Properties Questions in English

(D) $1$. Ozonolysis of $\text{but-2-ene}$ gives acetaldehyde $(CH_3CHO)$ as $X$.
$2$. Reaction of $X$ $(CH_3CHO)$ with $CH_3MgBr$ followed by hydrolysis gives propan$-2-$ol $(CH_3CH(OH)CH_3)$ as $Y$.
$3$. Dehydrogenation of $Y$ (a secondary alcohol) with $Cu$ at $573 \ K$ gives acetone $(CH_3COCH_3)$ as $Z$.
$4$. $Z$ is acetone,which is a methyl ketone. It gives a positive iodoform test (yellow precipitate with $I_2$ and $NaOH$),so statement $I$ is correct.
$5$. Acetone does not have $\alpha$-hydrogens that can lead to Cannizzaro reaction,so it does not undergo disproportionation with concentrated $NaOH$,making $II$ incorrect.
$6$. Acetone undergoes Wolff-Kishner reduction to form propane,so statement $III$ is correct.
$7$. Acetone is a ketone and does not reduce Fehling's reagent,so statement $IV$ is incorrect.
Therefore,statements $I$ and $III$ are correct.

(D) For reagent $X$: The starting material is a $\beta$-bromo ketone. Treatment with alcoholic $KOH$ (alc. $KOH$) leads to dehydrohalogenation via an $E2$ mechanism,resulting in the formation of an $\alpha,\beta$-unsaturated ketone.
For reagent $Y$: The product contains a carboxylate ion and a yellow precipitate $(CHI_3)$,which is characteristic of the iodoform reaction. This reaction is specific to methyl ketones (or alcohols that can be oxidized to methyl ketones) and is performed using $NaOI$ (sodium hypoiodite).

(A) . Tollen's reagent is ammoniacal silver nitrate.
$B$. Schiff's reagent is a solution of para-rosaniline hydrochloride decolorized by $SO_2$.
$C$. Rosenmund reduction uses $Pd$ supported on $BaSO_4$ as a catalyst.
$D$. Fehling solution '$B$' is an aqueous solution of Rochelle salt (potassium sodium tartrate) in $NaOH$.
Thus,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) $1$. The compound '$A$' has the molecular formula $C_8H_8O$ and reacts with $I_2/KOH$ (haloform reaction),which indicates the presence of a methyl ketone group $(-COCH_3)$.
$2$. Acetophenone $(C_6H_5COCH_3)$ fits this description. It reacts with $I_2/KOH$ to form potassium benzoate ($B$,$C_6H_5COOK$) and iodoform ($C$,$CHI_3$).
$3$. Iodoform $(CHI_3)$ on heating with silver powder $(Ag)$ undergoes dehalogenation to form acetylene ($D$,$HC\equiv CH$).
$4$. The reaction is: $2CHI_3 + 6Ag \rightarrow HC\equiv CH + 6AgI$.
$5$. Therefore,'$A$' is acetophenone and '$D$' is acetylene $(HC\equiv CH)$.

(A) The compound $A$ is $CH_3COCH_3$ (acetone),which has the molecular formula $C_3H_6O$.
$1$. Reaction with $I_2/NaOH$ (Iodoform test): Acetone reacts with $I_2$ in the presence of $NaOH$ to give a yellow precipitate of iodoform $(CHI_3)$.
$CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$. Here,$B$ is $CHI_3$.
$2$. Reaction with $Zn-Hg/HCl$ (Clemmensen reduction): Acetone undergoes reduction to form propane $(CH_3CH_2CH_3)$.
$CH_3COCH_3 \xrightarrow{Zn-Hg/HCl} CH_3CH_2CH_3$. Here,$C$ is $CH_3CH_2CH_3$.

(A) When acetaldehyde $(CH_3CHO)$ is heated with Fehling solution,it undergoes oxidation to form acetate ions,while the $Cu^{2+}$ ions in the Fehling solution are reduced to $Cu^+$ ions.
This results in the formation of a red precipitate of cuprous oxide $(Cu_2O)$.
The chemical reaction is:
$CH_3CHO + 2Cu(OH)_2 + NaOH \longrightarrow CH_3COONa + Cu_2O \downarrow (\text{Red}) + 3H_2O$

(B) Statement $I$: Aldehydes react with $HCN$ to form cyanohydrins. The reaction is a nucleophilic addition reaction where the $CN^-$ ion attacks the carbonyl carbon,followed by protonation to yield the cyanohydrin product.
Statement $II$: $A$ cyanohydrin is defined as an organic compound that contains both a hydroxyl group $(-OH)$ and a cyano group $(-CN)$ attached to the same carbon atom. This is consistent with the structure $R-CH(OH)(CN)$.
Both statements are correct.

(C) When ethanal $(CH_3CHO)$ reacts with semicarbazide $(NH_2NHCONH_2)$,a condensation reaction occurs where a molecule of water is eliminated to form a semicarbazone. The terminal $-NH_2$ group of semicarbazide is involved in the reaction with the carbonyl group of the aldehyde.
The reaction is as follows:
$CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2 + H_2O$
The correct structure of the product,acetaldehyde semicarbazone,is $H_3C-CH=N-NH-C(=O)-NH_2$.

(B) In reaction $(I)$,benzene reacts with $CO$ and $HCl$ in the presence of anhydrous $AlCl_3/CuCl$ to form benzaldehyde (Gattermann-Koch reaction). No carboxylate ion is formed.
In reaction $(II)$,acetaldehyde $(CH_3CHO)$ reacts with Tollen's reagent $([Ag(NH_3)_2]^+)$ in a basic medium to form an acetate ion $(CH_3COO^-)$,which is a carboxylate ion.
In reaction $(III)$,benzaldehyde reacts with Fehling's solution $(Cu^{2+}/OH^-)$ to form a benzoate ion $(C_6H_5COO^-)$,which is a carboxylate ion.
In reaction $(IV)$,aniline is treated with $Cl_2/hv$ followed by $H_2O$ at $373 \ K$,which does not produce a carboxylate ion.
Therefore,reactions $(II)$ and $(III)$ produce a carboxylate ion in their reaction mixture.

(A) An acetal is an organic compound formed by the reaction of an aldehyde with two equivalents of an alcohol in the presence of an acid catalyst. The general structure of an acetal is $R-CH(OR')_2$,where a single carbon atom is bonded to two alkoxy groups $(-OR')$,one hydrogen atom,and one alkyl group $(R)$.

(B) Acetaldoxime is formed by the reaction of acetaldehyde $(CH_3CHO)$ with hydroxylamine $(NH_2OH)$.
The reaction is as follows:
$CH_3CHO + NH_2OH \rightarrow CH_3CH=NOH + H_2O$
Thus,the structure of acetaldoxime is $CH_3CH=NOH$.

(D) The haloform reaction is given by compounds containing the methyl keto group $(CH_3-CO-)$ or compounds that can be oxidized to a methyl keto group,such as secondary alcohols with the structure $(R-CH(OH)-CH_3)$.
$1$. $CH_3CHO$ (acetaldehyde) contains the $CH_3-CO-$ group.
$2$. $CH_3CH_2OH$ (ethanol) can be oxidized to $CH_3CHO$,which contains the $CH_3-CO-$ group.
$3$. $CH_3COCH_3$ (acetone) contains the $CH_3-CO-$ group.
$4$. $C_2H_5COCH_2CH_3$ (pentan$-3-$one) does not contain a methyl keto group $(CH_3-CO-)$ and cannot be oxidized to one.
Therefore,$C_2H_5COCH_2CH_3$ does not undergo the haloform reaction.

(A) The reaction proceeds as follows:
$(i)$ $CH_3OH$ reacts with $KI$ to form $CH_3I$.
(ii) $CH_3I$ reacts with $Mg$ in dry ether to form the Grignard reagent $CH_3MgI$.
(iii) $CH_3MgI$ undergoes nucleophilic addition to the carbonyl group of $1-$indanone ($2$,$3$-dihydro-1H-inden$-1-$one).
(iv) Subsequent hydrolysis with $H_2O$ yields $1-$methyl$-2,3-$dihydro-1H-inden$-1-$ol.
$(v)$ Dehydration of the alcohol using $20\% H_3PO_4$ at $358 \ K$ results in the formation of the more stable conjugated alkene,$1$-methyl-1H-indene,as the major product.

(D) Step $1$: Conversion of toluene to $(P)$. Toluene reacts with $KMnO_4 / KOH$ followed by $H_3O^+$ to form $p$-methylbenzoic acid. This acid then reacts with $PCl_5$ to form $p$-methylbenzoyl chloride,which is $(P)$.
Step $2$: Conversion of bromobenzene to $(Q)$. Bromobenzene reacts with $Mg$ to form phenylmagnesium bromide,which then reacts with $CdCl_2$ to form diphenylcadmium. This organocadmium compound reacts with $(P)$ ($p$-methylbenzoyl chloride) to form the final product $(Q)$,which is $4$-methylbenzophenone.

(D) The Clemmensen reduction is a chemical reaction that reduces carbonyl groups (aldehydes and ketones) to methylene groups $(-CH_2-)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
Option $A$ represents the Rosenmund reduction.
Option $B$ represents the Wolff-Kishner reduction.
Option $C$ represents the reduction of an ester to an aldehyde using $DIBAL-H$.
Option $D$ represents the Clemmensen reduction,where the ketone $R-COCH_3$ is reduced to the alkane $R-CH_2-CH_3$.

(A) The given reaction is an example of the $Wolff-Kishner$ reduction.
This reaction involves the reduction of a carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
First,the ketone reacts with hydrazine $(NH_2NH_2)$ to form a hydrazone intermediate.
Then,in the presence of a strong base like $KOH$ and a high-boiling solvent like ethylene glycol,the hydrazone undergoes base-catalyzed decomposition to release nitrogen gas $(N_2)$ and form the corresponding alkane.
In this specific case,$3,3-dimethylcyclopentanone$ is reduced to $3,3-dimethylcyclopentane$.

(A) The Wolff-Kishner reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into methylene groups $(-CH_2-)$.
It involves heating the carbonyl compound with hydrazine $(NH_2NH_2)$ and a strong base like potassium hydroxide $(KOH)$ in a high-boiling solvent such as ethylene glycol.
The reaction proceeds via the formation of a hydrazone intermediate,which then undergoes base-catalyzed decomposition to release nitrogen gas $(N_2)$ and form the corresponding alkane.

(A) The starting material is $2,3$-dimethylbut-$2$-ene.
Step $1$: Ozonolysis of $2,3$-dimethylbut-$2$-ene with $(1) \ O_3$ and $(2) \ Zn/H_2O$ leads to the cleavage of the double bond,producing two molecules of acetone $(CH_3COCH_3)$.
Step $2$: The product ' $A$ ' is acetone. When acetone is treated with $Ba(OH)_2$ (a base),it undergoes an aldol condensation reaction.
Two molecules of acetone react to form $4$-hydroxy-$4$-methylpentan-$2$-one (diacetone alcohol).
Therefore,the product ' $P$ ' is $4$-hydroxy-$4$-methylpentan-$2$-one.

(C) $1$. Reaction of $CH_3CH_2COCH_3$ (Butan$-2-$one) with $CH_3MgBr$ followed by hydrolysis gives $A$,which is $CH_3CH_2C(OH)(CH_3)_2$ ($2$-methylbutan$-2-$ol).
$2$. Dehydration of $A$ with $20\% H_3PO_4$ at $358 \ K$ follows Saytzeff's rule to give the major product $B$,which is $CH_3CH=C(CH_3)_2$ ($2$-methylbut$-2-$ene).
$3$. Ozonolysis of $B$ $(CH_3CH=C(CH_3)_2)$ gives $CH_3CHO$ (Ethanal) and $(CH_3)_2CO$ (Propanone) as products $C$ and $D$.

(B) $1$. The starting material is propene $(CH_3CH=CH_2)$.
$2$. Hydroboration-oxidation of propene with $(i) B_2H_6$ and $(ii) H_2O_2/NaOH$ gives propan$-1-$ol $(CH_3CH_2CH_2OH)$ as $X$.
$3$. Dehydrogenation of propan$-1-$ol $(X)$ using $Cu$ at $573 \ K$ gives propanal $(CH_3CH_2CHO)$ as $Y$.
$4$. Propanal $(Y)$ undergoes aldol condensation in the presence of dilute $NaOH$ followed by heating $(\Delta)$ to give the $\alpha,\beta$-unsaturated aldehyde,$2$-methylpent-$2$-enal $(CH_3CH_2CH=C(CH_3)CHO)$ as $Z$.

(A) Step $1$: Ozonolysis of $but-2-ene$ $(CH_3-CH=CH-CH_3)$ yields two molecules of acetaldehyde $(CH_3CHO)$. Thus,$X = CH_3CHO$.
Step $2$: Aldol condensation of $2$ molecules of acetaldehyde $(2CH_3CHO)$ in the presence of dilute $NaOH$ followed by heating $(\Delta)$ gives $but-2-enal$ $(CH_3-CH=CH-CHO)$,which is $Z$.
Step $3$: Ozonolysis of $but-2-enal$ $(CH_3-CH=CH-CHO)$ breaks the double bond to form acetaldehyde $(CH_3CHO)$ and glyoxal $(CHO-CHO)$.
Therefore,$A$ and $B$ are $CH_3CHO$ and $CHO-CHO$.

(A) $1$. Propene undergoes anti-Markovnikov addition with $HBr$ in the presence of peroxide to form $1-$bromopropane: $CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2Br$.
$2$. $1-$bromopropane reacts with $Mg$ in the presence of dry ether to form the Grignard reagent,propylmagnesium bromide: $CH_3-CH_2-CH_2Br + Mg \xrightarrow{\text{dry ether}} CH_3-CH_2-CH_2MgBr$.
$3$. Propanal $(CH_3-CH_2-CHO)$ undergoes nucleophilic addition with propylmagnesium bromide followed by hydrolysis to form hexan$-3-$ol: $CH_3-CH_2-CHO + CH_3-CH_2-CH_2MgBr \rightarrow CH_3-CH_2-CH(OH)-CH_2-CH_2-CH_3$.
$4$. Hexan$-3-$ol on oxidation with $CrO_3$ forms hexan$-3-$one: $CH_3-CH_2-CH(OH)-CH_2-CH_2-CH_3 \xrightarrow{CrO_3} CH_3-CH_2-CO-CH_2-CH_2-CH_3$.

(D) $1$. The alkene $A$ is $but-2-ene$ $(CH_3CH=CHCH_3)$,which shows cis/trans isomerism.
$2$. Ozonolysis of $but-2-ene$ yields two molecules of acetaldehyde $(CH_3CHO)$,so $B = CH_3CHO$.
$3$. Acetaldehyde undergoes aldol condensation in the presence of $NaOH$ to form $but-2-enal$ $(CH_3CH=CHCHO)$.
$4$. Reaction of $but-2-enal$ with hydroxylamine $(NH_2OH)$ forms the corresponding oxime,$CH_3CH=CHCH=NOH$,which is $C$.

(B) The reaction of $m$-chlorobenzaldehyde with concentrated $NaOH$ is a Cannizzaro reaction.
Aldehydes that do not contain any $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) in the presence of a concentrated base.
In this reaction,one molecule of $m$-chlorobenzaldehyde is oxidized to the corresponding salt of the carboxylic acid (sodium $3$-chlorobenzoate),and another molecule is reduced to the corresponding alcohol ($3$-chlorobenzyl alcohol).
The reaction is: $2 \text{ } m\text{-Cl-C}_6\text{H}_4\text{CHO} + \text{NaOH (conc.)}$ $\rightarrow m\text{-Cl-C}_6\text{H}_4\text{CH}_2\text{OH} + m\text{-Cl-C}_6\text{H}_4\text{COONa}$.

(A) The reaction proceeds as follows:
$1$. Hydroboration-oxidation of methylenecyclopentane with $(i) B_2H_6$ and $(ii) H_2O_2/NaOH$ yields cyclopentylmethanol.
$2$. Oxidation of the primary alcohol with $PCC$ (Pyridinium chlorochromate) yields cyclopentanecarbaldehyde.
$3$. Nucleophilic addition of the Grignard reagent $PhMgBr$ to the aldehyde,followed by hydrolysis with $(v) H_2O$,yields the secondary alcohol,cyclopentylphenylmethanol.

(C) The reaction is a Cannizzaro reaction. In the mechanism,the nucleophilic attack of $OH^-$ on the carbonyl carbon of $Ph-CDO$ forms a tetrahedral intermediate. In the rate-determining step,a hydride (or deuteride) ion is transferred from this intermediate to another molecule of $Ph-CDO$. Since the intermediate contains a $D$ atom attached to the carbon,a deuteride ion $(D^-)$ is transferred to the second $Ph-CDO$ molecule. This results in the formation of $Ph-CD_2-O^-$ and $Ph-COO^-$. Upon workup (or proton exchange with the solvent),the alkoxide $Ph-CD_2-O^-$ becomes $Ph-CD_2-OH$.

(C) lactol exists in equilibrium with its open-chain hydroxy-aldehyde form.
In the presence of $NaBH_4$,the aldehyde group is reduced to a primary alcohol.
The lactol $S$ ($2$-hydroxytetrahydropyran) opens to form $5-$hydroxypentanal,which is then reduced by $NaBH_4$ to pentane$-1,5-$diol.

(D) The ease of dehydration of alcohols under acidic conditions depends on the stability of the carbocation formed after the loss of the water molecule.
In the given options,the dehydration of $4$-hydroxyhexan-$2$-one (Option $D$) leads to the formation of a carbocation that is stabilized by hyperconjugation with $5$ $\alpha$-hydrogen atoms.
Additionally,the resulting alkene product is conjugated with the carbonyl group,which provides extra stability through resonance.
Therefore,$4$-hydroxyhexan-$2$-one undergoes dehydration most readily.

(B) In an alkaline medium,the dehydration of $\beta$-hydroxy carbonyl compounds (aldol products) occurs via an $E1cB$ mechanism.
First,the base abstracts the acidic $\alpha$-proton to form an enolate ion.
Then,the hydroxide ion $(-OH)$ at the $\beta$-position is eliminated to form an $\alpha,\beta$-unsaturated carbonyl compound.
For this to occur,the molecule must have both a carbonyl group and a hydroxyl group at the $\beta$-position relative to each other,and an $\alpha$-hydrogen must be available.
Option $(B)$,$4$-hydroxy$-2-$pentanone,is a $\beta$-hydroxy ketone with an $\alpha$-hydrogen,which readily undergoes dehydration to form pent$-3-$en$-2-$one.

(D) The iodoform test is given by compounds containing the $CH_{3}CO-$ group or compounds that can be oxidized to this group,such as secondary alcohols with the $CH_{3}CH(OH)-$ group.
Option $A$: $CH_{3} CH_{2} CH_{2} CH_{2} CHO$ is a pentanal,which does not contain the $CH_{3}CO-$ group.
Option $B$: $CH_{3} CH_{2} CO CH_{2} CH_{3}$ is pentan-$3$-one,which lacks the $CH_{3}CO-$ group.
Option $C$: $CH_{3} CH_{2} CH_{2} CH_{2} CH_{2} OH$ is pentan-$1$-ol,a primary alcohol that does not form the required methyl ketone upon oxidation.
Option $D$: $CH_{3} CH_{2} CH_{2} CH(OH) CH_{3}$ is pentan-$2$-ol. It contains the $CH_{3}CH(OH)-$ group,which is oxidized by $I_{2}/NaOH$ to pentan-$2$-one $(CH_{3} CH_{2} CH_{2} CO CH_{3})$,a methyl ketone that subsequently undergoes the iodoform reaction to produce $CHI_{3}$ (iodoform).

(A) In the Cannizzaro reaction,an aldehyde lacking an $\alpha$-hydrogen undergoes disproportionation in the presence of a concentrated base to form an alcohol and a carboxylic acid salt. No new $C-C$ bond is formed in this process.
For example: $2HCHO + 50\% NaOH \rightarrow CH_3OH + HCOONa$.

(A) The reaction shows that compound $M$ reacts with $CH_3MgBr$ (a Grignard reagent) to release $CH_4$ gas. This indicates that $M$ contains an acidic hydrogen atom.
Upon treatment with $H^+$,the intermediate $N$ yields $CH_3COCH_2COCH_3$ (acetylacetone).
Acetylacetone has an active methylene group ($-CH_2-$ between two carbonyl groups) with acidic protons.
Therefore,$M$ must be $CH_3COCH_2COCH_3$ itself,as the Grignard reagent acts as a base to abstract the acidic proton from the active methylene group,forming the enolate $N$,which is then protonated back to the original ketone upon treatment with $H^+$.

(A) The reaction proceeds as follows:
$1$. $3$-bromo-$1$-chlorobenzene reacts with $Mg$ in dry $Et_2O$ to form the Grignard reagent,$3$-chlorophenylmagnesium bromide.
$2$. This Grignard reagent then undergoes a nucleophilic addition reaction with $4$-chlorobenzaldehyde.
$3$. Finally,aqueous $NH_4Cl$ workup protonates the alkoxide intermediate to yield the secondary alcohol,$3$-chlorophenyl-$4$-chlorophenylmethanol.

(A) The reaction of a carbonyl compound $R-CO-R'$ with methyl magnesium iodide $(CH_3MgI)$ followed by hydrolysis yields a tertiary or secondary alcohol with a chiral center if the four groups attached to the central carbon are different.
For the product to be chiral (enantiomeric),the central carbon must be bonded to four distinct groups: $R$,$R'$,$CH_3$,and $H$ (or $OH$ group).
$1$. Benzaldehyde $(C_6H_5CHO)$: Reacts to form $1$-phenylethanol,which has a chiral center $(C_6H_5, CH_3, H, OH)$. Thus,it produces enantiomers.
$2$. Propiophenone $(C_6H_5COCH_2CH_3)$: Reacts to form $2$-phenylbutan-$2$-ol. Here,the groups are $C_6H_5, CH_3, CH_2CH_3, OH$. All four are different,so it forms enantiomers.
$3$. Acetone $(CH_3COCH_3)$: Forms $2$-methylpropan-$2$-ol,which is achiral.
$4$. Acetaldehyde $(CH_3CHO)$: Forms propan-$2$-ol,which is achiral.
Given the options,Benzaldehyde is the standard example for producing a chiral secondary alcohol.

(A, C, D) The given compound is ethyl acetoacetate $(CH_3COCH_2COOC_2H_5)$.
$(A)$ It contains active methylene group,so it exhibits keto-enol tautomerism.
$(B)$ The enol form has an acidic hydroxyl group,so it reacts with metallic sodium to release $H_2$ gas.
$(C)$ The enol form gives a characteristic reddish-violet coloration with neutral $FeCl_3$ solution.
$(D)$ It contains a keto group,so it reacts with $2,4-$dinitrophenyl hydrazine $(2,4-DNP)$ to form a precipitate.
Therefore,statements $(A)$,$(C)$,and $(D)$ are correct.

(D) $1$. Reaction of benzaldehyde with methanol in the presence of dry $HCl$ gas leads to the formation of an acetal,$A$,which is benzaldehyde dimethyl acetal,$Ph-CH(OMe)_2$.
$2$. Treatment of the acetal $A$ with dilute $HCl$ results in its hydrolysis back to benzaldehyde $(Ph-CHO)$ and methanol.
$3$. The subsequent reaction of benzaldehyde with acetic anhydride $((CH_3CO)_2O)$ in the presence of sodium acetate $(CH_3COONa)$ is the Perkin condensation reaction,which yields cinnamic acid $(Ph-CH=CH-COOH)$ as the final product $B$.

(C) $NaBH_4$ is a selective reducing agent that reduces ketones and aldehydes to alcohols but does not reduce esters under mild conditions.
Ethyl $3$-oxobutanoate contains both a ketone group and an ester group.
Therefore,$NaBH_4$ will selectively reduce the ketone group at the $C-3$ position to a hydroxyl group,while the ester group at the $C-1$ position remains unaffected.
The reaction is:
$CH_3COCH_2COOC_2H_5 \xrightarrow{NaBH_4, CH_3OH} CH_3CH(OH)CH_2COOC_2H_5$
Thus,the product is ethyl $3$-hydroxybutanoate.

(A) The correct order of reactivity for the nucleophilic addition reaction of the given carbonyl compounds with ethylmagnesium iodide is $I > III > II > IV$.
This order is determined by two primary factors:
$(i)$ Inductive effect: Alkyl groups are electron-releasing. As the number of alkyl groups attached to the carbonyl carbon increases,the electron density on the carbonyl carbon increases,which reduces its electrophilicity and thus decreases reactivity towards nucleophilic attack.
$(ii)$ Steric effect: As the number and size of alkyl groups attached to the carbonyl carbon increase,the steric hindrance around the carbonyl carbon increases,making it more difficult for the nucleophile to attack.
Therefore,formaldehyde $(I)$ is the most reactive,followed by acetaldehyde $(III)$,acetone $(II)$,and finally di-tert-butyl ketone $(IV)$.