP$ derivative: $C_6H_5COCH_3 + H_2NNHC_6H_3(NO_2)_2 \rightarrow C_6H_5C(CH_3)=NNHC_6H_3(NO_2)_2 + H_2O$
$2$. Iodoform test: $C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 + 3NaI + 3H_2O$
$3$. Oxidation: $C_6H_5COCH_3 \xrightarrow{H_2CrO_4} C_6H_5COOH$

(N/A) $(i)$ The $+I$ effect of the alkyl group increases in the order: $\text{Ethanal} < \text{Propanal} < \text{Propanone} < \text{Butanone}$.
The electron density at the carbonyl carbon increases with the increase in the $+I$ effect. As a result,the chances of attack by a nucleophile decrease. Hence,the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is: $\text{Butanone} < \text{Propanone} < \text{Propanal} < \text{Ethanal}$.
$(ii)$ The $+I$ effect is more in ketone than in aldehyde. Hence,acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes,the $+I$ effect is the highest in $p-$Tolualdehyde because of the presence of the electron-donating $-CH_3$ group and the lowest in $p-$Nitrobenzaldehyde because of the presence of the electron-withdrawing $-NO_2$ group. The electron-withdrawing group increases the electrophilicity of the carbonyl carbon.
Hence,the increasing order of the reactivities of the given compounds is: $\text{Acetophenone} < p-\text{Tolualdehyde} < \text{Benzaldehyde} < p-\text{Nitrobenzaldehyde}$.

(N/A) The reactions involve the nucleophilic addition of ammonia derivatives to carbonyl compounds followed by the elimination of a water molecule to form imine derivatives.
$(i)$ Cyclopentanone + $NH_2OH \xrightarrow{H^+} \text{Cyclopentanone oxime} + H_2O$
(ii) Cyclohexanone + $2,4-DNP \xrightarrow{H^+} \text{Cyclohexanone-2,4-dinitrophenylhydrazone} + H_2O$
(iii) $R-CH=CH-CHO + NH_2CONHNH_2 \xrightarrow{H^+} R-CH=CH-CH=NNHCONH_2 + H_2O$
(iv) Acetophenone + $CH_3CH_2NH_2 \xrightarrow{H^+} \text{N-ethylacetophenone imine} + H_2O$

(N/A) $(i)$ $\text{Cyanohydrin}$: Formed by the addition of $HCN$ to aldehydes or ketones. Example: $CH_3CHO + HCN \rightarrow CH_3CH(OH)CN$ (Acetaldehyde cyanohydrin).
$(ii)$ $\text{Acetal}$: Formed when an aldehyde reacts with two equivalents of a monohydric alcohol in the presence of dry $HCl$. Example: $CH_3CHO + 2C_2H_5OH \xrightarrow{dry \ HCl} CH_3CH(OC_2H_5)_2 + H_2O$.
$(iii)$ $\text{Semicarbazone}$: Formed by the reaction of an aldehyde or ketone with semicarbazide. Example: $CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2 + H_2O$.
$(iv)$ $\text{Aldol}$: $A$ $\beta$-hydroxy aldehyde or ketone formed by the condensation of two molecules of an aldehyde or ketone containing $\alpha$-hydrogen. Example: $2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO$.
$(v)$ $\text{Hemiacetal}$: Formed by the addition of one molecule of alcohol to an aldehyde. Example: $CH_3CHO + C_2H_5OH \xrightarrow{HCl} CH_3CH(OH)(OC_2H_5)$.
$(vi)$ $\text{Oxime}$: Formed by the reaction of aldehydes or ketones with hydroxylamine $(NH_2OH)$. Example: $CH_3CHO + NH_2OH \rightarrow CH_3CH=NOH + H_2O$.
$(vii)$ $\text{Ketal}$: Formed by the reaction of a ketone with two equivalents of alcohol or a diol. Example: $CH_3COCH_3 + HOCH_2CH_2OH \xrightarrow{HCl} \text{Ethylene glycol ketal}$.
$(viii)$ $\text{Imine}$: Compounds containing a carbon-nitrogen double bond $(>C=NH)$,formed by reaction with ammonia. Example: $CH_3CHO + NH_3 \rightarrow CH_3CH=NH + H_2O$.
$(ix)$ $2,4-DNP$ $\text{derivative}$: Formed by the reaction of carbonyl compounds with $2,4$-dinitrophenylhydrazine. Example: $R_2C=O + H_2NNHC_6H_3(NO_2)_2 \rightarrow R_2C=NNHC_6H_3(NO_2)_2 + H_2O$.
$(x)$ $\text{Schiff's base}$: An imine formed by the reaction of an aldehyde or ketone with a primary amine $(RNH_2)$. Example: $CH_3CHO + CH_3NH_2 \rightarrow CH_3CH=NCH_3 + H_2O$.

(N/A) $(i)$ Benzaldehyde $2,4-$dinitrophenylhydrazone: $C_6H_5CH=NNHC_6H_3(NO_2)_2$
$(ii)$ Cyclopropanone oxime: $(CH_2)_2C=NOH$
$(iii)$ Acetaldehyde dimethyl acetal: $CH_3CH(OCH_3)_2$
$(iv)$ Cyclobutanone semicarbazone: $(CH_2)_3C=NNHCONH_2$
$(v)$ Ethylene ketal of hexan$-3-$one: $CH_3CH_2C(OCH_2CH_2O)CH_2CH_2CH_3$
$(vi)$ Methyl hemiacetal of formaldehyde: $HCH(OH)(OCH_3)$

(N/A) $(i)$ Cyclohexanecarbaldehyde reacts with $PhMgBr$ (a Grignard reagent) followed by acid hydrolysis to form cyclohexylphenylcarbinol.
$(ii)$ Cyclohexanecarbaldehyde reacts with Tollens' reagent $([Ag(NH_3)_2]^+)$ to form the cyclohexanecarboxylate ion and a silver mirror ($Ag$ precipitate).
$(iii)$ Cyclohexanecarbaldehyde reacts with semicarbazide in the presence of a weak acid to form cyclohexanecarbaldehyde semicarbazone.
$(iv)$ Cyclohexanecarbaldehyde reacts with excess ethanol in the presence of dry $HCl$ gas to form cyclohexanecarbaldehyde diethyl acetal.
$(v)$ Cyclohexanecarbaldehyde undergoes Clemmensen reduction with zinc amalgam $(Zn-Hg)$ and dilute hydrochloric acid $(HCl)$ to form methylcyclohexane.

(N/A) Aldehydes and ketones having at least one $\alpha-$hydrogen undergo aldol condensation. The compounds $(ii)$ $2-$methylpentanal,$(v)$ cyclohexanone,$(vi)$ $1-$phenylpropanone,and $(vii)$ phenylacetaldehyde contain one or more $\alpha-$hydrogen atoms. Therefore,these undergo aldol condensation.
Aldehydes having no $\alpha-$hydrogen atoms undergo Cannizzaro reactions. The compounds $(i)$ Methanal,$(iii)$ Benzaldehyde,and $(ix)$ $2,2-$dimethylbutanal do not have any $\alpha-$hydrogen. Therefore,these undergo Cannizzaro reactions.
Compound $(iv)$ Benzophenone is a ketone having no $\alpha-$hydrogen atom and compound $(viii)$ Butan$-1-$ol is an alcohol. Hence,these compounds do not undergo either aldol condensation or Cannizzaro reactions.
Aldol condensation $(ii)$: $2CH_3CH_2CH_2-CH(CH_3)-CHO \xrightarrow{dil. NaOH} CH_3CH_2CH_2-CH(CH_3)-CH(OH)-CH(CHO)-CH_2CH_2CH_3$ (product: $3-$Hydroxy$-2,4-$dimethyl$-2-$propylheptanal).
Cannizzaro reaction $(i)$: $2HCHO + conc. KOH \to CH_3OH + HCOOK$.
Cannizzaro reaction $(ix)$: $2CH_3CH_2-C(CH_3)_2-CHO \xrightarrow{conc. NaOH} CH_3CH_2-C(CH_3)_2-CH_2OH + CH_3CH_2-C(CH_3)_2-COONa$.

(N/A) $(i)$ Ethanal undergoes aldol condensation with dilute alkali to form $3-$hydroxybutanal,which upon reduction with $NaBH_4$ yields butane$-1,3-$diol.
$CH_3CHO$ $\xrightarrow{dil.NaOH} CH_3-CH(OH)-CH_2-CHO$ $\xrightarrow{NaBH_4} CH_3-CH(OH)-CH_2-CH_2OH$
$(ii)$ Ethanal undergoes aldol condensation with dilute alkali to form $3-$hydroxybutanal,which upon heating undergoes dehydration to yield but$-2-$enal.
$CH_3CHO$ $\xrightarrow{dil.NaOH} CH_3-CH(OH)-CH_2-CHO$ $\xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO$
$(iii)$ But$-2-$enal is oxidized using Tollen's reagent to yield but$-2-$enoic acid.
$CH_3-CH=CH-CHO \xrightarrow{[Ag(NH_3)_2]^+OH^-} CH_3-CH=CH-COOH$

(N/A) Propanal $(CH_3CH_2CHO)$ and butanal $(CH_3CH_2CH_2CHO)$ both contain $\alpha$-hydrogen atoms. Therefore,they undergo both self-aldol and cross-aldol condensation to give a mixture of four products. The products are formed as follows:

$1$. Propanal (Electrophile) + Propanal (Nucleophile)	$CH_3CH_2CH=C(CH_3)CHO$ ($2$-methylpent$-2-$enal)
$2$. Butanal (Electrophile) + Butanal (Nucleophile)	$CH_3CH_2CH_2CH=C(C_2H_5)CHO$ ($2$-ethylhex$-2-$enal)
$3$. Propanal (Electrophile) + Butanal (Nucleophile)	$CH_3CH_2CH=C(C_2H_5)CHO$ ($2$-ethylpent$-2-$enal)
$4$. Butanal (Electrophile) + Propanal (Nucleophile)	$CH_3CH_2CH_2CH=C(CH_3)CHO$ ($2$-methylhex$-2-$enal)

(A) The compound with molecular formula $C_9H_{10}O$ forms a $2,4-DNP$ derivative and reduces Tollens' reagent,which confirms it is an aldehyde.
Since it undergoes the Cannizzaro reaction,it must not have any $\alpha$-hydrogen atoms attached to the carbonyl carbon. However,in aromatic aldehydes,the $-CHO$ group is attached to the benzene ring,and the Cannizzaro reaction occurs if there are no $\alpha$-hydrogens on the carbon adjacent to the carbonyl group (though aromatic aldehydes generally undergo Cannizzaro regardless of ring substituents).
On vigorous oxidation,it yields $1,2-benzenedicarboxylic$ acid (phthalic acid). This indicates that the original compound is an ortho-substituted benzene derivative with an ethyl group at the ortho position relative to the $-CHO$ group.
Thus,the compound is $2-ethylbenzaldehyde$.

(N/A) $(i)$ Ethylbenzene + $KMnO_4/KOH, \Delta \rightarrow$ Benzoic acid $(C_6H_5COOH)$
$(ii)$ Phthalic acid + $SOCl_2, \Delta \rightarrow$ Phthaloyl chloride $(C_6H_4(COCl)_2)$
$(iii)$ Benzaldehyde + Semicarbazide $(H_2NCONHNH_2)$ $\rightarrow$ Benzaldehyde semicarbazone $(C_6H_5CH=NNHCONH_2)$
$(iv)$ Benzene + Benzoyl chloride $(C_6H_5COCl)$ + $AlCl_3 \rightarrow$ Benzophenone $(C_6H_5COC_6H_5)$
$(v)$ $4$-Oxocyclohexanecarbaldehyde + Tollens' reagent $[Ag(NH_3)_2]^+$ $\rightarrow$ $4$-Oxocyclohexanecarboxylic acid
$(vi)$ $2$-Formylbenzoic acid + $NaCN/HCl \rightarrow$ $2$-($1$-hydroxy-$1$-cyanomethyl)benzoic acid
$(vii)$ Benzaldehyde + Propanal + $dil. NaOH, \Delta \rightarrow$ $2$-Methyl-$3$-phenylprop-$2$-enal (Cross-Aldol product)
$(viii)$ Ethyl acetoacetate + $(i) NaBH_4, (ii) H^+ \rightarrow$ Ethyl $3$-hydroxybutanoate $(CH_3CH(OH)CH_2COOC_2H_5)$
$(ix)$ Cyclohexanol + $CrO_3 \rightarrow$ Cyclohexanone
$(x)$ Methylenecyclohexane + $(i) B_2H_6, (ii) H_2O_2/OH^- \rightarrow$ Cyclohexylmethanol (Hydroboration-oxidation)
$(xi)$ $1,1'$-Bicyclohexylidene + $(i) O_3, (ii) Zn-H_2O \rightarrow$ $2$ Cyclohexanone

(A) $1$. Calculate the empirical formula:
$C:H:O = \frac{69.77}{12} : \frac{11.63}{1} : \frac{18.6}{16} = 5.81 : 11.63 : 1.16 = 5 : 10 : 1$.
Empirical formula is $C_5H_{10}O$.
$2$. Determine molecular formula:
Empirical formula mass $= 5(12) + 10(1) + 16 = 86$.
Since molecular mass is $86$,the molecular formula is $C_5H_{10}O$.
$3$. Analyze chemical properties:
- Does not reduce Tollens' reagent: Not an aldehyde.
- Forms addition compound with $NaHSO_3$: Contains a carbonyl group.
- Positive iodoform test: Contains a $CH_3CO-$ group.
- Oxidation yields ethanoic acid $(CH_3COOH)$ and propanoic acid $(CH_3CH_2COOH)$: The cleavage occurs at the $C-C$ bond adjacent to the carbonyl group.
$4$. Conclusion:
The compound is $Pentan-2-one$ $(CH_3COCH_2CH_2CH_3)$.

(A) The reaction of Grignard reagent $(CH_3MgBr)$ with carbonyl compounds followed by hydrolysis yields alcohols:
$1$. $HCHO + CH_3MgBr$ $\rightarrow CH_3CH_2OMgBr$ $\xrightarrow{H_2O/H^+} CH_3CH_2OH$ (Primary alcohol).
$2$. $CH_3CHO + CH_3MgBr$ $\rightarrow CH_3CH(CH_3)OMgBr$ $\xrightarrow{H_2O/H^+} CH_3CH(OH)CH_3$ (Secondary alcohol).
$3$. $CH_3COCH_3 + CH_3MgBr$ $\rightarrow (CH_3)_3COMgBr$ $\xrightarrow{H_2O/H^+} (CH_3)_3COH$ (Tertiary alcohol).