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Chemistry MCQs for NEET — Practice Questions with Answers

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An organic compound X (mol. formula C₆H₅O₂N) has 6 carbons in a ring system, three double bonds and also a nitro group as substituent X is:

homocyclic but not aromatic

aromatic but not homocyclic

homocyclic and aromatic

heterocyclic

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Explanation

The given organic compound with the molecular formula C6H5O2N has 6 carbon atoms in a ring system, three double bonds, and a nitro group (-NO2) as a substituent. This compound is nitrobenzene, which is both homocyclic (contains only carbon atoms in the ring) and aromatic.

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Which of the following reactions is appropriate for converting acetamide to methanamine ?

Carbylamine reaction

Hofmann hypobromamide reaction

Stephens reaction

Gabriels phthalimide synthesis

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Explanation

The Hofmann bromamide reaction is a method for converting an amide into an alkylamine with one more carbon atom than the starting material. It involves treating the amide with bromine and then an alkoxide base, and is the appropriate reaction for converting acetamide to methanamine.

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The correct statement regarding the basicity of arylamines is

Arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring p-electron system.

Arylamines are generally more basic than alkylamines because of aryl group.

Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized

Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring p-electron system.

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Explanation

Arylamines are generally less basic than alkylamines because the nitrogen lone pair electrons in arylamines are delocalized by interaction with the aromatic ring Ï€-electron system, making the nitrogen atom less available for protonation, which reduces the basicity.

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The electrolytic reduction of nitrobenzene in strongly acidic medium produces

p-aminophenol

azoxybenzene

azobenzene

aniline

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Method by which aniline cannot be prepared is

hydrolysis phenyl isocyanide with an acidic solution

degradation of benzamide with bromine in alkaline solution

reduction of nitrobenzene with H₂/Pd in ethanol

potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous NaOH solution

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Explanation

Due to resonance in chlorobenzene C-Cl bond acquires double bond character hence, C-Cl bond is inert towards nucleophile (phthalimide ion).

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Which of the following will be most stable diazonium salt $R N_{2}^{+} X^{-}$ ?

${CH}_{3} N_{2}^{+} X^{-}$

$C_{6} H_{5} N_{2}^{+} X^{-}$

${CH}_{3} {CH}_{2} N_{2}^{+} X^{-}$

$C_{6} H_{5} {CH}_{2} N_{2}^{+} X^{-}$

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Explanation

The stability of diazonium salts increases with an increase in the delocalization of the positive charge over the aromatic ring. The benzene diazonium salt (o2) is more stable than the aliphatic diazonium salts (o1 and o3) due to better delocalization of the positive charge.

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Which one of the following is an amine hormone ?

Thyroxine

Oxypurin

Insulin

Progesterone

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Explanation

The hormone which is proteinous in nature are said to be amine hormone because protein are the polymers of polypeptides and polypeptides are made up of amino acids. Amino acid contains amino group.

Thyroxin is an amine hormone. It is secreted by thyroid gland. Its function is to control the metabolism of carbohydrates, proteind and lipids.

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Which of the following is more basic than aniline ?

Diphenylamine

Triphenylamine

p-nitroaniline

Benzylamine

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0.2 g of an organic compound on complete combustion produces 0.44 of CO₂, then the percentage of carbon in it is :-

50%

60%

70%

80%

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Explanation

Mass of the compound = 0.2

Mass of CO₂ formed = 0.44

Percentage formed = 0.44

Percentage of

$C = \frac{12}{44} \times \frac{Mass of {CO}_{2} formed}{Mass of QC} \times 100 = \frac{12}{44} \times \frac{0.44}{0.2} \times 100 = 60 %$

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Glucose and Mannose are

C₁ epimers

Anomers

C₂ epimers

C₄ epimers

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Explanation

D-Mannose is an epimer of D-glucose because the two sugars differ only in the configuration at C-2 .

When a molecule such as glucose converts to a cyclic form, it generates a new chiral centre at C-1 .

Anomers are special cases — they are epimers that differ in configuration only at the anomeric carbon

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