(a) Which halogen compound in the following pair will react faster in $S_N2$ reactions and why ? $CH_3-CH_2-I$ OR $CH_3-CH_2-Br$ (b) Why is chloroform stored in closed dark coloured bottles ?

(a) Which halogen compound in the following pair will react faster in SN2 reactions and why ? CH₃-CH₂-I OR CH₃

Question

S_N2

(OR)

CH_3-CH_2-Br

(b) Why is chloroform stored in closed dark coloured bottles ?

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Part (a)Concept understanding — SN1 and SN2 Mechanism

The Core Idea: Two Ways to Swap a Group

Imagine you have a molecule with a leaving group (like a halogen) attached to a carbon. You want to replace that leaving group with a nucleophile (something that loves positive charge). There are two fundamentally different ways this can happen — like two different ways to replace a lightbulb.

SN2 is like unscrewing the old bulb and screwing in the new one in one smooth motion. SN1 is like first pulling the old bulb out completely, leaving an empty socket, and then putting the new bulb in.

That empty socket — the carbocation — is the key difference.

SN2: One Step, Backside Attack

The name says it all: Substitution, Nucleophilic, Bimolecular. "Bimolecular" means two molecules (the nucleophile and the substrate) are involved in the rate-determining step.

The Mechanism

The nucleophile attacks the carbon from the backside — directly opposite the leaving group. As the nucleophile approaches, the leaving group starts to leave. At the transition state, the nucleophile is partially bonded and the leaving group is partially detached. Then the leaving group departs completely, and the nucleophile is fully bonded.

SN2 mechanism in wedge-dash notation: hydroxide attacks the carbon of CH3Cl from the back side, passing through a five-coordinate transition state (partial HO and Cl bonds, both delta-minus) before chloride leaves

All of this happens in one step — no intermediate.

The Stereochemistry: Inversion

Because the nucleophile attacks from the back, the configuration at the carbon inverts — like an umbrella turning inside out in a strong wind. If you start with an R configuration, you get S (and vice versa). This is called Walden inversion.

Backside attack in the SN2 reaction (NCERT Fig 6.2): the incoming hydroxide (red) approaches the carbon opposite the outgoing halide (green), and the three hydrogens flip through like an umbrella inverting

What Favours SN2?

Primary carbon (least steric hindrance — the backside is wide open)
Strong nucleophile (needs to push its way in)
Good leaving group (but not too good — it needs to wait for the nucleophile)
Polar aprotic solvent (doesn't solvate the nucleophile too tightly)

Watch out

SN2 is impossible on tertiary carbons — the three bulky groups block the backside completely. The nucleophile simply cannot get close enough.

Steric effects in the SN2 reaction (NCERT Fig 6.3): relative rates fall from methyl (30) to ethyl (1) to isopropyl (0.02) to tert-butyl (0) as alkyl groups crowd the backside approach of the nucleophile

SN1: Two Steps, Carbocation Intermediate

Substitution, Nucleophilic, Unimolecular. "Unimolecular" means only one molecule (the substrate) is involved in the rate-determining step.

The Mechanism

Step 1 (slow, rate-determining): The leaving group leaves on its own, forming a carbocation (a carbon with only six electrons — positively charged and very unstable).

Step 2 (fast): The nucleophile attacks the carbocation. Since the carbocation is flat (trigonal planar), the nucleophile can attack from either side with equal probability.

The two steps of the SN1 mechanism for 2-bromo-2-methylpropane: slow reversible ionisation to the planar tert-butyl carbocation, then fast attack by hydroxide to give 2-methylpropan-2-ol

The Stereochemistry: Racemisation

Because the nucleophile can attack from either face of the flat carbocation, you get a racemic mixture — equal amounts of R and S. If the starting material is optically pure, the product will be optically inactive.

SN1 racemisation of 2-bromobutane: the planar carbocation is attacked by hydroxide on either face, giving equal amounts of (+)- and (-)-butan-2-ol — an optically inactive racemic mixture

Note

In practice, you often get slightly more inversion than retention (about 60:40) because the leaving group can partially block one face as it departs. But the key idea is loss of stereochemistry.

What Favours SN1?

Tertiary carbon (the carbocation is stabilised by three alkyl groups — hyperconjugation and inductive effect)
Weak nucleophile (doesn't need to force its way in — the carbocation is desperate for electrons)
Excellent leaving group (must be able to leave on its own)
Polar protic solvent (stabilises the carbocation and the leaving group)

Watch out

SN1 is impossible on primary carbons — a primary carbocation is so unstable it effectively doesn't exist. The leaving group would never leave on its own.

The Big Comparison Table …

Part (b)Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
KMnO₄: similar to dichromate, but stronger — can over-oxidize. …

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

Oxidation reactions are fundamental to chemistry, and understanding why they work the way they do is essential for mastering Indian board exams (Class 11, 12, JEE, NEET). Let's break down the core ideas from first principles.

1. The Core Definition: What Does "Oxidation" Really Mean?

Historically, oxidation meant "adding oxygen." But that's too narrow. The modern, exam-correct definition is:

Oxidation is the loss of electrons by a species.

This is the electronic concept (given by the ionic theory). The why behind this definition comes from the behavior of atoms during chemical reactions.

Why do atoms lose electrons?

Atoms seek stability. They achieve this by having a full outer electron shell (octet, duplet, or pseudo-inert gas configuration).

Metals (like Na, Mg, Fe) have few valence electrons (1, 2, or 3). It's energetically easier for them to lose these electrons than to gain 5, 6, or 7.
Non-metals (like O, Cl, F) have many valence electrons (5, 6, or 7). It's energetically easier for them to gain electrons.

So, when a metal reacts with a non-metal, the metal loses electrons (gets oxidized), and the non-metal gains electrons (gets reduced).

Example:

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}

Na loses 1 electron: $\text{Na} \rightarrow \text{Na}^+ + e^-$ (Oxidation)
Cl gains 1 electron: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ (Reduction)

Key takeaway: Oxidation and reduction always happen together (Redox reactions). You cannot have one without the other.

2. The Key Formula(e): Oxidation Number Rules

The oxidation number (O.N.) is a bookkeeping tool. It's not a real charge (except in ionic compounds), but it helps track electron flow.

Why do we assign oxidation numbers?

Because in covalent compounds (like $\text{CH}_4$ or $\text{H}_2\text{O}$ ), electrons are shared, not transferred. We need a way to pretend they are transferred to see which atom "owns" the electrons more.

The Rules (and why they exist)

Rule	Statement	Why this rule?
1	O.N. of an element in its free state = 0	No electron transfer has occurred.
2	O.N. of a monatomic ion = its charge	The atom has actually lost/gained that many electrons.
3	O.N. of H = +1 (except in metal hydrides where it's -1)	H is less electronegative than O, F, Cl, but more electronegative than metals.
4	O.N. of O = -2 (except in peroxides where it's -1, superoxides -1/2, and with F where it's +2)	O is highly electronegative (3.44 on Pauling scale). It "pulls" electrons toward itself.
5	Sum of O.N. in a neutral compound = 0	The compound has no net charge.
6	Sum of O.N. in a polyatomic ion = charge of the ion	The ion's overall charge must be accounted for.

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

Suppose you need to find the O.N. of S in $\text{H}_2\text{SO}_4$ .

Step 1: Write known O.N.s:

H: +1 (rule 3)
O: -2 (rule 4)
S: let it be $x$ (unknown)

Step 2: Apply rule 5 (neutral compound sum = 0):

2(+1) + x + 4(-2) = 0

Step 3: Solve:

2 + x - 8 = 0

x - 6 = 0

x = +6

Why this works: The oxidation number is a mathematical consequence of the electronegativity hierarchy. Oxygen is more electronegative than sulfur, so it "takes" the electrons. Hydrogen is less electronegative than sulfur, so it "gives" electrons to sulfur. The net result is that sulfur appears to have lost 6 electrons.

3. The Key Formula(e): Balancing Redox Equations

Two methods are exam-critical: Oxidation Number Method and Ion-Electron Method (Half-Reaction Method).

Why do we need these methods?

Because in a redox reaction, the total number of electrons lost (oxidation) must equal the total number of electrons gained (reduction). This is the Law of Conservation of Charge.

The Ion-Electron Method (for acidic medium) — Step-by-step why

Example: Balance $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$ (acidic)

Step 1: Write half-reactions.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Why? Fe loses 1 electron (O.N. goes from +2 to +3).
Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$

Why? Mn gains electrons (O.N. goes from +7 to +2).

Step 2: Balance atoms other than H and O.

Mn is already balanced (1 on each side).

Step 3: Balance O by adding $\text{H}_2\text{O}$ .

Left: 4 O atoms. Right: 0 O atoms.
Add 4 $\text{H}_2\text{O}$ to the right:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 4: Balance H by adding $\text{H}^+$ (because acidic medium).

Right: 8 H atoms (from 4 $\text{H}_2\text{O}$ ). Left: 0 H atoms.
Add 8 $\text{H}^+$ to the left:

8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 5: Balance charge by adding electrons.

Left: $8(+1) + (-1) = +7$ charge.
Right: $+2$ charge.
To make left = right, add 5 electrons to the left:

8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 6: Multiply half-reactions to equalize electrons.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5)
Reduction: $8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (×1)

Step 7: Add them:

5\text{Fe}^{2+} + 8\text{H}^+ + \text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}

Why this works: Every step is driven by conservation laws:

Mass balance: Same number of each atom on both sides.
Charge balance: Net charge on left = net charge on right.
Electron balance: Electrons lost = electrons gained.

4. The Key Formula(e): Electrochemical Series and Cell Potential

For a galvanic cell (voltaic cell), the cell potential $E^\circ_{\text{cell}}$ is:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

Why this formula? …

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$CH_3CH_2I$ reacts faster in $S_N2$ because $I^-$ is a better leaving group than $Br^-$ .
Chloroform is stored in closed dark bottles because light and air slowly oxidise it to poisonous phosgene, $COCl_2$ .

Part (a) — Which reacts faster in $S_N2$ ?

An $S_N2$ reaction is a single-step, back-side attack in which the nucleophile bonds as the leaving group departs. Its rate is strongly influenced by how good the leaving group is — i.e. how weak the C–X bond is and how stable the departing halide ion is.

Comparing the two ethyl halides (same alkyl group, so sterics are identical):

Bond strength: C–I (~240 kJ mol⁻¹) is weaker than C–Br (~285 kJ mol⁻¹), so C–I breaks more easily.
Leaving-group stability: $I^-$ is the larger, more polarisable ion; its charge is spread over a bigger volume, making it more stable and a better leaving group.

\text{Leaving-group ability: } I^- > Br^- > Cl^- > F^-

Therefore ethyl iodide, $CH_3CH_2I$ , reacts faster than ethyl bromide. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study] Depending on the molecule involved in controlling the rate of reaction, nucleophilic substitution reaction can be divided in two categories: nucleophilic unimolecular (SN1) and nucleophilic bimolecular (SN2). Alkyl halide reactivity towards SN1 and SN2 reactions depends on a number of variables, including steric hindrance, stability of the intermediate or transition state and solvent polarity. Primary alkyl halides, followed by secondary and tertiary alkyl halide are most favourable to the SN2 reaction mechanism. In the case of SN1 reactions, this order is reversible.$
(i) Which of the following is most reactive towards nucleophilic substitution reaction ?
(a) CHCl3
(b) CH2=CHCl
(c) ClCH2CH=CH2
(d) CH2CH=CHCl
›Reveal solutionSolution
Allyl chloride is most reactive because ionisation of the C–Cl bond gives an allylic carbocation stabilised by resonance with the adjacent C=C double bond, favouring rapid SN1 substitution.

Comparing the four halides:

CHCl₃ (chloroform): a trihalomethane; it is not a typical alkyl/allyl/vinyl halide substrate for facile nucleophilic substitution under these conditions — its C–H (not a good leaving-group carbon) and its multiple Cl on the same very electron-poor carbon make it comparatively unreactive here.

CH₂=CHCl (vinyl chloride): the C–Cl bond is attached directly to an sp² carbon of the double bond. This bond has partial double-bond character (due to conjugation of a chlorine lone pair with the π system) and is both shorter and stronger than a normal C–Cl bond; also, any positive charge generated at that carbon cannot be stabilised by the adjacent π system (it would need an empty orbital where the π bond already sits). Vinylic and aryl halides are therefore very unreactive towards nucleophilic substitution. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study, continued]$
(ii) Isopropyl chloride undergoes hydrolysis by
(a) SN1 and SN2 mechanism
(b) SN1 mechanism
(c) SN2 mechanism
(d) None of the above
›Reveal solutionSolution
Isopropyl chloride is a secondary alkyl halide; secondary halides sit in the intermediate reactivity zone and can react by either SN1 (via a moderately stable secondary carbocation) or SN2 (moderate steric hindrance still allows backside attack), so both mechanisms operate, often simultaneously depending on conditions.

Reactivity trend for nucleophilic substitution:

Primary halides are sterically unhindered, favouring SN2 (backside attack is easy) but their carbocation would be unstable, disfavouring SN1.

Tertiary halides are too sterically hindered for backside attack (SN2 is disfavoured) but readily form a stable tertiary carbocation, strongly favouring SN1. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following on heating with aqueous KOH, produces acetaldehyde?$
(a) $CH_3CH_2Cl$
(b) $CH_2ClCH_2Cl$
(c) $CH_3CHCl_2$
(d) $CH_3COCl$
›Reveal solutionSolution
Only the geminal dihalide $CH_3CHCl_2$ hydrolyses to an unstable gem-diol that collapses to the aldehyde; the other substrates give an alcohol, a diol, or a carboxylate instead.

$CH_3CHCl_2$ (c), a gem-dihalide (both $Cl$ on the same carbon):

$CH_3CHCl_2 + 2KOH(aq) \rightarrow CH_3CH(OH)_2 + 2KCl$

The geminal diol $CH_3CH(OH)_2$ is unstable (two $-OH$ groups on the same carbon) and spontaneously eliminates water:

$CH_3CH(OH)_2 \rightarrow CH_3CHO + H_2O$

Net: $CH_3CHCl_2 + 2KOH \rightarrow CH_3CHO + 2KCl + H_2O$ — acetaldehyde.

Why the others are wrong:

(a) $CH_3CH_2Cl$ + aq. $KOH$ undergoes simple nucleophilic substitution to give ethanol, $CH_3CH_2OH$ — not an aldehyde. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which one of the following is the correct statement?$
(a) Alkyl halides are more reactive than aryl halides towards nucleophilic substitution reaction
(b) Alkyl halides are less reactive than aryl halides towards nucleophilic substitution reaction
(c) Nucleophilic substitution reaction proceeds through carbocation
(d) Aryl halides cannot be prepared by electrophilic substitution to arenes
›Reveal solutionSolution
Resonance donation of the halogen's lone pair into the aromatic ring strengthens and shortens the aryl $C-X$ bond and makes the ring electron-rich, so alkyl halides are far more reactive than aryl halides toward nucleophilic substitution.

Why aryl halides resist substitution: the halogen's lone pair delocalises into the benzene ring by resonance, giving the $C-X$ bond partial double-bond character — it becomes shorter and stronger than a normal $C-X$ single bond, and harder to break. The electron-rich ring also repels an incoming nucleophile, and backside ( $S_N2$ -type) attack on the $sp^2$ carbon is sterically/geometrically hindered by the planar ring; an $S_N1$ -type pathway would require a highly unstable phenyl cation, which does not form under normal conditions.

Why the other statements are wrong:

(b) is the exact reverse of the truth. …
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CBSE 2025Set ANNUAL1 mark
$Q. How can the following conversion be carried out? (Give chemical equation only) — Bromoethane to propane nitrile$
›Reveal solutionSolution

Nucleophilic substitution of bromide by cyanide ion (alcoholic KCN) gives the nitrile.

Bromoethane is heated with alcoholic potassium cyanide (KCN); the cyanide ion (a good nucleophile, attacking through carbon) displaces bromide in an $S_N2$ reaction:

$C_2H_5Br + KCN \xrightarrow{\text{ethanolic}} C_2H_5CN + KBr$

…

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CBSE 2024Set D1 markMCQ
$Q. C2H5Br + NaOH -> C2H5OH + NaBr is an example of which of the following types of reaction?$
(a) Electrophilic substitution
(b) Nucleophilic substitution
(c) Both (A) and (B)
(d) None of these
›Reveal solutionSolution

The nucleophile OH- attacks the electrophilic carbon of C2H5Br and displaces the leaving group Br-, giving C2H5OH. This is nucleophilic substitution.

In C2H5Br the C-Br bond is polar; carbon carries a partial positive charge. The hydroxide ion (OH-) from NaOH is an electron-rich nucleophile that attacks this carbon and expels bromide as the leaving group:

C2H5Br + OH- -> C2H5OH + Br-

…

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CBSE 2024Set D1 markMCQ
$Q. An aldehyde on oxidation gives$
(a) an alcohol
(b) a ketone
(c) an ether
(d) an acid
›Reveal solutionSolution

Oxidation of an aldehyde gives a carboxylic acid.

Aldehydes carry an H on the carbonyl carbon and are easily oxidised. With oxidising agents (or even mild reagents such as Tollen's or Fehling's), an aldehyde is converted to the corresponding carboxylic acid:

R-CHO + [O] -> R-COOH

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Arylhalides are less reactive towards nucleophilic substitution reaction as compared to alkylhalides due to :$
(a) The formation of less stable carbonium ion
(b) Resonance stabilization
(c) Longer carbon-halogen bond
(d) Inductive effect
›Reveal solutionSolution
Aryl halides resist nucleophilic substitution because resonance between the halogen's lone pair and the ring strengthens the C-X bond.

In an aryl halide such as chlorobenzene, a lone pair on the halogen conjugates with the pi-electron system of the ring. This delocalisation:

Gives the carbon-halogen bond partial double-bond character, making it shorter and stronger than a normal C-X single bond, so it resists heterolytic cleavage (needed for both SN1 and SN2 pathways).

Increases electron density on the ring (especially ortho/para to X), which repels an approaching nucleophile.

The carbon bearing the halogen is sp2 hybridised, holding the bonding electrons closer to carbon and further strengthening the bond. …
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CBSE 2024Set ANNUAL1 mark
$Q. How would you obtain the following? Benzoic acid from ethyl benzene$
›Reveal solutionSolution

Vigorous oxidation (hot alkaline $KMnO_4$ ) of any alkylbenzene side chain, regardless of its length, converts it entirely to a single $-COOH$ group attached directly to the ring.

Ethylbenzene, $C_6H_5{-}CH_2CH_3$ , has a two-carbon side chain with benzylic hydrogens. Strong oxidising agents like hot alkaline potassium permanganate attack the side chain at the benzylic position and progressively oxidise it, cleaving off the extra carbon(s) and leaving only the ring-attached carbon as a carboxyl group — the exact chain length beyond the first carbon does not matter, the product is always benzoic acid:

…

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CBSE 2023Set 56/1/11 markMCQ
$CH_3CONH_2$
›Reveal solutionSolution
This is the Hofmann bromamide degradation reaction. An amide ( $CH_3CONH_2$ ) reacts with bromine and a base to give a primary amine with one fewer carbon atom. The product here is methylamine ( $CH_3NH_2$ ), which corresponds to option (B).

The reaction you're looking at is a classic name reaction in organic chemistry — the Hofmann bromamide degradation. It's one of the most reliable ways to convert an amide into a primary amine, and it always involves a loss of one carbon from the chain. Let's understand why.

The key idea: the amide group ( $-CONH_2$ ) gets "chopped" by bromine in the presence of a strong base. The carbonyl carbon (the one attached to oxygen) is lost as carbon dioxide, and the nitrogen ends up attached to the alkyl group that was originally next to the carbonyl. So the product has one carbon fewer than the starting amide.

Now let's walk through the reaction step by step for your specific compound, acetamide ( $CH_3CONH_2$ ).

Identify the starting material.

Acetamide has the structure $CH_3-CO-NH_2$ . The alkyl group attached to the carbonyl is a methyl group ( $CH_3-$ ). The amide carbon is the carbonyl carbon.

Recall the general outcome of Hofmann degradation.

The reaction is:

$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$
Notice that the product $R-NH_2$ has the same $R$ group as the starting amide, but the carbonyl carbon is gone (it becomes carbonate). So the amine has one less carbon than the amide.

Apply to acetamide.

Here $R = CH_3-$ . So the amine formed is $CH_3-NH_2$ , which is methylamine.

Check the options.

(A) $CH_3COONa$ — this is sodium acetate, not an amine.

(B) $CH_3NH_2$ — methylamine, matches our prediction. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following will react faster in SN2 reaction?$
(a) 1-bromopentane
(b) 2-bromopentane
(c) 3-bromopentane
(d) 2-bromo-2-methyl butane
›Reveal solutionSolution

SN2 rate depends inversely on steric hindrance around the carbon bearing the leaving group; the primary halide has the least hindrance and reacts fastest.

In an SN2 reaction, the nucleophile attacks the carbon from the side opposite the leaving group in a single concerted step, so the rate is very sensitive to steric crowding at that carbon: primary (1 degree) > secondary (2 degree) > tertiary (3 degree), and tertiary halides essentially do not undergo SN2. Among the choices, 1-bromopentane has its Br on a pr …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. In Benzaldehyde + [O] --(Air)--> A, A is$
(a) Benzene
(b) Benzoic acid
(c) Benzyl alcohol
(d) None of these
›Reveal solutionSolution

Aromatic aldehydes like benzaldehyde undergo slow autoxidation in air, converting -CHO to -COOH.

C6H5CHO + [O] --(air)--> C6H5COOH

On exposure to air, benzaldehyde is slowly autoxidised at the aldehydic hydrogen, converting the -CHO group into a -COOH group and giving benzoic acid. (Th …

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Q.(a) Which halogen compound in the following pair will react faster in SN2S_N2SN​2 reactions and why ? CH3−CH2−ICH_3-CH_2-ICH3​−CH2​−I

Part (a) — Which reacts faster in SN2S_N2SN​2?

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Q.(a) Which halogen compound in the following pair will react faster in $S_N2$ reactions and why ? $CH_3-CH_2-I$

Part (a) — Which reacts faster in $S_N2$ ?

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