2-methylbutan-2-ol is reacted with HCl to produce 2-chloro-2-methylbutane explain with mechanism
Author: Sarvan Kumar
Order of reactivity of alcohols (Allyl, benzyl, primary, secondary and tertiary) with HX
The order of reactivity of alcohols with hydrogen halides (HX, like HCl, HBr, HI) depends mainly on how easily the C–OH bond can be broken and replaced by halogen.
Key points:
- The reaction proceeds either via carbocation formation (SN1) or direct displacement (SN2) depending on the substrate.
- Carbocation stability (tertiary > secondary > primary) governs the rate in SN1 cases.
- Resonance-stabilized carbocations (benzylic, allylic) are even more reactive than tertiary.
Reactivity order:
Benzylic alcohol > Allylic alcohol > Tertiary alcohol > Secondary alcohol > Primary alcohol
Explanation:
- Benzylic alcohols → form benzylic carbocations stabilized by resonance with aromatic ring → very fast.
- Allylic alcohols → form allylic carbocations stabilized by resonance with adjacent double bond → also very fast.
- Tertiary alcohols → form stable 3° carbocations → faster than 2° or 1°.
- Secondary alcohols → less stable carbocation, slower.
- Primary alcohols → do not form stable carbocations easily, react only by SN2 mechanism with HX, so they are slowest.
Final order:
Benzyl > Allyl > 3° > 2° > 1°
Write IUPAC Name of ethylidene dibromide, Isopropylidene dichloride, ethylene dibromide, isobutene dibromide and tetramethylene dibromide
Ethylidene dibromide
Structure: CH₃–CHBr₂
IUPAC name: 1,1-dibromoethane
Isopropylidene dichloride
Structure: (CH₃)₂CCl₂
IUPAC name: 2,2-dichloropropane
Ethylene dibromide
Structure: Br–CH₂–CH₂–Br
IUPAC name: 1,2-dibromoethane
Isobutene dibromide (addition product of Br₂ across 2-methylprop-1-ene)
Structure: CH₂Br–CBr(CH₃)₂ (derived from CH₂=C(CH₃)₂ + Br₂)
IUPAC name: 1,2-dibromo-2-methylpropane
Tetramethylene dibromide
Structure: Br–(CH₂)₄–Br
IUPAC name: 1,4-dibromobutane
Nucleophilicity Is Affected by Steric Effects explain with example
Nucleophilicity is the ability of a species (usually an ion or molecule with a lone pair or π electrons) to attack an electrophilic center, such as a carbon atom in an alkyl halide.
While several factors affect nucleophilicity (like charge, solvent, and electronegativity), one major factor is steric hindrance — the crowding around the reactive site.
What Are Steric Effects?
Steric effects occur when bulky groups around the nucleophile block or hinder its approach toward the electrophilic atom.
Even if the nucleophile is strongly basic and has high electron density, its reactivity (nucleophilicity) decreases if it is physically hindered.
General Trend
Smaller, less hindered nucleophiles are more nucleophilic because they can easily reach and attack the electrophile.
Less steric hindrance → higher nucleophilicity
Example 1: Alkoxides
Let’s compare methoxide ion (CH₃O⁻) and tert-butoxide ion [(CH₃)₃CO⁻].
- CH₃O⁻: small, unhindered → can easily attack → strong nucleophile
- (CH₃)₃CO⁻: bulky tert-butyl group shields oxygen → difficult to attack → weak nucleophile
Result:
CH₃O⁻ > (CH₃)₃CO⁻ in nucleophilicity
But the reverse can be true for basicity, since tert-butoxide is a strong base.
Example 2: Amines
Compare NH₃, CH₃NH₂, and (CH₃)₃N:
| Compound | Steric hindrance | Nucleophilicity |
|---|---|---|
| NH₃ | None | Highest |
| CH₃NH₂ | Mild | High |
| (CH₃)₃N | Very bulky | Low |
Nucleophilicity Of RO⁻ and RS⁻ in polar protic and polar aprotic solvent
Which is a better nucleophile in aqueous solution?
- In water (a protic solvent), small, highly basic ions like RO⁻ are strongly solvated by hydrogen bonding.
- Larger, less basic ions like RS⁻ are less solvated and therefore more available to attack electrophiles.
Answer: RS⁻ is the better nucleophile in aqueous solution.
Which is a better nucleophile in DMSO?
- DMSO is a polar aprotic solvent — it does not strongly solvate anions.
- Hence, nucleophilicity parallels basicity in aprotic solvents.
- Since RO⁻ is the stronger base, it is also the stronger nucleophile in DMSO.
Answer: RO⁻ is the better nucleophile in DMSO.
a. Br⁻ or Cl⁻ in H₂O → Br⁻
- In protic solvents larger, more polarizable anions are less strongly solvated and therefore better nucleophiles: I⁻ > Br⁻ > Cl⁻ > F⁻.
b. Br⁻ or Cl⁻ in DMSO → Cl⁻
- In polar aprotic solvents nucleophilicity follows basicity (smaller, harder = better): Cl⁻ is a better nucleophile than Br⁻ in DMSO.
c. CH₃O⁻ or CH₃OH in H₂O → CH₃O⁻
- The anion (methoxide) is far more nucleophilic than neutral methanol, even though it is somewhat solvated in water.
d. CH₃O⁻ or CH₃OH in DMSO → CH₃O⁻
- In a polar aprotic solvent the anionic nucleophile is even more reactive (less solvated), so methoxide wins by a larger margin.
e. HO⁻ or ⁻NH₂ in H₂O → HO⁻
- Practical point: ⁻NH₂ is so basic that it is protonated by water (it doesn’t exist appreciably in aqueous solution), so hydroxide is the available nucleophile.
f. HO⁻ or ⁻NH₂ in DMSO → ⁻NH₂
- In an aprotic medium where ⁻NH₂ can exist, the stronger base (⁻NH₂) is the stronger nucleophile.
g. I⁻ or Br⁻ in H₂O → I⁻
- Same protic-solvent rule: the larger, more polarizable I⁻ is less solvated and therefore the better nucleophile in water.
h. I⁻ or Br⁻ in DMSO → Br⁻
- In polar aprotic solvents nucleophilicity follows basicity/charge density: the smaller Br⁻ is the stronger nucleophile than I⁻.
Why Thiols (RSH) Are Stronger Acids Than Alcohols (ROH)
When you first look at alcohols (ROH) and thiols (RSH), they seem quite similar — both contain a hydrogen attached to a group 16 element (oxygen or sulfur).
Yet, when it comes to acidity, thiols clearly win.
In this post, we’ll explore why thiols (RSH) are stronger acids than alcohols (ROH) — step by step.
Step 1: Acid strength depends on conjugate base stability
The strength of an acid depends on how stable its conjugate base is after losing a proton (H⁺).
- Alcohols (ROH) lose H⁺ to form alkoxide ions (RO⁻)
- Thiols (RSH) lose H⁺ to form thiolate ions (RS⁻)
The more stable the conjugate base, the stronger the acid.
So, we must compare the stability of RO⁻ and RS⁻.
Step 2: The role of atom size and polarizability
Sulfur (S) is larger and more polarizable than oxygen (O).
That means the negative charge on sulfur in RS⁻ is spread over a larger volume, making it more stable.
In contrast, oxygen is smaller and holds the negative charge tightly on a small area, creating stronger charge density and less stability.
Result: RS⁻ is more stable than RO⁻ → RSH is a stronger acid.
Step 3: Bond strength difference
The S–H bond is weaker than the O–H bond.
It takes less energy to break the S–H bond and release a proton.
Weaker bond → easier H⁺ release → stronger acid
Step 4: Electronegativity vs Polarizability
Although oxygen is more electronegative than sulfur, electronegativity isn’t the main factor here.
For acid strength, charge delocalization and bond strength dominate — and sulfur’s large, polarizable nature stabilizes the anion far better than oxygen.
tep 5: pKa values tell the story
| Compound | Approx. pKa | Acid Strength |
|---|---|---|
| Methanol (CH₃OH) | ~16 | Weaker acid |
| Methanethiol (CH₃SH) | ~10 | Stronger acid |
A lower pKa means a stronger acid — so thiols are clearly more acidic than alcohols.
✅ Final Summary
| Factor | Alcohol (ROH) | Thiol (RSH) | Stronger Acid? |
|---|---|---|---|
| Atom size | Small (O) | Large (S) | ✅ RSH |
| Charge delocalization | Less | More | ✅ RSH |
| Bond strength | Strong O–H | Weak S–H | ✅ RSH |
| pKa | ~16 | ~10 | ✅ RSH |
Conclusion
Thiols (RSH) are stronger acids than alcohols (ROH) because their conjugate bases (RS⁻) are more stable.
Sulfur’s larger size and greater polarizability allow the negative charge to spread out, and the weaker S–H bond makes proton loss easier.
So, while oxygen wins in electronegativity, sulfur wins in stability, making RSH the stronger acid overall.
What is oleum? Difference between oleum and H2SO4
Oleum, also known as fuming sulfuric acid, is a highly reactive compound consisting of dissolved sulfur trioxide gas (SO₃) in 100% sulfuric acid.
What do you mean by fuming?
When we say “fuming” (like fuming sulfuric acid = oleum), we mean:
- The liquid gives off visible vapors (“fumes”) when exposed to air.
- In the case of oleum, those vapors are actually SO₃ gas, which escapes easily because SO₃ is volatile.
- When this SO₃ gas meets moisture in the air, it reacts immediately to form tiny droplets of H₂SO₄, which look like white, smoky fumes.
That’s why:
- Concentrated H₂SO₄ does not “fume” much in air.
- Oleum does, because of the escaping SO₃.
So fuming = releasing visible vapors (SO₃ + H₂SO₄ mist) into the air
Fumes of H2SO4 dangerous?
Yes ✅ the fumes of H₂SO₄ (sulfuric acid) are very dangerous.
Here’s why:
- What the fumes are:
Pure H₂SO₄ itself doesn’t evaporate much, but when it’s very concentrated (or as oleum) it releases SO₃ vapors.
These vapors react with moisture in the air → form fine droplets of H₂SO₄ mist. - Effects on health:
- Extremely corrosive to skin, eyes, and mucous membranes.
- Causes severe burns on contact.
- If inhaled, can damage the nose, throat, lungs (leading to breathing difficulty, pulmonary edema).
- Even short exposure to strong fumes is harmful.
That’s why handling concentrated H₂SO₄ / oleum requires fume hoods, protective gloves, goggles, and masks.
H₂S₂O₇ is the chemical formula of oleum?
Oleum is also called fuming sulfuric acid.
It is not a single pure compound but rather a solution of SO₃ dissolved in H₂SO₄.
H₂S₂O₇ = pyrosulfuric acid, which represents one particular composition of oleum (when there is exactly one mole of SO₃ combined with one mole of H₂SO₄).
Oleum in general can have different amounts of SO₃, not just the 1:1 case.
If it is mixture, so why it is represented by molecular formula H2S2O7?
Oleum is really a mixture (H₂SO₄ + SO₃), but in chemistry we often like to represent it in a molecular way. That’s why one of its possible compositions is written as H₂S₂O₇ (pyrosulfuric acid).
Convenience in Representation
- Writing H₂S₂O₇ makes it easier to handle in equations and discussions, instead of always writing “H₂SO₄·SO₃”.
- For example, when balancing reactions in textbooks, H₂S₂O₇ is simpler to use.
But in Reality
- Oleum is not just H₂S₂O₇ — it may have more (or less) SO₃ dissolved in H₂SO₄.
- So H₂S₂O₇ is only one specific case of oleum.
Why Sulphonation of OF BENZENE IS Carried out with Oleum not pure Conc. H2SO4?
leum already contains a large amount of free SO₃ dissolved in H₂SO₄.
That SO₃ directly acts as the electrophile in the sulphonation of benzene.
The H₂SO₄ in oleum helps by protonating SO₃ to form the stronger electrophile
MCQs – Biomolecules
1. The basic unit of proteins is:
a) Nucleotides
b) Amino acids
c) Fatty acids
d) Monosaccharides
2. Which of the following is a disaccharide?
a) Glucose
b) Fructose
c) Maltose
d) Galactose
3. The number of chiral carbons in glucose is:
a) 2
b) 3
c) 4
d) 5
4. Which of the following is a reducing sugar?
a) Sucrose
b) Maltose
c) Starch
d) Cellulose
5. Which vitamin is water-soluble?
a) Vitamin A
b) Vitamin D
c) Vitamin K
d) Vitamin C
6. Which of the following is a storage polysaccharide in animals?
a) Starch
b) Glycogen
c) Cellulose
d) Chitin
7. The prosthetic group of haemoglobin is:
a) Chlorophyll
b) Haem
c) Biotin
d) Cyanocobalamin
8. DNA differs from RNA in having:
a) Ribose sugar
b) Thymine instead of uracil
c) Single strand only
d) Guanine
9. The linkage between two amino acids is:
a) Glycosidic bond
b) Peptide bond
c) Hydrogen bond
d) Ester bond
10. Which of the following is a fibrous protein?
a) Hemoglobin
b) Insulin
c) Keratin
d) Albumin
11. In DNA, complementary base pairing occurs between:
a) A–G and C–T
b) A–T and G–C
c) A–C and T–G
d) A–U and G–C
12. The process of conversion of glucose to ethanol by yeast is:
a) Oxidation
b) Fermentation
c) Reduction
d) Hydrolysis
13. The deficiency of Vitamin D causes:
a) Night blindness
b) Rickets
c) Beriberi
d) Scurvy
14. Which of the following carbohydrates is a non-reducing sugar?
a) Glucose
b) Maltose
c) Sucrose
d) Lactose
15. Insulin is a:
a) Steroid
b) Vitamin
c) Protein
d) Carbohydrate
16. The sugar present in RNA is:
a) Deoxyribose
b) Ribose
c) Fructose
d) Glucose
17. The secondary structure of proteins is stabilized by:
a) Hydrogen bonds
b) Ionic bonds
c) Disulfide bonds
d) Peptide bonds
18. The genetic material in most viruses is:
a) DNA only
b) RNA only
c) Both DNA and RNA
d) Protein
19. Which one is a sulphur-containing amino acid?
a) Glycine
b) Serine
c) Cysteine
d) Alanine
20. Starch is composed of:
a) Amylose only
b) Amylopectin only
c) Amylose and amylopectin
d) Glucose and fructose
✅ Answer Key
- b
- c
- c
- b
- d
- b
- b
- b
- b
- c
- b
- b
- b
- c
- c
- b
- a
- b
- c
- c
MCQs – Amines
1. The general formula of primary amines is:
a) R₂NH
b) R₃N
c) R–NH₂
d) Ar–NO₂
2. Aniline is less basic than ethylamine because:
a) –NH₂ group is electron withdrawing in aniline
b) Lone pair of nitrogen in aniline is delocalised into the benzene ring
c) Aniline is aromatic
d) Resonance decreases stability of aniline
3. Which one gives carbylamine test?
a) Primary amine
b) Secondary amine
c) Tertiary amine
d) Amide
4. The IUPAC name of CH₃–NH₂ is:
a) Methyl amine
b) Aminomethane
c) Methanamine
d) Both (a) and (c)
5. The Hinsberg reagent is:
a) Benzenesulphonyl chloride
b) Chlorobenzene
c) Benzoyl chloride
d) Acetyl chloride
6. Which amine gives a nitrosoamine with nitrous acid at low temperature?
a) Primary aliphatic amine
b) Secondary amine
c) Tertiary amine
d) Aromatic amine
7. Which amine reacts with nitrous acid to give nitrogen gas?
a) Primary aliphatic amine
b) Secondary amine
c) Tertiary amine
d) Aromatic amine
8. The product formed when aniline reacts with bromine water is:
a) Monobromoaniline
b) 2,4,6–Tribromoaniline
c) 3–Bromoaniline
d) 4–Bromoaniline
9. Which one is most basic in aqueous solution?
a) Aniline
b) Ethylamine
c) p–Nitroaniline
d) Benzylamine
10. Which of the following is used as a starting material for Hofmann bromamide reaction?
a) Amide
b) Amine
c) Nitro compound
d) Ammonium salt
11. In Gabriel phthalimide synthesis, the product obtained is:
a) Primary amine
b) Secondary amine
c) Tertiary amine
d) Quaternary salt
12. Which of the following will not respond to carbylamine test?
a) CH₃NH₂
b) (CH₃)₂NH
c) C₂H₅NH₂
d) C₆H₅NH₂
13. Acetanilide is obtained by reaction of aniline with:
a) Acetaldehyde
b) Acetic anhydride
c) Acetyl chloride
d) Both (b) and (c)
14. The basic strength of amines in aqueous medium decreases in the order:
a) NH₃ > RNH₂ > R₂NH > R₃N
b) R₂NH > RNH₂ > R₃N > NH₃
c) R₃N > R₂NH > RNH₂ > NH₃
d) NH₃ > R₃N > R₂NH > RNH₂
15. Diazonium salts are stable:
a) At room temperature
b) Only in solid state
c) Below 5 °C
d) In presence of HCl only
16. Reaction of an amine with aryl diazonium salt to form azo compounds is called:
a) Coupling reaction
b) Condensation reaction
c) Neutralisation reaction
d) Elimination reaction
17. The Hinsberg test is used to distinguish:
a) Primary and secondary alcohols
b) Primary, secondary and tertiary amines
c) Aliphatic and aromatic amines
d) Aldehydes and ketones
18. In the Hofmann bromamide reaction, the major product formed from ethanamide is:
a) Ethylamine
b) Methylamine
c) Aniline
d) Ammonia
19. The reaction of nitrous acid with aniline at 0–5 °C gives:
a) p–Nitroaniline
b) Benzenediazonium chloride
c) Anisole
d) Chlorobenzene
20. The reagent used to convert nitrobenzene to aniline is:
a) Zn + HCl
b) Sn + HCl
c) Fe + HCl
d) All of these
✅ Answer Key
- c
- b
- a
- d
- a
- b
- a
- b
- b
- a
- a
- b
- d
- b
- c
- a
- b
- b
- b
- d
MCQs – Aldehydes, Ketones & Carboxylic Acids
1. Which of the following gives a positive silver mirror test (Tollen’s test)?
a) Formaldehyde
b) Acetone
c) Benzophenone
d) Acetophenone
2. Which of the following does not give Fehling’s test?
a) Formaldehyde
b) Acetaldehyde
c) Glucose
d) Benzaldehyde
3. The most reactive aldehyde towards nucleophilic addition is:
a) HCHO
b) CH₃CHO
c) C₆H₅CHO
d) CH₃COCH₃
4. Acetone reacts with HCN to form:
a) Cyanohydrin
b) Alcohol
c) Ketal
d) Ether
5. Which test is used to distinguish aldehyde from ketone?
a) Tollen’s test
b) Iodoform test
c) Baeyer’s test
d) Benedict’s test
6. The functional group –CHO is present in:
a) Ketones
b) Aldehydes
c) Carboxylic acids
d) Esters
7. The general formula of carboxylic acid is:
a) CnH₂nO
b) CnH₂nO₂
c) CnH₂n+1O₂
d) CnH₂n−1O₂
8. Which one of the following acids is strongest?
a) Acetic acid
b) Formic acid
c) Benzoic acid
d) Propanoic acid
9. The reduction of aldehydes and ketones with NaBH₄ gives:
a) Alcohols
b) Acids
c) Esters
d) Ethers
10. Acetophenone gives positive iodoform test because it contains:
a) –CHO group
b) –COOH group
c) CH₃CO– group
d) –OH group
11. The product formed when acetaldehyde is oxidized with acidified K₂Cr₂O₇ is:
a) Acetic acid
b) Methanol
c) Ethanol
d) Ethanoic anhydride
12. Which of the following does not give iodoform test?
a) Ethanol
b) Acetaldehyde
c) Methanol
d) Acetone
13. Which reaction converts aldehyde into primary alcohol?
a) Oxidation
b) Reduction
c) Decarboxylation
d) Hydrolysis
14. Acetone on ozonolysis gives:
a) Methanol + Formic acid
b) Acetic acid + Formic acid
c) Acetic acid only
d) Formaldehyde only
15. The carboxylic acid having maximum boiling point is:
a) Formic acid
b) Acetic acid
c) Propanoic acid
d) Butanoic acid
16. Benzaldehyde does not respond to:
a) Tollen’s test
b) Fehling’s test
c) Oxidation test
d) Cannizzaro reaction
17. Which one gives effervescence with NaHCO₃ solution?
a) CH₃OH
b) CH₃CHO
c) CH₃COOH
d) C₂H₆
18. In the Rosenmund reduction, acid chlorides are reduced to:
a) Alcohols
b) Aldehydes
c) Alkanes
d) Ketones
19. Which acid is present in vinegar?
a) Formic acid
b) Acetic acid
c) Oxalic acid
d) Citric acid
20. Hell–Volhard–Zelinsky reaction is shown by:
a) Aldehydes
b) Ketones
c) Carboxylic acids
d) Esters
✅ Answer Key
- a
- d
- a
- a
- a
- b
- b
- b
- a
- c
- a
- c
- b
- b
- a (highest due to lowest molar mass & strongest H-bonding)
- b
- c
- b
- b
- c
Sarvan sir- Chemistry For All 