Organic Reactions

Step 1: Concept
Amino acids exist as Zwitterions at their isoelectric point, but change form in acidic or basic media.
Step 2: Analysis
At (acidic medium), the carboxylate group ( ) of the Zwitterion accepts a proton.
Step 3: Conclusion
The structure becomes a cation: .
Final Answer: (B)

Step 1: Concept
Phenolphthalein is a phthalein dye synthesized via a condensation reaction.
Step 2: Analysis
Heating phthalic anhydride with phenol in the presence of concentrated (as a dehydrating agent) yields phenolphthalein.
Step 3: Conclusion
The required reagent is phenol.
Final Answer: (C)

Step 1: Concept
Identify the attacking species and the leaving group.
Step 2: Analysis
The Bromine atom (leaving group) is replaced by the hydroxyl group ( ), which acts as a nucleophile.
Step 3: Conclusion
This is a unimolecular nucleophilic substitution ( ) reaction.
Final Answer: (C)

Step 1: Concept
Identify the attacking species and the leaving group.
Step 2: Analysis
The Bromine atom (leaving group) is replaced by the hydroxyl group ( ), which acts as a nucleophile.
Step 3: Conclusion
This is a unimolecular nucleophilic substitution ( ) reaction.
Final Answer: (C)

Step 1: Concept
Hydrolysis of gem-dihalides initially produces an unstable gem-diol.
Step 2: Analysis
Hydrolysis replaces two Cl atoms with two OH groups. The resulting 2,2-propane diol is unstable and loses a water molecule.
Step 3: Conclusion
The dehydration of the gem-diol results in the formation of acetone.
Final Answer: (A)

Step 1: Concept
Carbonyl groups ( ) in ketones typically undergo addition reactions initiated by nucleophiles.
Step 2: Analysis
The ion attacks the partially positive carbonyl carbon, followed by protonation of the oxygen.
Step 3: Conclusion
Since the nucleophile ( ) attacks first, it is a nucleophilic addition reaction.
Final Answer: (C)

Step 1: Concept
Carboxylic acids react with ammonia to form salts, which dehydrate to amides and then to nitriles.
Step 2: Analysis
* (A: Ammonium acetate). * (B: Acetamide). * (C: Acetonitrile).
Step 3: Conclusion
The end product C is methyl cyanide (acetonitrile).
Final Answer: (D)

Step 1: Find Concentration
Initial . Diluted 100 times, .
Step 2: Account for Water dissociation
Total .
Step 3: Find pOH and pH
. . This value is between 7 and 8.
Final Answer: (a)

Step 1: Stefan's Law
Energy radiated ( ) is proportional to the fourth power of absolute temperature ( ): .
Step 2: Substitution
New energy .
Step 3: Calculation
.
Final Answer: (a)

Step 1: Given Data
For an equilateral prism, the angle of prism . The problem states the angle of minimum deviation .
Step 2: Formula
The refractive index is given by .
Step 3: Calculation
.
Final Answer: (b)

Step 1: Concept
Potential .
Step 2: Analysis
On the axial line, or , so .
Step 3: Conclusion
This yields the maximum magnitude of potential for a given distance.
Final Answer: (A)

Step 1: Formula
Magnetic dipole moment .
Step 2: Variables
From the formula, depends on the number of turns ( ), current ( ), and area ( ).
Step 3: Conclusion
The external magnetic field does not appear in the definition of the dipole moment itself; it only determines the torque acting on that moment.
Final Answer: (a)

Step 1: Setup
Consider a small element at distance . Turns per unit length .
Step 2: Field Element
Field due to element : .
Step 3: Integration
.
Final Answer: (d)

Step 1: Compare Equations
Standard wave equation: . The given equation is .
Step 2: Find Velocity
From comparison, and . Wave velocity .
Step 3: Distance
Distance .
Final Answer: (b)

Step 1: Reaction
Ethyl bromide ( ) reacts with alcoholic silver nitrite ( ).
Step 2: Analysis
is a covalent compound. The nucleophilic attack occurs through the Nitrogen atom rather than the Oxygen atom.
Step 3: Equation

Step 4: Conclusion
The major product formed is nitro ethane.
Final Answer: (D)

Step 1: Concept
Iso-propyl chloride is a secondary ( ) alkyl halide.
Step 2: Analysis
Secondary halides are at the border of nucleophilic substitution mechanisms.
Step 3: Conclusion
They can undergo hydrolysis via either or pathways, depending on factors like the nature of the solvent and the strength of the nucleophile.
Final Answer: (C)

Step 1: Concept
Aldehydes that do not contain an -hydrogen atom undergo the Cannizzaro reaction when treated with concentrated alkali (50% NaOH).
Step 2: Analysis
Benzaldehyde ( ) lacks an -hydrogen atom.
Step 3: Reaction
In this disproportionation reaction, one molecule of benzaldehyde is reduced to benzyl alcohol and another is oxidized to sodium benzoate. $ $
Step 4: Conclusion
Thus, benzaldehyde provides the corresponding alcohol and acid salt.
Final Answer: (B)

Step 1: Reaction
The reduction of nitrobenzene by and is a neutral reduction.
Step 2: Calculation
.
Step 3: Conclusion
The product formed is N-phenyl hydroxylamine.
Final Answer: (C)

Step 1: Understanding nucleophilicity.
Nucleophilicity refers to the ability of a species to donate a pair of electrons to form a bond with an electrophile. It depends on the species' electron density and size. More negatively charged and less electronegative species are generally better nucleophiles.

Step 2: Analyze the options.
- : Fluoride ion is highly electronegative and its electron density is tightly held, making it a weaker nucleophile.
- : Hydroxide ion is a good nucleophile, but not the strongest among the given options.
- : The methyl anion ( ) is a very strong nucleophile because the negative charge is localized on a small, less electronegative carbon atom.
- : The amide anion ( ) is a good nucleophile, but not as good as the methyl anion.

Step 3: Conclusion.
The correct answer is (3), , as it is the most nucleophilic species among the options.

Step 1: Understanding the chloral-chlorobenzene reaction.
Chloral (trichloroacetaldehyde) reacts with chlorobenzene to produce DDT (dichlorodiphenyltrichloroethane), which is an insecticide.

Step 2: Explanation of the options.
- (1) BHC: BHC (benzene hexachloride) is another insecticide, but it is not the product of chloral and chlorobenzene.
- (2) DDT: DDT is the correct product. The reaction between chloral and chlorobenzene results in DDT.
- (3) Gammexene: This is another name for BHC, so it is not the correct answer.
- (4) Lindane: Lindane is a specific isomer of BHC, but not the result of this reaction.

Step 3: Conclusion.
The correct answer is (2) DDT.

Step 1: Understanding the Wurtz reaction.
The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal to form alkanes. When two different alkyl iodides react, multiple products can be formed.

Step 2: Explanation of the options.
- (1) Butane: Butane is formed when two ethyl radicals combine.
- (2) Ethane: Ethane is formed when two methyl radicals combine.
- (3) Propane: Propane is formed when one methyl and one ethyl radical combine.
- (4) A mixture of above three: Since different combinations of methyl and ethyl groups are possible, the reaction gives a mixture of butane, ethane, and propane.

Step 3: Conclusion.
The correct answer is (4) A mixture of above three.

Step 1: Understanding the Concept:
Anhydride = compound that reacts with water to form an acid. Mixed anhydride gives two different acids.
Step 2: Detailed Explanation:
ClO + H O HClO + HClO (disproportionation).
2ClO + H O HClO + HClO .
Thus, ClO is mixed anhydride of chlorous (HClO ) and chloric (HClO ) acids.
Step 3: Final Answer:
Thus, ClO is mixed anhydride of HClO and HClO .

Step 1: Understanding the Concept:}
Wurtz reaction with dihalide gives cyclic product via intramolecular coupling.
Step 2: Detailed Explanation:
1-bromo-3-chlorocyclobutane has halogens on carbons 1 and 3. Reaction with Na (Wurtz-type) causes elimination of NaBr and NaCl, forming a bond between C1 and C3, giving bicyclo[1.1.0]butane.
Step 3: Final Answer:
Thus, product is bicyclobutane.

Step 1: Understanding the Concept:
Sequence involves substitution, followed by oxidative cleavage and further reaction leading to ring contraction.
Step 2: Detailed Explanation:
Cyclohexyl bromide reacts with NH to form amine (A).
On treatment with O /H O, oxidative cleavage occurs forming a diketone or dialdehyde (B).
Further reaction with BaO leads to rearrangement and ring contraction forming a smaller ring ketone.
Step 3: Final Answer:
Thus, the final product corresponds to option (C).

Step 1: Understanding the Concept:
This reaction involves phenol treated with and , leading to formation of a methylene-bridged cyclic ether.

Step 2: Detailed Explanation:
In basic medium, phenol forms phenoxide ion. The acts as a methylene donor.
Since two groups are present in ortho-position, intramolecular cyclization occurs forming a five-membered ring with linkage.
This results in a benzodioxole structure (cyclic acetal type).

Step 3: Final Answer:
Option (a)

Step 1: Understanding the Concept:
Based on the reaction sequence shown.

Step 2: Detailed Explanation:

Step 3: Final Answer:
As per the figure. reduced to 1st order alcohol

Step 1: Understanding the Concept:
Based on the reaction shown.

Step 2: Detailed Explanation:

Step 3: Final Answer:
As per the figure.

Step 1: Understanding the Concept:
C H is toluene. Side-chain chlorination, then ring bromination, then reduction of Cl to H.

Step 2: Detailed Explanation:
1. Toluene + 3Cl , heat → side-chain chlorination: C H CCl (benzotrichloride, A).
2. A + Fe/Br → ring bromination (meta-directing due to -CCl group): 3-bromo-benzotrichloride (B).
3. B + Zn/HCl → reduction of -CCl to -CH : m-bromo toluene (C).

Step 3: Final Answer:
m-bromo toluene

Step 1: Concept:
Identify each reaction type:

• → Nucleophilic substitution (chain extension)
• Partial hydrolysis of nitrile → Amide
• → Hofmann bromamide reaction (gives amine with one less carbon)

Step 2: Detailed Explanation:
First Reaction:
Product is Propanenitrile.

Second Reaction (Partial Hydrolysis):
Product is Propanamide.

Third Reaction (Hofmann Bromamide Reaction):
In Hofmann bromamide reaction: - Amide loses one carbon - Forms a primary amine
So, product is Ethylamine.

Step 3: Final Answer:

Step 1: Understanding the Concept:
This is a reaction sequence starting from formaldehyde.
Step 2: Detailed Explanation:
\begin{enumerate} \item HCHO A: Reduction of formaldehyde gives methanol (CH OH). A = CH OH. \item CH OH B: P + I produces PI , which converts alcohol to alkyl iodide. B = CH I (methyl iodide). \item CH I C: Nucleophilic substitution gives methyl cyanide (acetonitrile). C = CH CN. \item CH CN D: Hydrolysis of nitrile gives carboxylic acid. D = CH COOH (acetic acid). \end{enumerate}
Step 3: Final Answer:
D is acetic acid, option (A).

Step 1: Understanding the Concept:
Carbylamine reaction is a specific test for primary amines.
Step 2: Detailed Explanation:
In the carbylamine test, a primary amine is heated with chloroform (trihalogenated methane) and alcoholic KOH. The reaction produces an isocyanide (carbylamine) which has a foul smell.
Step 3: Final Answer:
The mixture is trihalogenated methane and a primary amine, option (B).

Step 1: Understanding the Concept:
Acidity depends on the stability of the conjugate base formed after losing a proton.
Step 2: Detailed Explanation:
\begin{itemize} \item RCOOH (Carboxylic acids): Strongest among these due to resonance stabilization of carboxylate ion. pKa ≈ 4-5. \item H CO (Carbonic acid): Weak acid, pKa ≈ 6.4. \item C H OH (Phenol): Weak acid, pKa ≈ 10. \item H O (Water): Very weak acid, pKa ≈ 15.7. \item ROH (Alcohols): Very weak acid, pKa ≈ 16-18. \end{itemize}
Step 3: Final Answer:
The correct order is RCOOH>H CO >C H OH>H O>ROH, option (A).

Step 1: Understanding the Concept:
Lucas reagent (ZnCl /HCl) reacts with alcohols to form alkyl chlorides. The rate depends on the stability of the carbocation intermediate.
Step 2: Detailed Explanation:}}
\begin{itemize} \item (A) butan-1-ol: Primary alcohol, forms primary carbocation (least stable), no reaction at room temperature. \item (B) butan-2-ol: Secondary alcohol, forms secondary carbocation, reacts slowly (turbidity in 5-10 min). \item (C) 2-methyl propan-1-ol: Primary alcohol (neopentyl type), forms primary carbocation, no reaction at room temperature. \item (D) 2-methyl propan-2-ol: Tertiary alcohol, forms tertiary carbocation (most stable), reacts immediately (turbidity instantly). \end{itemize}
Step 3: Final Answer:
2-methyl propan-2-ol (tert-butyl alcohol) reacts fastest, option (D).

Step 1: Understanding the Concept:
Basicity depends on the availability of the lone pair on nitrogen. Electron-donating groups increase basicity, while electron-withdrawing groups decrease basicity.
Step 2: Detailed Explanation:
\begin{itemize} \item (A) (ethylamine): Aliphatic amine with electron-donating ethyl group, strong base. \item (B) (acetamide): The carbonyl group is electron-withdrawing, delocalizing the lone pair on nitrogen through resonance, reducing basicity. \item (C) (benzamide): Similar to acetamide, but the phenyl ring further reduces basicity due to conjugation. \item (D) (hydrazine): Less basic than ethylamine due to the electron-withdrawing effect of the additional group. \end{itemize}
Step 3: Final Answer:
Ethylamine is the most basic, option (A).

Step 1: Understanding the Concept:
Decarboxylation is the removal of CO from a carboxylic acid or its salt. Sodium salts of carboxylic acids on decarboxylation with soda lime (NaOH + CaO) yield alkanes.
Step 2: Detailed Explanation:
Sodium propionate is . Upon heating with soda lime, it undergoes decarboxylation: The product is ethane.
Step 3: Final Answer:
Decarboxylation of sodium propionate gives ethane, option (D).

Step 1: Understanding the Concept:
Formic acid (HCOOH) is unique because it has a hydrogen atom directly attached to the carboxyl group.
Step 2: Detailed Explanation:
When reacts with carboxylic acids, it generally replaces the -OH group with -Cl to form an acyl chloride ( ). However, formic acid, being the simplest carboxylic acid, behaves differently. The initial product, formyl chloride (HCOCl), is unstable and decomposes immediately at room temperature to carbon monoxide (CO) and hydrogen chloride (HCl). The overall reaction is:
Step 3: Final Answer:
Formic acid reacts with to give CO and HCl, option (D).

Step 1: Understanding the Concept:
Vinegar is a common household product used for cooking and preservation. It is derived from the fermentation of ethanol.
Step 2: Detailed Explanation:
Vinegar typically contains 5-8% of acetic acid by volume. Acetic acid's IUPAC name is ethanoic acid (CH COOH). Other acids like citric acid (in citrus fruits) and oxalic acid (in tomatoes, spinach) are not the primary component of vinegar.
Step 3: Final Answer:
Vinegar is a dilute solution of ethanoic acid, option (A).

Step 1: Understanding the Concept:
Acidity of phenols is enhanced by electron-withdrawing groups, especially when they are ortho or para to the OH group. This is because these groups stabilize the conjugate base (phenoxide ion) through resonance and inductive effects.
Step 2: Detailed Explanation:
The aldehyde group ( ) is a strong electron-withdrawing group. When it is ortho or para to the OH group, it can participate in resonance with the phenoxide ion, delocalizing the negative charge further onto the oxygen of the aldehyde group, thus stabilizing it. Meta-substitution does not allow this resonance stabilization. A simple alcohol ( ) is not acidic.
Step 3: Final Answer:
Therefore, the ortho- or para-hydroxybenzaldehyde is the most acidic among the options.

Step 1: Understanding the Concept:
The Wurtz reaction is a coupling reaction used to form alkanes from alkyl halides.
Step 2: Detailed Explanation:
In the Wurtz reaction, two molecules of an alkyl halide (usually primary) react with sodium metal in the presence of dry ether to form a higher alkane. . Sodium acts as a reducing agent, and the reaction proceeds via a free radical mechanism.
Step 3: Final Answer:
Thus, the correct reagent is sodium in dry ether, option (D).

Step 1: Understanding the Concept:
Dry distillation or heating of calcium salts of carboxylic acids yields ketones. This is a general method for ketone synthesis.
Step 2: Detailed Explanation:
Calcium benzoate is the calcium salt of benzoic acid . Upon dry distillation, it undergoes decomposition to form a ketone. The product is diphenylmethanone, commonly known as benzophenone.
Step 3: Final Answer:
Thus, the correct option is (D) benzophenone.

Step 1: Understanding the Concept:
Grignard reagents ( ) act as strong nucleophiles. They react with various electrophiles to form new carbon-carbon bonds.
Step 2: Detailed Explanation:
When a Grignard reagent reacts with carbon dioxide ( ), it forms a carboxylate salt. This salt upon acid hydrolysis yields a carboxylic acid. Other options like or give different products like alcohols or sulfinic acids.
Step 3: Final Answer:
Thus, the correct answer is (B) .

Step 1: Understanding the Concept:
Diazomethane (CH N ) decomposes upon heating to produce a carbene (:CH ).
Step 2: Detailed Explanation:}}
The reaction of ethene with diazomethane in the presence of heat or light is a carbene addition reaction. Diazomethane decomposes to form methylene carbene (:CH ) which then adds to the double bond of ethene to form cyclopropane. The intermediate is the methylene carbene.
Step 3: Final Answer:
The intermediate is :CH (methylene carbene), option (D).

Step 1: Understanding the Concept:
S 2 reactions are favored by polar aprotic solvents. These solvents solvate the cation but not the anion, leaving the nucleophile highly reactive.
Step 2: Detailed Explanation:
\begin{itemize} \item CH OH and H O are polar protic solvents. They solvate the nucleophile through hydrogen bonding, reducing its nucleophilicity and thus slowing the S 2 reaction. \item DMSO (dimethyl sulfoxide) is a polar aprotic solvent. It solvates the cation but not the anion, leaving the nucleophile "naked" and highly reactive, maximizing the S 2 rate. \item Benzene is non-polar and does not stabilize the charged transition state, so the reaction is very slow. \end{itemize}
Step 3: Final Answer:
The rate of S 2 reaction is maximum in DMSO, option (C).

Step 1: Understanding the Concept:
Hoffmann bromamide degradation.
Step 2: Detailed Explanation:
R-CONH + Br + 4KOH R-NH + 2KBr + K CO + 2H O
CH CONH CH NH
Step 3: Final Answer:
Product is CH NH (methylamine).

Step 1: Understanding the Concept:
Find longest chain containing -OH group.
Step 2: Detailed Explanation:
Longest chain: 5 carbons (pentane)
-OH at C2, methyl at C4
Name: 4-methylpentan-2-ol
Step 3: Final Answer:
4-methyl pentane-2-ol.

Step 1: Understanding the Concept:
Reaction with NH gives mixture of amines.
Step 2: Detailed Explanation:
R-X + NH R-NH (1°), then R NH (2°), then R N (3°)
Step 3: Final Answer:
All types of amines are formed.

Step 1: Understanding the Concept:
Addition follows Markovnikov's rule.
Step 2: Detailed Explanation:
HOCl adds such that OH goes to more substituted carbon, Cl to less substituted.
CH -CH=CH + HOCl CH -CH(OH)-CH Cl
Step 3: Final Answer:
Product is CH -CH(OH)-CH Cl.

Step 1: Understanding the Concept:
This is Wurtz reaction.
Step 2: Detailed Explanation:
2C H I + 2Na C H + 2NaI
Step 3: Final Answer:
Product is butane (C H ).

Step 1: Understanding the Concept:
KMnO oxidizes alkyl side chain to carboxylic acid.
Step 2: Detailed Explanation:
C H -CH -CH C H -COOH (benzoic acid)
Complete oxidation of side chain.
Step 3: Final Answer:
Product is benzoic acid.

Step 1: Understanding the Concept:
Cannizzaro reaction occurs for aldehydes without -H.
Step 2: Detailed Explanation:
HCHO and C H CHO have no -H, undergo Cannizzaro.
CH CHO has -H, undergoes aldol condensation.
Step 3: Final Answer:
C H CHO and HCHO.

Step 1: Understanding the Concept:
Acetylation replaces H with acetyl group.
Step 2: Detailed Explanation:
R NH + CH COCl R NCOCH + HCl
Product is N,N-dialkyl acetamide.
Step 3: Final Answer:
N,N-dialkyl acetamide.

Step 1: Understanding the Concept:
Nitrobenzene reduction to aniline.
Step 2: Detailed Explanation:
Sn/HCl is commonly used for reduction of nitro group to amine in acidic medium.
Step 3: Final Answer:
Reducing agent is Sn/HCl.

Step 1: Understanding the Concept:
AgNO gives nitroalkanes.
Step 2: Detailed Explanation:
C H Br + AgNO C H NO + AgBr
Step 3: Final Answer:
Product is nitro ethane.