Free BCHET-147 Solved Assignment | July 2025 | BSCG, BSCM | English & Hindi Medium | IGNOU

Give the IUPAC nomenclature of potassium dichromate. Also give its reaction with potassium iodide in acidic medium.

Answer:

1. IUPAC Nomenclature of Potassium Dichromate

• The cation is potassium (K⁺).
• The anion is the dichromate ion (Cr₂O₇²⁻).
• In the Stock notation (which is part of IUPAC nomenclature), the oxidation state of the metal is indicated by a Roman numeral in parentheses. Since the two chromium atoms are in the +6 oxidation state, the name becomes dichromate(VI).
• Therefore, the full systematic name is Potassium dichromate(VI).

2. Reaction with Potassium Iodide in Acidic Medium

Step-by-Step Explanation:

• Color Change: The solution changes color from orange (the color of dichromate ions, Cr₂O₇²⁻) to a dark brown or reddish-brown (the color of iodine, I₂, in solution). The final solution may also have a greenish tinge due to the formation of chromium(III) ions (Cr³⁺).
• Formation of a Precipitate: In some cases, the formed iodine may appear as a grayish-black solid if the concentration is high enough.

Give the structure of methyl lithium.

Answer:

Predominant Structure (in solid state and common solvents like diethyl ether)

• Visualization: Imagine a pyramid with four lithium atoms at the corners.
• Bonding: Each methyl group (carbon atom) sits above the center of a triangular face of the Li₄ pyramid and forms a multi-center bond with the three lithium atoms that make up that face.
• Formula: This cluster is represented as (CH₃Li)₄ or more accurately, (CH₃)₄Li₄.

In the Gas Phase or at Very High Temperatures

With the help of suitable equations, give any two methods for preparation of metal carbonyls by reductive carbonylation.

Answer:

1. Preparation from Metal Halides by Reductive Carbonylation

2. Preparation from Metal Oxides by Reductive Carbonylation

Explain the structure of Fe(CO)₅ based on valence bond approach.

Answer:

Electronic Configuration and Oxidation State

• The oxidation state of Fe in Fe(CO)₅ is zero.
• The ground state electronic configuration of Fe is $$[Ar]3d^{6}4s^2$$$$[Ar]3d^{6}4s^2$$[Ar]3d^(6)4s^(2)[Ar] 3d^6 4s^2$$[Ar]3d^{6}4s^2$$.
  - CO is a strong field ligand and causes pairing of electrons in the 3d orbitals of Fe, resulting in electron rearrangement that creates vacant orbitals suitable for bonding.

Hybridization and Geometry

• According to valence bond theory, Fe in Fe(CO)₅ undergoes $$ds{p}^3$$$$ds{p}^3$$dsp^(3)dsp^3$$ds{p}^3$$ hybridization.
  - This involves the mixing of one 3d orbital, one 4s orbital, and three 4p orbitals to form five hybrid orbitals that are directed towards the five CO ligands.
  - The geometry of Fe(CO)₅ is trigonal bipyramidal: three CO ligands occupy the equatorial positions, and two occupy the axial positions.

Bonding

• The CO ligands act as σ-donors, donating lone pair electrons to the vacant hybrid orbitals on Fe.
• CO is also a π-acceptor ligand, which leads to back donation of electron density from filled d orbitals of Fe to empty π* orbitals of CO, stabilizing the complex.
• This bonding leads to a diamagnetic complex due to paired electrons.

Summary

• Fe(CO)₅ adopts a trigonal bipyramidal structure based on $$ds{p}^3$$$$ds{p}^3$$dsp^(3)dsp^3$$ds{p}^3$$ hybridization of Fe orbitals.
  - The three equatorial and two axial positions of CO ligands correspond to the geometry explained by the valence bond approach.
  - The bonding involves σ-donation from CO and π-back donation from Fe, promoting stability.

With suitable reactions (any two) explain how carbonyl hydrides are formed by reaction of metal carbonyls with hydrogen.

Answer:

1. Formation of Cobalt Carbonyl Hydride

2. Formation of Manganese Carbonyl Hydride

Explanation

• In these reactions, molecular hydrogen (H₂) acts as a reducing agent.
• The metal centers in the carbonyl clusters are reduced, and hydride ions (H⁻) are introduced as ligands.
• The product complexes are called metal carbonyl hydrides, characterized by a direct metal-hydrogen bond.
• These hydrides serve as important intermediates or catalysts in several homogeneous catalytic processes.

With suitable diagrams explain the polarity of the free CO molecule as well as when it binds with a transition metal cation.

Answer:

Polarity of Free CO Molecule

• The CO molecule has a triple bond between carbon and oxygen, consisting of one sigma bond and two pi bonds.
• Oxygen is more electronegative than carbon; hence, electrons tend to be drawn toward oxygen.
• However, the polarity of the CO molecule is unusual because the molecular dipole moment points from oxygen (which carries a slight positive charge) toward carbon (which carries a slight negative charge).
• This is because the electron density from the pi bonds shifts somewhat from oxygen to carbon, creating a dipole with negative charge near carbon.
• Therefore, the dipole moment direction is $${\text{O}}^{\delta }+\to {\text{C}}^{\delta }-$$$${\text{O}}^{\delta }+\to {\text{C}}^{\delta }-$$"O"^(delta)+rarr"C"^(delta)-\text{O}^\delta+ \rightarrow \text{C}^\delta-$${\text{O}}^{\delta }+\to {\text{C}}^{\delta }-$$, opposite to what electronegativity alone would predict.
  - In summary, CO has a small net dipole moment with negative character on carbon and positive character on oxygen.

Polarity of CO When Bound to a Transition Metal Cation

• When CO binds to a transition metal cation, it acts as a ligand through the lone pair on carbon donating electrons to the metal (σ-donation).
• The metal can back-donate electron density from filled d orbitals into the $${\pi }^{\ast }$$$${\pi }^{\ast }$$pi^(**)\pi^*$${\pi }^{\ast }$$ antibonding orbitals of CO (π-back donation).
  - This back donation increases the electron density on the oxygen end and reduces the dipole moment, sometimes reversing the polarity compared to free CO.
  - The metal-CO bond often has a dipole pointing from metal (positive) toward carbon (negative), with significant polarization toward the CO ligand.
  - This results in a shift of electron density emphasizing the metal-to-CO interaction and modifies the CO bond strength and overall polarity.

Diagrams :

What information is obtained from the IR spectrum of Fe₂(CO)₉?

Answer:

Give a short account of the essential and non essential elements occurring in biological systems.

Answer:

Essential Elements

• These are elements required by organisms to carry out vital biochemical and physiological functions.
• Essential elements are further divided into:
  • Macronutrients: Needed in large amounts, such as carbon (C), hydrogen (H), oxygen (O), nitrogen (N), phosphorus (P), sulfur (S), potassium (K), calcium (Ca), magnesium (Mg), and sodium (Na). They form the basic building blocks of biomolecules and help in processes like energy transfer, signaling, and structural integrity.
  • Micronutrients (Trace elements): Required in very small amounts but are critical for enzyme function and metabolic pathways. These include iron (Fe), zinc (Zn), copper (Cu), manganese (Mn), cobalt (Co), molybdenum (Mo), selenium (Se), and iodine (I).

Non-Essential Elements

• Non-essential elements are those which may be present in organisms but are not necessary for normal growth or survival.
• Some non-essential elements may have no biological role or may be toxic at high concentrations.
• Occasionally, certain non-essential elements can substitute for essential ones or have beneficial effects but are not strictly required.

Summary

• Essential elements are indispensable for life and enzymatic functions.
• Non-essential elements may be incidental or potentially toxic and do not play a fundamental role in biological systems.

Give the different ways by which lead affects the human body.

Answer:

1. Neurological and Neurodevelopmental Effects (Most Sensitive Target)

• In Children:
  • Reduced IQ and Cognitive Deficits: Lead interferes with neurotransmitter release and brain development, leading to permanent learning disabilities and lower intellectual capacity.
  • Behavioral Problems: Causes increased incidence of attention-deficit/hyperactivity disorder (ADHD), aggression, irritability, and antisocial behavior.
  • Impaired Development: Delays in reaching developmental milestones (e.g., talking, walking).
  • Poor Academic Performance: Difficulty concentrating and shortened attention span.
• In Adults:
  • Neuropathy: Damage to peripheral nerves, leading to weakness, pain, and tingling in the extremities (e.g., "wrist drop" is a classic sign).
  • Cognitive Effects: Impaired memory, poor concentration, and mood disorders.
  • Headaches and Seizures: In cases of high exposure.

2. Hematological Effects (Effects on Blood)

• Anemia: By inhibiting key enzymes (δ-aminolevulinic acid dehydratase and ferrochelatase), lead prevents the formation of hemoglobin, leading to hemolytic anemia (where red blood cells are destroyed prematurely).
• Porphyrinuria: The disruption of heme synthesis causes precursors like aminolevulinic acid (ALA) and porphyrins to build up and be excreted in urine.
• Reduced Oxygen Carrying Capacity: Anemia results in fatigue, weakness, and pallor.

3. Cardiovascular Effects

• Hypertension (High Blood Pressure): Chronic low-level lead exposure is a recognized risk factor for high blood pressure, a major cause of heart disease and stroke.
• Increased Risk of Heart Disease: Lead can cause damage to the heart muscle and blood vessels through oxidative stress and inflammation.

4. Renal (Kidney) Effects

• Chronic Kidney Disease: Long-term exposure damages the proximal tubular cells of the kidney, leading to impaired kidney function and potentially kidney failure.
• Gout: Lead-induced kidney damage can reduce the excretion of uric acid, leading to hyperuricemia and gout ("saturnine gout").

5. Gastrointestinal Effects

• "Lead Colic": A classic symptom of lead poisoning, characterized by severe abdominal cramping, pain, constipation, and nausea.
• Loss of Appetite and Weight Loss.

6. Reproductive and Developmental Effects

• In Men: Reduced sperm count, increased abnormal sperm, and potential infertility.
• In Women: Increased risk of miscarriage, stillbirth, and premature birth. Lead can cross the placental barrier.
• Effects on the Fetus: Because it crosses the placenta, lead exposure in the mother can cause serious harm to the developing fetus, including reduced growth, premature birth, and neurological damage.

7. Skeletal System

• Bone Storage: Approximately 95% of the lead in an adult's body is stored in the bones and teeth. It replaces calcium in the bone matrix.
• Endogenous Exposure: During periods of physiological stress (e.g., pregnancy, lactation, osteoporosis), lead can be released from bones back into the bloodstream, acting as a source of internal exposure long after external exposure has ended.

8. Other Systems

• Auditory: Hearing loss, especially in children.
• Dental: Lead lines (blue-black lines) on gums in cases of extreme exposure.
• Immune System: May impair immune function, making the body more susceptible to disease.

Key Characteristics of Lead Toxicity:

• Cumulative: Lead accumulates in the body over a lifetime.
• No Safe Level: The World Health Organization (WHO) states there is no known level of lead exposure that is considered safe.
• Children are Most Vulnerable: They absorb 4-5 times more ingested lead than adults, and their developing nervous systems are exquisitely sensitive.

Explain the conversion of heme to hemin along with a suitable diagram.

Answer:

Reaction Summary- Heme: Contains iron in the ferrous state (Fe²⁺)

• Hemin: Contains iron in the ferric state (Fe³⁺), with one chloride ion bound as an axial ligand to iron, making it more stable in aqueous solution.

Explanation- In this oxidation process, the central Fe in heme loses one electron to oxygen, converting from Fe(II) to Fe(III).

• A chloride ion from HCl coordinates axially to Fe(III) in hemin, stabilizing the complex.
• Hemin is more stable and less prone to oxidation than heme and is often used for physiological and experimental studies related to heme.

Diagram - Porphyrin ring shown as a planar macrocycle.

• Central iron atom labeled Fe²⁺ in heme.
• Arrow indicating oxidation to Fe³⁺.
• A chloride ion (Cl⁻) shown coordinating axially in hemin.

How will you prepare pentan-2-one from ethyl 3-oxobutanoate. Give the reactions involved.

Answer:

Step 1: Hydrolysis and Decarboxylation of Ethyl 3-oxobutanoate

• Ethyl 3-oxobutanoate undergoes alkaline hydrolysis (saponification) followed by acidification, resulting in the β-keto acid.
• Upon heating, the β-keto acid undergoes decarboxylation, losing CO₂ to give 3-oxobutanoic acid, which tautomerizes to pentan-2-one.

Detailed Reactions:

1. Hydrolysis:

2. Acidification:

3. Decarboxylation:

Step 2: Reduction (if necessary)

• The product pentan-2-one (a ketone) is obtained directly after decarboxylation.
• No further reduction is necessary if pentan-2-one itself is the desired compound.

Summary

Give IUPAC name the following compounds.

Answer:

Compound 1 (Top Structure)

• Features:
  • Naphthalene core
  • Substituents: OH (hydroxy), CH₃ (methyl), NH₂ (amino)
  • Positioning: OH and CH₃ on the same ring, NH₂ adjacent to OH
• IUPAC Name:
  2-Amino-6-methyl-1-naphthol

Compound 2 (Bottom Structure)

• Features:
  • Naphthalene core
  • Substituents: NO₂ (nitro), CH₃ (methyl), Cl (chloro)
  • Positioning: NO₂ and methyl on one ring, Cl on the adjoining ring
• IUPAC Name:
  4-Chloro-5-methyl-1-nitronaphthalene

Is anthracene aromatic compound? Explain.

Answer:

Explanation:

• Anthracene is composed of three fused benzene rings in a linear arrangement, giving it the molecular formula $$C_{14}{H}_{10}$$$$C_{14}{H}_{10}$$C_(14)H_(10)C_{14}H_{10}$$C_{14}{H}_{10}$$.
  - Each benzene ring is aromatic, and the conjugation extends over all three rings, creating a larger conjugated π-electron system.
  - Anthracene follows the Hückel rule for aromaticity, having $$4n+2$$$$4n+2$$4n+24n + 2$$4n+2$$ π electrons (where $$n$$$$n$$nn$$n$$ = 4 for 14 π-electrons over the system).
• The molecule is planar, allowing the p-orbitals from each ring to overlap and delocalize electrons across the entire system.
• This extensive conjugation and electron delocalization contribute to anthracene’s aromatic stability.
• Anthracene exhibits typical aromatic behavior like electrophilic substitution reactions predominantly at the 9-position.

Among pyridine and pyrrole which one is more basic? Explain

Answer:

Explanation:

• Pyridine:
  • The nitrogen atom in pyridine has a lone pair of electrons located in an sp² hybrid orbital perpendicular to the aromatic π system.
  • This lone pair is not involved in the aromaticity of the ring, so it is available to accept a proton (H⁺), making pyridine a good Lewis base.
  • Due to this, pyridine readily donates its lone pair and exhibits relatively strong basicity (pKa of conjugate acid ≈ 5.14).
• Pyrrole:
  • The nitrogen lone pair in pyrrole is part of the aromatic sextet (it contributes two electrons to the aromatic π system).
  • Because the lone pair is delocalized into the ring for maintaining aromaticity, it is less available to accept a proton.
  • Hence, pyrrole is much less basic than pyridine (pKa of conjugate acid ≈ -3.8).
  • Protonation of pyrrole would disrupt aromaticity, which is energetically unfavorable.

Summary:

In pyridine substitution occur which at the 3 position. Explain.

Answer:

Why Substitution in Pyridine Occurs at the 3-Position

• Pyridine is a heteroaromatic compound with a nitrogen atom in the ring, which is more electronegative than carbon.
• The nitrogen atom withdraws electron density from the ring via its inductive effect, making the ring electron-deficient overall, especially at the 2- and 4-positions.
• During electrophilic substitution, the intermediate formed is a resonance-stabilized carbocation (sigma complex).

Explanation for Preference of 3-Position:

• When substitution occurs at the 3-position, the positive charge in the intermediate does not localize on the nitrogen atom.
• Substitution at the 2- or 4-position would result in intermediates where the positive charge is directly adjacent to the nitrogen, which is destabilizing due to the electronegative nitrogen withdrawing electron density.
• Hence, the intermediate formed from substitution at the 3-position is more stable compared to those at 2- and 4-positions.
• This increased stability makes substitution at the 3-position more favorable.

Summary:

Conclusion:

Which are the two main parameters on which the absorption bands in ultraviolet and visible spectra are measured?

Answer:

1. Wavelength (λ):
   • The position of the absorption band is measured in terms of the wavelength of light absorbed, usually expressed in nanometers (nm) or sometimes in wavenumbers (cm⁻¹).
   • The wavelength corresponds to the energy difference between electronic energy levels involved in the transition.
2. Absorbance or Transmittance (Intensity):
   • The strength or intensity of the absorption band is measured by absorbance (A) or sometimes as transmittance (%T).
   • Absorbance is related to the concentration of the absorbing species, path length, and molar absorptivity by Beer-Lambert law.
   • It indicates how much light is absorbed at a particular wavelength.

Explain Woodward rules for predicting π-π* absorption in dienes.

Answer:

Woodward Rules for Predicting π-π* Absorption in Dienes

Key Points of Woodward Rules for Dienes:

1. Base Values for λ_max:
   • Acyclic Dienes: 214 nm
   • Homoannular Dienes (both double bonds in the same ring): 253 nm
   • Heteroannular Dienes (double bonds in different rings or positions): 214 nm
2. Additive Effects of Substituents and Structural Features:
   • Each alkyl substituent or ring residue attached to the conjugated system: +5 nm
   • Each exocyclic double bond: +5 nm
   • Each additional double bond conjugated (extension of π-system): +30 nm
3. Auxochromes (electron-donating groups) also cause shifts:
   • Examples: –OR (alkoxy), –Cl, –Br groups add a few nm to λ_max.

How to Use Woodward Rules:

• Start with the base value depending on the diene type.
• Add increments to the base value for each substituent, exocyclic double bond, or extra conjugation.
• The total gives an approximate λ_max for the π-π* electronic transition.

Summary Table:

Explain with the help of a diagram the orbital energy relationship between isolated and conjugated C=C and C=O groups.

Answer:

Isolated C=C and C=O Groups

• For isolated double bonds such as C=C and C=O, each has its own set of π and π* orbitals separated by a certain energy gap.
• The π orbital is the bonding orbital, and the π* orbital is the antibonding orbital.
• Due to the difference in electronegativity, the C=O π orbital is lower in energy than the C=C π orbital, and the π* orbital of C=O is also lower in energy than the π* orbital of C=C.

Conjugated C=C and C=O Groups

• When C=C and C=O groups are conjugated (i.e., alternating double bonds), their π orbitals interact and combine to form molecular orbitals delocalized over both groups.
• This interaction causes splitting of the original π and π* orbitals into new bonding and antibonding orbitals in the conjugated system.
• The conjugation lowers the energy of the bonding π orbital (stabilization) and raises the energy of the antibonding π orbital* (destabilization) leading to a smaller HOMO-LUMO gap.
• Consequently, conjugated systems absorb light at longer wavelengths (lower energy) compared to isolated groups due to this reduced energy gap.

Diagram Explanation

• Interaction between the π orbitals of C=C and C=O in conjugation causes the formation of new molecular orbitals with altered energies.

Summary

• The orbital interaction in conjugated systems reduces the HOMO-LUMO gap compared to isolated double bonds.
• This leads to characteristic absorption shifts in UV-Vis spectroscopy for conjugated compounds.
• The C=O group generally has lower energy π and π* orbitals than C=C due to the higher electronegativity of oxygen.

For polyatomic molecule, what are the different types of normal modes of vibration? Give the normal modes of vibrations for the symmetric stretching of a triatomic linear and triatomic angular molecule with suitable diagrams.

Answer:

Types of Normal Modes of Vibration in Polyatomic Molecules

1. Stretching Vibrations:
   • Change in bond length between two atoms.
   • Can be symmetric or asymmetric.
2. Bending Vibrations:
   • Change in the angle between two bonds.
   • Types of bending: scissoring, rocking, wagging, and twisting.

Normal Modes for Triatomic Molecules

1. Triatomic Linear Molecule (e.g., CO₂)

• There are 3 atoms → $$3N=9$$$$3N=9$$3N=93N = 9$$3N=9$$ degrees of freedom
  - Translational and rotational motions = 5 (for linear)
  - Number of vibrational modes = $$3N-5=4$$$$3N-5=4$$3N-5=43N - 5 = 4$$3N-5=4$$

Vibrational Modes:

• Symmetric stretching: Both bond lengths increase or decrease simultaneously.
• Asymmetric stretching: One bond lengthens while the other shortens.
• Bending (degenerate): Two modes where the molecule bends in and out of the plane (perpendicular directions).

Symmetric stretching diagram:

• Both $$C=O$$$$C=O$$C=OC=O$$C=O$$ bonds stretch or compress together symmetrically.

2. Triatomic Angular (Bent) Molecule (e.g., H₂O)

• 3 atoms → 9 degrees of freedom
• Translational + rotational = 6 (for nonlinear)
• Number of vibrational modes = $$3N-6=3$$$$3N-6=3$$3N-6=33N - 6 = 3$$3N-6=3$$

Vibrational Modes:

• Symmetric stretching: Both bonds stretch or compress simultaneously.
• Asymmetric stretching: One bond lengthens while the other shortens.
• Bending: Change in bond angle.

Symmetric stretching diagram:

• Both $$O–H$$$$O–H$$O–HO–H$$O–H$$ bonds stretch or compress in phase.

Summary Table:

Give the characteristic frequencies of the ketones and aldehydes. How are they different from esters?

Answer:

Characteristic Frequencies of Ketones and Aldehydes

• Both ketones and aldehydes contain the $$\text{C}=\text{O}$$$$\text{C}=\text{O}$$"C"="O"\text{C}= \text{O}$$\text{C}=\text{O}$$ (carbonyl) group which shows a strong and sharp absorption in the IR spectrum.
  - Ketones:
  - $$\text{C}=\text{O}$$$$\text{C}=\text{O}$$"C"="O"\text{C}= \text{O}$$\text{C}=\text{O}$$ stretching frequency appears typically around 1715 cm⁻¹.
• Aldehydes:
  • $$\text{C}=\text{O}$$$$\text{C}=\text{O}$$"C"="O"\text{C}= \text{O}$$\text{C}=\text{O}$$ stretching frequency appears slightly higher, around 1725–1740 cm⁻¹.
  • Additionally, aldehydes show characteristic $$\text{C–H}$$$$\text{C–H}$$"C–H"\text{C–H}$$\text{C–H}$$ stretching bands due to aldehydic hydrogen near 2700–2800 cm⁻¹ (often appears as two weak bands).

Characteristic Frequencies of Esters

• Esters also have a carbonyl group attached to an oxygen atom (–COOR).
• The $$\text{C}=\text{O}$$$$\text{C}=\text{O}$$"C"="O"\text{C}= \text{O}$$\text{C}=\text{O}$$ stretching frequency of esters usually appears at 1735–1750 cm⁻¹ (higher than ketones and aldehydes).
  - Esters also show a strong absorption due to the C–O stretching vibration in the range 1050–1300 cm⁻¹.

Key Differences Between Ketones/Aldehydes and Esters in IR Spectrum

Summary

• Esters have a higher carbonyl stretching frequency than ketones and aldehydes due to the electron-withdrawing effect of the oxygen atom attached to the carbonyl carbon.
• Aldehydes uniquely exhibit $$\text{C–H}$$$$\text{C–H}$$"C–H"\text{C–H}$$\text{C–H}$$ stretching bands around 2700–2800 cm⁻¹, which are absent in ketones and esters.

List the factors affecting the position and intensity of IR bonds in the IR spectrum.

Answer:

Factors Affecting the Position (Frequency) of IR Absorption Bands:

1. Bond Strength (Force Constant):
   Stronger bonds vibrate at higher frequencies (wavenumbers). For example, triple bonds absorb at higher frequencies than double bonds, which absorb higher than single bonds.
2. Reduced Mass of the Atoms:
   Heavier atoms vibrate at lower frequencies because the vibrational frequency depends inversely on the square root of the reduced mass of the bonded atoms.
3. Electronegativity and Electronic Effects:
   • Electron-withdrawing groups increase bond polarity and often increase the frequency.
   • Resonance or conjugation lowers the frequency by delocalizing the electron density, weakening the bond (e.g., in conjugated carbonyls).
   • Inductive effects also influence the frequency.
4. Hydrogen Bonding:
   Hydrogen bonding lowers the frequency of the involved bond (e.g., O–H stretch) and broadens the absorption band due to bond weakening and increased dipole moment fluctuations.
5. Molecular Environment and Physical State:
   • Phase (solid, liquid, gas) can affect frequency and band shape.
   • Ring strain can shift frequencies by affecting bond angles and strengths.
6. Fermi Resonance and Coupling Effects:
   Interaction between fundamental vibration and overtone or combination bands can shift frequencies and intensities.

Factors Affecting the Intensity of IR Absorption Bands:

1. Change in Dipole Moment:
   The more the dipole moment changes during vibration, the more intense the absorption. Non-polar bonds show weak or no IR absorption.
2. Concentration of the Sample:
   Higher concentration leads to greater intensity since more molecules absorb IR radiation.
3. Number of Bonds Vibrating:
   Molecules with more identical bonds show stronger absorption (e.g., many C–H bonds lead to intense C–H stretch).
4. Symmetry of the Molecule:
   IR active vibrations require a change in dipole moment. Symmetric molecules may have IR inactive vibrations (e.g., symmetric stretching in nonpolar molecules).

Summary

Explain the bands appearing in the IR spectrum of an alkane giving a suitable example.

Answer:

Bands Appearing in the IR Spectrum of an Alkane

Key IR Bands for Alkanes (Example: Octane)

1. C–H Stretching Vibrations:
   • Appear as strong bands in the region 3000–2850 cm⁻¹.
   • These correspond to the stretching of sp³ hybridized C–H bonds.
   • Methyl ($$–C{H}_3$$$$–C{H}_3$$–CH_(3)–CH_3$$–C{H}_3$$) and methylene ($$–C{H}_2–$$$$–C{H}_2–$$–CH_(2)––CH_2–$$–C{H}_2–$$) groups contribute to these bands.
2. C–H Bending Vibrations:
   • Scissoring (bending) motion: Appears around 1465–1450 cm⁻¹.
   • Methyl rock: Appears near 1375–1350 cm⁻¹.
   • Methylene rock: Appears near 720–730 cm⁻¹ (for long-chain alkanes).
3. C–C Bond Vibrations:
   • Generally weak and appear in the 800–1300 cm⁻¹ region but are often overlapped by other vibrations.

Summary of Alkane IR Bands