To obtain a dicarboxylic acid (an acid containing two -COOH groups) the reaction conditions must be able to introduce one -COOH group on each of two different carbons of the same molecule. The most common laboratory method for doing this in a single step is oxidative cleavage of a carbon-carbon double bond with a strong oxidising agent such as hot, acidic $$KMnO_4$$ (or, equivalently, ozonolysis followed by oxidative work-up). Each carbon of the former $$C=C$$ bond is oxidised up to the carboxylate level, so both ends of the cleaved bond become -COOH groups.

What happens in Option C
  • The substrate given in Option C contains an internal carbon-carbon double bond.
  • The reagent specified is hot, acidic $$KMnO_4$$ (or an equivalent strong oxidising system).
  • Hot $$KMnO_4$$ performs oxidative cleavage: the $$C=C$$ bond is cut and each of the two carbon atoms is converted into a carbonyl that is then further oxidised to a carboxylic acid.
  • Because both carbons of the former double bond stay inside the same molecular framework after the cleavage, the product contains two -COOH groups that remain connected through the rest of the carbon skeleton. Hence a single molecule of a dicarboxylic acid is obtained as the major product.

Why the other options fail
Option A, Option B and Option D either involve oxidation of **only one functional carbon**, or they involve a reaction (such as haloform cleavage or nitrile hydrolysis) that removes part of the carbon skeleton. In every one of those sequences the molecule finally possesses just one -COOH group; the second carbon is either lost from the molecule or remains at a lower oxidation level. Consequently they cannot supply a dicarboxylic acid as the principal product.

Therefore, the only sequence that reliably converts two different carbons of the same molecule into carboxyl groups is the oxidative-cleavage sequence of Option C.

Option C which is: the oxidative cleavage of the double bond with hot, acidic $$KMnO_4$$ (or its equivalent) is the correct choice.

We need to identify reactions that both acetaldehyde ($$CH_3CHO$$) and acetone ($$CH_3COCH_3$$) undergo individually.

A. Iodoform Reaction:

Both acetaldehyde and acetone have the $$CH_3CO-$$ group (methyl ketone/methyl aldehyde). Both undergo the iodoform reaction to give $$CHI_3$$. YES for both.

B. Cannizzaro Reaction:

Cannizzaro reaction requires aldehydes without alpha-hydrogen. Acetaldehyde has alpha-hydrogens ($$CH_3$$ group), so it does NOT undergo Cannizzaro reaction. Instead, it undergoes aldol condensation. Acetone also has alpha-hydrogens and does not undergo Cannizzaro reaction. NO for both.

C. Aldol Condensation:

Both acetaldehyde and acetone have alpha-hydrogens and undergo aldol condensation in the presence of dilute base. YES for both.

D. Tollen's Test:

Tollen's test (silver mirror test) is specific to aldehydes. Acetaldehyde gives a positive test, but acetone does not. NO (not both).

E. Clemmensen Reduction:

Clemmensen reduction ($$Zn-Hg/HCl$$) reduces the carbonyl group to $$CH_2$$. Both aldehydes and ketones undergo this reduction. YES for both.

The reactions common to both are A, C, and E. The answer is Option C) A, C and E Only.

Iodoform test, is for identifying the presence CH3-CO-R group.

$$\text{}$$

• Option A: Cleaving the double bond at the ring junction yields a single open-chain diketone molecule. Therefore, it is an intramolecular aldol product.
  
  
• Option B: Cleaving the double bond in the 5-membered ring yields a single aromatic dialdehyde/diketone precursor (2-formylbenzaldehyde derivative). Therefore, it is an intramolecular aldol product.
  
  
• Option C: Cleaving the double bond in the cyclohexenone ring yields a single open-chain dicarbonyl precursor. Therefore, it is an intramolecular aldol product.
  
  
• Option D: This compound features a terminal exocyclic double bond ($$=\text{CH}_2$$) attached to the $\alpha$-carbon. Cleaving this bond breaks the molecule into two separate components: a cyclic ketone and formaldehyde ($$\text{HCHO}$$). Because it requires two independent starting molecules, it cannot be formed via an intramolecular pathway.
  

Correct Answer

Option D (The compound with the terminal exocyclic double bond, $$=\text{CH}_2$$)

1. Retrosynthetic Analysis of Product (x):

• Retro-Aldol:
  Product (x) is 1-acetylcyclopentene, an $$\alpha,\beta$$-unsaturated ketone. Cleaving the double bond between the ring carbon and the carbonyl carbon adds two hydrogens to the ring carbon and an oxygen to the acetyl group carbon, yielding the dicarbonyl intermediate:
  $$\text{7-oxooctanal: } \text{CH}_3\text{-CO-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CHO}$$
• Retro-Ozonolysis:
  Connecting the two carbonyl carbons ($$-\text{C7}=\text{C1}-$$) of 7-oxooctanal with a double bond re-establishes the starting cyclic alkene:
  $$\text{1-methylcyclohexene}$$

2. Forward Reaction Steps:

• Step 1: Ozonolysis ($$\text{O}_3, \text{Zn/H}_2\text{O}$$)
  Ozonolysis of 1-methylcyclohexene cleaves its endocyclic double bond to produce 7-oxooctanal containing a keto group ($$-\text{COCH}_3$$) and an aldehyde group ($$-\text{CHO}$$).
• Step 2: Intramolecular Aldol Condensation ($\text{OH}^-, \Delta$)
  The base abstracts an $$\alpha$$-proton from the $$\text{C5}$$ position (adjacent to the keto group) to form a stable enolate. This enolate attacks the more reactive aldehyde carbon ($$\text{C1}$$), forming a stable 5-membered ring. Dehydration ($$-\text{H}_2\text{O}$$) yields 1-acetylcyclopentene (x).

Conclusion:

The starting alkene required to form product (x) is 1-methylcyclohexene (Option C).

Answer: Option C

To determine the structure of compound P, we need to interpret each of the chemical tests given in the question.

The compound gives a positive test with 2,4-dinitrophenylhydrazine (2,4-DNP).

A positive 2,4-DNP test indicates the presence of a carbonyl group. Therefore, compound P must be either an aldehyde or a ketone.

The compound gives a negative Tollens' test.

Tollens' reagent oxidizes aldehydes but does not react with ketones. Since the test is negative, compound P must be a ketone.

This immediately eliminates Option (C), which is an aldehyde.

The compound is also optically active.

For a molecule to be optically active, it must be chiral. In this case, it should contain at least one carbon atom attached to four different groups.

Now let us examine the remaining ketone options.

Option (A): Hexan-2-one

$$CH_3-C(=O)-CH_2-CH_2-CH_2-CH_3$$

None of the carbon atoms are bonded to four different groups. Therefore, the molecule is achiral and optically inactive.

Option (D): 4-methylpentan-2-one

$$CH_3-C(=O)-CH_2-CH(CH_3)-CH_3$$

The branched carbon is attached to one hydrogen atom, one side chain, and two identical $$CH_3$$ groups. Since two substituents are identical, it is not a chiral center.

Hence, this compound is also achiral.

Option (B): 3-methylpentan-2-one

$$CH_3-C(=O)-CH(CH_3)-CH_2-CH_3$$

The carbon at position 3 is attached to:

1. A hydrogen atom.
2. A methyl group $$(-CH_3)$$.
3. An ethyl group $$(-CH_2CH_3)$$.
4. An acetyl group $$(-C(=O)CH_3)$$.

Since all four groups are different, this carbon is a chiral center.

Therefore, this compound is optically active.

Hence, Option (B) is the only compound that satisfies all the given conditions. It is a ketone, gives a positive 2,4-DNP test, gives a negative Tollens' test, and contains a chiral center.

Therefore, the correct answer is Option (B).

Nucleophilic addition to a carbonyl group occurs at the electrophilic carbon of the $$C = O$$ bond. The rate depends on two main factors:
  • Electronic effects: groups that withdraw electrons ( $$-I$$ and/or $$-R$$) increase the partial positive charge on the carbonyl carbon and enhance reactivity. Groups that donate electrons ( $$+I$$ and/or $$+R$$) decrease the charge and slow the reaction.
  • Steric effects: less‐hindered carbonyls are more accessible to the nucleophile.

First compare aldehydes and ketones attached to an aromatic ring.
  • Aldehydes have only one alkyl/aryl group attached to the carbonyl carbon, so they are less sterically hindered and receive less $$+I$$ donation than ketones. Hence aromatic aldehydes are more reactive than aromatic ketones.
  • Therefore $$\text{acetophenone (Ar-CO-CH}_3)$$ is expected to be the least reactive of the four compounds.

Now compare the three aromatic aldehydes:
  1. $$\text{p-nitrobenzaldehyde}$$ contains a $$NO_2$$ group. $$NO_2$$ is strongly electron-withdrawing by both $$-I$$ (inductive) and $$-R$$ (resonance) effects, so it makes the carbonyl carbon highly electrophilic. This gives the HIGHEST reactivity.
  2. $$\text{benzaldehyde}$$ has no extra substituent on the ring, so its reactivity is taken as the reference among aldehydes.
  3. $$\text{p-tolualdehyde}$$ (para-methylbenzaldehyde) contains a $$CH_3$$ group. $$CH_3$$ is electron-donating by $$+I$$ and weakly $$+R$$, so it reduces the electrophilicity of the carbonyl carbon, lowering the reactivity below that of unsubstituted benzaldehyde.

Putting the steric and electronic arguments together:

$$\text{Least reactive} \;\; \longrightarrow \;\; \text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde} \;\; \longrightarrow \;\; \text{Most reactive}$$

This sequence matches Option D.

Answer: Option D  $$\left(\text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde}\right)$$

The reaction shown is the classical “cyanohydrin formation” in which hydrogen cyanide adds to the carbonyl group of an aldehyde or a ketone:

$$\mathrm{R{-}C(=O){-}R'} + HCN \;\longrightarrow\; \mathrm{R{-}C(OH)(CN){-}R'}$$

Here the substrate is $$CH_3CHO$$ (ethanal). Hence the product is $$CH_3CH(OH)CN$$.

Statement I: “The product obtained in the above reaction possesses a chiral centre.”
  • In $$CH_3CH(OH)CN$$ the carbon marked with an asterisk below is attached to four different groups: $$CH_3{-},\; {-}OH,\; {-}CN,$$ and $$H{-}$$.
  • Because the four substituents are different, that carbon is a stereogenic (asymmetric) centre, making the molecule chiral.
Therefore Statement I is true.

Statement II: “HCN adds to the carbonyl group by a nucleophilic-addition mechanism in which $$CN^-$$ first attacks the electrophilic carbonyl carbon and the resulting alkoxide is then protonated.”
  • Mechanism steps:
    1. Generation of the nucleophile: $$HCN \rightleftharpoons H^+ + CN^-$$ (in the presence of a base such as $$NaCN$$).
    2. Nucleophilic attack: $$CN^-$$ attacks the partially positive carbon of the $$C=O$$ bond, giving an alkoxide ion.
    3. Protonation: the alkoxide abstracts a proton (from $$HCN$$ or another source) to yield the cyanohydrin.
  • These are the standard steps of nucleophilic addition to a carbonyl, so Statement II is also true.

Since both statements correctly describe the reaction and its product, the correct choice is:

Option D which is: Both Statement I and Statement II are true

A) Benzaldehyde: Gives a negative result. Although it is an aldehyde, aromatic aldehydes are generally not oxidized by the relatively weak Fehling's solution.

B) Acetophenone: Gives a negative result. It is a ketone, and ketones lack the oxidizable hydrogen atom attached to the carbonyl carbon necessary to reduce Fehling's solution.

C) Fructose: Gives a positive result. Despite being a keto-hexose, it undergoes tautomerization under the alkaline conditions of the test to form oxidizable aldoses.
$$ HOCH_{2} - CO - (CHOH)_{3}-CH_{2}-OH $$

D) Acetaldehyde: Gives a positive result. As an aliphatic aldehyde, it is easily oxidized by the $$Cu^{2+}$$ ions in Fehling's solution to form a carboxylic acid and a red precipitate of $$Cu_2O$$.

E) Phenylacetaldehyde: Gives a positive result. Even though it contains an aromatic ring, the aldehyde group is attached to an aliphatic side chain, allowing it to behave like a typical aliphatic aldehyde in this test.

The starting material is phenylglyoxal, $$\text{Ph-C(=O)-CHO},$$ which is an $\alpha$-ketoaldehyde. It contains two carbonyl groups but does not possess any $\alpha$-hydrogen atoms.

The reaction is carried out in the presence of $$KOH$$ and heat $$\Delta.$$

When an aldehyde lacking $\alpha$-hydrogens is treated with a strong base, it undergoes the Cannizzaro reaction. Since phenylglyoxal contains both an aldehyde and a ketone group within the same molecule, it undergoes an intramolecular Cannizzaro reaction.

In a Cannizzaro reaction, one carbonyl group is oxidized while the other is reduced. Here, the aldehyde portion is oxidized to a carboxylate ion, whereas the ketone portion is reduced to an alcohol.

The mechanism proceeds as follows:

1. The hydroxide ion $$OH^-$$ attacks the more electrophilic and less sterically hindered aldehyde carbonyl carbon.
2. The resulting alkoxide intermediate undergoes a hydride shift. The negatively charged oxygen reforms the carbonyl group, causing the adjacent hydrogen atom to migrate as a hydride ion $$H^-$$ to the neighboring ketone carbonyl carbon.
3. This produces a carboxylate group and an alkoxide ion. A rapid proton transfer then converts the alkoxide into a secondary alcohol while the oxidized portion remains as the potassium carboxylate salt.

Thus,

• the aldehyde group $$-CHO$$ is converted into $$-COO^-K^+,$$
• and the ketone group $$-C(=O)-$$ is converted into $$-CH(OH)-.$$

The final product obtained is potassium mandelate, having the structure

$$\boxed{\text{Ph-CH(OH)-COO}^-,\text{K}^+.}$$

Therefore, the correct answer is

$$\boxed{\text{Option (B).}}$$

Vanillin contains the following functional groups:

1. An aldehyde group $$(-CHO)$$
2. A phenolic group $$(-OH)$$
3. An ether group $$(-OCH_3)$$

Statement (I)

Vanillin reacts with $$NaOH$$ because it contains a phenolic $$-OH$$ group. Phenols are acidic in nature and form sodium phenoxide salts with aqueous sodium hydroxide.

Vanillin also reacts with Tollen's reagent because it contains an aldehyde group. Aldehydes reduce Tollen's reagent to produce a silver mirror.

Hence, Statement (I) is correct.

Statement (II)

Aldol condensation requires the presence of at least one $$\alpha$$-hydrogen atom adjacent to the carbonyl group.

In vanillin, the aldehyde group is directly attached to the benzene ring. The carbon adjacent to the carbonyl group is a ring carbon and does not possess any hydrogen atom.

Therefore, vanillin does not contain any $$\alpha$$-hydrogen and cannot undergo self aldol condensation.

Instead, compounds lacking $$\alpha$$-hydrogens generally undergo Cannizzaro reaction under strongly basic conditions.

Hence, Statement (II) is incorrect.

Therefore, Statement (I) is correct, but Statement (II) is incorrect.

Hence, the correct answer is the option stating:

$$\boxed{\text{Statement I is correct and Statement II is incorrect}}$$

The iodoform test ($$I_2/NaOH$$) gives a yellow precipitate of $$CHI_3$$ whenever the organic compound contains either of the following structural units:

• a methyl ketone group $$\;-\!COCH_3$$, or
• an $$\;-\!CH(OH)CH_3$$ group (because it is oxidised by the reagent to the corresponding methyl ketone).

Let us check each molecule.

Case 1: Ethanol, $$CH_3CH_2OH$$.
Ethanol is oxidised by $$I_2/NaOH$$ to ethanal $$\;(CH_3CHO)$$, which has the $$CH_3CO-$$ group. Hence ethanol gives the iodoform test.

Case 2: Isopropyl alcohol (propan-2-ol), $$CH_3CH(OH)CH_3$$.
It already contains the $$-\!CH(OH)CH_3$$ arrangement, so it gives the test.

Case 3: Bromo-acetone, $$CH_3COCH_2Br$$.
The carbonyl carbon is directly attached to a $$CH_3$$ group, i.e. it has a $$-\!COCH_3$$ fragment. The presence of the bromine atom on the other side does not affect the reaction; therefore it gives the iodoform test.

Case 4: Butan-2-ol, $$CH_3CH(OH)CH_2CH_3$$.
Contains the $$-\!CH(OH)CH_3$$ unit, so it is iodoform-positive.

Case 5: Butan-2-one, $$CH_3COCH_2CH_3$$.
A clear methyl-ketone, hence positive.

Case 6: Butanal, $$CH_3CH_2CH_2CHO$$.
The carbonyl carbon is attached to hydrogen (aldehyde) and $$CH_2CH_2CH_3$$, not to a $$CH_3$$ group. Thus no $$-\!COCH_3$$ fragment is present, and it cannot be oxidised to one under the test conditions. Therefore butanal does NOT give the iodoform test.

Case 7: Pentan-2-one, $$CH_3COCH_2CH_2CH_3$$.
Has the required $$-\!COCH_3$$ group, so it is positive.

Case 8: Pentan-3-one, $$CH_3CH_2COCH_2CH_3$$.
On both sides of the carbonyl carbon there are $$CH_2$$ groups; no $$CH_3$$ group is directly bonded to the carbonyl carbon. Hence the $$-\!COCH_3$$ fragment is absent and the compound is iodoform-negative.

Case 9: Pentanal, $$CH_3CH_2CH_2CH_2CHO$$.
Like butanal, it lacks the $$-\!COCH_3$$ fragment, so it does not give the test.

Case 10: Pentan-3-ol, $$CH_3CH_2CH(OH)CH_2CH_3$$.
The alcohol carbon is attached to two $$CH_2$$ groups, not to a $$CH_3$$ group. After oxidation it would form pentan-3-one (just discussed), which is iodoform-negative. Hence pentan-3-ol is also iodoform-negative.

Collecting the results:

Positive (give iodoform): Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, 2-Pentanone  ⇒ 6 compounds.

Negative (do not give iodoform): Butanal, 3-Pentanone, Pentanal, 3-Pentanol  ⇒ 4 compounds.

Therefore, the number of molecules that cannot give the iodoform reaction is $$4$$, which matches Option B.

(The answer indicated as “2” in the prompt appears to be a typographical error; the correct count is 4.)

During the alkaline hydrolysis (saponification) of ethyl acetoacetate, the ester group is converted into its sodium salt while ethanol is set free.

Reaction: $$CH_3COCH_2COOCH_2CH_3 + NaOH \rightarrow CH_3COCH_2COONa + C_2H_5OH$$

After the reaction mixture is boiled, the volatile ethanol distils off. On evaporating the remaining liquid to dryness, the solid that remains is the sodium salt of acetoacetic acid.

Hence $$\text{Residue (A)} = CH_3COCH_2COONa$$    (sodium acetoacetate)

When this residue is subsequently treated with dilute hydrochloric acid, simple proton exchange takes place to give the free β-keto-acid (acetoacetic acid) and sodium chloride.

$$CH_3COCH_2COONa + HCl_{(dil)} \rightarrow CH_3COCH_2COOH + NaCl$$

Therefore $$\text{Compound (B)} = CH_3COCH_2COOH$$    (acetoacetic acid)

Both structures together correspond to Option D.

Final answer: Option D which is: Residue (A) - sodium acetoacetate ($$CH_3COCH_2COONa$$) and Compound (B) - acetoacetic acid ($$CH_3COCH_2COOH$$).

The test with aqueous $$NaHCO_{3}$$ is an acid-base reaction. It is positive (effervescence of $$CO_{2}$$) only for those organic compounds whose acidity is stronger than that of carbonic acid $$\bigl($$ $$pK_{a}\,\approx\,6.37$$ $$\bigr)$$. In the syllabus this essentially restricts the list to:

• Carboxylic acids $$\left( \,\ce{R-COOH}\, \right)$$
• Sulfonic acids $$\left( \,\ce{R-SO_{3}H}\, \right)$$ (rare in JEE questions)
Phenols, enols, alcohols, aldehydes and ketones are all too weakly acidic to react with $$NaHCO_{3}$$.

Now inspect the five compounds given in the question:

Thus only compounds A, C and D satisfy the condition.

Option A which is: A, C and D Only

To determine the major product of the given intramolecular aldol condensation, we analyze the structure of the starting material and the possible cyclization pathways available under basic conditions.

The starting compound is 6-oxoheptanal, which contains both an aldehyde and a ketone functional group. Its structure may be represented as

$$CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO.$$

Under basic conditions, an $$\alpha$$-hydrogen is abstracted to generate an enolate ion. The resulting enolate then undergoes an intramolecular nucleophilic attack on the other carbonyl group, leading to ring formation.

Three possible enolates can be generated:

• Formation of an enolate at the methyl carbon adjacent to the ketone would cyclize to give a seven-membered ring, which is both kinetically and thermodynamically less favourable.
• Formation of an enolate at the carbon adjacent to the aldehyde would attack the ketone carbonyl, producing a five-membered ring.
• Formation of an enolate at the carbon adjacent to the ketone would attack the aldehyde carbonyl, also producing a five-membered ring.

Since five-membered rings are considerably more stable than seven-membered rings, the reaction proceeds through one of the two five-membered ring pathways.

To determine the major pathway, the relative reactivity of the carbonyl groups must be considered. Aldehydes are more electrophilic than ketones because they possess only one electron-donating alkyl group and experience less steric hindrance. Consequently, the enolate generated adjacent to the ketone preferentially attacks the aldehyde carbonyl, making this pathway the most favourable.

The cyclization initially forms a five-membered ring containing a $$\beta$$-hydroxy ketone intermediate. Under the reaction conditions, dehydration occurs readily to give the corresponding $$\alpha,\beta$$-unsaturated ketone. The resulting product is 1-acetylcyclopent-1-ene, in which the acetyl group is directly attached to the double bond of the cyclopentene ring.

Therefore, the major product is 1-acetylcyclopent-1-ene, corresponding to Option A.

Note: The product represented by Option B would arise from cyclization involving attack on the less reactive ketone carbonyl, making it a minor product. The seven-membered ring represented by Option D is not favoured due to the lower stability and slower formation of larger rings.

The substrate supplied in the problem (two vicinal bromo-substituents in an aliphatic chain bearing a phenyl group) is treated with hot alcoholic $$\text{KOH}$$. Alcoholic $$\text{KOH}$$ is a strong base and therefore promotes bimolecular dehydro-halogenation (E2). Because two bromine atoms are present, two consecutive E2 eliminations can occur.

Step 1 : First E2 elimination
The base abstracts a β-hydrogen that is anti-periplanar to one of the C-Br bonds. The departure of Br⁻ in the same transition state furnishes an alkene. Among the several hydrogens that can be removed, the one that gives the more substituted alkene and simultaneously places the double bond conjugated with the benzene ring is preferred (Saytzeff rule + conjugative stabilisation). Hence the first elimination produces the intermediate

$$\mathrm{Ph\!-\!CH=CH\!-\!CH_2\!-\!CH(Br)\!-\!CH_2\!-\!CH_3}$$

This alkene is allylic to the phenyl ring and therefore stabilised.

Step 2 : Second E2 elimination
The remaining bromo group is still β to at least one hydrogen. A second molecule of base removes the β-H that is anti-periplanar to the C-Br bond. The double bond generated in this step can appear in two possible positions, but the product in which the two C=C bonds are conjugated with each other as well as with the phenyl ring is vastly more stable. Therefore the elimination gives

$$\mathrm{Ph\!-\!CH=CH\!-\!CH=CH\!-\!CH_2\!-\!CH_3}$$

This molecule is named as 2-phenylhepta-2,4-diene.

Why the other isomeric dienes are not formed in major amounts
A. 2-Phenylhepta-2,5-diene and C. 6-Phenylhepta-3,5-diene are non-conjugated; they lack the extended π-system and are therefore less stable.
B. 6-Phenylhepta-2,4-diene keeps the conjugation between the two C=C bonds but places the phenyl ring out of the conjugated segment; again it is less stabilised than the product obtained above.
D. 2-Phenylhepta-2,4-diene enjoys full conjugation over the phenyl ring and both double bonds, so it is the thermodynamically favoured (major) product.

Hence the major product is
Option D which is : 2-Phenylhepta-2,4-diene

The reagent $$H_2/Pd$$ adds across a carbon-carbon multiple bond. Therefore, when a compound absorbs 2 moles of $$H_2$$, it must contain exactly two carbon-carbon double bonds (each $$C=C$$ consumes one mole of $$H_2$$).

Hot, acidic $$KMnO_4$$ cleaves every $$C=C$$ completely, converting each carbon of the double bond into a carbonyl carbon that finally appears as $$-COOH$$. If a carbon of the original double bond was $$\;CH_2=$$, that terminal carbon is oxidised further to $$CO_2$$.

Let the oxidation product obtained from ‘X’ be $$HOOC\,(CH_2)_6\,COOH$$ (suberic acid).
Tracing the process backwards:
  • Replace each carboxyl carbon of the acid by the corresponding alkene carbon (because each $$-COOH$$ arose from a double-bond carbon).
  • Add back the $$CH_2$$ that was lost as $$CO_2$$ at both ends.

Working in this way we recover the precursor:

$$CH_2=CH\;-\;(CH_2)_6\;-\;CH=CH_2$$

This structure is $$\textbf{deca-1,9-diene}$$. It indeed contains two isolated $$C=C$$ bonds and therefore absorbs exactly two moles of $$H_2$$, matching the data given.

Now count the $$\sigma$$ bonds in deca-1,9-diene.

Number of carbon atoms, $$n = 10$$, hence the carbon skeleton has $$n-1 = 9$$ $$C-C$$ $$\sigma$$ bonds.

Hydrogen count (obtained from the structure):
  • Two terminal $$CH_2=$$ groups ⇒ $$2 \times 2 = 4$$ H
  • Two $$CH=$$ carbons (one in each $$CH=CH_2$$) ⇒ $$2 \times 1 = 2$$ H
  • Six internal $$CH_2$$ groups ⇒ $$6 \times 2 = 12$$ H
Total hydrogens $$= 4 + 2 + 12 = 18$$.

Therefore, $$C-H$$ $$\sigma$$ bonds $$= 18$$.

Total $$\sigma$$ bonds in the molecule:
$$C-C: 9 \;+\; C-H: 18 = 27$$.

Hence, the compound ‘X’ has $$\boxed{27}$$ $$\sigma$$ bonds.

In the Claisen-Schmidt reaction between benzaldehyde and acetone, represented by $$ 2C_6H_5CHO + CH_3COCH_3 \to C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O $$, if 5.3 g of benzaldehyde are used and 3.51 g of dibenzalacetone are obtained, the percentage yield can be calculated as follows.

The molar mass of benzaldehyde is $$6(12) + 5(1) + 12 + 16 + 1 = 106$$ g/mol, and the molar mass of dibenzalacetone is $$17(12) + 14(1) + 16 = 204 + 14 + 16 = 234$$ g/mol.

The number of moles of benzaldehyde is $$\frac{5.3}{106} = 0.05$$ mol. Since 2 mol of benzaldehyde produce 1 mol of dibenzalacetone, the theoretical yield in moles is $$0.05/2 = 0.025$$ mol.

Multiplying by the molar mass of dibenzalacetone gives the theoretical mass as $$0.025 \times 234 = 5.85$$ g.

Therefore, the percentage yield is $$\frac{3.51}{5.85} \times 100 = 60\%$$, so the yield of the reaction is 60%.

We need to find the ratio $$K_P/K_C$$ for the reaction $$CO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons CO_{2(g)}$$. The relationship between $$K_P$$ and $$K_C$$ is given by $$K_P = K_C(RT)^{\Delta n_g}$$ where $$\Delta n_g$$ is the change in the number of moles of gaseous species (moles of gaseous products minus moles of gaseous reactants).

Moles of gaseous products = 1 (CO$$_2$$) and moles of gaseous reactants = 1 (CO) + 1/2 (O$$_2$$) = 3/2, so $$\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$$.

Substituting into the formula gives $$\frac{K_P}{K_C} = (RT)^{\Delta n_g} = (RT)^{-1/2} = \frac{1}{\sqrt{RT}}$$, therefore the correct answer is Option (1): $$\frac{1}{\sqrt{RT}}$$.

We need to evaluate two statements about buffer solutions.

Analysis of Statement (I): "A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities."

This statement is FALSE. While a buffer solution is indeed a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid), it does not work with just "any" quantities. For a buffer to function effectively, the components must be mixed in appropriate proportions. Specifically:

- The buffer capacity depends on the ratio of the concentrations of the conjugate pair.

- An effective buffer typically has a pH within one unit of the $$pK_a$$ of the weak acid, which requires the ratio $$[salt]/[acid]$$ to be between 0.1 and 10.

- If the ratio is too extreme, the solution will not resist pH changes effectively.

Analysis of Statement (II): "Blood is naturally occurring buffer solution whose pH is maintained by $$H_2CO_3/HCO_3^-$$ concentrations."

This statement is TRUE. Blood is a naturally occurring buffer that maintains its pH around 7.4. The primary buffer system in blood is the carbonic acid-bicarbonate buffer system:

$$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$

When excess acid ($$H^+$$) enters the blood, bicarbonate ions ($$HCO_3^-$$) neutralise it. When excess base ($$OH^-$$) enters, carbonic acid ($$H_2CO_3$$) neutralises it. This maintains the blood pH within a narrow, life-sustaining range.

The correct answer is Option (3): Statement I is false but Statement II is true.

Find the equilibrium constant for $$X \rightleftharpoons W$$ given intermediate steps.

$$X \rightleftharpoons Y$$, $$K_1 = 1.0$$

$$Y \rightleftharpoons Z$$, $$K_2 = 2.0$$

$$Z \rightleftharpoons W$$, $$K_3 = 4.0$$

When reactions are added, their equilibrium constants are multiplied. Adding all three:

$$X \rightleftharpoons Y \rightleftharpoons Z \rightleftharpoons W$$

$$K = K_1 \times K_2 \times K_3 = 1.0 \times 2.0 \times 4.0 = 8.0$$

The correct answer is Option (4): 8.0.

The reaction: $$Fe_2O_{3(s)} + 3CO_{(g)} \rightleftharpoons Fe_{(l)} + 3CO_{2(g)}$$

Le Chatelier's Principle: The equilibrium shifts in response to changes in concentration, pressure, or temperature.

Key point: Pure solids and pure liquids do NOT affect the equilibrium position because their concentrations (activities) are constant. They do not appear in the equilibrium expression.

$$Fe_2O_3$$ is a solid. Adding more $$Fe_2O_3(s)$$ will not disturb the equilibrium because its activity remains 1.

Adding/removing $$CO_2$$ or removing $$CO$$ would shift the equilibrium as they are gaseous species that appear in the equilibrium expression.

The correct answer is Option (3): Addition of $$Fe_2O_3$$.

The reaction is $$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$$.

The cylinder is filled with equal number of molecules of $$H_2$$, $$I_2$$, and $$HI$$ at $$-20°C$$ (253 K) and 1 atm.

Since the number of moles of gaseous reactants equals the number of moles of gaseous products ($$\Delta n_g = 0$$), $$K_p = K_c$$ for this reaction.

Let the number of moles of each gas be $$n$$. The partial pressures are all equal (since equal moles at same T, V): each gas has partial pressure $$P/3$$ where $$P = 1$$ atm.

$$K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot P_{I_2}} = \frac{(P/3)^2}{(P/3)(P/3)} = \frac{P^2/9}{P^2/9} = 1$$

So $$K_p = 1 = x \times 10^{-1}$$.

This gives $$x = 10$$.

The correct answer is Option B: 10.

The extent of enolization depends on the stability of the enol form. Enol content increases when the enol is stabilized by conjugation, intramolecular hydrogen bonding, or aromaticity.

In 1,3-cyclohexanedione, enolization produces a conjugated system stabilized by intramolecular hydrogen bonding. Hence, it has a significant enol content.

In 1,4-cyclohexanedione, enolization does not generate an extended conjugated system and lacks substantial stabilization. Therefore, its enol content is relatively low.

In 1,3,5-cyclohexanetrione, complete enolization converts the molecule into 1,3,5-trihydroxybenzene (phloroglucinol). The resulting structure contains a fully conjugated aromatic benzene ring.

The aromatic enol form is exceptionally stable due to aromatic resonance stabilization, which is much greater than the stabilization provided by simple conjugation or intramolecular hydrogen bonding. As a result, the keto-enol equilibrium is shifted strongly toward the enol form.

Therefore, 1,3,5-cyclohexanetrione exhibits the highest enol content among the given compounds.

Hence, the correct answer is $$\text{1,3,5-cyclohexanetrione}$$

The reaction sequence is a standard industrial route to the poly-alcohol pentaerythritol and its ketal derivatives. We analyse every step.

Step 1 : Reaction of acetaldehyde with excess formaldehyde in conc. NaOH
• Formaldehyde has no $$\alpha$$-hydrogen, so in strong alkali it undergoes the Cannizzaro reaction: $$2\,HCHO + OH^- \;\longrightarrow\; HCOO^- + CH_3OH$$

• In the presence of acetaldehyde, a cross-Cannizzaro/Aldol sequence occurs.   1 mol acetaldehyde first gives an enolate which adds to two molecules of formaldehyde (cross-aldol).   The aldehydic group thus produced is then reduced by hydride transfer from a third molecule of formaldehyde (Cannizzaro step).
The overall stoichiometry is

$$CH_3CHO + 3\,HCHO + 3\,NaOH \; \xrightarrow{\;\;\text{heat}\;\;} C(CH_2OH)_4 \;+\; 3\,NaHCOO$$

• The poly-alcohol obtained is pentaerythritol, $$P = C(CH_2OH)_4$$.   It possesses no -CHO group, hence it does not respond to Tollens’ reagent.

• The salt $$Q = NaHCOO$$ on acidification furnishes formic acid, $$HCOOH$$, which does reduce Tollens’ reagent. Thus the statements regarding $$P$$ and $$Q$$ are consistent.

Step 2 : Ketal formation between $$P$$ and excess cyclohexanone in the presence of catalytic p-toluenesulfonic acid (PTSA)
Acid catalyses the reaction of a ketone with two alcohol functions to give a ketal. Each molecule of cyclohexanone requires two -OH groups:

$$\text{cyclohexanone}+2\,ROH \;\xrightarrow{\;\;H^+\;\;} \text{cyclohexanone ketal}\;(\;C=O \text{ is replaced by }C(OR)_2\;)$$

Pentaerythritol has four -OH groups, therefore it can furnish the necessary two alcohol units twice. Consequently, with excess cyclohexanone it forms a diketal:

$$R = \bigl[\;C(CH_2O{-})_4\bigr]$$ Each pair $$\;{-}OCH_2{-}$$ originates from one molecule of cyclohexanone, so $$R$$ contains two cyclohexanone rings attached through ketal linkages.

Counting -CH2- groups in $$R$$
• Central pentaerythritol carbon: four $$CH_2$$ groups → $$4$$
• Each cyclohexanone ring (after ketal formation) supplies five $$CH_2$$ groups   (the carbonyl carbon becomes quaternary, carries no hydrogens).
  Two rings ⇒ $$2 \times 5 = 10$$
Total methylene groups $$= 4 + 10 = 14$$

Counting oxygen atoms in $$R$$
Every ketal linkage contains the two original alcohol oxygens; thus the four -OH groups of pentaerythritol become four ether oxygens. (The carbonyl oxygens of the ketones are lost as water during ketalisation).
Total oxygen atoms $$= 4$$

Required sum
$$\text{(number of }{-}CH_2{-}\text{ groups)} + \text{(number of O atoms)} = 14 + 4 = 18$$

Therefore, the asked sum is 18.

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We need to identify the given organic reduction reaction where benzoyl chloride ($$C_6H_5COCl$$) is converted into benzaldehyde ($$C_6H_5CHO$$).

Statement (A): Rosenmund reduction. The Rosenmund reduction is a hydrogenation process where an acyl chloride is selectively reduced to an aldehyde using hydrogen gas ($$H_2$$) over palladium on barium sulfate ($$Pd\text{-}BaSO_4$$). The barium sulfate acts as a catalyst poison to prevent further reduction of the aldehyde into a primary alcohol. This matches the given reaction perfectly. This is TRUE.

Statement (B): Wolff-Kishner reduction. The Wolff-Kishner reduction is used to convert a carbonyl group (aldehyde or ketone) into a methylene group ($$\text{-}CH_2\text{-}$$) using hydrazine ($$NH_2NH_2$$) and a strong base like $$KOH$$. This is FALSE.

Statement (C): Stephen reduction. The Stephen reduction involves the preparation of aldehydes from nitrites ($$R\text{-}CN$$) using tin(II) chloride ($$SnCl_2$$) and hydrochloric acid ($$HCl$$), followed by hydrolysis. This is FALSE.

Statement (D): Etard reduction. The Etard reduction is a reaction that oxidizes an aromatic methyl group to an aldehyde group using chromyl chloride ($$CrO_2Cl_2$$). This is FALSE.

The given reaction represents the Rosenmund reduction.

Answer: Option A — Rosenmund reduction

We need to evaluate the truth of two statements about the Aldol reaction.

Statement I: "Acidity of alpha-hydrogens of aldehydes and ketones is responsible for the Aldol reaction."

The Aldol reaction proceeds through the following mechanism:

1. A base abstracts an $$\alpha$$-hydrogen from one carbonyl compound, forming a resonance-stabilised enolate ion.

2. This enolate ion (a nucleophile) attacks the electrophilic carbonyl carbon of another molecule.

3. The resulting alkoxide is protonated to give the $$\beta$$-hydroxy aldehyde (aldol product).

The entire reaction depends on the ability to form the enolate, which in turn requires that the $$\alpha$$-hydrogen be sufficiently acidic ($$pK_a \approx 17{-}20$$ for typical aldehydes and ketones). Without acidic $$\alpha$$-hydrogens, the enolate cannot form and the Aldol reaction cannot proceed. Therefore, Statement I is correct.

Statement II: "Reaction between benzaldehyde and ethanal will NOT give a Cross-Aldol product."

Let us analyse the two reactants:

- Benzaldehyde ($$C_6H_5CHO$$): This aldehyde has no $$\alpha$$-hydrogen (the carbon adjacent to $$C=O$$ is part of the benzene ring). Therefore, benzaldehyde cannot form an enolate and cannot act as the donor component.

- Ethanal ($$CH_3CHO$$): This aldehyde has three $$\alpha$$-hydrogens on the methyl group. It can readily form an enolate ion.

In a cross-Aldol reaction between these two, ethanal provides the enolate (nucleophile), which attacks the carbonyl carbon of benzaldehyde (electrophile). This produces 3-hydroxy-3-phenylpropanal, which upon dehydration gives cinnamaldehyde ($$C_6H_5CH=CHCHO$$). This is a well-known reaction called the Claisen-Schmidt reaction.

Since the cross-Aldol product is formed, Statement II is incorrect.

The correct answer is Option 1: Statement I is correct but Statement II is incorrect.

The reaction involves benzene reacting with $$CO$$ and $$HCl$$ in the presence of anhydrous $$AlCl_3$$ and $$CuCl$$.

Under these conditions, $$CO$$ and $$HCl$$ generate a formyl electrophile in situ. This electrophile undergoes electrophilic aromatic substitution with benzene, introducing a $$-CHO$$ group onto the aromatic ring.

As a result, benzene is converted into benzaldehyde.

This method of direct formylation of an aromatic ring using $$CO$$ and $$HCl$$ in the presence of $$AlCl_3$$ and $$CuCl$$ is known as the Gatterman-Koch reaction.

Stephen reaction converts nitriles into aldehydes and does not involve direct formylation of benzene.

Etard reaction oxidizes the methyl group of toluene to form an aldehyde using chromyl chloride.

Rosenmund reduction converts acid chlorides into aldehydes using hydrogen over a poisoned palladium catalyst.

Hence, the given reaction is the $$\text{Gatterman-Koch reaction}$$

The iodoform test is highly specific for compounds containing either a methyl ketone group ($$\text{CH}_3\text{--C}(=\text{O})--$$) or a secondary methyl alcohol group ($$\text{CH}_3\text{--CH(OH)}--$$) that can easily oxidize to a methyl ketone.

When starting with a geminal dihalide complex, treatment with a strong base like aqueous $$\text{KOH}$$ causes nucleophilic substitution to replace the halogen atoms with hydroxyl ($$-\text{OH}$$) groups. If this intermediate step generates a stable or unstable gem-diol that eliminates water to form a methyl ketone structure, it will give a positive iodoform reaction upon adding $$\text{KOI}$$.

Step-by-Step Analysis of Compound A:

1. Nucleophilic Substitution with Base:

   Compound A is 2,2-dichlorobutane ($$\text{CH}_3\text{--CH}_2\text{--C(Cl)}_2\text{--CH}_3$$). When treated with aqueous $$\text{KOH}$$, the two chlorine atoms on carbon-2 are substituted by two hydroxyl groups, forming an unstable gem-diol intermediate:

   $$\text{CH}_3\text{--CH}_2\text{--C(Cl)}_2\text{--CH}_3 \xrightarrow{\text{aq. KOH}} \text{CH}_3\text{--CH}_2\text{--C(OH)}_2\text{--CH}_3$$

2. Dehydration to Form a Functional Carbonyl:

   Since two hydroxyl groups cannot comfortably sit on the same carbon atom, the intermediate immediately loses a water molecule ($$-\text{H}_2\text{O}$$) to form 2-butanone:

   $$\text{CH}_3\text{--CH}_2\text{--C(OH)}_2\text{--CH}_3 \xrightarrow{-\text{H}_2\text{O}} \text{CH}_3\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_3$$

3. Iodoform Conversion:

   The resulting 2-butanone contains the required methyl ketone ($$\text{--C}(=\text{O})\text{CH}_3$$) structural component. When potassium hypoiodite ($$\text{KOI}$$) is added, it undergoes oxidative cleavage to produce a bright yellow precipitate of iodoform ($$\text{CHI}_3$$) along with a potassium carboxylate salt:

   $$\text{CH}_3\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_3 \xrightarrow{\text{KOI}} \text{CH}_3\text{--CH}_2\text{--COOK} + \text{CHI}_3 \downarrow$$

Conclusion:

As clearly depicted in the reaction scheme, 2,2-dichlorobutane converts directly into a methyl ketone intermediate under basic conditions, making it perfectly suited to deliver a positive iodoform response.

Answer: Option A

The four reagents that are routinely used to identify and distinguish aldehydes and ketones are Fehling’s solution, the Iodoform reagent, 2,4-DNP (Brady’s reagent) and Schiff’s reagent. Each of them gives a very characteristic observation that the examiner places in List-II. We recall the basic chemistry of every reagent first, then match.

Case A : Fehling’s solution
Fehling’s solution contains $$Cu^{2+}$$ complexed with tartrate ion in an alkaline medium. Only aliphatic aldehydes can reduce $$Cu^{2+}$$ to $$Cu_2O$$, giving a brick-red precipitate. Neither ketones nor aromatic aldehydes respond.
⇒ Observation matched with Fehling’s solution is “formation of a red (brick-red) precipitate”. This statement is numbered IV.

Case B : Iodoform test ($$I_2/NaOH$$)
The haloform reaction is given by carbonyl compounds that contain the $$CH_3CO$$- group (i.e. methyl ketones) and by acetaldehyde (which is readily oxidised to that group). The tell-tale observation is a pale-yellow precipitate having the characteristic smell of iodoform, $$CHI_3$$.
⇒ “Formation of yellow precipitate of $$CHI_3$$ (iodoform)” is statement II.

Case C : 2,4-DNP (Brady’s reagent)
Both aldehydes and ketones add to 2,4-dinitrophenylhydrazine giving intensely coloured yellow to orange crystalline hydrazones. This test is used merely to confirm the presence of the carbonyl (>C=O) group.
⇒ “Formation of yellow/orange crystalline precipitate” is statement I.

Case D : Schiff’s reagent
Schiff’s reagent is a decolourised solution of fuchsin dye in $$SO_2$$. With aldehydes the magenta colour reappears almost instantaneously, whereas ketones do not restore the colour under ordinary conditions.
⇒ “Solution turns colourless to magenta (positive with aldehydes)” is statement III.

Collecting all the matches:
A - IV (B - II) C - I D - III

Therefore the correct choice is
Option B which is: A - IV, B - II, C - I, D - III.

An aldehyde that possesses at least one $$\alpha$$-hydrogen undergoes the base-catalysed aldol reaction. In the given problem the reactant is ethanal (acetaldehyde) and the reagent is dilute aqueous $$NaOH$$ kept below about $$40^{\circ}\text{C}$$, so dehydration does not take place. Only the aldol addition is observed.

Step 1 - Enolate formation
$$CH_3CHO + OH^- \;\longrightarrow\; CH_2CHO^- + H_2O$$

Step 2 - Nucleophilic attack on a second carbonyl molecule
The enolate ion formed above attacks the carbonyl carbon of another acetaldehyde molecule:
$$CH_2CHO^- + CH_3CHO \;\longrightarrow\; CH_3CH(OH)CH_2CHO + OH^-$$

Step 3 - Protonation
The alkoxide produced in step 2 is protonated by water, regenerating the hydroxide ion used in step 1. The overall stoichiometry is therefore
$$2CH_3CHO \xrightarrow[\;dil.\;]{NaOH,\; 295\,\text{K}} CH_3CH(OH)CH_2CHO$$

The product is $$\beta$$-hydroxyaldehyde, specifically 3-hydroxybutanal (often called “aldol”). Among the options given, this structure corresponds to Option A.

Hence the correct answer is:
Option A which is: 3-hydroxybutanal (CH3CH(OH)CH2CHO).

We need to identify which compounds give a silver mirror test (Tollens' test) with ammoniacal silver nitrate ($$[Ag(NH_3)_2]^+$$).

Tollens' reagent (ammoniacal silver nitrate) acts as a mild oxidising agent that oxidises aldehydes and certain other reducing compounds while itself being reduced. In this process silver ions ($$Ag^+$$) are reduced to metallic silver ($$Ag$$), which deposits as a mirror on the test tube wall:
$$RCHO + 2[Ag(NH_3)_2]^+ + 2OH^- \rightarrow RCOO^- + 2Ag \downarrow + 4NH_3 + H_2O$$

Formic acid has the structure $$H-\underset{\displaystyle \|}{\overset{\displaystyle O}{C}}-OH$$. Unlike other carboxylic acids, it contains an aldehyde-like $$H-C=O$$ group in addition to the $$-OH$$ group. This structural feature allows it to act as a reducing agent, so formic acid gives a positive silver mirror test.

Formaldehyde (HCHO) is an aldehyde and readily reduces Tollens' reagent. It is oxidised all the way to formate or carbonate because both hydrogen atoms on the carbon can be oxidised, and thus formaldehyde gives a positive silver mirror test.

Benzaldehyde ($$C_6H_5CHO$$) is an aromatic aldehyde containing the $$-CHO$$ group, and all aldehydes give a positive Tollens' test whether they are aliphatic or aromatic. Consequently, benzaldehyde gives a positive silver mirror test.

Acetone ($$CH_3COCH_3$$) is a ketone that lacks the $$-CHO$$ group and is not easily oxidised by the mild oxidising agent in Tollens' reagent. Therefore, acetone does not give a silver mirror test.

Consequently, formic acid, formaldehyde, and benzaldehyde give a positive silver mirror test, whereas acetone does not. The correct answer is Option 1: A, B and C only.

The Claisen-Schmidt reaction to prepare dibenzalacetone involves the condensation of acetone with benzaldehyde:

$$ CH_3COCH_3 + 2C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O $$

The molar mass of acetone (CH_3COCH_3) is 12×3 + 6 + 16 = 58 g/mol, that of benzaldehyde (C_6H_5CHO) is 12×7 + 6 + 16 = 106 g/mol, and that of dibenzalacetone (C_{17}H_{14}O) is 12×17 + 14 + 16 = 234 g/mol.

To prepare 351 g of dibenzalacetone:

$$ \text{Moles of dibenzalacetone} = \frac{351}{234} = 1.5 \text{ mol} $$

The available 87 g of acetone corresponds to:

$$ \text{Moles of acetone} = \frac{87}{58} = 1.5 \text{ mol} $$

Since the reaction requires 2 mol of benzaldehyde per mole of acetone:

$$ \text{Moles of benzaldehyde} = 2 \times 1.5 = 3 \text{ mol} $$

Thus, the mass of benzaldehyde needed is:

$$ \text{Mass} = 3 \times 106 = 318 \text{ g} $$

Therefore, 318 g of benzaldehyde is required.

Formation of Product A (Elimination)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in alcoholic medium ($$\text{C}_2\text{H}_5\text{OH}$$).
• Mechanism: Alcoholic $$\text{NaOH}$$ acts as a strong base, favoring $$\text{E}2$$ dehydrohalogenation.
• Product A: Ethene ($$\text{CH}_2=\text{CH}_2$$)
• Hydrogen Count in A: 4

2. Formation of Product B (Substitution)

• Reaction: Bromoethane ($$\text{CH}_3\text{CH}_2\text{Br}$$) reacting with $$\text{NaOH}$$ in aqueous medium ($$\text{H}_2\text{O}$$).
• Mechanism: Aqueous $$\text{NaOH}$$ provides strong nucleophilic hydroxyl ions ($$\text{OH}^-$$), favoring $$\text{S}_\text{N}2$$ substitution.
• Product B: Ethanol ($$\text{CH}_3\text{CH}_2\text{OH}$$)
• Hydrogen Count in B: 6

We need to count the number of compounds that give a positive Fehling's test from: Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, Cyclohexane carbaldehyde.

Recall what Fehling's test detects.

Fehling's test uses Fehling's reagent (an alkaline solution of cupric tartrate, deep blue in colour) to detect aliphatic aldehydes. Aldehydes reduce $$Cu^{2+}$$ (blue) to $$Cu_2O$$ (red-brown precipitate). Important: aromatic aldehydes do NOT give a positive Fehling's test because the benzene ring reduces the reactivity of the aldehyde group. Ketones also do not give a positive test.

Analyse each compound.

- Benzaldehyde ($$C_6H_5CHO$$): Aromatic aldehyde. Does NOT give positive Fehling's test.

- Acetaldehyde ($$CH_3CHO$$): Aliphatic aldehyde. Positive.

- Acetone ($$CH_3COCH_3$$): Ketone. Does NOT give positive test.

- Acetophenone ($$C_6H_5COCH_3$$): Aromatic ketone. Does NOT give positive test.

- Methanal ($$HCHO$$): Aliphatic aldehyde (simplest one). Positive.

- 4-Nitrobenzaldehyde ($$O_2N-C_6H_4-CHO$$): Aromatic aldehyde. Does NOT give positive Fehling's test.

- Cyclohexane carbaldehyde ($$C_6H_{11}CHO$$): Aliphatic aldehyde (cyclohexane ring is not aromatic). Positive.

Count.

Positive Fehling's test: Acetaldehyde, Methanal, Cyclohexane carbaldehyde = 3.

The answer is 3.

Ethanal ($$CH_3CHO$$) reacts with semicarbazide ($$NH_2NHCONH_2$$) to form a semicarbazone through a condensation reaction.

The reaction is:

$$ CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2 + H_2O $$

The product is acetaldehyde semicarbazone: $$CH_3CH=N-NH-CO-NH_2$$.

Counting the nitrogen atoms in the product:

1. The $$=N-$$ nitrogen

2. The $$-NH-$$ nitrogen

3. The $$-NH_2$$ nitrogen

Total nitrogen atoms = $$3$$.

Therefore, the answer is $$\boxed{3}$$.

Assertion A: Acyl chlorides can be subjected to Wolff-Kishner reduction.

This is False. Wolff-Kishner reduction requires treatment with hydrazine (NH$$_2$$NH$$_2$$) and strong base (KOH/NaOH) at high temperature. Acyl chlorides would react with hydrazine to form acid hydrazides rather than undergo the intended carbonyl reduction.

Reason R: Wolff-Kishner reduction converts C=O to CH$$_2$$.

This is True. The Wolff-Kishner reduction specifically reduces the carbonyl group to a methylene group (-CH$$_2$$-) through formation and decomposition of a hydrazone intermediate.

A is false but R is true.

The conversion of n-heptane into benzoic acid and benzyl alcohol proceeds through aromatization, controlled oxidation, disproportionation, and acidification.

n-Heptane undergoes cyclization and dehydrogenation in the presence of chromium oxide at high temperature and pressure to form toluene:

$$\mathrm{C_7H_{16} \xrightarrow[20\,atm]{Cr_2O_3,\;770\,K} C_6H_5CH_3}$$

Toluene is then subjected to Étard oxidation using chromyl chloride followed by hydrolysis to give benzaldehyde:

$$\mathrm{C_6H_5CH_3 \xrightarrow{CrO_2Cl_2,\;H_3O^+} C_6H_5CHO}$$

Benzaldehyde contains no α-hydrogen and therefore undergoes the Cannizzaro reaction with concentrated sodium hydroxide:

$$\mathrm{2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa}$$

The sodium benzoate formed is acidified to obtain benzoic acid:

$$\mathrm{C_6H_5COONa + H_3O^+ \rightarrow C_6H_5COOH + Na^+ + H_2O}$$

Thus,

$$Q = \mathrm{C_6H_5COOH}$$

$$R = \mathrm{C_6H_5CH_2OH}$$

Hence, the correct sequence of reagents is

$$\boxed{(i)\ Cr_2O_3,\ 770\,K,\ 20\,atm \;\rightarrow\; (ii)\ CrO_2Cl_2,\ H_3O^+ \;\rightarrow\; (iii)\ NaOH \;\rightarrow\; (iv)\ H_3O^+}$$

Therefore, the correct answer is $$\boxed{\text{Option A}}$$.

Explanation:

The reagents are:

$$Zn(Hg),\ \mathrm{conc.}\ HCl,\ \Delta$$

These are the conditions for Clemmensen reduction.

Clemmensen reduction converts an aldehyde or ketone carbonyl group into a methylene group.

The given compound is:

$$C_6H_5-CH(OH)-CH(CH_3)-CH_2-CHO$$

The aldehyde group:

$$-CHO$$

is reduced to:

$$-CH_3$$

Hence,

$$-CH_2-CHO\longrightarrow-CH_2-CH_3$$

The benzylic $$-OH$$ group is not affected under the expected transformation.

Therefore, the product formed is:

$$C_6H_5-CH(OH)-CH(CH_3)-C_2H_5$$

This corresponds to option B.

Statement I: True

• Reactant: A keto-amide ($$\text{H}_2\text{N-C}(=\text{O})-\text{CH}_2-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3$$).
• Reagent: Clemmensen reduction utilizes zinc amalgam in concentrated hydrochloric acid ($$\text{Zn(Hg)} / \text{HCl}$$).
• Reaction:
  1. The strongly acidic medium ($$\text{HCl}$$) hydrolyzes the primary amide group ($$\text{-CONH}_2$$) into a carboxylic acid group ($$\text{-COOH}$$).
  2. The ketone group ($$\text{-C=O-}$$) is reduced to a methylene group ($$\text{-CH}_2-$$).
• Product: Pentanoic acid ($$\text{HOOC-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$$), which matches the product shown in Statement I.

Statement II: False

• Reactant: A halo-ketone ($$\text{Cl-CH}_2-\text{CH}_2-\text{CO-CH}_3$$).
• Reagent: Wolff-Kishner reduction uses hydrazine and a strong base ($$\text{NH}_2\text{NH}_2 / \text{KOH}$$) at high temperatures.
• Reaction: While the carbonyl group is targeted for reduction to an alkane, the presence of a strong base ($$\text{KOH}$$) induces a highly favorable competing side reaction—dehydrohalogenation (elimination of $$\text{HCl}$$). Instead of obtaining the simple reduced haloalkane ($$\text{Cl-CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$$), elimination reactions dominate, making the statement false.
  

• Statement I is true.
• Statement II is false.

Correct Answer: [C] Statement I is true but Statement II is false

Assertion A: Acetal/Ketal is stable in basic medium. — TRUE. Acetals and ketals are stable under basic conditions but hydrolyze under acidic conditions.

Reason R: The high leaving tendency of alkoxide ion gives stability to acetal/ketal in basic medium. — FALSE. Alkoxide ions are actually poor leaving groups (high pKa ~16). The stability in basic medium is because the first step of hydrolysis requires protonation of the oxygen (acid catalysis), which doesn't occur in basic conditions.

Since A is true and R is false, the correct answer is Option A: $$\boxed{\text{A is true but R is false}}$$.

1. Fehling's Test (Positive):

   A positive Fehling's test indicates the presence of an aliphatic aldehyde group ($$\text{--CHO}$$). This eliminates any option that contains a ketone, ester, or other non-reducing functional groups, focusing our attention on compounds with an active formyl functionality.




2. Sodium Fusion Extract (Lassaigne's Test) with Sodium Nitroprusside:

   When an organic compound containing both Nitrogen ($$\text{N}$$) and Sulfur ($$\text{S}$$) is fused with sodium metal, they react together with carbon to form sodium thiocyanate ($$\text{NaSCN}$$):

   $$\text{Na} + \text{C} + \text{N} + \text{S} \rightarrow \text{NaSCN}$$
   • Blood-Red Coloration: When ferric ions ($$\text{Fe}^{3+}$$) or specific complexing reagents like sodium nitroprusside are introduced into an acidified thiocyanate extract solution, it yields a characteristic blood-red color due to the formation of ferric thiocyanate complexes:
   $$\text{Fe}^{3+} + \text{SCN}^\ominus \rightarrow [\text{Fe}(\text{SCN})]^{2+} \quad \text{(Blood-Red Solution)}$$
   • Absence of Prussian Blue: Because both nitrogen and sulfur are fixed together as thiocyanate ($$\text{SCN}^\ominus$$) rather than isolated cyanide ($$\text{CN}^\ominus$$) ions, there are no free cyanide ions available to form the classic Prussian blue complex ($$\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$$).

   Therefore, the compound must contain both nitrogen and sulfur atoms alongside its aldehyde group.

Evaluating the Structural Candidates:

• Oxazole derivatives: Contain Nitrogen ($$\text{N}$$) and Oxygen ($$\text{O}$$), but lack Sulfur ($$\text{S}$$). They do not give a blood-red color during Lassaigne's element detection.
• Pyrrole derivatives: Contain Nitrogen ($$\text{N}$$), but lack Sulfur ($$\text{S}$$), giving a Prussian blue test rather than blood-red thiocyanate formations.
• 2-Thiazolecarboxaldehyde ($$\text{C}_4\text{H}_3\text{NOS}$$): The thiazole ring features both an aromatic nitrogen atom and a sulfur heteroatom in its cyclic core, along with a directly attached formyl group ($$\text{--CHO}$$). This perfectly satisfies both the structural and chemical criteria.

Conclusion:

The simultaneous presence of the aldehyde moiety and both nitrogen and sulfur within the thiazole ring framework satisfies all requirements for a positive Fehling's reaction and a blood-red Lassaigne's outcome.

Answer: Option C (2-Thiazolecarboxaldehyde)

1. Step 1: Cyanohydrin Formation and Alcoholysis ($$\text{1) NaCN}$$, $$\text{2) EtOH, H}_2\text{O, H}^\oplus$$)

   The starting material is a ketone, 4-chloroacetophenone ($$\text{Cl--C}_6\text{H}_4\text{--COCH}_3$$).

   • First, nucleophilic addition of cyanide ($$\text{CN}^\ominus$$) to the carbonyl group yields a cyanohydrin intermediate: $$\text{Cl--C}_6\text{H}_4\text{--C(OH)(CN)(CH}_3)$$.
   • Second, treating the cyanohydrin with ethanol ($$\text{EtOH}$$) under acidic aqueous conditions hydrolyzes and esterifies the nitrile group ($$\text{--CN}$$) into an ethyl ester group ($$\text{--CO}_2\text{Et}$$).

   This gives the first intermediate shown in the scheme: ethyl 2-(4-chlorophenyl)-2-hydroxypropanoate.




2. Step 2: Grignard Reaction with Excess Methylmagnesium Bromide ($$\text{MeMgBr (excess)}$$ followed by $$\text{H}_3\text{O}^\oplus$$)

   The intermediate contains two key functional groups: a tertiary alcohol ($$\text{--OH}$$) and an ester ($$\text{--CO}_2\text{Et}$$).

   • The first equivalent of the Grignard reagent ($$\text{MeMgBr}$$) acts as a base and deprotonates the acidic hydroxyl proton ($$\text{--OH}$$), which is later regenerated during acidic workup.
   • Another equivalent of $$\text{MeMgBr}$$ undergoes nucleophilic acyl substitution at the ester carbonyl carbon, displacing the ethoxide leaving group ($$\text{--OEt}$$) to form a methyl ketone intermediate ($$\text{--COCH}_3$$).
   • Because the Grignard reagent is added in excess, a final equivalent immediately attacks this newly formed ketone group to yield a tertiary alkoxide.

   After acidic workup ($$\text{H}_3\text{O}^\oplus$$), the final product contains a highly branched tertiary alcohol unit attached to the original aromatic framework.

Structure of the Final Product:

The carbon atom alpha to the benzene ring preserves its original hydroxyl group ($$\text{--OH}$$) and methyl group ($$\text{--Me}$$). The ester group ($$\text{--CO}_2\text{Et}$$) is converted completely into a 2-hydroxypropan-2-yl group ($$\text{--C(OH)Me}_2$$).

Therefore, the structural formula of the final major product is:

$$\text{Cl--C}_6\text{H}_4\text{--C(OH)(Me)--C(OH)(Me)}_2$$

Conclusion:

The reaction sequence effectively introduces multiple methyl groups by transforming the ester functionality into a tertiary alcohol while maintaining the structural integrity of the chloro-substituted benzene ring.

Answer: The molecular structure labeled as 'C' in the reaction scheme, which corresponds to $$\text{Cl--C}_6\text{H}_4\text{--C(OH)(Me)--C(OH)(Me)}_2$$.

Step 1: Conjugate Addition (1,4-addition)

• The combination of MeMgBr and CuI forms a Gilman reagent (organocopper species), which selectively undergoes 1,4-addition (conjugate addition) rather than 1,2-addition.
• The methyl group $$(\text{Me}^{-})$$ attacks the $$\beta$$-carbon of the 3-methylcyclohex-2-en-1-one.
• This creates a quaternary center at the 3-position and generates a nucleophilic enolate intermediate.

Step 2: Alpha-Alkylation

• Instead of quenching with water, the reaction adds n-PrI (n-propyl iodide).
• The enolate oxygen-carbon double bond reforms, and the $$\alpha $$-carbon acts as a nucleophile to attack the n-propyl iodide via an $$S_{N}2$$mechanism.
• The n-propyl group is added to the 2-position (the $$\alpha $$-carbon).

The major product of this reaction sequence is 3,3-dimethyl-2-propylcyclohexanone.

Propanal undergoes crossed aldol condensation with methanal in the presence of dilute $$NaOH$$ to form a $$\beta$$-hydroxy aldehyde.

On heating, dehydration occurs to give the $$\alpha,\beta$$-unsaturated aldehyde:

$$CH_2=C(CH_3)CH_2CHO$$

Addition of $$HCN$$ generated from $$NaCN/H_3O^+$$ across the carbonyl group forms a cyanohydrin.

Acidic hydrolysis converts the $$-CN$$ group into a $$-COOH$$ group, giving an $$\alpha$$-hydroxy carboxylic acid with molecular formula:

$$C_5H_8O_3$$

The carbon bearing the $$-OH$$ group becomes a new chiral center.

Since the carbonyl group is planar, cyanide attacks from both faces with equal probability.

Hence, the product is obtained as a racemic mixture.

The presence of the $$-COOH$$ group causes the compound to react with saturated $$NaHCO_3$$, liberating carbon dioxide.

$$RCOOH+NaHCO_3\rightarrow RCOONa+CO_2+H_2O$$

Hence, the correct answer is:

$$\boxed{\text{c}}$$

Step 1: Reduction with $$\text{LiAlH}_4$$ (excess)

The starting material is $$1,1,2,2\text{-tetracyanoethane}$$:

$$\text{(NC)}_2\text{CH}-\text{CH(CN)}_2$$

Lithium aluminum hydride ($$\text{LiAlH}_4$$) is a strong reducing agent that reduces all four nitrile ($$-\text{C}\equiv\text{N}$$) groups completely into primary amine ($$-\text{CH}_2\text{NH}_2$$) groups.

The intermediate tetra-amine formed after workup with $$\text{H}_2\text{O}$$ is:

$$\text{(H}_2\text{NCH}_2)_2\text{CH}-\text{CH}(\text{CH}_2\text{NH}_2)_2$$

Step 2: Reaction with Acetophenone (excess)

Acetophenone ($$\text{Ph}-\text{CO}-\text{CH}_3$$) reacts with primary amines ($$-\text{NH}_2$$) via a nucleophilic addition-elimination reaction to form imines ($$-\text{N}=\text{C}\langle$$):

$$\text{R}-\text{NH}_2 + \text{O}=\text{C}(\text{CH}_3)\text{Ph} \longrightarrow \text{R}-\text{N}=\text{C}(\text{CH}_3)\text{Ph} + \text{H}_2\text{O}$$

Since acetophenone is in excess, all four primary amine groups will react to form four imine groups in the final product $$P$$.

Step 3: Counting $$\text{sp}^2$$ Hybridised Carbon Atoms

Let's analyze the hybridization of carbons in one attached acetophenone-derived unit ($$-\text{N}=\text{C}(\text{CH}_3)\text{Ph}$$):

1. Imine Carbon ($$-\text{N}=\underline{\text{C}}-$$): $$1\text{ sp}^2$$ carbon
2. Methyl Carbon ($$-\text{CH}_3$$): $$0\text{ sp}^2$$ carbons ($$\text{sp}^3$$)
3. Phenyl Ring ($$-\text{C}_6\text{H}_5$$): All $$6$$ carbons in the benzene ring are $$\text{sp}^2$$ hybridised.

$$\text{Number of sp}^2\text{ carbons per unit} = 1 + 6 = 7$$

$$\text{Total number of sp}^2\text{ hybridised carbons} = 4 \times 7 = \mathbf{28}$$

The starting material is a 2-alkylcyclohexanone (a cyclic ketone with an alkyl group $$R$$ at the $$\alpha$$-position).

When treated with a strong oxidizing agent like $$\text{KMnO}_4$$, cyclic ketones undergo oxidative cleavage of the $$\text{C}-\text{C}$$ bond adjacent to the carbonyl group.

Regioselectivity: The bond cleavage preferentially occurs between the carbonyl carbon ($$\text{C}_1$$) and the more substituted $$\alpha$$-carbon ($$\text{C}_2$$, which bears the alkyl group $$R$$).

The carbonyl carbon ($$\text{C}_1$$) is oxidized to a carboxylic acid ($$-\text{CO}_2\text{H}$$), and the substituted $$\alpha$$-carbon ($$\text{C}_2$$) is oxidized to a ketone ($$-\text{C}(=\text{O})R$$) group because it holds the alkyl chain.

Thus, the ring opens up to form a keto-acid.

Product 'A' is then treated with hydrazine and a strong base ($$\text{NH}_2\text{NH}_2, \text{KOH}$$) followed by acidic workup ($$\text{H}_3\text{O}^+$$). This reagent combination describes a Wolff-Kishner Reduction.

The Wolff-Kishner reduction selectively converts carbonyl groups ($$\text{C}=\text{O}$$) of aldehydes and ketones into methylene groups ($$\text{CH}_2$$).

Selectivity: It does not reduce carboxylic acid ($$-\text{CO}_2\text{H}$$) groups.

The ketone carbonyl group ($$\text{C}=\text{O}$$) attached to $$R$$ is completely reduced to a $$-\text{CH}_2-$$ group.

Correct Option B.

$$LiAlH_4$$ is a powerful reducing agent so, it will completely reduces primary amides down to primary amines by converting the carbonyl carbon into a methylene group ($$-CH2-$$). AND The ketone carbonyl group is cleanly reduced to a secondary alcohol group.

$$LiAlH_4$$ is a nucleophilic reducing agent, meaning it attacks electron-deficient carbon atoms (like carbonyls). It does not reduce ordinary isolated carbon-carbon double bonds because alkenes are electron-rich and repel nucleophilic attack. Therefore, the central double bond remains completely unaffected.

Hence Final structure would be C

Match the reactions with their reagents.

(A) Hoffmann Degradation: Conversion of amide to amine using $$Br_2/NaOH$$ — matches (III).

(B) Clemmensen Reduction: Reduction of $$C=O$$ to $$CH_2$$ using $$Zn-Hg/HCl$$ — matches (IV).

(C) Cannizzaro Reaction: Disproportionation of aldehydes (without $$\alpha$$-H) using concentrated KOH — matches (I).

(D) Reimer-Tiemann Reaction: Introduction of $$CHO$$ group to phenol using $$CHCl_3/NaOH$$ — matches (II).

Matching: A-III, B-IV, C-I, D-II

This matches Option 3.

The answer is $$\boxed{A-III, B-IV, C-I, D-II}$$ (Option 3).

1. Formation of Product [A]

The starting material is 2-methylpentanal. It is treated with $$\text{NaHSO}_3$$, $$\text{dil. HCl}$$, $$\text{NaCN}$$, and $$\text{H}_2\text{O}$$, which are the standard reagents for cyanohydrin formation via the bisulfite addition pathway.

• Mechanism: The aldehyde group ($$-\text{CHO}$$) undergoes nucleophilic addition by the cyanide ion ($$\text{CN}^-$$), followed by protonation to yield a cyanohydrin.
• Structure of [A]:
  $$A = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)CN}$$
  

2. Formation of Product [B]

Product [A] (the cyanohydrin) is treated with Lithium Aluminum Hydride ($$\text{LiAlH}_4$$), a powerful reducing agent.

• Mechanism: $$\text{LiAlH}_4$$ reduces the nitrile group ($$-\text{CN}$$) down to a primary amine group ($$-\text{CH}_2\text{NH}_2$$). The alcohol group ($$-\text{OH}$$) remains unaffected after workup.
• Structure of [B]:
  $$B = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)CH}_2\text{NH}_2$$

3. Formation of Product [C]

Product [A] is treated with $$\text{HCl}/\text{H}_2\text{O}$$ and heated ($$\Delta$$).

• Mechanism: Acidic hydrolysis of a nitrile group ($$-\text{CN}$$) converts it into a carboxylic acid group ($$-\text{COOH}$$). Because it is a cyanohydrin, this yields a $$\alpha$$-hydroxy acid.
• Structure of [C]:$$C = \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(OH)COOH}$$

The iodoform test is positive for compounds containing either a $$CH_3CO-$$ group (methyl ketone) or a $$CH_3CH(OH)-$$ group (secondary alcohol with methyl group adjacent to OH).

Analyzing each compound:

(a) 1-Phenylbutan-2-one (C₆H₅CH₂COCH₂CH₃): The carbonyl has CH₂ groups on both sides, not a methyl group. Negative ✗

(b) 2-Methylbutan-2-ol ((CH₃)₂C(OH)CH₂CH₃): This is a tertiary alcohol with no CH₃CHOH pattern. Negative ✗

(c) 3-Methylbutan-2-ol (CH₃CH(OH)CH(CH₃)₂): Has a CH₃CHOH group. Positive ✓

(d) 1-Phenylethanol (C₆H₅CH(OH)CH₃): Has a CH₃CHOH group. Positive ✓

(e) 3,3-Dimethylbutan-2-one (CH₃COC(CH₃)₃): Has a CH₃CO group. Positive ✓

(f) 1-Phenylpropan-2-ol (C₆H₅CH₂CH(OH)CH₃): Has a CH₃CHOH group. Positive ✓

Number of compounds giving positive iodoform test = 4 (c, d, e, f).

Molecules having 

give positive iodoform test

Glyoxylic acid has the structure: OHC-COOH (an aldehyde and a carboxylic acid group).

When glyoxylic acid reacts with excess MeMgBr (Grignard reagent):

The first equivalent of MeMgBr reacts with the acidic -COOH group (acid-base reaction):

$$-\text{COOH} + \text{CH}_3\text{MgBr} \rightarrow -\text{COO}^-\text{MgBr}^+ + \text{CH}_4$$

The second equivalent of MeMgBr performs nucleophilic addition to the -CHO group:

$$-\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow -\text{CH(OMgBr)CH}_3$$

After hydrolysis with H$$_3$$O$$^+$$, the product is 2-hydroxypropionic acid (lactic acid type).

So for $$y$$ moles of glyoxylic acid, $$x = 2y$$ moles of MeMgBr are needed.

$$\frac{x}{y} = 2$$

C is the only option with a di-ketone being in conjugation with double bonds, the other options are not conjugated. Therefore, C is the correct option.

For a carbonyl compound the fraction present as its enol tautomer increases when the enol form enjoys extra stabilisation. Three effects are especially important:

(i) Resonance of the enolate/enol with more than one carbonyl group.
(ii) Intramolecular hydrogen bonding that locks the six-membered chelate ring.
(iii) Extended conjugation of the C=C bond formed in the enol.

Let us analyse each option in this light.

A simple aldehyde/ketone contains only one C=O group. The enol that could form is stabilised only by ordinary $$\pi$$-conjugation with that single carbonyl oxygen, so its mole-fraction is very small (typically <0.1 %). No intramolecular H-bonding is possible.

A simple ketone with no electron-withdrawing group on either side behaves almost exactly like Case A. The enol percentage is likewise very low.

Ethyl acetoacetate is a 1,3-dicarbonyl compound, so its enolate is resonance-stabilised by two carbonyls. However the ester carbonyl cannot engage in effective intramolecular hydrogen bonding with the -OH formed in the enol, and conjugation is interrupted by the intervening -O- atom of the ester. Hence the enol content is larger than in A or B but still moderate (≈7-8 %).

Pentane-2,4-dione (acetylacetone) is a true 1,3-diketone. In its enol form the molecule forms a six-membered ring via a strong intramolecular O-H···O hydrogen bond, and the $$\ce{C=C}$$ bond is conjugated simultaneously with both carbonyl groups:

$$\ce{CH3-C(=O)-CH2-C(=O)-CH3 \;\rightleftharpoons\; CH3-C(=O)-CH= C(OH)-CH3}$$

This dual resonance plus chelation lowers the energy of the enol dramatically, so the equilibrium lies far toward the enol side (≈70 % at room temperature).

Comparing all four, the stabilisation is greatest in Case D, therefore the enol content is the highest for that compound.

Option D which is: Pentane-2,4-dione (acetylacetone) has the highest enol content.

1. Identify the Major Product R

When 4-bromobenzyl alcohol is treated with red phosphorus and bromine (red P / Br₂), it acts as a brominating agent (PBr₃ formed in situ). This reagent selectively replaces the aliphatic hydroxyl group (-OH) with a bromine atom (-Br). The aromatic ring-bound bromine remains unaffected.

• Starting material: 4-bromobenzyl alcohol (C₇H₇OBr)
• Major product R: 4-bromobenzyl bromide
• Molecular formula of R: C₇H₆Br₂

2. Calculate the Molar Mass of Compound R (C₇H₆Br₂)

Using the given atomic masses:

• Carbon (C): 7 atoms × 12 = 84
• Hydrogen (H): 6 atoms × 1 = 6
• Bromine (Br): 2 atoms × 80 = 160

Molar mass of R = 84 + 6 + 160 = 250 g/mol

3. Analysis of Carius Method for Estimation of Bromine

In the Carius method, all bromine atoms present in the organic compound are converted quantitatively into Silver Bromide (AgBr) precipitates.

1 mole of C₇H₆Br₂ contains 2 moles of Br atoms, which will yield 2 moles of AgBr.

• Molar mass of AgBr = Mass of Ag + Mass of Br = 108 + 80 = 188 g/mol

4. Stoichiometric Calculation

From the relationship established above:

• 1 mole of R (250 g) gives 2 moles of AgBr (2 × 188 g = 376 g)

Now, we calculate the amount of AgBr formed from 1.00 g of R:

• Mass of AgBr formed = (376 g / 250 g) × 1.00 g
• Mass of AgBr formed = 1.504 g

Rounding off to two decimal places as shown in standard numerical formats:

• Mass of AgBr formed = 1.50 g

Direct oxidation of toluene usually produces benzoic acid because strong oxidizing agents over-oxidize the methyl group.

Controlled oxidation using $$\mathrm{CrO_3}$$ in acetic anhydride $$\mathrm{((CH_3CO)_2O)}$$ prevents this over-oxidation.

The methyl group is first converted into the stable intermediate benzylidene diacetate.

This intermediate resists further oxidation.

Acid hydrolysis using $$\mathrm{H_3O^+}$$ then converts the intermediate into benzaldehyde $$\mathrm{(C_6H_5CHO)}$$.

Reagents such as $$\mathrm{KMnO_4}$$ directly oxidize toluene into benzoic acid $$\mathrm{(C_6H_5COOH)}$$.

The Gattermann-Koch reaction using $$\mathrm{CO/HCl/AlCl_3}$$ introduces a formyl group onto the aromatic ring rather than oxidizing the side chain.

Hence, the correct reagent sequence for conversion of toluene to benzaldehyde is Option $$\mathrm{(B)}$$.

Thus, D is the right option.

We need to find the final product A in the reaction sequence where butanone reacts with HCN, followed by treatment with 95% $$H_2SO_4$$ and heat.

Butanone ($$CH_3CH_2COCH_3$$) undergoes nucleophilic addition with HCN to form a cyanohydrin:

$$CH_3CH_2COCH_3 + HCN \rightarrow CH_3CH_2C(OH)(CN)CH_3$$

The product is 2-hydroxy-2-methylbutanenitrile.

Concentrated $$H_2SO_4$$ (95%) acts as both a dehydrating agent and a hydrolysis catalyst:

First, the nitrile group ($$-CN$$) is hydrolyzed to a carboxylic acid group ($$-COOH$$) in the acidic medium:

$$CH_3CH_2C(OH)(CN)CH_3 \xrightarrow{H_2SO_4} CH_3CH_2C(OH)(COOH)CH_3$$

Then, dehydration occurs — the hydroxyl group and a hydrogen from the adjacent carbon are eliminated:

$$CH_3CH_2C(OH)(COOH)CH_3 \xrightarrow{-H_2O} CH_3CH=C(CH_3)COOH$$

The final product is $$CH_3-CH=C(CH_3)-COOH$$, which is 2-methylbut-2-enoic acid (also known as tiglic acid or $$\alpha,\beta$$-unsaturated acid).

Therefore, the correct answer is Option A.

In the first step, the alkene reacts with bromine in the presence of methanol.

$$\mathrm{Br_2/CH_3OH}$$

The $$\mathrm{\pi}$$ electrons of the double bond attack $$\mathrm{Br_2}$$ to form a cyclic bromonium ion intermediate.

Since methanol is present in excess, it acts as the nucleophile instead of $$\mathrm{Br^-}$$.

Methanol attacks the more substituted carbon because it carries greater partial positive character.

After deprotonation, intermediate $$\mathrm{A}$$ is formed.

$$\mathrm{Alkene \xrightarrow{Br_2/CH_3OH} Bromomethoxy\ Compound\ (A)}$$

Intermediate $$\mathrm{A}$$ contains:

1. A methoxy group $$\mathrm{(-OCH_3)}$$ on the tertiary carbon

2. A bromine atom on the adjacent secondary carbon

In the second step, compound $$\mathrm{A}$$ reacts with hydroiodic acid.

$$\mathrm{HI}$$ protonates the ether oxygen, converting it into a good leaving group.

$$\mathrm{R-OCH_3 + H^+ \longrightarrow R-O^+HCH_3}$$

Since the oxygen is attached to a tertiary carbon, cleavage occurs through an $$\mathrm{S_N1}$$ mechanism.

Methanol leaves, generating a stable tertiary carbocation.

$$\mathrm{R-O^+HCH_3 \longrightarrow R^+ + CH_3OH}$$

The iodide ion then attacks the tertiary carbocation.

$$\mathrm{R^+ + I^- \longrightarrow R-I}$$

The bromine atom on the neighbouring carbon remains unchanged throughout the reaction.

Thus, in the final product $$\mathrm{B}$$:

1. The bromine atom remains on the secondary carbon

2. The methoxy group is replaced by iodine

Therefore, the correct structure is:

$$\boxed{\mathrm{Option\ containing\ Br\ and\ I\ on\ adjacent\ carbons}}$$

To determine which reactions produce benzaldehyde, the reagents and reaction conditions must be analysed.

In reaction $$\mathrm{(A)}$$, benzoic acid first reacts with $$\mathrm{SOCl_2}$$ to form benzoyl chloride.

Subsequent treatment with $$\mathrm{H_2/Pd/BaSO_4}$$ performs the:

$$\mathrm{Rosenmund\ Reduction}$$

which selectively converts the acid chloride into benzaldehyde.

Hence, reaction $$\mathrm{(A)}$$ gives benzaldehyde.

In reaction $$\mathrm{(B)}$$, benzyl alcohol reacts with $$\mathrm{CrO_3/H_2SO_4}$$ (Jones reagent).

This strong oxidising agent converts the primary alcohol completely into benzoic acid.

Hence, benzaldehyde is not obtained.

In reaction $$\mathrm{(C)}$$, methyl benzoate is treated with $$\mathrm{NaBH_4}$$.

$$\mathrm{NaBH_4}$$ is too mild to reduce esters.

Therefore, no aldehyde is formed and PCC has no effect.

Hence, benzaldehyde is not obtained.

In reaction $$\mathrm{(D)}$$, toluene reacts with $$\mathrm{CrO_3/(CH_3CO)_2O}$$ to form a protected gem-diacetate intermediate.

Acid hydrolysis using $$\mathrm{H_3O^+,\ \Delta}$$ converts this intermediate into benzaldehyde.

Hence, reaction $$\mathrm{(D)}$$ gives benzaldehyde.

Therefore, the reactions producing benzaldehyde are: $$\mathrm{(A)\ and\ (D)}$$
Thus, the right option is C.

Explanation: The reaction begins with iodolactonization.

The carboxylic acid is converted into a carboxylate ion by $$NaHCO_3$$.

The alkene reacts with $$I_2$$ to form an iodonium ion intermediate.

The carboxylate ion attacks intramolecularly to form a stable five-membered lactone ring, placing iodine on the more substituted carbon.

In the next step, pyridine and heat promote dehydrohalogenation $$-HI$$ is eliminated to form a double bond.

Elimination occurs preferentially at the $$\alpha$$-carbon adjacent to the carbonyl group because it produces an $$\alpha,\beta$$-unsaturated lactone. The resulting conjugated system is thermodynamically more stable than non-conjugated alternatives.

Therefore, the major product is the conjugated $$\alpha,\beta$$-unsaturated lactone corresponding to Option C.

We need to follow the sequence of reactions starting from hex-4-en-2-ol, which has the structure: $$CH_3-CH(OH)-CH_2-CH=CH-CH_3$$.

PCC (Pyridinium Chlorochromate) is a mild oxidizing agent that converts a secondary alcohol to a ketone without further oxidation. $$CH_3-CH(OH)-CH_2-CH=CH-CH_3 \xrightarrow{PCC} CH_3-CO-CH_2-CH=CH-CH_3$$ Compound A is hex-4-en-2-one.

Sodium hypoiodite is the iodoform reagent. It reacts with methyl ketones ($$R-CO-CH_3$$) to give a carboxylate salt and iodoform ($$CHI_3$$): $$CH_3-CO-CH_2-CH=CH-CH_3 + 3NaOI \to CHI_3 + CH_3-CH=CH-CH_2-COONa$$. Compound B is sodium pent-3-enoate ($$CH_3CH=CHCH_2COONa$$).

Heating sodium pent-3-enoate with soda lime ($$NaOH + CaO$$) causes decarboxylation, removing $$CO_2$$ and replacing the $$-COONa$$ group with $$-H$$: $$CH_3-CH=CH-CH_2-COONa \xrightarrow{NaOH/CaO} CH_3-CH=CH-CH_3 + Na_2CO_3$$. Compound C is 2-butene ($$CH_3CH=CHCH_3$$).

The correct answer is Option C: 2-butene.

Reaction $$\mathrm{A}$$ shows the conversion of a nitrile to an aldehyde using:

$$\mathrm{SnCl_2/HCl}$$

followed by hydrolysis.

This reaction is known as the Stephen reaction.

$$\mathrm{R-CN \xrightarrow[\ H_2O\ ]{SnCl_2/HCl} R-CHO}$$

Therefore:

$$\mathrm{A \rightarrow III}$$

Reaction $$\mathrm{B}$$ shows oxidation of toluene using chromyl chloride:

$$\mathrm{CrO_2Cl_2}$$

to form benzaldehyde.

This reaction is known as the Etard reaction.

$$\mathrm{C_6H_5CH_3 \xrightarrow{CrO_2Cl_2} C_6H_5CHO}$$

Therefore:

$$\mathrm{B \rightarrow II}$$

Reaction $$\mathrm{C}$$ shows reduction of an acid chloride to an aldehyde using:

$$\mathrm{H_2/Pd-BaSO_4}$$

This reaction is known as the Rosenmund reduction.

$$\mathrm{R-COCl \xrightarrow{H_2/Pd-BaSO_4} R-CHO}$$

Therefore:

$$\mathrm{C \rightarrow IV}$$

Reaction $$\mathrm{D}$$ shows formylation of benzene using:

$$\mathrm{CO + HCl/AlCl_3}$$

This reaction is known as the Gatterman-Koch reaction.

$$\mathrm{C_6H_6 \xrightarrow{CO/HCl,\ AlCl_3} C_6H_5CHO}$$

Therefore:

$$\mathrm{D \rightarrow I}$$

Correct matching:

$$\boxed{\mathrm{A-III,\ B-II,\ C-IV,\ D-I}}$$

Thus, the right option is A.

Benzaldehyde $$\mathrm{(C_6H_5CHO)}$$ reacts with $$\mathrm{CH_3MgBr}$$ to form the secondary alcohol $$\mathrm{C_6H_5CH(OH)CH_3}$$, which on oxidation gives acetophenone $$\mathrm{(C_6H_5COCH_3)}$$.

Acetaldehyde $$\mathrm{(CH_3CHO)}$$ reacts with $$\mathrm{C_6H_5MgBr}$$ to form the same secondary alcohol $$\mathrm{C_6H_5CH(OH)CH_3}$$, which on oxidation again gives acetophenone $$\mathrm{(C_6H_5COCH_3)}$$.

In the reaction of ethyl benzoate $$\mathrm{(C_6H_5COOC_2H_5)}$$ with excess $$\mathrm{CH_3MgBr}$$, the first addition temporarily forms acetophenone $$\mathrm{(C_6H_5COCH_3)}$$.

However, ketones readily react with Grignard reagents. The second equivalent of $$\mathrm{CH_3MgBr}$$ attacks the ketone immediately to form a tertiary alcohol $$\mathrm{C_6H_5C(OH)(CH_3)_2}$$.

Thus, acetophenone is not obtained as the final major product.

Reaction of benzoyl chloride $$\mathrm{(C_6H_5COCl)}$$ with dialkylcadmium $$\mathrm{((CH_3)_2Cd)}$$ stops at the ketone stage because dialkylcadmium reagents are less reactive than Grignard reagents.

Hence, the sequence that does not yield acetophenone as the major product is Option $$\mathrm{(C)}$$.

Option $$\mathrm{C}$$ is incorrect.

In reaction $$\mathrm{(C)}$$, the starting compound is a primary alcohol.

Treatment with $$\mathrm{Na_2Cr_2O_7/H_2SO_4}$$ forms a strong oxidising medium.

Primary alcohols under these conditions are completely oxidised into carboxylic acids.

Hence, the actual product formed is:

$$\mathrm{p\text{-}Toluic\ Acid}$$

and not $$\mathrm{p\text{-}Tolualdehyde}$$.

Reaction $$\mathrm{(A)}$$ is the $$\mathrm{Rosenmund\ Reduction}$$, where $$\mathrm{H_2/Pd-BaSO_4}$$ selectively reduces an acid chloride into an aldehyde.

Reaction $$\mathrm{(B)}$$ uses $$\mathrm{DIBAL\text{-}H}$$, which reduces a nitrile into an aldehyde after hydrolysis.

Reaction $$\mathrm{(D)}$$ also uses $$\mathrm{DIBAL\text{-}H}$$, which selectively reduces an ester into an aldehyde.

Therefore, the incorrect reaction is:

$$\mathrm{C}$$

The reaction sequence begins with propanenitrile (CH₃-CH₂-CN) reacting with methylmagnesium bromide (CH₃MgBr), a Grignard reagent, in the presence of dry ether. The nucleophilic methyl group from the Grignard reagent attacks the electrophilic carbon of the nitrile group, forming an intermediate imine salt designated as compound A (CH₃-CH₂-C(CH₃)=NMgBr).

This intermediate is then immediately subjected to acidic hydrolysis (H₃O⁺), which cleaves the carbon-nitrogen double bond to yield a ketone. Consequently, compound B is butan-2-one (CH₃-CH₂-CO-CH₃).

In the final step, butan-2-one undergoes a Clemmensen reduction using zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). This reduction explicitly targets the carbonyl group, completely deoxygenating it and converting the >C=O group into a methylene (-CH₂-) group. Therefore, the final product, the correct structure of C, is the straight-chain alkane butane (CH₃-CH₂-CH₂-CH₃).

We need to identify the products A and B formed in the given reaction.

This question involves ozonolysis of a suitable alkene. Ozonolysis cleaves the carbon-carbon double bond and produces aldehydes and/or ketones depending on the substituents.

When an alkene undergoes ozonolysis (treatment with $$O_3$$ followed by reductive workup with Zn/$$H_2O$$), the double bond is cleaved, and each carbon of the double bond gets an oxygen (=O) to form carbonyl compounds.

For propene ($$CH_3CH=CH_2$$):

$$CH_3CH=CH_2 \xrightarrow{O_3/Zn, H_2O} CH_3CHO + HCHO$$

The products are:

$$HCHO$$ (formaldehyde) — from the $$=CH_2$$ end

$$CH_3CHO$$ (acetaldehyde) — from the $$CH_3CH=$$ end

Based on the given options and the convention in the question:

$$A = HCHO$$ (Formaldehyde)

$$B = CH_3CHO$$ (Acetaldehyde)

Option A: Acrolein and formaldehyde — Incorrect

Option B: Acetaldehyde and acrolein — Incorrect

Option C: Formaldehyde and acetaldehyde — Correct

Option D: Acrolein and acetaldehyde — Incorrect

Hence, the correct answer is Option C: Formaldehyde and acetaldehyde.

Explanation: Enamine formation requires removal of an $$\mathrm{\alpha}$$-hydrogen from the carbon adjacent to the carbonyl group in the final dehydration step.

Therefore, a ketone must contain at least one $$\mathrm{\alpha}$$-hydrogen to form an enamine.

In di-$$\mathrm{tert}$$-butyl ketone, the $$\mathrm{\alpha}$$-carbon is bonded to three methyl groups and the carbonyl carbon, $$\mathrm{-C(CH_3)_3}$$.

Thus, the $$\mathrm{\alpha}$$-carbon contains no hydrogen atoms.

Because no $$\mathrm{\alpha}$$-hydrogen is available, formation of the required $$\mathrm{C=C}$$ bond is impossible.

Hence, di-$$\mathrm{tert}$$-butyl ketone cannot form an enamine.

Ketones such as diethyl ketone and cyclohexanone possess $$\mathrm{\alpha}$$-hydrogens and therefore readily form enamines.

Correct Option: $$\mathrm{(B)}$$

We need to find which reactant gives the target alcohol on reaction with one mole of phenyl magnesium bromide ($$PhMgBr$$) followed by acidic hydrolysis.

Grignard reagents ($$RMgBr$$) react with various carbonyl compounds as follows:

(i) With formaldehyde ($$HCHO$$): Gives a primary alcohol ($$RCH_2OH$$)

(ii) With other aldehydes ($$R'CHO$$): Gives a secondary alcohol ($$RR'CHOH$$)

(iii) With ketones ($$R'COR''$$): Gives a tertiary alcohol

(iv) With nitriles ($$R'C \equiv N$$): After hydrolysis, gives a ketone

Based on the question context, the target alcohol is a secondary alcohol with one phenyl group. When $$PhMgBr$$ reacts with acetaldehyde ($$CH_3CHO$$):

$$PhMgBr + CH_3CHO \rightarrow CH_3CH(OMgBr)Ph \xrightarrow{H_3O^+} CH_3CH(OH)Ph$$

This gives 1-phenylethanol, which is a secondary alcohol with the structure $$C_6H_5CH(OH)CH_3$$.

Option A: $$CH_3C \equiv N$$ (acetonitrile) + $$PhMgBr$$ gives an imine intermediate, which on hydrolysis gives acetophenone ($$CH_3COC_6H_5$$), a ketone — not the target alcohol.

Option B: $$PhC \equiv N$$ (benzonitrile) + $$PhMgBr$$ would give benzophenone imine, which on hydrolysis gives benzophenone — not the target alcohol.

Option C: Formaldehyde ($$HCHO$$) + $$PhMgBr$$ gives benzyl alcohol ($$C_6H_5CH_2OH$$), a primary alcohol — not a secondary alcohol.

Option D: Acetaldehyde ($$CH_3CHO$$) + $$PhMgBr$$ gives 1-phenylethanol — this matches the target alcohol.

Hence, the correct answer is Option D: Acetaldehyde.

Toluene:

$$\mathrm{C_7H_8}$$

Molar mass of toluene:

$$\mathrm{= (7\times12) + (8\times1)}$$

$$\mathrm{= 84 + 8 = 92\ g\ mol^{-1}}$$

Benzaldehyde:

$$\mathrm{C_7H_6O}$$

Molar mass of benzaldehyde:

$$\mathrm{= (7\times12) + (6\times1) + 16}$$

$$\mathrm{= 84 + 6 + 16 = 106\ g\ mol^{-1}}$$

Moles of toluene used:

$$\mathrm{= \frac{5}{92}\ mol}$$

Reaction stoichiometry:

$$\mathrm{1\ mol\ Toluene \rightarrow 1\ mol\ Benzaldehyde}$$

Therefore, theoretical moles of benzaldehyde:

$$\mathrm{= \frac{5}{92}\ mol}$$

Theoretical mass of benzaldehyde:

$$\mathrm{= \frac{5}{92}\times106\ g}$$

Given yield:

$$\mathrm{92\%}$$

Actual mass produced:

$$\mathrm{= \left(\frac{5}{92}\times106\right)\times\frac{92}{100}}$$

$$\mathrm{= \frac{5\times106}{100}}$$

$$\mathrm{= \frac{530}{100}}$$

$$\mathrm{= 5.3\ g}$$

Expressing in the form $$\mathrm{\times10^{-2}\ g}$$:

$$\mathrm{5.3\ g = 530\times10^{-2}\ g}$$

$$\boxed{\mathrm{530}}$$

Cyclohexanone contains $$\alpha$$-hydrogens. In the presence of $$OH^-$$ followed by heating, it undergoes self-aldol condensation.

In the first step, hydroxide ion removes an $$\alpha$$-hydrogen from cyclohexanone to form an enolate ion.

The enolate ion then attacks the carbonyl carbon of another cyclohexanone molecule, forming a new carbon-carbon bond and producing a $$\beta$$-hydroxy ketone (aldol).

On heating, the $$\beta$$-hydroxy ketone undergoes dehydration to form an $$\alpha,\beta$$-unsaturated ketone.

The major product $$P$$ is 2-(cyclohexylidene)cyclohexan-1-one.

The product contains:

1. One carbonyl double bond, $$C=O$$

   $$1\ \pi\text{-bond} = 2\ \pi\text{-electrons}$$

2. One carbon-carbon double bond, $$C=C$$

   $$1\ \pi\text{-bond} = 2\ \pi\text{-electrons}$$

Therefore, the total number of $$\pi$$ electrons in product $$P$$ is

$$2+2=4$$

Hence, the number of $$\pi$$ electrons present in the major product is

$$\boxed{4}$$

Step 1: Identify product A from the fermentation of glucose.

$$\text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{Zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$$

Product A is ethanol ($$\text{C}_2\text{H}_5\text{OH}$$).

Step 2: Identify product B from the iodoform reaction.

NaOI is sodium hypoiodite, the reagent for the iodoform reaction. Ethanol undergoes the iodoform reaction:

$$\text{CH}_3\text{CH}_2\text{OH} + 4\text{NaOI} \xrightarrow{\Delta} \text{CHI}_3 + \text{HCOONa} + 3\text{NaOH}$$

The products are iodoform ($$\text{CHI}_3$$) and sodium formate ($$\text{HCOONa}$$).

Therefore, product B is sodium formate ($$\text{HCOONa}$$), which contains 1 carbon atom.

The number of carbon atoms in product B is 1.

We have the molecular formula $$\mathrm{C_3H_6}$$. In open-chain form, the only alkene possible is propene, written structurally as $$\mathrm{CH_3-CH=CH_2}$$.

The first reagent is aqueous acid, written as $$\mathrm{H^+/H_2O}$$. In the presence of dilute acid, an alkene undergoes electrophilic addition of water. The governing principle is Markovnikov’s rule: “When HX or H-OH adds to an unsymmetrical alkene, the hydrogen attaches to the doubly-bonded carbon already holding the larger number of hydrogens, and the other part (here -OH) goes to the other carbon.” Applying that rule to propene, the -OH group attaches to the middle carbon while the extra hydrogen attaches to the terminal carbon.

Writing every step symbolically,

$$$\mathrm{CH_3-CH=CH_2 \;+\; H^+ \xrightarrow{\;\;}\; CH_3-CH^+-CH_3}$$$ $$$\mathrm{CH_3-CH^+-CH_3 \;+\; H_2O \longrightarrow CH_3-CH(OH)-CH_3 + H^+}$$$

Thus the product after the first step is $$\mathrm{CH_3-CH(OH)-CH_3}$$, i.e. propan-2-ol (commonly called isopropyl alcohol). Let us designate it as compound $$A$$.

Now $$A$$ is treated with “$$\mathrm{KIO/dil\;KOH}$$”. In the laboratory the iodoform reaction is normally carried out with iodine and alkali ($$\mathrm{I_2/KOH}$$); $$\mathrm{I_2}$$ in base actually produces the active species hypoiodite $$\mathrm{IO^-}$$, so writing the reagent as $$\mathrm{KIO}$$ in dilute $$\mathrm{KOH}$$ conveys the same chemistry. The iodoform reaction occurs whenever the substrate contains either

(i) a methyl ketone group $$\mathrm{ -CO-CH_3}$$, or
(ii) an alcohol that can be oxidised in situ to such a ketone, specifically $$\mathrm{R-CH(OH)-CH_3}$$.

Propan-2-ol fits criterion (ii) because it has the fragment $$\mathrm{ -CH(OH)-CH_3}$$. The reaction sequence is:

1. Oxidation of the secondary alcohol to the corresponding methyl ketone (propan-2-one, acetone).

$$$\mathrm{CH_3-CH(OH)-CH_3 + [O] \longrightarrow CH_3-CO-CH_3 + H_2O}$$$

2. Successive iodination of the $$\mathrm{ -CO-CH_3}$$ group and base-promoted cleavage, giving yellow iodoform and a carboxylate anion whose carbon skeleton is the original ketone minus the iodine-bearing methyl group.

For acetone the detailed stoichiometry can be summarised as

$$$\mathrm{CH_3-CO-CH_3 + 3\,I_2 + 4\,KOH \longrightarrow CHI_3 \downarrow + CH_3COOK + 3\,KI + 3\,H_2O}$$$

Therefore the two organic products obtained from the iodoform reaction of propan-2-ol are

$$$\mathrm{CHI_3} \quad\text{(iodoform, a yellow solid)}$$$

and

$$$\mathrm{CH_3COOK} \quad\text{(potassium acetate)}$$$

Denoting them in the order given in the problem statement, we can identify

$$B = \mathrm{CHI_3}, \qquad C = \mathrm{CH_3COOK}$$

Among the alternatives, only Option D lists the pair “$$\mathrm{CHI_3}$$, $$\mathrm{CH_3COOK}$$”.

Hence, the correct answer is Option D.

Mesityl oxide is a well-known organic compound formed by the aldol condensation of acetone. When two molecules of acetone undergo aldol condensation, they first form diacetone alcohol, which then undergoes dehydration to give mesityl oxide.

The product has the molecular formula $$C_6H_{10}O$$ and its structure is $$CH_3\text{-}CO\text{-}CH=C(CH_3)_2$$. To name it by IUPAC rules, we identify the longest carbon chain containing both the carbonyl group and the double bond. The longest such chain has 5 carbons (pent-), so the parent name is pentanone.

Numbering from the end nearer to the carbonyl group: carbon 1 is $$CH_3$$, carbon 2 carries the $$C=O$$ (keto group), carbon 3 is $$CH=$$, carbon 4 is $$=C(CH_3)$$, and carbon 5 is the terminal $$CH_3$$ attached to carbon 4. The double bond is between carbons 3 and 4, and there is a methyl branch at carbon 4.

Putting it together, the IUPAC name is 4-methylpent-3-en-2-one. This matches option (D).

Option A represents only the intermediate tertiary alcohol formed after Grignard addition and ignores the subsequent reaction with $$HCl$$.

Option B incorrectly assumes hydrogenation of the double bond, which cannot occur with the given reagents.

Option C represents the correct final product formed after electrophilic addition of $$HCl$$ across the double bond according to Markovnikov’s rule.

The proton adds to the terminal carbon, while $$Cl^-$$ attaches to the more substituted carbon.

Option D represents the anti-Markovnikov product, which is not formed because $$HCl$$ normally follows Markovnikov addition in the absence of peroxide conditions.

Therefore, Option C is the major product.

The 2,4-DNP (2,4-dinitrophenylhydrazine) test is a standard qualitative test used to detect the presence of the carbonyl group ($$C=O$$) in aldehydes and ketones.

When 2,4-dinitrophenylhydrazine reacts with an aldehyde or ketone, it forms a yellow, orange, or red precipitate of 2,4-dinitrophenylhydrazone. This confirms the presence of the carbonyl functional group.

Among the given options — aldehyde, amine, halogens, and ether — the 2,4-DNP test is specifically used to identify aldehydes (and ketones). Amines, halogens, and ethers do not contain a carbonyl group and will not give a positive 2,4-DNP test.

Therefore, the correct answer is aldehyde.

The reaction sequence involves a crossed aldol addition followed by the haloform reaction.

First, $$\mathrm{NaOH}$$ deprotonates the only available $$\mathrm{\alpha}$$-position of $$\mathrm{2,2\text{-}Dimethylcyclopentan\text{-}1\text{-}one}$$ to form an enolate ion.

This enolate attacks the carbonyl carbon of acetaldehyde, producing intermediate $$\mathrm{P}$$, a $$\mathrm{\beta}$$-hydroxy ketone containing the group $$\mathrm{(-CH(OH)-CH_3)}$$.

Intermediate $$\mathrm{P}$$ is then treated with $$\mathrm{I_2/NaOH}$$, initiating the iodoform reaction.

Sodium hypoiodite first oxidises the secondary alcohol into a methyl ketone.

The methyl group undergoes tri-iodination followed by cleavage to form yellow precipitate:

$$\mathrm{CHI_3}$$

and a sodium carboxylate salt.

Filtration removes the iodoform precipitate.

Acidification with $$\mathrm{HCl}$$ protonates the carboxylate ion to form the corresponding carboxylic acid.

The final major product $$\mathrm{X}$$ is:

$$\mathrm{2,2\text{-}Dimethyl\text{-}5\text{-}carboxycyclopentan\text{-}1\text{-}one}$$

First we write the full structure of the given ester. Pent-4-ynoic acid contains five carbon atoms in the main chain and the triple bond starts at carbon-4. Converting the -COOH group into an ethyl ester gives

$$\mathrm{CH_3CH_2C\!\equiv\!C-CO_2CH_2CH_3}$$

Now we examine the nature of each functional group. The molecule has

(i) an ester carbonyl $$-CO_2-$$, and

(ii) a terminal alkyne hydrogen $$\mathrm{\;-C\!\equiv\!C-H}$$ at carbon-5. A terminal alkyne hydrogen is appreciably acidic (pK$$_a\approx25$$) and is removed instantaneously by a Grignard reagent, because every Grignard reagent behaves as a very strong base as well as a nucleophile.

We next recall the two standard facts we shall need.

• Acid-base reaction with a terminal alkyne $$\mathrm{R-C\!\equiv\!C-H + R'MgX \;\longrightarrow\; R-C\!\equiv\!C^{\;-}MgX^{+} + R'H}$$

• Nucleophilic addition of a Grignard reagent to an ester For every mole of an ester, two moles of a Grignard reagent are required to reach the tertiary-alcohol stage:

$$\mathrm{RCOOR' + RMgX \;\xrightarrow{\;(1)\;}\; R-CO-R \;(\text{ketone})}$$

and then

$$\mathrm{R-CO-R + RMgX \;\xrightarrow{\;(2)\;}\; R-C(OMgX)R_2 \;\xrightarrow{\;H_2O\;/\;H^+\;} R-C(OH)R_2}$$

Keeping these two rules in mind, we follow the reaction of ethyl pent-4-yn-oate with methylmagnesium bromide step by step.

Step 1 - deprotonation of the terminal alkyne

$$\mathrm{CH_3CH_2C\!\equiv\!C-H + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+} + CH_4\uparrow}$$

This consumes one mole of $$\mathrm{CH_3MgBr}$$ and produces methane gas. The alkyne is now converted into its magnesium acetylide; the ester function is still intact.

Step 2 - first nucleophilic attack on the ester carbonyl

$$\mathrm{CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+}\;-CO_2CH_2CH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3CH_2OMgBr}$$

After the first nucleophilic addition and expulsion of the ethoxide ion, a ketone $$\mathrm{RC(=O)CH_3}$$ is formed. A second mole of the Grignard reagent has therefore been used.

Step 3 - second nucleophilic attack on the ketone

$$\mathrm{CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-C(OMgBr)(CH_3)_2}$$

Hydrolysing the adduct finally gives the alcohol:

$$\mathrm{CH_3CH_2C\!\equiv\!C-C(OH)(CH_3)_2}$$

The carbon bearing the -OH group now carries three alkyl substituents (two $$\mathrm{CH_3}$$ groups and one $$\mathrm{CH_2C\!\equiv\!CCH_2CH_3}$$ group). By definition, such an alcohol is a tertiary (3°) alcohol. Hence Statement I is perfectly correct.

Let us count the number of moles of $$\mathrm{CH_3MgBr}$$ that were consumed overall:

• one mole in Step 1 (acid-base reaction), • one mole in Step 2 (first nucleophilic addition), • one mole in Step 3 (second nucleophilic addition).

Total $$= 3$$ moles of $$\mathrm{CH_3MgBr}$$ per mole of the ester. Statement II, which claims that only two moles are utilised, is therefore incorrect.

We thus conclude that Statement I is true whereas Statement II is false.

Hence, the correct answer is Option B.

We need an oxidising agent, thus the option is C.

Step 1 : Formation of the acyl chloride (compound A)
Acetic acid reacts with thionyl chloride as
$$CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$$
Thus $$A = CH_3COCl$$ (ethanoyl chloride or acetyl chloride).