NEET Chemistry - MCQ Practice Questions

Understanding:

We need to identify which alkyl bromide undergoes ${\text{S}}_{\text{N}}1$ substitution most readily.

Formula:

The rate of ${\text{S}}_{\text{N}}1$ depends on the stability of the carbocation intermediate:

Carbocation stability order:

Step 1: Classify each substrate

Step 2: Identify the most reactive

The $3^{\circ }$ carbocation formed from $({\text{CH}}_3{)}_3\text{CBr}$ is stabilised by three alkyl groups donating electron density via hyperconjugation and induction, making it the most stable and easiest to form.

Answer:

The compound $({\text{CH}}_3{)}_3\text{CBr}$ undergoes ${\text{S}}_{\text{N}}1$ reaction most readily.

Quick Tip:

Always look for the most substituted carbon bearing the leaving group when predicting ${\text{S}}_{\text{N}}1$ reactivity.

Understanding:

We need to identify the final product of aldol condensation of ${\text{CH}}_3\text{CHO}$ followed by dehydration.

Formula:

Aldol condensation:

Step 1: Aldol addition

Two molecules of ${\text{CH}}_3\text{CHO}$ react in the presence of dilute base. The $\alpha$-carbon of one molecule attacks the carbonyl carbon of another:

This gives $3$-hydroxybutanal (aldol product).

Step 2: Dehydration

On heating, the $\beta$-hydroxy aldehyde loses water to form an $\alpha ,\beta$-unsaturated aldehyde:

The product is but-2-enal (crotonaldehyde).

Answer:

The final product of aldol condensation followed by dehydration of acetaldehyde is but-2-enal.

Quick Tip:

The aldol addition product is a $\beta$-hydroxy carbonyl compound; dehydration always removes ${\text{H}}_2\text{O}$ from the $\alpha$ and $\beta$ positions to give the conjugated $\alpha ,\beta$-unsaturated product.

Understanding:

We need to count all $\sigma$ and $\pi$ bonds in ${\text{CH}}_3{\text{CH}}_2\text{C}\equiv \text{CH}$.

Formula:

For any bond:

Step 1: Draw the structure

The connectivity of ${\text{CH}}_3-{\text{CH}}_2-\text{C}\equiv \text{CH}$ gives:

Step 2: Add up

Answer:

But-1-yne contains $9$ sigma bonds and $2$ pi bonds.

Quick Tip:

A quick formula: for ${\text{C}}_n{\text{H}}_m$ with one triple bond, $\sigma =(n-1)+m$ carbon-carbon and carbon-hydrogen $\sigma$ bonds; count $\pi$ bonds separately from the degree of unsaturation.

Understanding:

We need to identify the reagent that converts ${\text{RCONH}}_2$ to ${\text{RNH}}_2$, specifically with loss of one carbon.

Formula:

Hofmann bromamide degradation:

Step 1: Analyse each option

${\text{LiAlH}}_4$ reduces ${\text{RCONH}}_2$ to ${\text{RCH}}_2{\text{NH}}_2$ — the carbon count is retained, giving a primary amine with the same number of carbons.

Step 2: Identify the degradation reaction

In the Hofmann bromamide reaction, ${\text{Br}}_2/\text{NaOH}$ acts on the amide nitrogen, forming an isocyanate intermediate which hydrolyses. The carbonyl carbon is lost as ${\text{CO}}_2$:

The product ${\text{RNH}}_2$ has one carbon fewer than the starting amide.

Step 3: Eliminate other options

${\text{HNO}}_2$ reacts with primary amines (not amides to give a degraded amine), and $\text{Sn}/\text{HCl}$ reduces nitro groups. Neither removes a carbon.

Answer:

The Hofmann bromamide reaction using ${\text{Br}}_2/\text{NaOH}$ converts a primary amide to a primary amine with one fewer carbon.

Quick Tip:

Whenever you see a reaction that decreases the carbon chain by one while converting an amide to an amine, it is always the Hofmann degradation.

Understanding:

We need to assign the correct IUPAC name to ${\text{CH}}_3{\text{CH(OH)CH(CH}}_3){\text{CH}}_2\text{Cl}$.

Formula:

IUPAC rules:

1. Identify the longest continuous carbon chain containing the principal characteristic group ($-\text{OH}$ has higher seniority than $-\text{Cl}$).

2. Number from the end that gives the $-\text{OH}$ group the lowest locant.

Step 1: Identify the parent chain

The longest continuous chain is 4 carbons: $\text{C1}-\text{C2}-\text{C3}-\text{C4}$.

Step 2: Number to give $-\text{OH}$ the lowest locant

Numbering from the ${\text{CH}}_3$ end gives $\text{OH}$ at position $2$ and $\text{Cl}$ at position $4$.

Numbering from the ${\text{CH}}_2\text{Cl}$ end gives $\text{OH}$ at position $3$ — higher, so not preferred.

Step 3: Name substituents

Full name (substituents in alphabetical order): $4$-chloro-$3$-methylbutan-$2$-ol.

Answer:

The IUPAC name of the given compound is $4$-chloro-$3$-methylbutan-$2$-ol.

Quick Tip:

Always number the chain to give the principal characteristic group ($-\text{OH}$ over $-\text{Cl}$) the lowest possible locant, not the halogen.

Understanding:

We need to apply Markovnikov's rule to the addition of $\text{HBr}$ to ${\text{CH}}_3\text{CH}={\text{CH}}_2$ (propene).

Formula:

Markovnikov's rule (mechanistic statement):

The nucleophile (${\text{Br}}^{-}$) then attacks the carbocation.

Step 1: Protonation step

${\text{H}}^{+}$ from $\text{HBr}$ adds to $\text{C1}$ (which has more H atoms):

or to $\text{C2}$:

The $2^{\circ }$ carbocation is more stable, so ${\text{H}}^{+}$ goes to $\text{C1}$.

Step 2: Attack by ${\text{Br}}^{-}$

${\text{Br}}^{-}$ attacks the $2^{\circ }$ carbocation at $\text{C2}$:

Major product: $2$-bromopropane.

Answer:

According to Markovnikov's rule, bromine adds to the secondary carbon $\text{C2}$, giving $2$-bromopropane as the major product.

Quick Tip:

Markovnikov's rule in one line: the electrophile (${\text{H}}^{+}$) goes to the carbon that already has more hydrogens, because this generates the more stable (more substituted) carbocation.

Understanding:

We need to arrange phenol, ethanol, and water in order of decreasing ${\text{pK}}_{\text{a}}$ (i.e., increasing acidity).

Formula:

Step 1: Recall approximate ${\text{pK}}_{\text{a}}$ values

Step 2: Explain the trend

Phenol is most acidic because its conjugate base (phenoxide ion) is stabilised by resonance with the aromatic ring — the negative charge is delocalised over the ring. Water is slightly more acidic than ethanol because the alkyl group in ethanol donates electron density (induction), destabilising the ethoxide ion. Ethanol's conjugate base (ethoxide) is less stable than hydroxide, so ethanol is the weakest acid.

Step 3: Arrange

Answer:

The correct order is ethanol $>$ water $>$ phenol in terms of ${\text{pK}}_{\text{a}}$.

Quick Tip:

Resonance stabilisation of the conjugate base is a much stronger effect than induction; phenoxide's resonance makes phenol thousands of times more acidic than alcohols.

Understanding:

We need to identify the product of catalytic hydrogenation of a nitrile ($\text{R-CN}$) with ${\text{H}}_2$ over nickel.

Formula:

Step 1: Analyse the reaction

A nitrile contains a $\text{C}\equiv \text{N}$ triple bond (one $\sigma$ + two $\pi$ bonds). Catalytic hydrogenation with $2$ moles of ${\text{H}}_2$ reduces the $\text{C}\equiv \text{N}$ completely:

Step 2: Eliminate other options

Answer:

The catalytic hydrogenation of a nitrile produces a primary amine.

Quick Tip:

Reduction of a nitrile always gives a primary amine with the same number of carbons as the nitrile — useful for synthesising amines from halides via $\text{R-X}\to \text{R-CN}\to {\text{R-CH}}_2{\text{NH}}_2$.

Understanding:

We need to identify the pair that are enantiomers (non-superimposable mirror images of each other).

Formula:

Enantiomers are stereoisomers that are non-superimposable mirror images, differing only in the configuration at every chiral centre:

Step 1: Analyse option A

$(R)\text{-}2\text{-bromobutane}$ and $(S)\text{-}2\text{-bromobutane}$ have the same molecular formula ${\text{C}}_4{\text{H}}_9\text{Br}$, the same connectivity, and opposite configurations ($R$ vs $S$) at the single chiral centre (C2). They are non-superimposable mirror images — the definition of enantiomers.

Step 2: Analyse other options

Answer:

$(R)\text{-}2\text{-bromobutane}$ and $(S)\text{-}2\text{-bromobutane}$ are enantiomers.

Quick Tip:

Enantiomers must have the same molecular formula and connectivity but opposite $(R/S)$ configuration at every chiral centre. If any substituent differs, they are not enantiomers.

Understanding:

We need to explain why aniline is less basic than cyclohexylamine (an aliphatic amine).

Formula:

Step 1: Lone pair in cyclohexylamine

In cyclohexylamine, nitrogen is ${\text{sp}}^3$ hybridised. Its lone pair is in a pure ${\text{sp}}^3$ orbital and is completely available to accept a proton.

Step 2: Lone pair in aniline

In aniline, the nitrogen lone pair overlaps with the $\pi$ system of the benzene ring (resonance):

This delocalisation reduces the electron density on nitrogen, making it much less available for protonation. The conjugate acid (anilinium ion) loses this resonance stabilisation, so aniline does not gain as much from protonation.

Step 3: Eliminate option D

Although nitrogen in aniline has partial ${\text{sp}}^2$ character and ${\text{sp}}^2$ orbitals are more electronegative, the dominant reason is resonance delocalisation, not orbital electronegativity. Option D is partially true but not the best or primary explanation taught at this level.

Answer:

Aniline is less basic because resonance delocalisation of the nitrogen lone pair into the benzene ring reduces its availability for accepting a proton.

Quick Tip:

Any time $-{\text{NH}}_2$ is directly attached to an aromatic ring, expect reduced basicity due to resonance. The same effect makes aromatic amines much weaker bases than aliphatic amines.