Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. Write down the decreasing o

NCERT Exemplar · Q30

Q. Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal with primary, secondary and tertiary alcohols.

CBSEShort· 2mImportance★★★★★

61% · 82/135 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

IUPAC nomenclature is a systematic way to name a compound so that its name alone tells you its exact structure, with no ambiguity. Every organic name follows the same underlying recipe, whatever the functional group.

The Recipe

Identify the principal characteristic group. If the molecule has a functional group senior enough to be named as a suffix (carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine, and so on down the seniority order), that group decides the suffix and must be included in the parent chain. A halogen is never senior enough to be a suffix — it is always named as a prefix ("halo-"), whatever else is present.
Choose the parent chain. The parent is the longest continuous carbon chain that contains the principal characteristic group (if there is one). Among chains of the same length, the one with the most substituents wins.
Number the chain. Number from whichever end gives the LOWEST LOCANT to the principal characteristic group first. If there is no principal group (e.g. a simple haloalkane, or an alkene with a halogen substituent), lowest locant goes to the site of unsaturation (double/triple bond) first, then to substituents as a set.

Watch out

When two numbering directions give the SAME locant for the principal group/unsaturation (a genuine tie), the tie-break is the lowest locant SET for the substituents as a group — compare the two sets at their first point of difference. Only if the sets are themselves tied does the alphabetically-first substituent get the lower number.
Name and cite the substituents as prefixes, in alphabetical order (ignoring multiplying prefixes like di-/tri- but not ignoring structural prefixes like iso-/cyclo-), each with its own locant.
Assemble the name: locants + substituent prefixes (alphabetical) + parent chain name + suffix (if any).

Worked Example

$CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3$ : the longest chain is 5 carbons (pentane), no principal characteristic group (just a halogen substituent), so number for the lowest locant set. From the left: Cl at C2, methyl at C3 → set {2,3}. From the right: methyl at C3, Cl at C4 → set {3,4}. {2,3} is lower, so numbering from the left wins: 2-chloro-3-methylpentane. …

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

IUPAC nomenclature is not a single formula, but a system of rules designed to give every organic compound a unique, unambiguous name. The "why" behind these rules lies in clarity, consistency, and communication — ensuring that a chemist in Tokyo and one in Toronto draw the same structure from the same name.

1. The Core Principle: The Longest Carbon Chain

Rule: Identify the longest continuous chain of carbon atoms. This becomes the parent chain (e.g., pentane, hexane).

Why?

The longest chain represents the backbone of the molecule.
It gives the most stable, fundamental name — shorter chains would be branches, not the main structure.
Example: In a molecule with 5 carbons in a row and a 2-carbon branch, calling it "pentane" (not "ethane") tells you the core skeleton is 5 carbons long.

Key idea: The parent chain is the maximum continuous path — not necessarily the one that looks "straight" on paper.

2. Numbering: Lowest Locants (The "First Point of Difference" Rule)

Rule: Number the parent chain so that substituents get the smallest possible numbers. When there's a tie, compare the first point of difference.

Why?

This ensures reproducibility — two chemists will always number the same way.
It avoids ambiguity: "2-methylpentane" is unambiguous; "3-methylpentane" would be a different compound.
The first point of difference rule: If you have substituents at positions 2,4 and 3,5, choose 2,4 because 2 < 3 (the first number is smaller).

Example:

For a methyl group on carbon 2 vs. carbon 4 of a 5-carbon chain:
- 2-methylpentane (correct)
- 4-methylpentane (wrong — higher number)

3. Alphabetical Order of Substituents

Rule: List substituents in alphabetical order (ignoring prefixes like di-, tri-, sec-, tert- but not iso-).

Why?

Alphabetical order is a universal sorting convention — no need to remember priority based on size or complexity.
It makes names searchable and predictable.
Example: "3-ethyl-2-methylpentane" (e before m) — not "2-methyl-3-ethylpentane".

Exception: Prefixes like iso- and neo- are considered part of the name (e.g., isopropyl comes before methyl because "i" < "m").

4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes

Rule: Number the chain so that the double or triple bond gets the lowest possible number, even if it means giving a substituent a higher number.

Why?

The functional group (alkene/alkyne) is more important than alkyl substituents.
The bond position defines the compound's reactivity and geometry.
Example: In pent-2-ene (not pent-3-ene), the double bond is between carbons 2 and 3 — the lower number (2) is used.

Priority order:

Principal functional group (e.g., -OH, -COOH, C=C)
Multiple bonds
Substituents (alkyl, halo, etc.)

5. The "Suffix" and "Prefix" System …

Start your 14-day free trial to unlock the full solution →

The reactivity of sodium metal with alcohols depends on the acidity of the O–H bond. Since the acidity decreases as the alkyl group becomes more bulky and electron-donating, the order of reactivity is: primary > secondary > tertiary.

Why does sodium react with alcohols at all?

Sodium metal is a strong reducing agent. When it meets an alcohol, it donates an electron to the proton of the hydroxyl group. This breaks the O–H bond, releasing hydrogen gas and leaving behind the alkoxide ion ( $RO^-$ ) paired with $Na^+$ .

The key point: the reaction rate depends on how easily the O–H bond breaks. That ease is directly linked to the acidity of the alcohol — the more acidic the O–H hydrogen, the faster it reacts with sodium.

What controls the acidity of an alcohol?

In the gas phase, the acidity order is actually: tertiary > secondary > primary. But in the liquid phase (which is what we deal with in a lab), the order flips. Why? Because solvation effects dominate.

The alkoxide ion ( $RO^-$ ) that forms after deprotonation is stabilised by the solvent (usually the alcohol itself, or any polar medium). A smaller, less bulky alkoxide ion can be better solvated — the solvent molecules can pack around it more tightly, stabilising the negative charge more effectively.

Primary alkoxide ( $CH_3CH_2O^-$ ): small, compact, easily solvated → very stable → reaction is fast.
Secondary alkoxide ( $(CH_3)_2CHO^-$ ): bulkier, solvation is less efficient → less stable → reaction is slower.
Tertiary alkoxide ( $(CH_3)_3CO^-$ ): very bulky, solvation is poor → least stable → reaction is slowest.

So the reactivity order with sodium metal in the liquid phase is:

Primary > Secondary > Tertiary

Step-by-step reasoning

Identify the reaction type.

This is an acid–base reaction where sodium acts as a base (it accepts a proton). The rate depends on the acidity of the O–H bond in the liquid phase.
Recall the liquid-phase acidity trend for alcohols.

Due to solvation effects, the acidity decreases as the alkyl group becomes more substituted: …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make when answering this question, along with how to avoid each.

Mistake 1: Getting the Reactivity Order Backwards

The Error:

Students often assume that because tertiary carbocations are more stable, tertiary alcohols must be more reactive with sodium. They write:

Tertiary > Secondary > Primary (✗)

Why it’s wrong:

Reaction with sodium is an acid-base reaction, not a carbocation-forming reaction. The alcohol acts as a weak acid ( $\text{ROH}$ ), and sodium acts as a base. The key factor is the stability of the alkoxide ion ( $\text{RO}^-$ ) formed.

Primary alkoxides are the most stable (least steric hindrance, negative charge is more exposed and better solvated).
Tertiary alkoxides are the least stable (bulky alkyl groups hinder solvation and destabilize the negative charge).

Correct Order:

\text{Primary} > \text{Secondary} > \text{Tertiary}

How to Avoid:

Always ask: “Which alkoxide ion is most stable?” Not “Which carbocation is most stable?” Draw the alkoxide ions and compare steric hindrance around the negative oxygen.

Mistake 2: Confusing This with Dehydration or SN1 Reactivity

The Error:

Students mix up the reactivity order for:

Reaction with Na (acid-base, alkoxide stability)
Acid-catalyzed dehydration (carbocation stability)
SN1 reactions (carbocation stability)

They apply the tertiary > secondary > primary order from dehydration to this reaction.

How to Avoid:

Make a mental checklist:

Reaction Type	Key Intermediate	Reactivity Order
With Na metal	Alkoxide ion stability	1° > 2° > 3°
Dehydration (H⁺/heat)	Carbocation stability	3° > 2° > 1°
SN1 (with HX)	Carbocation stability	3° > 2° > 1°

Tip: If the reaction involves a metal (Na, K, Mg), think alkoxide stability → primary first.

Mistake 3: Ignoring the Role of Steric Hindrance

The Error:

Students only think about electronic effects (inductive effect) and forget that steric hindrance matters.

Why it matters:

In a tertiary alcohol, the bulky alkyl groups physically block the Na atom from approaching the O–H bond.
This slows down the reaction even if the O–H bond were equally polar.

How to Avoid:

Visualize the reaction: Na atom needs to get close to the O–H hydrogen. In a tertiary alcohol, three alkyl groups crowd the oxygen. In a primary alcohol, only one small alkyl group is present. Less crowding = faster reaction.

Mistake 4: Forgetting That This Is an Acid-Base Reaction

The Error:

Students treat it as a redox or substitution reaction and look for leaving groups or oxidation states. …

Start your 14-day free trial to unlock the full solution →

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 96 on this concept.

CBSE 2026Set A1 markMCQ
$Q. The IUPAC name of CH3CH2COCH2CH3 is$
(a) Diethyl ketone
(b) 3-Pentanone
(c) 2-Pentanone
(d) Methyl propyl ketone
›Reveal solutionSolution

Number the five-carbon chain to give the C=O the lowest locant: the ketone carbon is C-3, so the IUPAC name is pentan-3-one (3-pentanone).

Structure: CH3-CH2-CO-CH2-CH3. The longest chain containing the carbonyl has 5 carbons -> pentanone. Numbering from either end places the carbonyl carbon at position 3. Hence the IUPAC na …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. IUPAC name of [a benzene ring with an -OC2H5 (ethoxy) group attached] is$
(a) Benzyl ethoxide
(b) Benzene ethoxide
(c) Ethoxybenzene
(d) Ethoxybenzoyl
›Reveal solutionSolution

Simple ethers with one aromatic and one alkyl group are IUPAC-named as 'alkoxy' + 'benzene', treating the -OR group as a substituent on the ring.

The compound is a benzene ring bearing an ethoxy substituent (-O-C2H5). Following IUPAC substitutive nomenclature for ethers, the smaller/simpler group's oxygen chain is named as an 'alkoxy' prefix attached to the parent (here benzene):

-OC2H5 = ethoxy group

So the compound is Ethoxybenzene (common name: phenetole).

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. IUPAC name of CH3COOH is:$
(a) Ethanol
(b) ethanoic acid
(c) Ethanal
(d) ethane
›Reveal solutionSolution

CH₃COOH is acetic acid, IUPAC name ethanoic acid.

CH₃COOH is a two-carbon carboxylic acid, commonly called acetic acid. In IUPAC nomenclature, carboxylic acids are named by replacing the "-e" of the corresponding alkane name with "-oic …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of Formic acid is :$
(a) Methanoic acid
(b) Ethanoic acid
(c) Ethandioic acid
(d) Methandioic acid
›Reveal solutionSolution
Formic acid, HCOOH, is the one-carbon carboxylic acid, named methanoic acid by IUPAC rules.

Formic acid has the structure H–COOH (a single carbon bearing the carboxyl group). To name a carboxylic acid by IUPAC nomenclature:

Select the longest carbon chain including the –COOH carbon.

Replace the terminal '-e' of the corresponding alkane name with '-oic acid'.

Here the chain has just one carbon (methane), so the name becomes methanoic acid.

…
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of isobutyl chloride is$
(a) 2-chlorobutane
(b) 1-chloro-2-methyl propane
(c) 2-chloro-2-methyl propane
(d) 1-chlorobutane
›Reveal solutionSolution

Isobutyl chloride has the structure (CH3)2CH-CH2-Cl. Number the longest chain (propane, 3 carbons) so the substituent (Cl) gets the lowest locant, then name the methyl branch.

Structure of isobutyl chloride: (CH3)2CH-CH2-Cl

This can be redrawn as: Cl-CH2-CH(CH3)-CH3

Step 1: Identify the longest carbon chain containing the point of attachment of Cl. That's a 3-carbon (propane) chain: C1(CH2Cl)-C2(CH, bearing a methyl branch)-C3(CH3).

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Write IUPAC name of the following compound:$
›Reveal solutionSolution
Numbering the 6-carbon chain from the -COOH carbon (C1) puts the ketone at C4 and the terminal bromine at C6, giving 6-bromo-4-oxohexanoic acid.

The drawn structure is a straight 6-carbon chain: BrCH2-CH2-CO-CH2-CH2-COOH, i.e. a terminal -CH2Br at one end and -COOH at the other, with a ketone in between.

The carboxylic acid (-COOH) is the senior group and is numbered C1.

Numbering from that end: C1 (COOH), C2 (CH2), C3 (CH2), C4 (the ketone C=O), C5 (CH2), C6 (the terminal CH2Br). …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of CH3COOH is$
(a) Acetic acid
(b) Formic acid
(c) Ethanoic acid
(d) Methanoic acid
›Reveal solutionSolution

CH3COOH = ethanoic acid (IUPAC).

CH3COOH has two carbon atoms and a −COOH group. In IUPAC nomenclature, a two-carbon carboxylic acid is 'ethan' (two carbons) + 'oic acid' = ethanoic acid. Its common (trivial) …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The IUPAC name of CH3-CH(OH)-CH3 is$
(a) Propan-1-ol
(b) Propan-2-ol
(c) n-propyl alcohol
(d) Isopropyl alcohol
›Reveal solutionSolution

CH3−CH(OH)−CH3 = propan-2-ol.

The chain has three carbons (propane). The −OH group is attached to the central (second) carbon. Numbering to give the OH the lowest locant places it at position 2, so the …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The structure of a valuable organic compound used as solvent in many chemical industries is shown below (a central carbon bearing a CH₃ group above, a CH₃ group to the left, an OH group below, and a -CH₂-OCH₃ group to the right). The IUPAC name of the organic compound is$
(a) 1-methoxy-2-methylpropan-2-ol
(b) 3-methoxy-2-methylpropan-2-ol
(c) 1-methoxy-3-methylpropan-3-ol
(d) 2-methoxy-2-methylpropan-2-ol
›Reveal solutionSolution
The structure is (CH₃)₂C(OH)CH₂OCH₃; the parent is propan-2-ol with a 2-methyl and a 1-methoxy substituent, giving 1-methoxy-2-methylpropan-2-ol — option (A).

The central carbon bears two $CH_3$ groups, an $OH$ , and a $-CH_2-OCH_3$ group, so the molecule is

$(CH_3)_2C(OH)-CH_2-OCH_3.$

Naming:

The longest carbon chain containing the –OH is three carbons (propane): $C1 = CH_2(OCH_3)$ , $C2 = C(OH)(CH_3)$ , $C3 = CH_3$ . …
Start your 14-day free trial to see the full solution →
CBSE 2025Set 56/5/11 markMCQ
$CH_3 - CH_2 - N(CH_3) - CH_2 - CH_2 - CH_3$
›Reveal solutionSolution
The compound is a secondary amine with an ethyl and a methyl group on the nitrogen, and a three-carbon chain as the parent. The correct IUPAC name is N-ethyl-N-methylpropan-1-amine, which corresponds to option (B).

The key to naming amines under IUPAC rules is to identify the longest continuous carbon chain attached to the nitrogen — that becomes the parent alkane name, with the suffix "-amine". The other groups on the nitrogen are treated as substituents, prefixed with "N-" to show they are attached to the nitrogen atom, not to the carbon chain.

Let's break this down step by step.

Draw the structure from the given condensed formula.

The formula is $CH_3 - CH_2 - N(CH_3) - CH_2 - CH_2 - CH_3$ .

The nitrogen has three bonds: one to an ethyl group ( $-CH_2CH_3$ ), one to a methyl group ( $-CH_3$ ), and one to a propyl group ( $-CH_2CH_2CH_3$ ). So the molecule is:

CH3 | CH3-CH2-N-CH2-CH2-CH3

The nitrogen is connected to three different alkyl groups: ethyl, methyl, and propyl.

Identify the parent chain.

The IUPAC rule for amines: the parent chain is the longest continuous carbon chain attached directly to the nitrogen. Here, the chain on the right is $CH_2-CH_2-CH_3$ — that's three carbons, a propyl group. The chain on the left is $CH_2-CH_3$ — two carbons, an ethyl group. The methyl group ( $CH_3$ ) is just one carbon.

So the longest chain is the three-carbon chain (propyl). The parent name becomes propan-1-amine (since the nitrogen is at the end of the chain, carbon 1).

Name the substituents on the nitrogen.

The other two groups attached to the nitrogen are an ethyl group ( $-CH_2CH_3$ ) and a methyl group ( $-CH_3$ ). Since they are on the nitrogen, not on the carbon chain, they are prefixed with "N-".

Alphabetically, ethyl comes before methyl. So we write N-ethyl-N-methyl.

Combine the parts.

The full name is: N-ethyl-N-methylpropan-1-amine. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$\text{H}_3\text{C}-\underset{\underset{\text{Br}}{|}}{\text{CH}}-\text{CH}_2-\underset{\overset{\text{O}}{||}}{\text{C}}-\text{H}$
(a) 3–bromobutyraldehyde
(b) 2–bromopropanaldehyde
(c) 3–bromobutanal
(d) 2–bromobutanal
›Reveal solutionSolution
The compound is $\text{CH}_3\text{-CH(Br)-CH}_2\text{-CHO}$ , a 4-carbon aldehyde with bromine on C-3 → 3-bromobutanal.

The structure is $\text{H}_3\text{C-CH(Br)-CH}_2\text{-CHO}$ .

The principal functional group is the aldehyde (–CHO), so it gets the lowest locant. Numbering starts at the carbonyl carbon: C1 = CHO, C2 = $\text{CH}_2$ , C3 = CH(Br), C4 = $\text{CH}_3$ .

Longest chain = 4 carbons → butanal. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. The IUPAC name of CH3COOC2H5 is$
(a) Methyl propanoate
(b) Ethyl ethanoate
(c) Acetoethane
(d) Ethoxyethane
›Reveal solutionSolution
CH3COOC2H5 = ethyl ethanoate (common name ethyl acetate).

The compound is an ester derived from ethanoic acid (CH3COOH) and ethanol (C2H5OH). An ester R-COO-R' is named as 'alkyl alkanoate':

The alkyl group from the alcohol part (C2H5) -> ethyl …
Start your 14-day free trial to see the full solution →

←Previous questionQ29 — Arrange the following compounds in increasing order of acidity and giv…→Next questionQ31 — What happens when benzene diazonium chloride is heated with water?

Q.Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal with primary, secondary and tertiary alcohols.

Concept understanding — IUPAC Nomenclature

IUPAC Nomenclature (Organic Compounds)

The Recipe

Worked Example

Why this formula?

IUPAC Nomenclature: Why the Rules Work the Way They Do

1. The Core Principle: The Longest Carbon Chain

2. Numbering: Lowest Locants (The "First Point of Difference" Rule)

3. Alphabetical Order of Substituents

4. Multiple Bonds: The "Lowest Locant" Rule for Alkenes/Alkynes

5. The "Suffix" and "Prefix" System …

Why does sodium react with alcohols at all?

What controls the acidity of an alcohol?

Step-by-step reasoning

Method: Steric Hindrance & Carbanion Stability Approach

Steps

Mistake 1: Getting the Reactivity Order Backwards

Mistake 2: Confusing This with Dehydration or SN1 Reactivity

Mistake 3: Ignoring the Role of Steric Hindrance

Mistake 4: Forgetting That This Is an Acid-Base Reaction

Q.Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal with primary, secondary and tertiary alcohols.

Why does sodium react with alcohols at all?

What controls the acidity of an alcohol?

Step-by-step reasoning

🎓Unlock everything free for 14 days

Unlock everything free for 14 days