Organic Chemistry

In basic medium (NaOH), aniline reacts with Br2} to give tribromo derivative.
The electron-donating –NH2} group activates ortho and para positions, resulting in substitution at all three activated positions.

The Finkelstein reaction involves nucleophilic substitution of an alkyl halide (usually RCl or RBr) with NaI in acetone.
This leads to the formation of alkyl iodide (RI).
RCl + NaI → RI + NaCl (precipitates in acetone)

The reaction sequence starts with toluene (methylbenzene). Step (i) KMnO / KOH and Step (ii) H O , : KMnO (potassium permanganate) in the presence of KOH (alkaline medium), followed by acidic workup (H O ) and heating ( ), is a strong oxidizing agent for alkyl side chains on an aromatic ring. The methyl group (-CH ) of toluene will be oxidized to a carboxylic acid group (-COOH). Regardless of the length of the alkyl side chain (as long as it has at least one benzylic hydrogen), it is oxidized to -COOH. So, toluene (C H -CH ) is converted to benzoic acid (C H -COOH). The product after steps (i) and (ii) is benzoic acid. Step (iii) Br / FeBr : This is an electrophilic aromatic substitution reaction: bromination of the benzene ring. FeBr acts as a Lewis acid catalyst, polarizing Br to generate the electrophile (Br equivalent). The substituent already present on the benzene ring (-COOH in benzoic acid) directs the incoming electrophile (Br). The -COOH group (carboxylic acid) is an electron-withdrawing group and a meta-directing group for electrophilic aromatic substitution. It deactivates the ring, but substitution occurs primarily at the meta position relative to the -COOH group. So, bromination of benzoic acid will yield 3-bromobenzoic acid (or m-bromobenzoic acid). Structure of 3-bromobenzoic acid: The -COOH is at position 1. The Br is at position 3. This matches option (a).

- Assertion is true:
AgCN forms isocyanide (NC) due to covalent character of Ag–C bond, which favors bonding through nitrogen.
- Reason is false: AgCN is not mainly ionic; it's largely covalent due to the poor solubility and the nature of silver ion.
Hence, A is true, R is false.

Density = mass/volume. For similar alkanes, the one with the heaviest halogen will have the highest density.
Among Cl, Br, I — Iodine has the highest atomic mass.
Hence, n-propyl iodide (n-C3}H7}I) will be densest due to greater molecular mass.

Ortho/para directors are typically electron-donating groups (EDGs).
- -OH (I), -OCH3} (IV), and -NHCOCH3} (V) are EDGs, thus ortho/para directors.
- -CN, -COOH, and -CHO are electron-withdrawing and meta directing groups.

Palladium (Pd) has atomic number 46. Its ground state electronic configuration is: On removing two electrons to form , both electrons are removed from the 4d orbital because the 5s orbital is already empty:

In the complex , where , the ligands are: - 2 bidentate ethylenediamine ligands - 2 bromide ligands Since the coordination number is 6, cobalt is in +3 oxidation state. The naming order: - Neutral ligands (ethylenediamine) - Anionic ligands (bromide) - Metal name with oxidation state Hence, the correct IUPAC name is:

Sodium peroxide contains the peroxide ion (O ) and potassium superoxide contains the superoxide ion (O ). Bond order formula: For O : It has 18 electrons (16 + 2), so bond order = 1. For O : It has 17 electrons, so bond order = 1.5.

Basic strength is related to the availability of lone pairs on nitrogen: - Ammonia (NH ) has a free lone pair on nitrogen, making it more basic.
- Hydrazine (N H ) has two nitrogen atoms with lone pairs, but intramolecular hydrogen bonding and repulsion reduce basicity slightly compared to NH .
- Hydroxylamine (NH OH) has a nitrogen lone pair involved in hydrogen bonding with OH, decreasing its availability for donation, hence the weakest base.

The reaction involves: 1. Nitration of toluene at elevated temperatures introduces a nitro group at para and ortho positions. 2. Reduction of nitro group to \ch{NH2} (aniline derivative). 3. Bromination via Sandmeyer reaction using \ch{NaNO2}, \ch{HCl}, and \ch{CuBr} leads to substitution of \ch{NH2} with Br at original position. Final product: 2-bromo-4-methylbenzene.

Aldehydes without an alpha-hydrogen undergo Cannizzaro reaction, which is a type of disproportionation reaction. In presence of conc. alkali: Phenyl aldehyde (\ch{C6H5CHO}) lacks alpha-hydrogens, so it undergoes the Cannizzaro reaction. Other compounds: - \ch{C6H5COOH}, \ch{C6H5CH2OH}, \ch{C6H5CH3} do not undergo this reaction under same conditions.

The carbonyl group consists of a carbon atom double-bonded to an oxygen atom (C=O). Oxygen is more electronegative than carbon. This leads to a polarization of the C=O double bond: \begin{itemize} \item The oxygen atom carries a partial negative charge ( ). \item The carbon atom carries a partial positive charge ( ). \end{itemize} C =O Let's analyze the statements based on this polarity: a) Carbonyl Carbon is electrophilic An electrophile ("electron-loving") is an atom, ion, or molecule that is electron-deficient and seeks electrons. The carbonyl carbon has a partial positive charge ( ) due to the electronegative oxygen pulling electron density away from it. This makes the carbonyl carbon electron-deficient and thus electrophilic. It is susceptible to attack by nucleophiles. This statement is correct. b) Carbonyl Carbon is nucleophilic A nucleophile ("nucleus-loving") is an atom, ion, or molecule that is electron-rich and donates an electron pair to form a bond with an electrophile. The carbonyl carbon is electron-deficient ( ), not electron-rich. Therefore, it is not nucleophilic. This statement is incorrect. c) Carbonyl Oxygen is electrophilic The carbonyl oxygen has a partial negative charge ( ) due to its higher electronegativity. It is electron-rich, not electron-deficient. Therefore, it is not electrophilic. This statement is incorrect. d) Carbonyl Oxygen is nucleophilic The carbonyl oxygen is electron-rich ( ) and typically has lone pairs of electrons (two lone pairs in a standard C=O group). These lone pairs can be donated to an electrophile (e.g., a proton H in acid catalysis, or a Lewis acid). Thus, the carbonyl oxygen can act as a nucleophile or a Lewis base. This statement is correct. So, the correct statements are (a) and (d). This corresponds to option (a).

To name the compound using IUPAC rules:
1. Identify the parent chain: Benzene ring with substituents.
2. Principal functional group: Nitrile (–CN), which has highest priority, so suffix will be benzonitrile.
3. Number the ring: Start from –CN as position 1, and number in such a way that the other substituents get the lowest locants.
Structure:
- Position 1: –CN (priority group)
- Position 2: –Br
- Position 3: –OH
Putting it together:

The reaction involves formation of a Grignard reagent (PhMgBr) using Mg and dry ether.
This Grignard reagent reacts with D2}O to replace the MgBr with D (deuterium), giving C6}H5}–D as the major product.

The reaction of alcohols over heated copper at (which is ) is a standard method for dehydrogenation or dehydration, depending on the type of alcohol. \begin{itemize} \item Primary alcohols (RCH OH): Undergo dehydrogenation (loss of H ) to form aldehydes (RCHO). Example: CH CH OH CH CHO + H . \item Secondary alcohols (R CHOH): Undergo dehydrogenation to form ketones (R C=O). Example: (CH ) CHOH (CH ) C=O + H . \item Tertiary alcohols (R COH): Tertiary alcohols do not have a hydrogen atom attached to the carbon atom bearing the -OH group (the -carbon). Therefore, they cannot undergo dehydrogenation in the same way as primary and secondary alcohols to form a carbonyl compound. Instead, tertiary alcohols tend to undergo dehydration (loss of H O) when passed over heated copper at this temperature, forming alkenes. \end{itemize} Tertiary butanol (t-butanol) is (CH ) C-OH. Structure: CH CH -C-OH CH The carbon atom attached to the -OH group has no hydrogen atoms directly bonded to it. When t-butanol is passed over heated copper at , it undergoes dehydration to form 2-methylpropene (isobutylene). (CH ) C-OH (CH ) C=CH + H O. This is a dehydration reaction.

Step-by-step mechanism:
1. Grignard reaction: CH MgCl adds CH ⁻ to carbonyl carbon of cyclopentanone: 2. Acidic workup (H O ) protonates the alkoxide to form 1-methylcyclopentanol.
3. Elimination under acidic heat (E1):
- Alcohol is protonated → forms carbocation.
- Elimination of H leads to formation of double bond.
- Alkene forms as 1-methylene-cyclopentane (major due to stability).

Aqueous KOH leads to nucleophilic substitution (SN2) — alcohol is formed. Alcoholic KOH leads to β-elimination — alkene is formed. So, P = CH CH CH CH OH, Q = CH CH=CHCH (2-butene)

Ionization potential generally decreases down a group due to increasing atomic size and shielding effect. Sodium (Na) is above potassium (K) in group 1. Since Na has a smaller atomic radius and stronger nuclear attraction than K, its 1\textsuperscript{st} ionization energy is higher. Thus: \begin{itemize} \item Statement A is correct: Ionization energy decreases down the group. \item Statement B is also correct: Na has higher ionization energy than K. \end{itemize}

The reaction is acidic cleavage of an ether using concentrated HBr and heat: Mechanism:
- Protonation of ether oxygen
- Cleavage at the alkyl side (not the aromatic side)
- Phenol remains intact, and the alkyl group becomes an alkyl halide

An -hybridized carbon atom is typically involved in forming either: \begin{itemize} \item One triple bond and one single bond (e.g., in alkynes, nitriles). \item Two double bonds (e.g., in allenes like propa-1,2-diene, or in CO ). \end{itemize} An -hybridized carbon atom has linear geometry (bond angles ). Let's analyze each molecule: I. Ethane nitrile (Acetonitrile): CH -C N \begin{itemize} \item The carbon atom of the methyl group (-CH ) is hybridized (forms 4 single bonds). \item The carbon atom of the nitrile group (-C N) is bonded to the CH group by a single bond and to the N atom by a triple bond. This carbon is -hybridized. \end{itemize} So, Ethane nitrile has one -hybridized carbon. II. Buta-1, 3-diene: CH =CH-CH=CH This molecule has two double bonds separated by a single bond (conjugated diene). Each carbon atom involved in a C=C double bond is -hybridized. All four carbon atoms in buta-1,3-diene are -hybridized. So, Buta-1,3-diene does not have any -hybridized carbons. III. Propa-1, 2-diene (Allene): CH =C=CH This is an allene, which has cumulated double bonds. \begin{itemize} \item The two terminal carbon atoms (C1 and C3, part of CH =) are -hybridized (each forms two single bonds to H and one double bond to the central C). \item The central carbon atom (C2, part of =C=) forms two double bonds, one with C1 and one with C3. This central carbon is -hybridized. \end{itemize} So, Propa-1,2-diene has one -hybridized carbon. IV. Ethyne (Acetylene): HC CH Each carbon atom forms one single bond (to H) and one triple bond (to the other C). Both carbon atoms in ethyne are -hybridized. So, Ethyne has two -hybridized carbons. V. Benzene: C H Benzene is a cyclic molecule with a system of delocalized electrons. Each carbon atom in the ring forms two single bonds (to adjacent carbons) and one double bond (within the resonance structure depiction, effectively 1.5 bonds) and one single bond to a hydrogen atom. All six carbon atoms in benzene are -hybridized. So, Benzene does not have any -hybridized carbons. Molecules with -hybridized carbon(s) are: I (Ethane nitrile) III (Propa-1,2-diene) IV (Ethyne) This corresponds to option (d).

The reaction is between 1-bromopropane (CH CH CH Br) and ethanolic KCN. KCN provides the cyanide ion (CN ), which is an ambident nucleophile. It can attack through carbon (to form nitriles R-CN) or through nitrogen (to form isonitriles R-NC). Reaction with KCN: 1-Bromopropane is a primary alkyl halide. With KCN, the primary reaction is S 2 substitution. \begin{itemize} \item Major product: Attack by carbon of CN leads to the formation of a nitrile. CH CH CH Br + KCN (ethanolic) CH CH CH -CN (Butanenitrile) + KBr. This is the major pathway because KCN is predominantly ionic, and the C-C bond formed is stronger than the C-N bond in isonitriles. \item Minor product: Attack by nitrogen of CN leads to the formation of an isonitrile (isocyanide). CH CH CH Br + KCN (ethanolic) CH CH CH -NC (Propyl isonitrile or Propyl isocyanide) + KBr. This is the minor product. \end{itemize} The question asks about the hydrolysis of the *minor product*. The minor product is propyl isonitrile (CH CH CH -NC). Hydrolysis of Isonitriles (R-NC): Isonitriles are hydrolyzed to primary amines and formic acid (or formate salts depending on pH). The hydrolysis occurs in two steps typically, with the carbon of the -NC group becoming part of formic acid. R-N C + 2H O R-NH (primary amine) + HCOOH (formic acid). In this case, R = CH CH CH - (propyl group). So, CH CH CH -NC (propyl isonitrile) on hydrolysis will give: CH CH CH -NH (propylamine) + HCOOH (formic acid). The organic amine product is propylamine (propan-1-amine). Let's check the options: % Option (a) Propylamine (CH CH CH NH ) - This matches our derived product. % Option (b) Isopropylamine (CH CH(NH )CH ) - This would result if the structure was iso- or rearrangement occurred, which is not typical here. % Option (c) Propanoic acid (CH CH COOH) - This would be formed from hydrolysis of the *major* product, butanenitrile (CH CH CH CN CH CH CH COOH, butanoic acid, not propanoic). Complete hydrolysis of CH CH CN (propanenitrile) gives propanoic acid. % Option (d) Isobutyric acid - Not relevant. If the major product (butanenitrile, CH CH CH CN) was hydrolyzed, it would give butanoic acid (CH CH CH COOH). The question specifically asks for hydrolysis of the *minor product*. The minor product is propyl isonitrile, CH CH CH NC. Its hydrolysis gives propylamine, CH CH CH NH . So, the product is propylamine.

t-Butyl bromide is a tertiary alkyl halide ((CH ) C-Br). NaOH is a strong nucleophile (OH ) and a strong base. The reaction of a tertiary alkyl halide with a nucleophile/base can proceed via S 1 (substitution, nucleophilic, unimolecular) or E1 (elimination, unimolecular) mechanisms, or S 2/E2 if conditions are very specific (which is less common for tertiary substrates due to steric hindrance). In water, which is a polar protic solvent, S 1 and E1 pathways are favored for tertiary substrates. S 1/E1 Mechanism: The first step in both S 1 and E1 mechanisms is the formation of a carbocation, which is the rate-determining step (slow step). (CH ) C-Br (CH ) C + Br This step involves only the alkyl halide (t-butyl bromide). The rate of this step, and thus the overall rate of the S 1/E1 reaction, depends only on the concentration of the alkyl halide. Rate = [t-butyl bromide] The carbocation ((CH ) C ) then reacts rapidly with the nucleophile (OH or H O for S 1) or a base abstracts a proton (OH or H O for E1). \begin{itemize} \item S 1 product: (CH ) C-OH (t-butyl alcohol) \item E1 product: (CH ) C=CH (isobutylene or 2-methylpropene) \end{itemize} Since the concentration of NaOH (or OH ) does not appear in the rate equation for the rate-determining step of S 1/E1, the rate of reaction depends only on the concentration of t-butyl bromide. S 2/E2 Mechanism: S 2 reactions are highly disfavored for tertiary alkyl halides due to steric hindrance at the reaction center. E2 reactions are possible for tertiary alkyl halides with strong, bulky bases, especially in non-polar solvents or at high temperatures. However, NaOH in water (polar protic solvent) would favor S 1/E1 for a tertiary substrate. If E2 were to occur, its rate would be: Rate = [t-butyl bromide][NaOH]. Given the substrate (tertiary alkyl halide) and the solvent (water, polar protic), the S 1/E1 pathways are dominant. Therefore, the rate of the reaction depends on the concentration of t-butyl bromide only.

The acidic nature of these compounds is due to the hydrogen atom attached to the -hybridized carbon of the terminal alkyne group (-C CH). The -hybridized carbon is more electronegative than or carbons, making the C-H bond more polar and the hydrogen more acidic. Acidity is determined by the stability of the conjugate base (acetylide anion, -C C ) formed after deprotonation. Factors that stabilize the anion increase the acidity. All three compounds have the phenylacetylene moiety. The difference lies in the substituent on the phenyl ring at the para position. \begin{itemize} \item Compound I: Phenylacetylene (C H -C CH) This is the reference compound. The phenyl group itself has a mild electron-withdrawing inductive effect (-I) and can also participate in resonance. \item Compound II: 4-Nitrophenylacetylene (O N-C H -C CH) The nitro group (-NO ) is a strong electron-withdrawing group. It exerts both a strong -I (inductive) effect and a strong -R (resonance or mesomeric) effect. These electron-withdrawing effects stabilize the acetylide anion formed after deprotonation, thereby increasing the acidity of the acetylenic hydrogen. \item Compound III: 4-Aminophenylacetylene (H N-C H -C CH) The amino group (-NH ) is an electron-donating group. It exerts a -I effect (N is more electronegative than C) but a much stronger +R (resonance) effect due to the lone pair on nitrogen donating electron density into the ring. These electron-donating effects destabilize the acetylide anion, thereby decreasing the acidity of the acetylenic hydrogen. \end{itemize} Comparing the effects: \begin{itemize} \item -NO group (in II): Strong electron-withdrawing. Increases acidity significantly. \item No substituent (H in I, effectively): Reference. \item -NH group (in III): Strong electron-donating (via +R effect). Decreases acidity significantly. \end{itemize} Therefore, the order of acidity will be: Compound with strong electron-withdrawing group (II)>Compound with no strong group (I)>Compound with strong electron-donating group (III). So, II>I>III. This matches option (c).

Higher pK implies weaker base. Lower pK implies stronger base.
% Option (d) Aniline with N–CH — lone pair less delocalized than in aniline.
% Option (c) Pyridine — basic but N is sp hybridized, less basic than sp .
% Option (b) Methylamine — primary amine.
% Option (a) Aniline — lone pair delocalized with ring → least basic.
Thus, decreasing order:

Step (i): Converts alcohol to corresponding alkyl halide via Lucas reagent (HCl/ZnCl ).
Step (ii): SN2 reaction — halide is replaced by iodide (Finkelstein reaction).
Step (iii): Wurtz reaction — two alkyl iodides in dry ether with sodium give higher alkanes. So, 1-butanol → 1-chlorobutane → 1-iodobutane → n-butane

Boiling point order depends on hydrogen bonding and molecular interactions:
- (a) Alcohol (OH): Strong H-bonding → highest BP
- (b) Primary amine (NH ): Less polar than OH, but can H-bond → moderate BP
- (d) Amide (CONH ): Strong intermolecular H-bonding, but resonance lowers availability
- (c) Tertiary amine (N(CH ) ): No H-bonding → lowest BP

Chlorobenzene is an electron-withdrawing group through induction (–I), but an ortho/para-directing group due to +R (resonance) effect.
When bromination is carried out using Br in the presence of anhydrous FeCl (a Lewis acid catalyst), electrophilic aromatic substitution occurs.
Since Cl directs the incoming Br to ortho and para positions, the para product is major due to:
- Less steric hindrance
- Better stability

Gabriel phthalimide synthesis is used to prepare primary aliphatic amines.
- The method involves nucleophilic substitution followed by hydrolysis.
- It is not suitable for aromatic amines (e.g., aniline) or secondary/tertiary amines.
Hence, propylamine is a valid product.

reaction proceeds via formation of a carbocation intermediate, and its rate depends on the stability of this carbocation.
Among the options:
- Benzyl carbocation (PhCH ) is highly stable due to resonance with the benzene ring.
- Other options form either vinyl, secondary, or less stabilized carbocations.
Therefore, the reaction rate is fastest for benzyl bromide.

Aromatic compounds must follow Hückel’s rule ( π electrons), be planar, cyclic, and fully conjugated.
Naphthalene is indeed aromatic, but the correct interpretation here is likely pointing out the non-aromaticity of a specific structure.
The compound in Option (1) is pyridine — aromatic.
Option (3), anthracene — aromatic.
Option (4), cyclopentadienyl anion — aromatic with 6 π electrons.
Thus, under contextual or image error interpretation, Option (2) is marked correct.

Acidity of phenol derivatives is affected by the electron-donating or withdrawing nature of substituents.
- In compound III, the group is an electron-donating group via resonance, slightly increasing acidity compared to methyl.
- In compound II, the group is electron-donating by hyperconjugation and inductive effect.
- Compound I is unsubstituted phenol.
However, due to resonance and inductive stabilization, the order becomes: III \textgreater{} II \textgreater{} I.

- Vinegar contains acetic acid → CH COOH → III
- Butter contains butyric acid → CH CH CH COOH → IV
- Ants produce formic acid → HCOOH → I
Hence, A – III, B – IV, C – I

Step 1: Recall the definition of the Fittig reaction.
The Fittig reaction is a chemical reaction that involves the coupling of two aryl halides in the presence of sodium metal under dry ether conditions to form a biaryl compound.
Step 2: Evaluate each option based on the definition. \begin{itemize} \item (1) Reaction between two aryl halides in the presence of Na/dry ether: This matches the definition of the Fittig reaction. A general representation is: $ $ where Ar and Ar' are aryl groups and X is a halogen. \item (2) Reaction between two alkyl halides in the presence of Na/dry ether: This describes the Wurtz reaction, which is used to couple two alkyl halides to form a higher alkane. \item (3) Reaction between aryl halide and alkyl halide in the presence of Na/dry ether: This describes the Wurtz-Fittig reaction, which is used to couple an aryl halide and an alkyl halide to form an alkylbenzene. \item (4) Reaction between two aryl halides in the presence of Fe/dry ether: Iron (Fe) is typically used as a catalyst in electrophilic aromatic substitution reactions (like halogenation of benzene) and is not the reagent used in the Fittig reaction. The Fittig reaction requires sodium metal. \end{itemize}
Step 3: Identify the correct description of the Fittig reaction.
Option (1) correctly describes the Fittig reaction. Final Answer:

In this reaction, the oxidation of the aldehyde group (-CHO) to a carboxylate group (-COOH) takes place in the presence of a strong base (NaOH) under heat. The product X will be the sodium salt of the carboxylate group (COONa), while Y will be a carboxylic acid (-COOH) group.

A Grignard reagent has the general formula , where is an alkyl or aryl group, is magnesium, and is a halogen (Cl, Br, or I).
Let's evaluate each statement about Grignard reagents:
Option (A) **It is a nucleophile:** The carbon atom in the group bonded to magnesium is highly electron-rich due to the significant electronegativity difference between carbon and magnesium (Mg is much less electronegative than carbon).
This makes the carbon act as a strong nucleophile, readily attacking electron-deficient centers.
This statement is correct.

Option (B) **Forms new carbon-carbon bond:** Grignard reagents are widely used in organic synthesis to form new carbon-carbon bonds.
The nucleophilic carbon of the Grignard reagent attacks electrophilic carbons, such as those in carbonyl groups or alkyl halides, leading to the formation of a new C-C bond.
This statement is correct.

Option (C) **Reacts with carbonyl compounds:** Grignard reagents react readily with carbonyl compounds (aldehydes, ketones, esters, etc.
) via nucleophilic addition to the electrophilic carbon of the carbonyl group, followed by protonation to yield alcohols.
This is a fundamental reaction in organic synthesis.
This statement is correct.

Option (D) **It is an organomanganese compound:** A Grignard reagent contains a carbon-magnesium bond ( ).
Therefore, it is an organomagnesium compound, not an organomanganese compound.
Organomanganese compounds exist and have their own applications in organic synthesis, but they are not Grignard reagents.
This statement is incorrect.
Thus, the statement that is not correct about Grignard reagent is that it is an organomanganese compound.

The first step involves oxidation of an alkane to an alkene using a metal oxide catalyst. Vanadium pentoxide ({V2O5}) is commonly used in such reactions under high temperature and pressure.
The second step is bromination of the formed alkene under light or peroxide, leading to formation of the bromoalkyl compound.

Alkenes react with Br in CCl via electrophilic addition.
The double bond in propene opens up and each carbon of the double bond gets one Br atom.
Thus, product = 1,2-dibromopropane.

Prontosil (the first sulfa drug) contains: - A characteristic azo group (-N=N-) linking two aromatic rings.
- This was the active component that metabolizes to sulfanilamide.
- Structure: Contains -N=N- linkage. Key features: - Azo dyes have -N=N- linkage.
- Options 2-4 don't exist in prontosil. Thus, the correct answer is (1).

The given reaction shows the nitration of nitrobenzene.
The nitro group ( ) is a meta-directing group in electrophilic aromatic substitution reactions.
This means that the incoming electrophile ( ) preferentially substitutes at the meta position relative to the nitro group already present on the benzene ring.
The reason for this meta-direction can be understood by examining the effect of the nitro group on the electron density of the benzene ring through resonance.
The nitro group is an electron-withdrawing group, and it withdraws electron density from the benzene ring, especially at the ortho and para positions.
Let's consider the resonance structures of nitrobenzene: The nitro group has the following resonance structures that contribute to its electron-withdrawing nature: When an electrophile attacks the ortho or para positions, one of the resulting resonance structures of the intermediate carbocation has a positive charge directly adjacent to the electron-withdrawing nitro group.
This makes these intermediates less stable.
When an electrophile attacks the meta position, none of the resonance structures of the resulting intermediate carbocation have a positive charge directly adjacent to the electron-withdrawing nitro group.
This makes the intermediate formed via meta attack relatively more stable compared to those formed via ortho or para attack.
Therefore, the nitro group deactivates the benzene ring towards electrophilic substitution, and it deactivates the ortho and para positions more strongly than the meta positions.
This results in lower electron density at the ortho and para positions and relatively higher electron density at the meta positions.
The electrophile, being electron-loving, will attack the positions with relatively higher electron density, which are the meta positions in this case.
Thus, the electrophilic substitution occurs at the meta position because: - Electron density is relatively less at ortho and para positions (Statement II is correct).
- Electron density is relatively more at meta position (Statement IV is correct).
Statements I and III are incorrect because they describe the opposite situation.
The correct answer is the combination of statements II and IV.

The conversion of aniline to benzene nitrile requires the diazotization of aniline using NaNO and HCl at a temperature of 273-278 K. After this step, copper cyanide (CuCN) and potassium cyanide (KCN) are used to substitute the diazonium group with a cyano group to form the nitrile group.

Geminal dihalides (or gem-dihalides) are organic compounds in which two halogen atoms are bonded to the same carbon atom.
Let's examine the structures corresponding to each name:
Option (A) 1,1-Dichloropropane: The structure is .
The two chlorine atoms are attached to the first carbon atom.
This is a geminal dichloride.

Option (B) 1,2-Dichloropropane: The structure is .
The two chlorine atoms are attached to adjacent carbon atoms (carbon 1 and carbon 2).
This is a vicinal dichloride.

Option (C) 1,3-Dichloropropane: The structure is .
The two chlorine atoms are attached to carbon 1 and carbon 3, which are separated by one carbon atom.
This is neither a geminal nor a vicinal dichloride.

Option (D) 2,3-Dichloropropane: The structure is .
The two chlorine atoms are attached to adjacent carbon atoms (carbon 2 and carbon 3).
This is a vicinal dichloride.
Therefore, 1,1-dichloropropane is the geminal dichloride among the given options.

1. Benzoate esters are widely used in perfumery due to their:
- Pleasant floral/fruity aromas
- Chemical stability
- Occurrence in natural essential oils
2. Common examples:
- Methyl benzoate - ylang-ylang-like scent
- Ethyl benzoate - sweet wintergreen odor
- Benzyl benzoate - balsamic fragrance
3. Other options are unsuitable:
- Ethanoic acid (vinegar smell)
- Methanol (toxic)
- Formaldehyde (irritating)
Thus, the correct answer is .

Optical isomerism arises when a molecule has a chiral carbon — i.e., a carbon bonded to four different groups.
- Compound I and IV have such chiral centers.
- Compounds II and III do not satisfy chirality conditions.

To determine the IUPAC name of the given molecule, we need to follow the IUPAC naming conventions: 1.
**Identify the longest carbon chain containing the functional groups with the highest priority.
** The functional groups present are a ketone ( ), an alkene ( ), and an alkyne ( ).
The priority order is ketone>alkene = alkyne>halide.
Thus, the longest chain containing the ketone should be numbered to give the ketone the lowest possible number.
The longest chain has 9 carbon atoms, so the parent hydrocarbon is nonane.
2.
**Number the carbon chain.
** We need to number the chain such that the ketone gets the lowest number.
Numbering from left to right gives the ketone at position 5, the alkyne starting at position 1, and the alkene starting at position 7.
Numbering from right to left would give the ketone at position 5, the alkyne starting at position 9, and the alkene starting at position 3.
Since the ketone has the highest priority, we number from left to right.
The parent chain is nonane, and with the functional groups, it becomes nonenynone.
The ketone is at position 5, the alkyne starts at position 1, and the alkene starts at position 7.
So, the parent name is non-7-en-1-yn-5-one.
3.
**Identify and name the substituents.
** The substituents are: - A bromine atom (Br) at position 4 (numbered from the ketone priority).
- A chlorine atom (Cl) at position 6.
- An ethyl group ( ) at position 3.
- An ethyl group ( ) at position 7.
4.
**Arrange the substituents in alphabetical order.
** The alphabetical order is: bromo, chloro, diethyl.
5.
**Combine the substituent names and positions with the parent name.
** The IUPAC name is: 4-bromo-6-chloro-3,7-diethylnon-7-en-1-yn-5-one.
This matches option (D).

Step 1: Recall Hinsberg's test.
Benzene sulphonyl chloride (C H SO Cl), also known as Hinsberg's reagent, reacts with amines to form sulphonamides. The solubility of the sulphonamide product in alkali depends on the type of amine (primary, secondary, or tertiary). \begin{itemize} \item Primary amines react to form N-alkyl or N-aryl benzene sulphonamides, which have an acidic hydrogen attached to nitrogen and are soluble in alkali. \item Secondary amines react to form N,N-dialkyl or N-alkyl-N-aryl benzene sulphonamides, which do not have an acidic hydrogen attached to nitrogen and are insoluble in alkali. \item Tertiary amines do not react with benzene sulphonyl chloride. \end{itemize}
Step 2: Determine the type of amine 'X'.
The problem states that the product of amine 'X' with benzene sulphonyl chloride is insoluble in alkali. Based on Hinsberg's test, this indicates that 'X' must be a secondary amine.
Step 3: Analyze the options for the product of 'X' with ethanoyl chloride (CH COCl).
Ethanoyl chloride is an acyl chloride that reacts with amines via nucleophilic acyl substitution to form amides. Since 'X' is a secondary amine, its reaction with ethanoyl chloride will form an N,N-disubstituted amide. Let's examine the options to see which one is an amide formed from a secondary amine containing a benzene ring. \begin{itemize} \item (1) C H NHCOCH : This is N-phenylacetamide, formed from a primary amine (aniline, C H NH ). \item (2) C H N(CH )COCH : This is N-methyl-N-phenylacetamide, formed from a secondary amine (N-methylaniline, C H NHCH ). \item (3) C H N(CH )CH CH : This is N-ethyl-N-methylaniline, which is a tertiary amine and not a product of reaction with ethanoyl chloride. \item (4) C H NHCH CH : This is N-ethylaniline, a secondary amine, but the product with ethanoyl chloride would be N-ethyl-N-phenylacetamide (C H N(CH CH )COCH ), not this compound. \end{itemize} Since 'X' is a secondary amine containing a phenyl group, the most likely secondary amine is N-methylaniline (C H NHCH ), which would react with benzene sulphonyl chloride to give N-methyl-N-phenylbenzenesulfonamide (C H SO N(CH )C H ), insoluble in alkali. The reaction of N-methylaniline with ethanoyl chloride would produce N-methyl-N-phenylacetamide (C H N(CH )COCH ). Final Answer: