Properties Questions in English

(A) The correct answer is $(A)$. In $NMR$ spectroscopy,the splitting of signals (spin-spin coupling) is caused by the interaction of the magnetic moments of protons attached to adjacent carbon atoms,provided these protons are not chemically equivalent to the absorbing proton.

(C) The correct answer is $(C)$.
Fehling's solution is a mixture of two solutions:
$1$. Fehling's solution $A$: Aqueous copper sulphate $(CuSO_4 \cdot 5H_2O)$.
$2$. Fehling's solution $B$: Alkaline sodium potassium tartrate (Rochelle salt).

(C) The dipole moments of the given compounds are as follows:
$(i)$ $Trans-2-butene$: The dipole moment is $0$ due to the cancellation of bond dipoles.
$(ii)$ $1, 3-Dimethylbenzene$: It has a small dipole moment due to the inductive effect of the two methyl groups.
$(iii)$ $Acetophenone$ $(C_6H_5COCH_3)$: It has a high dipole moment due to the strong electron-withdrawing nature of the carbonyl group $(C=O)$,which creates a significant charge separation.
$(iv)$ $Ethanol$ $(C_2H_5OH)$: It has a moderate dipole moment due to the electronegativity difference between oxygen and hydrogen.
Comparing these,$Acetophenone$ has the highest dipole moment among the given options.

(C) The reactivity of carbonyl compounds towards nucleophilic addition depends on the electrophilicity of the carbonyl carbon.
Alkyl groups are electron-donating due to the $+I$ effect,which decreases the positive charge on the carbonyl carbon and increases steric hindrance.
As the number of alkyl groups attached to the carbonyl carbon increases,the reactivity towards nucleophilic addition decreases.
Therefore,the order of reactivity is:
$HCHO (I) > CH_3CHO (II) > CH_3COCH_3 (III)$

(B) . The reaction of $C_{2}H_{5}CHO$ with $CH_{3}MgBr$ followed by hydrolysis produces $CH_{3}CH(OH)C_{2}H_{5}$ (butan-$2$-ol).
In this molecule,the carbon atom attached to the $-OH$ group is bonded to four different groups: $-H$,$-OH$,$-CH_{3}$,and $-C_{2}H_{5}$.
Since all four substituents are different,this carbon is a chiral center,making the molecule chiral.

(D) The reaction $CH_3CHO + HCN \to CH_3CH(OH)CN$ involves the nucleophilic addition of cyanide to the carbonyl carbon of acetaldehyde.
Since the carbonyl carbon is planar,the cyanide ion can attack from either side with equal probability.
This leads to the formation of both $d$ (dextro) and $l$ (laevo) enantiomers in equal amounts.
Therefore,the resulting product is a racemic mixture.

(B) In the Clemmensen reduction,aldehydes and ketones are reduced to their corresponding alkanes using amalgamated $Zn-Hg$ and $HCl$.

(D) The reaction described is the Wacker process,where ethylene $(CH_2=CH_2)$ is oxidized to acetaldehyde $(CH_3CHO)$ in the presence of $PdCl_2$ and $CuCl_2$.
Thus,$X$ is acetaldehyde $(CH_3CHO)$.
Acetaldehyde contains a carbonyl group $(C=O)$ and undergoes nucleophilic addition reactions.
Among the given options,$HCN$ (hydrogen cyanide) is a classic reagent that readily undergoes nucleophilic addition with the carbonyl group of aldehydes to form cyanohydrins.
Therefore,$HCN$ is the correct reagent.

(D) The reaction of $PCl_5$ with a ketone like propanone $(CH_3COCH_3)$ involves the replacement of the carbonyl oxygen atom with two chlorine atoms at the same carbon position.
This results in the formation of a $gem$-dichloride (specifically $2,2$-dichloropropane).
The reaction is: $CH_3COCH_3 + PCl_5 \to CH_3-CCl_2-CH_3 + POCl_3$.

(B) The oxidation of an aldehyde group $(-CHO)$ to a carboxylic acid group $(-COOH)$ while preserving the carbon-carbon double bond $(C=C)$ requires a selective oxidizing agent.
$CH_3CH=CHCHO$ is an $\alpha,\beta$-unsaturated aldehyde.
Using $KMnO_4$ in an acidic medium allows for the selective oxidation of the aldehyde group to the corresponding carboxylic acid,$CH_3CH=CHCOOH$.

(D) The presence of the $-CHO$ group exerts a strong electron-withdrawing effect due to resonance and inductive effects,which decreases the electron density at the $o-$ (ortho) and $p-$ (para) positions of the benzene ring.
As a result,the electrophilic substitution reaction is directed towards the $m-$ (meta) position,where the electron density is relatively higher compared to the $o-$ and $p-$ positions.
Therefore,$-CHO$ is a meta-directing group.

(B) Milk can be preserved by adding a few drops of $HCHO$ (formaldehyde) solution. $HCHO$ acts as a preservative by inhibiting the growth of bacteria in the milk.

(A) The process of silvering a mirror involves the deposition of a thin layer of metallic silver on the glass surface. This is typically achieved using $AgNO_3$ (silver nitrate) in the presence of a reducing agent (like glucose or formaldehyde) through the Tollens' test reaction. Therefore,$AgNO_3$ is the correct compound used for this purpose.

(A) Tollen's reagent is an ammoniacal silver nitrate solution.
It contains the diamminesilver$(I)$ complex ion,which is represented as $[Ag(NH_3)_2]^+$.
It is used to detect the presence of aldehyde groups in organic compounds.

(B) When chloral $(CCl_3CHO)$ is boiled with an aqueous solution of sodium hydroxide $(NaOH)$,it undergoes a haloform reaction to produce chloroform $(CHCl_3)$ and sodium formate $(HCOONa)$.
The chemical equation is: $CCl_3CHO + NaOH \xrightarrow{\text{Boil}} CHCl_3 + HCOONa$.

(D) The reaction of ethanol with excess chlorine proceeds in two steps:
$1$. Ethanol is first oxidized to acetaldehyde: $CH_3CH_2OH + Cl_2 \to CH_3CHO + 2HCl$
$2$. Acetaldehyde then undergoes chlorination to form chloral $(CCl_3CHO)$: $CH_3CHO + 3Cl_2 \to CCl_3CHO + 3HCl$
Therefore,the final product is chloral.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$Diethyl \ ketone$ $(CH_3CH_2-CO-CH_2CH_3)$ does not contain the $CH_3CO-$ group,therefore it does not form a yellow precipitate of iodoform $(CHI_3)$.

$CH_3COCH_3 \xrightarrow{NaOH/I_2} CHI_3$	$Yellow \ ppt$
$C_2H_5OH \xrightarrow{NaOH/I_2} CHI_3$
$CH_3-CH(OH)-CH_3 \xrightarrow{NaOH/I_2} CHI_3$

(B) The reaction of a compound with iodine $(I_2)$ and alkali $(NaOH)$ is known as the iodoform test.
Compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group give a yellow precipitate of iodoform $(CHI_3)$.
Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group.
The reaction is:
$C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 (\text{yellow ppt.}) + C_6H_5COONa + 3NaI + 3H_2O$.

(C) The reaction of acetone $(CH_3COCH_3)$ with iodine $(I_2)$ in the presence of sodium hydroxide $(NaOH)$ is a classic iodoform test.
$CH_3COCH_3 + 3I_2 + 4NaOH \to CHI_3 + 3NaI + CH_3COONa + 3H_2O$
In this reaction,acetone forms iodoform $(CHI_3)$,which is a yellow precipitate.

(D) The iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Among the given options,$Ethanol$ $(CH_3CH_2OH)$ contains the $CH_3CH(OH)-$ group,which is oxidized to acetaldehyde $(CH_3CHO)$ during the reaction,subsequently undergoing the iodoform test.
Formalin,methanol,and acetic acid do not contain these structural requirements.

(A) The molecular formula $C_2Cl_3OH$ can be written as $CCl_3CHO$ (Chloral).
Since $A$ reduces Fehling's solution,it must contain an aldehyde $(-CHO)$ group.
Chloral $(CCl_3CHO)$ is formed by the chlorination of ethyl alcohol $(C_2H_5OH)$:
$C_2H_5OH + 4Cl_2 \to CCl_3CHO + 5HCl$.
Upon oxidation,Chloral gives trichloroacetic acid $(CCl_3COOH)$,which is a monocarboxylic acid $(B)$.
Therefore,$A$ is Chloral.

(D) Grignard reagents $(RMgX)$ are nucleophilic and readily undergo addition reactions with polar multiple bonds such as $C = O$,$-C \equiv N$,and $C = S$ groups.
$1$. Reaction with $C = O$: $RMgX + >C = O$ $\rightarrow >C(R)-OMgX$ $\xrightarrow{H_2O} >C(R)-OH + Mg(OH)X$
$2$. Reaction with $-C \equiv N$: $RMgX + -C \equiv N$ $\rightarrow -C(R) = NMgX$ $\xrightarrow{H_2O} -C(R) = O + NH_3 + Mg(OH)X$
$3$. Reaction with $C = S$: $RMgX + >C = S$ $\rightarrow >C(R)-SMgX$ $\xrightarrow{H_2O} >C(R)-SH + Mg(OH)X$
Since Grignard reagents react with all these groups,the correct option is $(D)$.

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

(B) An acetal is a functional group with the general structure $R_2C(OR')_2$,where $R$ can be hydrogen or an alkyl/aryl group and $R'$ is an alkyl group.
Since an acetal contains two ether linkages $(-C-O-C-)$ attached to the same carbon atom,it is classified as a diether.

(B) The reaction of formaldehyde $(HCHO)$ with methyl magnesium iodide $(CH_3MgI)$ is a nucleophilic addition reaction.
Step $1$: Formaldehyde reacts with $CH_3MgI$ to form an addition product: $HCHO + CH_3MgI \rightarrow CH_3CH_2OMgI$.
Step $2$: Aqueous hydrolysis of the addition product yields ethyl alcohol: $CH_3CH_2OMgI + H_2O \rightarrow CH_3CH_2OH + Mg(OH)I$.
Thus,the final product is ethyl alcohol $(CH_3CH_2OH)$.

(B) Benzaldehyde $(C_6H_5CHO)$ lacks $\alpha$-hydrogen atoms.
When treated with concentrated alkali (like $NaOH$),it undergoes a self-oxidation and reduction reaction known as the Cannizzaro reaction.
In this reaction,one molecule of benzaldehyde is reduced to benzyl alcohol $(C_6H_5CH_2OH)$ and another is oxidized to sodium benzoate $(C_6H_5COONa)$.
The reaction is: $2C_6H_5CHO + conc. NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$.

(C) The reaction between formaldehyde $(HCHO)$ and concentrated potassium hydroxide $(KOH)$ is a classic example of the Cannizzaro reaction.
$2HCHO + KOH \rightarrow CH_3OH + HCOOK$
In this reaction,one molecule of formaldehyde is reduced to methyl alcohol $(CH_3OH)$,while the other is oxidized to potassium formate $(HCOOK)$.

(D) The Grignard reagent $(RMgX)$ is a strong nucleophile that reacts with compounds containing polar multiple bonds.
It readily undergoes nucleophilic addition reactions with carbonyl groups $(>C=O)$,thiocarbonyl groups $(>C=S)$,and cyano groups $(-C \equiv N)$.
Therefore,the correct option is $D$.

(D) Acetone $(CH_3COCH_3)$ reacts with Grignard reagent $(CH_3MgI)$ to form an addition product.
$CH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3C-OMgI$
Upon further hydrolysis,the addition product yields $2-$methyl$-2-$propanol (also known as $tert-$butyl alcohol).
$(CH_3)_3C-OMgI + H_2O \rightarrow (CH_3)_3C-OH + Mg(OH)I$

(C) When an ester $(RCOOR')$ reacts with a Grignard reagent $(R''MgX)$,the reaction proceeds in two steps.
First,the Grignard reagent attacks the carbonyl carbon of the ester to form a ketone intermediate $(R-CO-R'')$ and an alkoxide $(R'OMgX)$.
Since the Grignard reagent is in excess,it immediately reacts with the newly formed ketone to produce an alkoxide intermediate.
Finally,upon acidic hydrolysis,this intermediate yields a tertiary alcohol $(R-C(OH)(R'')_2)$.

(B) The reduction of benzaldehyde $(C_6H_5CHO)$ using a reducing agent like $LiAlH_4$ or $NaBH_4$ leads to the formation of a primary alcohol.
The reaction is as follows:
$C_6H_5CHO + 2[H] \xrightarrow{LiAlH_4} C_6H_5CH_2OH$
Thus,the product formed is benzyl alcohol.

(A) Acetyl bromide $(CH_3COBr)$ reacts with the first equivalent of $CH_3MgI$ to form acetone $(CH_3COCH_3)$.
Since $CH_3MgI$ is in excess,it further reacts with the formed acetone to produce a tertiary alkoxide intermediate $(CH_3-C(CH_3)_2-OMgI)$.
Finally,hydrolysis of this intermediate with a saturated solution of $NH_4Cl$ yields $2$-methyl-$2$-propanol $(CH_3-C(CH_3)_2-OH)$.
Reaction: $CH_3COBr + 2CH_3MgI$ $\rightarrow CH_3-C(CH_3)_2-OMgI$ $\xrightarrow{NH_4Cl} CH_3-C(CH_3)_2-OH$.

(A) $2C_6H_5CHO + NaOH \to C_6H_5CH_2OH + C_6H_5COONa$
Benzaldehyde undergoes the Cannizzaro reaction in the presence of concentrated alkali.
This is a disproportionation reaction where one molecule of benzaldehyde is reduced to benzyl alcohol and another is oxidized to sodium benzoate.

(C) $1$. Reaction with $CH_3MgBr$: Ethanal $(CH_3CHO)$ reacts with methylmagnesium bromide $(CH_3MgBr)$ followed by hydrolysis to form $2$-propanol $(CH_3CH(OH)CH_3)$.
$2$. Reaction with $C_2H_5OH$/dry $HCl$: Ethanal reacts with two equivalents of ethanol $(C_2H_5OH)$ in the presence of dry $HCl$ to form an acetal $(CH_3CH(OC_2H_5)_2)$.

(B) $NaBH_4$ is a selective reducing agent that reduces the carbonyl group $(-CHO)$ to a primary alcohol $(-CH_2OH)$ without affecting the carbon-carbon double bond $(C=C)$.
$H_2/Ni$ is a non-selective reducing agent that would reduce both the carbonyl group and the carbon-carbon double bond.
Therefore,the correct reagent is $NaBH_4$.
$C_6H_5-CH=CHCHO \xrightarrow{NaBH_4} C_6H_5-CH=CHCH_2OH$

(B) The Jones reagent ($CrO_3$ in $H_2SO_4/H_2O$) is a strong oxidizing agent used for the oxidation of alcohols.
It oxidizes primary alcohols to carboxylic acids and secondary alcohols to ketones.
The given substrate is $pent-3-en-2-ol$,which is an allylic secondary alcohol.
Jones reagent oxidizes the secondary alcohol group $(-CH(OH)-)$ to a ketone group $(-C=O)$ without affecting the carbon-carbon double bond $(C=C)$ under standard conditions.
Therefore,the reaction is:
$CH_3-CH=CH-CH(OH)-CH_3 \xrightarrow{Jones \ reagent} CH_3-CH=CH-COCH_3$
The product $X$ is $pent-3-en-2-one$ $(CH_3CH=CHCOCH_3)$.

(A) The iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
All the given options ($CH_3COCH_3$,$CH_3COC_2H_5$,$C_6H_5COCH_3$,and $CH_3COC_6H_5$) contain the $CH_3CO-$ group attached to a carbon or hydrogen atom.
Therefore,all these compounds are capable of undergoing the iodoform reaction.
However,in the context of standard chemistry problems where a single choice is required,it is noted that all these methyl ketones respond positively to the iodoform test.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. Ethyl alcohol $(C_2H_5OH)$ contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test: $C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$.
$2$. Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test: $CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$.
$3$. Methyl alcohol $(CH_3OH)$ does not contain either of these groups,so it does not give the iodoform test.
Therefore,only ethyl alcohol and acetone give the iodoform test.

(D) Aldehydes react with one equivalent of a monohydric alcohol in the presence of dry $HCl$ gas to yield alkoxyalcohol intermediate,known as a hemiacetal.
Further reaction with another molecule of alcohol in the presence of dry $HCl$ gives a gem-dialkoxy compound,which is known as an acetal.
The general reaction is:
$R-CHO + R'OH \xrightarrow{dry \ HCl} R-CH(OH)(OR')$ (Hemiacetal)
$R-CH(OH)(OR') + R'OH \xrightarrow{dry \ HCl} R-CH(OR')_2 + H_2O$ (Acetal)
Therefore,the correct option is $(D)$.

(C) The conversion of $pent-3-en-2-ol$ to $pent-3-yn-2-one$ involves the oxidation of the secondary alcohol group to a ketone while preserving the carbon-carbon double bond (or in this case,the triple bond formation requires specific conditions).
Chromic anhydride in glacial acetic acid (Jones reagent or similar conditions) is effective for the oxidation of alcohols to ketones without affecting the unsaturation in the chain.

(C) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on the magnitude of the positive charge on the carbonyl carbon atom.
Alkyl groups are electron-donating ($+I$ effect),which decreases the positive charge on the carbonyl carbon and reduces its reactivity.
In $HCHO$ (formaldehyde),there are no alkyl groups attached to the carbonyl carbon.
In $CH_3CHO$ (acetaldehyde),one methyl group is attached.
In $C_6H_5CHO$ (benzaldehyde),the phenyl group provides resonance stabilization,further reducing the electrophilicity of the carbonyl carbon.
Therefore,the order of reactivity is $HCHO > CH_3CHO > C_6H_5CHO$.
Thus,$HCHO$ is the most reactive.

(C) Step $1$: Hydration of ethyne $(HC \equiv CH)$ in the presence of $1\% \, HgSO_4$ and $20\% \, H_2SO_4$ yields acetaldehyde $(CH_3CHO)$ as product $A$.
Step $2$: Reaction of acetaldehyde $(CH_3CHO)$ with Grignard reagent $(CH_3MgX)$ followed by hydrolysis yields isopropyl alcohol $(CH_3CHOHCH_3)$ as product $B$.
Step $3$: Oxidation of isopropyl alcohol $(CH_3CHOHCH_3)$ using an oxidizing agent $([O])$ yields acetone $(CH_3COCH_3)$.

(D) Step $1$: Acetylene $(HC \equiv CH)$ undergoes hydration in the presence of $30\% \ H_2SO_4$ and $HgSO_4$ to form vinyl alcohol,which tautomerizes to acetaldehyde $(CH_3CHO)$. Thus,$A = CH_3CHO$.
Step $2$: Acetaldehyde contains $\alpha$-hydrogen atoms,so it undergoes aldol condensation in the presence of dilute $NaOH$ to form $3$-hydroxybutanal $(CH_3-CH(OH)-CH_2CHO)$. Thus,$B = CH_3-CH(OH)-CH_2CHO$.

(C) The Clemmensen reduction involves the reduction of carbonyl groups $(C=O)$ to methylene groups $(CH_2)$ using zinc amalgam $(Zn(Hg))$ and concentrated hydrochloric acid $(HCl)$.
Acetone is $CH_3COCH_3$.
When acetone undergoes Clemmensen reduction,the carbonyl group is reduced to a methylene group:
$CH_3COCH_3 + 4[H] \xrightarrow{Zn(Hg)/HCl} CH_3CH_2CH_3 + H_2O$.
The product formed is propane.

(A) $NH_3$ reacts with these aldehydes as follows:
$(i)$ $6HCHO + 4NH_3 \rightarrow (CH_2)_6N_4 + 6H_2O$
$(ii)$ $CH_3CHO + NH_3 \rightarrow CH_3CH(OH)NH_2$
$(iii)$ $C_6H_5CHO + NH_3 \rightarrow C_6H_5-CH(NCHC_6H_5)_2$
The product $(CH_2)_6N_4$ is known as urotropine (hexamethylenetetramine),which is used as a urinary antiseptic.

(D) Benzoin is an organic compound with the chemical formula $C_6H_5CH(OH)COC_6H_5$.
It is formed by the benzoin condensation of benzaldehyde.
The structure contains a ketone group $(C=O)$ with a hydroxyl group $(-OH)$ attached to the adjacent carbon atom (the $\alpha$-carbon).
Therefore,it is classified as an $\alpha$-hydroxy ketone.

(C) The reaction of acetone $(CH_3COCH_3)$ with methylmagnesium chloride $(CH_3MgCl)$ is a nucleophilic addition reaction.
Step $1$: $CH_3COCH_3 + CH_3MgCl \to (CH_3)_3COMgCl$.
Step $2$: Upon hydrolysis,the intermediate $(CH_3)_3COMgCl$ yields tert-butyl alcohol $(CH_3)_3COH$ as the final product $Y$.
Thus,$X = CH_3MgCl$ and $Y = (CH_3)_3COH$.

(B) The reaction of $m$-chlorobenzaldehyde with concentrated $KOH$ $(50\%)$ is a classic example of the Cannizzaro reaction.
Aldehydes that do not contain $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali.
$m$-chlorobenzaldehyde lacks $\alpha$-hydrogens,so it undergoes the Cannizzaro reaction to form $m$-chlorobenzoate and $m$-chlorobenzyl alcohol.

(A) The reaction of a ketone $(RR'C=O)$ with $HCN$ in the presence of $KCN$ (catalyst) leads to the formation of a cyanohydrin,where $A = RR'C(OH)(CN)$.
Next,the reduction of the cyanohydrin $(RR'C(OH)(CN))$ using a strong reducing agent like $LiAlH_4$ reduces the nitrile group $(-CN)$ to a primary amine group $(-CH_2NH_2)$.
Therefore,$A = RR'C(OH)(CN)$ and $B = LiAlH_4$.

(A) The reduction of aldehydes and ketones to hydrocarbons is known as the $Clemmensen$ reduction.
In this reaction,the carbonyl group is reduced to a methylene group $(-CH_2-)$ using $Zn-Hg$ (zinc amalgam) and concentrated $HCl$.