Solution: (i) is e (ii) is d (iii) is a (iv) is b (v) is f (vi) is c

Solution: (i) is c (ii) is d (iii) is a (iv) is b

Solution: (i) is b (ii) is e (iii) is d (iv) is a (v) is c

Solution: (i) is d (ii) is e (iii) is a (iv) is b (v) is c

Solution: Both reactions resemble each other. In Friedel Craft’s acylation reaction, an aryl group or benzene is treated with an acid chloride in the presence of anhydrous AlCl3 and corresponding...

Solution: This is due to the formation of polysubstituted products. To avoid the formation of polysubstituted products Friedel-craft’s alkylation reaction is not used for the preparation of...

Solution:

Solution: Due to the resonance in carboxylic acids, the negative charge is at the more electronegative oxygen whereas, in alcohols or phenols, the negative charge is on a less electronegative atom....

Solution: Compound A = CH3-MgBr Compound B = CH3-COOH Compound C = CH3COOCH3

Solution: The oxygen atom in carbonyl compound pull more shared pair of electron towards itself and so, carbon acquires partial positive charge and oxygen acquires partial negative charge in...

Solution: Both the compounds carbon atom is attached to the electronegative atom oxygen. Thus the oxygen pulls more shared pair of electron towards them and a partial positive charge will be...

Solution: NO2CH2COOH > FCH2COOH > C6H5COOH. NO2CH2COOH is most acidic among the given three compounds. Electron withdrawing groups like -NO2, increases the acidity of carboxylic acids by...

Solution: Compound ‘A’ belongs to the carboxylic acid. Compound ‘B’ belongs to alcohol. Compound ‘C’ belongs to an ester group.

Solution: When propanal reacts with 2-methyl propanal in the presence of NaOH, the mixture of aldehydes are formed. Both the reactants have an alpha-hydrogen and hence, can undergo cross aldol...

Solution: FCH2COOH > ClCH2COOH > C6H5CH2COOH > CH3COOH > CH3CH2OH. CH3CH2OH is least acidic among the given compounds. C6H5CH2COOH is more acidic than CH3COOH due to the resonance in...

Solution: Solution: The products formed on oxidation of 2, 5-dimethylhexan-3-one are the mixtures of ketone and carboxylic acids. Ketone is then further oxidized to carboxylic acids. Overall the...

Solution: The electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl3 is benzoylinium cation. The product formed in this reaction is benzophenone....

SOLUTION: Toluene is first converted to benzal chloride by side-chain chlorination, in presence of Chlorine gas and light. Benzal chloride on hydrolysis at 373K gives benzaldehyde.

Solution:

Solution: (i) Ethane-1,2-dial. (ii) Benzene-1, 4-dicarbaldehyde. (iii) 3-Bromobenzaldehyde.

Solution: (i) 3-Phenylprop-2-ene-1-al. (ii) Cyclohexanecarbaldehyde (iii) 3-Oxopentan-1-al (iv) IUPAC name: But-2-enal

Solution: One can differentiate between pentan-2-one and pentan-3-one by iodoform test. Pentan-2-one have a –CO-CH3 group and therefore forms a yellow precipitate of Iodoform. Pentan-2-one gives a...

Solution: Butanal has no intermolecular hydrogen bonding but butan-1-ol has intermolecular hydrogen bonding. This bonding in butan-1-ol makes it more stable at a higher temperature than butanal.

Solution: Option (A) and (B) are the answers. Reason:

Solution: Option (i) and (ii) are the answers Reason: Benzophenone can be obtained by the Friedel-Craft acylation reaction. The reaction is shown as

Solution: Option (i) and (iii) are the answers. Reason: Grigned reaction and aldol condensation is used to increase the number of carbon attom in the chain as follows:

Solution: Option (ii) and (iv) are the answers. Reason: The carbonyl group of aldehydes and ketones is reduced to CH2​ group on treatment with zinc amalgam and concentrated hydrochloric acid...

Solution: Option (ii) and (iii) are the answers. Reason: Treatment of compound with NaOH yields sodium phenoxide and sodium by means of nucleophilic substitution reaction as follows

Solution: Option (ii) and (iv) are the answers. reason: Aldehydes and ketones and having at least one alpha-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to beta-hydroxy...

Solution: Option (i) is the answer. Reason: From the above reaction carbonyl group is treated with Zn−Hg(Zinc Amalgum) and HCl

Solution: Option (ii) is the answer.  

Solution: Option (iii) is the answer. Reason: This reaction is called as lodoform reaction.  

Solution: Option (ii) is the answer. Reason:

Solution: Option (iv) is the answer. reason: Structure of A and the type of isomerism in the above reaction are Prop-1-en-2-ol, tautomerism respectively. Enol form tautomerises into keto...

Solution: Option (ii) is the answer. Reason: Benzaldhyde C6​H5​CHO on treatment with KOH yields the corresponding alcohol and acid. In this reaction, there is no alpha hydrogen atom present in...

Solution: Option (iv) is the answer. Reason: CH3CHO will not give Cannizzaro’s reaction because it contains a-hydrogen while other three compounds have no a-hydrogen. Hence, they will give...

Solution: Option (iii) is the answer. Reason: Aromatic aldehydes and ketones does not respond to Fehling's test. Sodium hydrogen sulphite,phenyl hydrazine, grignard reaction are common for carbonyl...

Solution: Option (ii) is the answer. Reason:

Solution: Option (iii) is the answer. Reason: The correct order of increasing acidic strength is Ethanol < Phenol < Acetic acid < Chloroacetic acid. Phenol is more acidic than ethanol...

Solution: Option (i) is the answer.    

Solution: Option (ii) is the answer. Reason: Addition of water to but-1-yne in the presence of H2​SO4​ and HgSO4​ gives 2-butaone. The addition takes place by markovnikoff's rule....

Solution: CH3CH2CH3 < CH3CH2NH2 < CH3CH2OH The dipole moment of CH3CH2OH is greater than that of CH3CH2NH2. CH3CH2CH3 has the least dipole moment among the three given compounds because it is...

Solution: The products formed in the reaction of aniline with bromine in a non-polar solvent such as CS2 are 4-Bromoaniline and 2-Bromoaniline where 4-Bromoaniline is the major product.

Solution: This reaction is carried out in a mild basic medium. This is an electrophilic substitution reaction. Aryldiazonium chloride reacts with aniline to form a yellow dye of p-Aminoazobenzene.

Solution: MeNH2 is a stronger base than MeOH because of the lower electronegativity and the presence of the lone pair of electrons on the nitrogen atom in MeNH2.

Solution: At high temperatures, benzene diazonium chloride is highly soluble in water, and at low temperatures, it is a very stable compound in water. Because it is unstable, it should be used as...

Solution: LiAlH4 and Sodium/Alcohol are the best reagents for converting nitrile to primary amine. The nitriles can be converted into a corresponding primary amine through reduction.

Solution: The NH2 group of aniline is acetylated before nitration to control the nitration reaction and the formation of tarry oxidation products and nitro derivatives. P-nitroaniline is the main...

Solution: Option (i) and (ii) are the answers. Reason:...

Solution: Option (i) and (ii) are the answers. Reason: In addition to the nitro derivatives, direct nitration of aniline produces tarry oxidation products. Furthermore, in a strongly acidic...

  Solution: Option (i) and (iii) are the answers. Reason:       

Solution: Option (i) and (ii) are the answers. Reason: Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium...

Solution: Option (i), (ii) and (iii) are the answers. Reason: Arenium ion involved in the bromination of aniline are as follows:

        Solution: Option (A) and (B) is the answer. Reason:

Solution: Option (ii) and (iii) are the answers. Reason: Hypophosphorous acid (phosphinic acid) and ethanol, for example, reduce diazonium salts to arenes, which are then oxidised to phosphorous...

Solution: Option (i) and (ii) are the answers. Reason: Only RNC and CHCl3​ are involved in carbylamine reaction.

Solution: Option (i), (ii), and (iii) are the answers. Reason: They are reducing agents.  

Solution: Option (iii) and (iv) are the answers. Reason: Sandmeyer's reaction is used for the preparation of chlorobenzene and bromobenzene.

Solution: Option (iv) is the answer. Reason: In Hoffmann Bromide degradation as the word, suggest, the amide is reduced to an amine with 1- carbonless, so this is the method in which we don't get...

Solution: Option (ii) is the answer. Reason: The order of boiling points of isomeric amines is 1 amine > 2 amines > 3 amines. 3 amines have no intermolecular association because there are no H...

Solution; Option (iv) is the answer. Reason: Option (iv) is the strongest Bronsted base as there is no delocalization of lone pair of electron of the atom which is not possible in aniline and in...

Solution: Option (i) is the answer. Reason: NH3​ is more basic than H2​O, therefore NH2−​ (Conjugate base of weak acid NH3​) is a stronger base than OH−.

Solution; Option (iv) is the answer. Reason: Option(iv)is the strongest Bronsted base as there is no delocalisation of lone pair of electron of N atom which is not possible in aniline and in...

Solution: Option (iii) is the answer. Reason: A Bronsted Lowry base is a proton acceptor or hydrogen ion acceptor. Amines have a stronger tendency to accept protons and are strong Bronsted bases....

Solution: Option (iv) is the answer. Reason: Diazonium cation is a weak electrophile and hence it reacts with electron-rich compounds containing electron-donating groups such as −OH, -$NH_2$ and...

Solution: Option (ii) is the answer. Reason: Best method for preparing primary aminos form alkyl halides without changing the number of carbon atoms in the chain is Gabriel synthesis. Because this...

Solution: Option (ii) is the answer. Reason: Diazonium salts in the presence of copper powder and halogen acid give aryl halide. Gattermann reaction is a variation of Sandmeyer reaction in which...

Solution: Option (i) is the answer Reason: When acid anhydrides react with primary amines, they produce amide. The H atom of the amino group is replaced with an acyl group in this nucleophilic...

      Solution: Option (ii) is the answer. Reason: The reactivity of amines is proportional to their basicity. If the R group is, the order of basicity is secondary amine ...

Solution: Option (iii) is the answer. Reason: Reduction of nitro aryl compounds in presence of Fe and HCl gives aromatic primary amines.

Solution: Option (iii) is the answer. Reason:

Solution: Option (ii) is the answer. Reason:

Solution: Option (iii) is the answer. Reason:

            Solution: Option (iv) is the answer. Reason: Electron donating:  group increases the basicity while electron-withdrawing group decreases the basicity of...

Solution: Option (ii) is the answer. Reason: Where the aryl amide is converted to arylamine in the presence of $Br_2$ and $NaOH$ .

Solution: Option (iv) is the answer. Reason:

Solution: Option (iii) is the answer. Reason:

Solution: Option (iv) is the answer. Reason: Gabriel synthesis :The reaction is given to the image.Source of nitrogen atom is Gabriel synthesis is Potassium phthalamide.    

Solution: Option (iii) is the answer. Reason:  

Solution: Option (ii) is the answer. Reason: LiAlH4​/ether reduces aryl nitro compounds to azo compounds 2C6H5NO2→LiAIH4C6H5N=N-C6H5