Test: JEE Previous Year Questions- Carbonyl Group - JEE MCQ

Through resonance both resonating structures are equivalent. That is why two C−0 bond lengths are found to be equal.

The correct answer is option C
LiAlH₄  can reduce the carboxylic acid group without affecting the double bond because alkene is an electron-rich species.

The correct answer is option D
The general formula C_n​H_2n​O₂​ could be for open chain carboxylic acids. Thus, butanoic acid CH₃​−CH₂​−CH₂​−COOH has formula C₄​H_8​O₂​. And, decanoic acid CH₃​−(CH​₂)₈​−COOH has formula C_10​H₂₀​O₂.​.

Option C is the correct answer.
So, acetonitrile does not have sp2 hybridized carbon.

The correct answer is option D

NO₂ group at any position shows electron withdrawing effect, thus acid strength is increased. But o -nitro benzoate ion is stabilised by intramolecular H-bonding like forces, hence its acid strength is maximum. Thus, the order of acid strength is (ii) > (iii) > (iv) > (i)
The effect is more pronounced at para position than meta.

The correct answer is Option D.
Cl⁻ is the best leaving group being the weakest nucleophile out of NH²⁻ , Cl− , OC₂H₅ and CH₃COO⁻. 
So, when Z is Cl⁻ , the ion leaves faster thus making the reaction the fastest when Z is Cl⁻ .

The correct answer is option A
2>3>4>1
Electron withdrawing groups increase the acidity of benzoic acid, o−o-isomer will have higher acidity than mm and pp isomer due to ortho effect. The NO₂NO₂ group at pp position has a more pronounced electron withdrawing effect than −NO₂-NO₂ group at mm position. Hence order of acidity is

 

aqueous NaCl is a neutral compound. Thus no reaction takes place between ethyl acetate and aqueous NaCl.

Thus resultant solution will be CH₃COOC₂​H_5​ + NaCl 

The correct answer is option A
Butan - 2 - one will get reduced into butane when treated with zinc and hydrochloric acid following Clemmensen reaction whereas Zn/HCl do not reduce ester, acid and amide

The correct answer is option C
The compound formed as a result of oxidation of ethyl benzene by alkaline KMnO4​ is benzoic acid. 
Other reagents that can be used are acidified potassium dichromate or dilute nitric acid.

The correct answer is Option D.

The fruity smell is evolved due to the formation of ester when ethanol and acid are treated in the presence of concentrated H2SO4.
The reaction is 
CH₃COOH + CH₃CH₂OH → CH₃COOC₂H₅

The correct answer is Option B.
When primary amine reacts with CHCl3 in alcoholic KOH, the product is isocyanide which is foul smelling.
R−NH₂+CHCl₃+3KOH -ΔR−NC+3KCl+3H₂O
This reaction is known as carbylamine test / Isocyande test. This test is not given by secondary or tertiary amines.

The correct answer is option A
It is ferri ferrocyanide
6NaCN+FeSO₄→Na₄[Fe(CN)₆]+Na₂SO₄
3Na₄[Fe(CN)₆]+2Fe₂(SO₄)₃→Fe₄[Fe(CN)₆]₃+6Na₂SO₄
Prussian blue
 

The correct answer is Option D.

Wurtz reaction is used to produce alkanes from alkyl halides.
2RX+2Na (in ether)→R−R+2NaX
Example :
2CH₃Cl+2Na (in ether)→C₂H₆+2NaCl
So, the Wurtz reaction has nothing to do with amines.

The correct answer is option A
The reaction of cyclohexanone with dimethylamine, in the presence of a catalytic amount of an acid forms a compound as shown in the reaction sequence. During the reaction, if water is continuously removed, the compound formed is generally known as an enamine.

The correct answer is option D
In the chemical reaction,
CH₃​CH₂​NH₂​+CHCl₃​+3KOH→(A)+(B)+3H₂​O
The compounds(A) and (B) are C₂​H₅​NC and 3KCl respectively. The given reaction is carbylamine test/Isocyanide test in which aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide give foul smelling alkyl isocyanides or carbylamines. Secondary or tertiary amines do not give this test.