2nd Year Chemistry Archives

Ethics and Values in Chemistry: Unit 1 Solved Exercises | 2nd Year Federal Board New Book

June 12, 2025 by samreen

Short Answer Questions:

i. Define cognitive bias.

Answer: Cognitive bias is a systematic error in thinking that influences judgments and decisions based on personal experiences, emotions, or subconscious preferences rather than objective evidence.
Key Concept: Subjective thinking patterns leading to irrational conclusions.
Tip/Trick: Remember: Bias = Brain’s shortcuts. Examples: Confirmation bias (favoring info that agrees with beliefs), availability bias (overweighting recent examples).

ii. What is a false cause fallacy?

Answer: False cause fallacy (or post hoc ergo propter hoc) assumes Event A caused Event B solely because A occurred before B, ignoring other factors.
Example: “I wore a lucky shirt, then passed my exam; the shirt caused my success.”
Key Concept: Confusing correlation with causation.
Tip/Trick: Ask: “Is there actual evidence linking cause and effect, or just timing?”

iii. Describe the straw man fallacy.

Answer: The straw man fallacy misrepresents an opponent’s argument as a weaker, distorted version to make it easier to attack.
Example: Original: “Reduce plastic use.” Straw man: “You want to ban all plastics and collapse industries!”
Key Concept: Intellectual dishonesty through distortion.
Tip/Trick: Spot phrases like “So you’re saying…” followed by an exaggerated claim.

iv. What does the fallacy of exclusion involve?

Answer: Fallacy of exclusion ignores crucial evidence that contradicts a claim, presenting an incomplete or biased view.
Example: “Chemical X is safe” (while hiding studies showing toxicity).
Key Concept: Cherry-picking data to support a narrative.
Tip/Trick: Always ask: “What evidence is missing?”

v. Give an example of a faulty analogy.

Answer: “Regulating chemicals is like banning cars because accidents happen – both stifle progress.”
Why faulty: Cars and chemical regulations aren’t comparable in risk scope or societal impact.
Key Concept: Weak comparisons that ignore critical differences.
Tip/Trick: Check if the analogy shares essential similarities. If not, it’s faulty.

vi. List one pro and one con of chemical substances.

Answer:

Pro: Pharmaceuticals save lives (e.g., antibiotics treat infections).
Con: Pesticides can contaminate ecosystems (e.g., DDT harming birds).
Key Concept: Dual nature of chemicals: lifesaving vs. harmful.
Tip/Trick: Pros relate to human progress; cons to unintended consequences.

vii. Responsibility of scientists/companies in chemical production.

Answer: Ensure safety through rigorous testing, transparent risk disclosure, ethical waste management, and compliance with regulations.
Example: Testing industrial solvents for long-term environmental impact.
Key Concept: Ethical duty to prioritize people and planet over profit.
Tip/Trick: Remember the precautionary principle: “Better safe than sorry.”

viii. Importance of regulations in the chemical industry.

Answer: Regulations enforce safety standards, prevent pollution, hold companies accountable, and protect public/environmental health.
Example: REACH laws in the EU ban carcinogenic chemicals.
Key Concept: Legal frameworks as safeguards against negligence.
Tip/Trick: Link to historical failures (e.g., Bhopal disaster → stricter laws).

ix. What is a claim in a scientific argument?

Answer: A claim is a statement asserting a fact or position, supported by evidence and reasoning.
Example: “Lead pollution reduces children’s IQ scores.”
Key Concept: Foundation of an argument requiring proof.
Tip/Trick: Claims answer “What are you trying to prove?”

x. Example of an assumption in renewable energy debates.

Answer: “Electric vehicles (EVs) are 100% eco-friendly.”
Hidden assumption: EV battery production/mining has no environmental cost.
Key Concept: Unverified premises weakening arguments.
Tip/Trick: Challenge assumptions with: “What evidence supports this?”

Summary Cheat Sheet

Concept	Core Idea	Red Flag
Cognitive Bias	Subjective thinking → errors	“I feel this is true”
False Cause	Sequence ≠ causation	“A happened, then B → A caused B”
Straw Man	Distort → attack	“So you’re saying [extreme]?”
Exclusion Fallacy	Hiding counter-evidence	Ignoring “inconvenient” data
Faulty Analogy	Weak comparisons	“It’s just like [unrelated]!”
Chemical Ethics	Balance innovation and safety	Absolute claims (“always safe”)
Scientific Claim	Assertion needing proof	Unsupported statements

Study Hack: Use real-world examples (e.g., Flint water crisis for regulation importance) to memorize concepts.

i. Explain confirmation bias and its potential impact on scientific research.

Statement:
Explain confirmation bias and its potential impact on scientific research.

Answer:
Confirmation bias is the tendency to favor, interpret, or recall information that confirms preexisting beliefs while ignoring or downplaying contradictory evidence. In scientific research, this manifests when researchers:

Selectively design experiments to yield expected results.
Overvalue data supporting their hypothesis and dismiss anomalies.
Cite only studies that align with their views during literature reviews.

Impact:

False Conclusions: Validates incorrect theories (e.g., linking vaccines to autism based on flawed data).
Wasted Resources: Diverts funding/time toward biased studies.
Erosion of Trust: Undermines scientific credibility when biases are exposed.

Key Concepts Used:

Cognitive psychology (information processing biases).
Scientific integrity (objectivity, reproducibility).
Research ethics (data transparency).

Tips and Tricks:

Red Flag: Ignoring “outlier” data.
Mitigation: Use blind trials, peer review, and preregistration of studies.
Example: In drug trials, confirmation bias might overstate efficacy by excluding non-responsive patients.

ii. Discuss ethical considerations in chemical production/use, balancing benefits and risks.

Statement:
Discuss the ethical considerations in the production and use of chemical substances, highlighting the balance between benefits and risks.

Answer:
Chemical ethics requires weighing societal benefits against health/environmental risks:

Benefits:

Life-saving drugs (e.g., antibiotics).
Food security (fertilizers/pesticides).

Risks:

Pollution (e.g., microplastics in oceans).
Health hazards (e.g., asbestos causing lung cancer).

Ethical Principles:

Precautionary Principle: Avoid chemicals with uncertain long-term effects.
Environmental Justice: Prevent disproportionate harm to marginalized communities (e.g., Flint water crisis).
Transparency: Disclose safety data (e.g., REACH regulations in the EU).

Balance Strategy:

Green Chemistry: Design biodegradable, low-toxicity alternatives.
Lifecycle Analysis: Assess environmental impact from production to disposal.

Key Concepts Used:

Risk-benefit analysis.
Stakeholder accountability (scientists, companies, regulators).
Sustainable development goals.

Tips and Tricks:

Framework: Use “People-Planet-Profit” to evaluate trade-offs.
Case Study: DDT boosted agriculture but caused ecological collapse → banned under Stockholm Convention.

iii. Deconstruct a scientific argument using EVs and air pollution.

Statement:
Deconstruct the structure of a scientific argument using the example of promoting electric vehicles to reduce air pollution.

Answer:

Component	Example for EVs
Claim	“Electric vehicles (EVs) reduce urban air pollution.”
Evidence	“EVs produce zero tailpipe emissions; gasoline cars emit CO₂, NOₓ, and particulates.”
Reasoning	“Transportation causes 60% of urban air pollution; replacing fossil-fuel vehicles cuts emissions.”
Counterclaim	“EV batteries require mining/power generation, shifting pollution elsewhere.”
Rebuttal	“Renewable-powered grids and battery recycling minimize net pollution (e.g., Tesla Gigafactories).”
Assumption	“EV adoption is scalable and infrastructure exists.”

Scientific Argument Structure:

Claim: Assertion requiring proof.
Evidence: Empirical data (emission studies).
Reasoning: Logical connection (pollution sources → solution).
Counter-Rebuttal: Addressing limitations.

Key Concepts Used:

Claim-evidence-reasoning framework.
Critical analysis of assumptions.
Holistic impact assessment.

Tips and Tricks:

Visualize: Create tables to map argument components.
Test Validity: Ask, “Does evidence directly support the claim?”

iv. Analyze regulations in the chemical industry with examples.

Statement:
Analyze the role of regulations and laws in ensuring ethical practices in the chemical industry, citing specific examples.

Answer:
Regulations enforce ethics through:

Risk Prevention:

REACH (EU): Requires safety assessments for chemicals; banned carcinogens like benzene in consumer products.

Accountability:

Bhopal Gas Tragedy (1984): Led to India’s Environment Protection Act, mandating disaster management plans.

Sustainability:

Montreal Protocol (1987): Phased out ozone-depleting CFCs, promoting safer refrigerants.

Impact:

Pros: Reduced pollution, safer workplaces, innovation in green chemistry.
Cons: Compliance costs; regulatory gaps in developing nations.

Key Concepts Used:

Regulatory frameworks (precautionary principle, polluter-pays).
Historical case studies.
Global vs. local governance.

Tips and Tricks:

Remember: Regulations = “Speed bumps for safety.”
Example Pairing:
Regulation: TSCA (USA)
Outcome: Banned PCBs in 1979 after proof of toxicity.

v. Evaluate pros/cons of pesticides in agriculture.

Statement:
Evaluate the pros and cons of using pesticides in agriculture, considering both human health and environmental impacts.

Answer:

Aspect	Pros	Cons
Human Health	↑ Food security (prevents famine)	Chronic diseases (cancer from glyphosate)
Environment	↑ Crop yields by 20-50%	Biodiversity loss (bees from neonicotinoids)
Economic	↓ Crop losses = affordable food	$15.6B/year in US health costs from exposure

Balanced Solutions:

Integrated Pest Management (IPM): Combine natural predators + minimal targeted pesticides.
Biodegradable Pesticides: e.g., Pyrethrins (plant-derived, low persistence).

Key Concepts Used:

Cost-benefit analysis.
Ecotoxicology (bioaccumulation, trophic cascades).
Sustainable agriculture.

Tips and Tricks:

Rule of Thumb: “No free lunch” – all chemicals have trade-offs.
Data Point: WHO estimates 3 million pesticide poisonings/year.

Summary Cheat Sheet

Question	Core Concept	Key Example
i	Bias → flawed research	Vaccine-autism retracted study
ii	Ethics = Benefit/Risk balance	DDT: malaria vs. ecosystem collapse
iii	Argument = Claim + Evidence + Reasoning	EV emissions reduction
iv	Regulations as safeguards	Bhopal → stricter laws
v	Pesticides: productivity vs. harm	Neonicotinoids and bee decline

Exam Strategy:

For ethics questions, always discuss trade-offs.
Use real-world examples to demonstrate depth (e.g., Flint for regulations, Silent Spring for pesticides).

Ethics and Values in Chemistry: Unit 1 MCQs Solved | Federal Board FSc Part 2 New Book

June 11, 2025 by samreen

Question i:

Statement:
What is cognitive bias?

Options:
a) The process of making decisions based on logical reasoning.
b) The tendency to make decisions based on personal characteristics and beliefs.
c) The evaluation of arguments using empirical evidence.
d) The method of constructing arguments using logical fallacies.

Correct Answer:
b

Explanation:
Cognitive bias refers to systematic errors in thinking that arise from personal beliefs, experiences, or emotions, leading to irrational judgments. Option b is correct because cognitive biases stem from subjective influences (e.g., stereotypes, emotions) rather than objective analysis.

a is incorrect: Logical reasoning opposes cognitive bias, which is inherently illogical.
c is incorrect: Cognitive biases often ignore empirical evidence in favor of preconceived notions.
d is incorrect: Logical fallacies are flawed argument structures, while cognitive biases are unconscious mental shortcuts.

Tips and Tricks:

Key phrase: “tendency” – Cognitive biases are unconscious tendencies, not deliberate methods.
Remember: If an option describes objectivity (e.g., “logical reasoning,” “empirical evidence”), it likely contradicts cognitive bias.

Question ii:

Statement:
What is a hasty generalization fallacy?

Options:
a) Drawing a conclusion based on a large, representative sample.
b) Drawing a conclusion based on insufficient evidence.
c) Drawing a conclusion by misrepresenting an opponent’s argument.
d) Drawing a conclusion based on a false cause-and-effect relationship.

Correct Answer:
b

Explanation:
Hasty generalization occurs when a broad conclusion is made from inadequate or unrepresentative data. Option b directly defines this fallacy (e.g., “All crows are black” after seeing two crows).

a is incorrect: Large samples support reliable generalizations, making this the opposite of hasty generalization.
c describes the straw man fallacy.
d describes the post hoc fallacy.

Tips and Tricks:

Spot keywords: “insufficient evidence,” “small sample size,” or “jumping to conclusions.”
Avoid confusion: If an option mentions “large sample” (a) or “cause-and-effect” (d), it refers to other fallacies.

Question iii:

Statement:
What does the post hoc ergo propter hoc fallacy assume?

Options:
a) One event is caused by another simply because it follows it.
b) One event is unrelated to another.
c) One event is the result of a thorough investigation.
d) One event is part of a larger series of events.

Correct Answer:
a

Explanation:
The post hoc fallacy (Latin: “after this, therefore because of this”) assumes Event A causes Event B solely because B occurs after A (e.g., “I wore a red shirt, then my team won; therefore, my shirt caused the win”).

b is incorrect: The fallacy assumes a false connection, not unrelatedness.
c is incorrect: This describes evidence-based reasoning, not a fallacy.
d is incorrect: While events may be part of a series, post hoc specifically confuses sequence with causation.

Tips and Tricks:

Translate the Latin: Post hoc = “after this.” If the answer highlights temporal sequence as causation, it’s correct.
Real-world example: Superstitions (e.g., “I broke a mirror, then got into an accident; the mirror caused the accident”).

Question iv:

Statement:
Which fallacy involves attacking a distorted version of an opponent’s claim?

Options:
a) Confirmation Bias
b) Hasty Generalizations
c) The Straw Man Fallacy
d) Redefinition

Correct Answer:
c

Explanation:
The straw man fallacy misrepresents an opponent’s argument to make it easier to attack (e.g., oversimplifying “Reduce carbon emissions” to “You want to destroy the economy!”).

a is incorrect: Confirmation bias is seeking evidence to support preexisting beliefs, not misrepresenting arguments.
b is incorrect: Hasty generalization involves flawed sampling, not distortion of claims.
d is incorrect: Redefinition changes word meanings but isn’t a formal fallacy.

Tips and Tricks:

Visualize the name: A “straw man” is a weak dummy substitute for the real argument.
Identify distortion: If an option describes twisting or exaggerating an opponent’s view, it’s likely straw man.

Summary of Correct Answers:

Question	Correct Answer
i	b
ii	b
iii	a
iv	c

Key Study Tip: Memorize fallacies using real-life examples. For instance:

Cognitive bias → Believing a news article aligns with your views without fact-checking (confirmation bias).
Hasty generalization → “My neighbor’s dog bit me; all dogs are aggressive!”
Post hoc → “It rained after I washed my car; my car wash caused the rain.”
Straw man → “You support renewable energy? So you’d leave us all in the dark!”

Question iv:

Statement:
Which fallacy involves attacking a distorted version of an opponent’s claim?

Options:
a) Confirmation Bias
b) Hasty Generalizations
c) The Straw Man Fallacy
d) Redefinition

Correct Answer:
c) The Straw Man Fallacy

Explanation:
The Straw Man Fallacy occurs when someone misrepresents or oversimplifies an opponent’s argument to make it easier to attack. For example:

Original argument: “We should reduce fossil fuel use.”
Straw man distortion: “You want to ban all cars and destroy the economy!”
Why others are incorrect:
a) Confirmation bias involves favoring information that confirms preexisting beliefs.
b) Hasty generalizations draw conclusions from insufficient evidence.
d) Redefinition changes word meanings but isn’t a formal fallacy.

Tips and Tricks:

Visualize a “straw man” dummy: Attacking a weak, fake version of the real argument.
Key phrase: “distorted version” = immediate link to Straw Man.

Question v:

Statement:
What does the principle of Occam’s Razor advocate?

Options:
a) Choosing the most complex explanation for an event
b) Choosing the simplest explanation for an event
c) Choosing an explanation based on tradition
d) Choosing an explanation based on authority

Correct Answer:
b) Choosing the simplest explanation for an event

Explanation:
Occam’s Razor states that when multiple explanations exist, the one with the fewest assumptions is most likely correct. Simplicity minimizes unnecessary complexity.
Why others are incorrect:

a) Complexity often introduces unwarranted assumptions.
c) Tradition may perpetuate errors (e.g., appeal to tradition fallacy).
d) Authority doesn’t guarantee truth (e.g., appeal to authority fallacy).

Tips and Tricks:

Remember: “The simplest solution is usually the best.”
Example: Preferring gravity over invisible angels to explain falling objects.

Question vi:

Statement:
What is an example of the appeal to tradition fallacy?

Options:
a) Believing a theory is valid because an expert supports it
b) Believing a practice is correct because it has always been done that way
c) Believing an argument because it cannot be tested
d) Believing a hypothesis because it is the simplest explanation

Correct Answer:
b) Believing a practice is correct because it has always been done that way

Explanation:
Appeal to tradition assumes longevity implies validity (e.g., “Bloodletting works because doctors did it for centuries”).
Why others are incorrect:

a) Describes appeal to authority.
c) Relates to untestable claims, not tradition.
d) Describes Occam’s Razor.

Tips and Tricks:

Spot phrases like “we’ve always done this” or “it’s traditional.”
Tradition ≠ correctness (e.g., historical use of lead in makeup).

Question vii:

Statement:
Which of the following is a pro of chemical substances?

Options:
a) They always have a negative impact on health
b) They reduce carbon footprints through clean energy technologies
c) They are always safe for workers
d) They never cause environmental pollution

Correct Answer:
b) They reduce carbon footprints through clean energy technologies

Explanation:
Chemicals enable green technologies (e.g., lithium-ion batteries for electric cars, catalytic converters).
Why others are incorrect:

a) and d) use absolutes (“always,” “never”) – chemicals have nuanced impacts.
c) Safety depends on handling/regulations, not inherent properties.

Tips and Tricks:

Reject absolute terms; focus on context-specific benefits.
Link “clean energy” to chemical innovation (solar panels, biofuels).

Question viii:

Statement:
What is a con of weak or outdated regulations regarding chemicals?

Options:
a) They protect public health effectively
b) They foster innovation in chemical processes
c) They fail to address current risks
d) They reduce compliance costs for companies

Correct Answer:
c) They fail to address current risks

Explanation:
Outdated regulations ignore modern threats (e.g., nanoplastics, PFAS “forever chemicals”) due to evolving science.
Why others are incorrect:

a) Weak regulations endanger health.
b) Innovation requires safety frameworks to avoid disasters (e.g., Bhopal gas tragedy).
d) Reduced costs for companies ≠ public benefit.

Tips and Tricks:

“Fail to address” signals unmanaged dangers.
Historical cases: Asbestos regulations lagged decades behind evidence.

Question ix:

Statement:
What role do premises play in a scientific argument?

Options:
a) They oppose the main claim
b) They provide the foundation for the claim
c) They represent the conclusion
d) They are unsupported assumptions

Correct Answer:
b) They provide the foundation for the claim

Explanation:
Premises are evidence-based statements supporting a conclusion (e.g., “CO₂ traps heat [premise 1]; human activity emits CO₂ [premise 2] → thus, humans cause warming [claim]”).
Why others are incorrect:

a) Counterclaims oppose; premises support.
c) Conclusions follow from premises.
d) Premises should be empirically validated.

Tips and Tricks:

Premises = “building blocks” of arguments.
Test: “Does this statement back up the main idea?”

Question x:

Statement:
In the context of written articles, what would be a counterclaim for promoting electric vehicles?

Options:
a) Electric cars produce no emissions
b) Urban areas suffer from high air pollution
c) Electric cars still have high environmental costs due to battery production

Correct Answer:
c) Electric cars still have high environmental costs due to battery production

Explanation:
A counterclaim challenges the main argument. Promoting EVs often highlights zero tailpipe emissions (a), but battery production involves mining, energy use, and waste (c).
Why others are incorrect:

a) Supports EV promotion.
b) Justifies EVs as a solution to pollution.

Tips and Tricks:

Counterclaims introduce drawbacks/limitations.
Key conflict: “Environmental costs” vs. perceived sustainability.

Answer Summary:

Q	Correct	Key Concept
iv	c	Straw Man Fallacy
v	b	Occam’s Razor
vi	b	Appeal to Tradition
vii	b	Pros of Chemicals
viii	c	Regulation Cons
ix	b	Premises in Arguments
x	c	Counterclaims

Free Radical Mechanism of Halogenation of Alkyl Halides Tips and Tricks

December 4, 2024 by samreen

free radical mechanism of halogenation of alkyl halides, a vital concept in organic chemistry. This post simplifies the three-step mechanism: initiation, propagation, and termination. Learn how different halogens react, their selectivity patterns, and tips for controlling reaction outcomes. Whether you’re a student or an enthusiast, these tips and tricks will help you understand and apply this mechanism effectively.

Main Reaction

General Reaction:
R−H+X₂→ R-X + HX (heat or UV)

Where:

R−H: Alkane
X₂: Halogen (e.g., Cl₂, Br₂)
R−X: Alkyl halide

Example: Chlorination of methane:
CH₄+Cl₂→ CH₃Cl + HCl

Mechanism: Free Radical Halogenation

This reaction proceeds through a free radical chain mechanism in three key steps:

1. Initiation

Halogen molecule absorbs energy (heat/UV), splitting into two halogen free radicals.
Cl₂→Cl^. + Cl^.
The bond cleavage is homolytic, producing highly reactive radicals.

2. Propagation

Step 1: A halogen radical abstracts a hydrogen atom from the alkane, forming an alkyl radical.
CH₄+Cl^⋅→CH₃^⋅+HCl
Step 2: The alkyl radical reacts with another halogen molecule, producing an alkyl halide and regenerating the halogen radical.
CH₃^⋅+Cl₂→CH₃Cl+Cl^⋅

3. Termination

Two radicals combine to terminate the chain reaction.
Cl^⋅+Cl^⋅→Cl2
CH₃^⋅+Cl^⋅→CH₃Cl
CH₃^⋅+CH₃^⋅→C₂H₆

Tips and Tricks for the Mechanism

Radical Stability Order:
Tertiary > Secondary > Primary > Methyl radicals.
- More substituted alkyl radicals are more stable, favoring halogenation at tertiary carbons.
Reactivity of Halogens:
- Cl₂: Reacts vigorously, less selective, forms multiple products.
- Br₂: Slower and more selective, prefers substitution at more stable radicals.
Controlling Products:
- Use excess alkane to minimize polyhalogenation.
- Use excess halogen for multiple substitutions.
Temperature Effect:
- Higher temperatures favor radical formation and increase reaction rate.
UV Light:
- Necessary for initiating the reaction by breaking the halogen bond.
Selectivity Rule:
- Bromination is more selective than chlorination due to differences in activation energy for hydrogen abstraction.

Detailed Insights on Selectivity in Halogenation

The selectivity of halogenation depends on the halogen used and the type of hydrogen being replaced. Here’s a detailed breakdown:

1. Reactivity and Selectivity of Halogens

Chlorination (Cl₂):
- Reactivity: High, leading to faster reactions.
- Selectivity: Poor; less discrimination between types of hydrogens (e.g., primary vs. secondary vs. tertiary).
- Tends to produce multiple products unless controlled conditions are applied.
Bromination (Br₂):
- Reactivity: Moderate, slower than chlorination.
- Selectivity: High; bromine prefers hydrogen attached to the most stable carbon radical (tertiary > secondary > primary).
Iodination (I₂):
- Generally not favored due to low reactivity and a highly endothermic nature.
- Needs special conditions, like oxidizing agents, to proceed.
Fluorination (F₂):
- Extremely reactive and often explosive. Rarely used in laboratory settings.

2. Factors Influencing Selectivity

Radical Stability

Tertiary radicals are most stable due to hyperconjugation and inductive effects, making them preferred sites for halogenation in bromination.

Energy Considerations

Chlorination:
- Lower activation energy differences among primary, secondary, and tertiary hydrogens.
- Produces a mix of products.
Bromination:
- Higher activation energy differences, favoring reactions at tertiary carbons due to stability.

4. Controlling Polyhalogenation

Polyhalogenation (multiple substitutions) can be controlled by:

Using an excess of alkane relative to the halogen.
Limiting the amount of halogen added.
Carefully controlling reaction conditions like temperature and UV exposure.

5. Selectivity Ratio

For bromination, the relative reactivity of hydrogens is much higher for tertiary hydrogens:

Tertiary:Secondary:Primary Reactivity Ratio (approx.):
- Chlorination: 5:4:1
- Bromination: 1700:80:1

Key Takeaways

Chlorination is suitable for fast, broad reactions but often yields multiple products.
Bromination is ideal when targeting specific products due to its high selectivity.
Reaction conditions and halogen choice dictate product formation and yield.

Environmental Chemistry

November 30, 2024October 9, 2024 by samreen

Enhance your understanding of Environmental Chemistry with comprehensive exercises and notes. Cover key topics like pollution, green chemistry, and sustainable practices to excel in exams and deepen your knowledge.

Q. 4 Discuss in detail the components of the environment.
The environment is composed of several key components:

Atmosphere: The layer of gases surrounding Earth, including oxygen, nitrogen, carbon dioxide, and others. It protects life on Earth by blocking harmful solar radiation.
Lithosphere: The solid outer part of the Earth, comprising rocks, soil, and minerals. It supports terrestrial ecosystems and human infrastructure.
Hydrosphere: All the water on Earth’s surface, including oceans, rivers, lakes, and groundwater. It is crucial for life, as it supports aquatic ecosystems and is essential for human consumption.
Biosphere: The global sum of all ecosystems, where living organisms interact with the other components of the environment. This includes all plants, animals, and microorganisms.

Q. 5 Describe the natural and human sources of carbon monoxide, nitrogen oxides, and sulphur oxides.

Carbon monoxide (CO):
Natural sources: Wildfires, volcanic eruptions, and the oxidation of methane in the atmosphere.
Human sources: Incomplete combustion of fossil fuels from vehicles, industrial processes, and residential heating.
Nitrogen oxides (NOx):
Natural sources: Lightning strikes, microbial activity in soils, and wildfires.
Human sources: Burning of fossil fuels in vehicles, power plants, and industrial activities.
Sulphur oxides (SOx):
Natural sources: Volcanic eruptions and the decomposition of organic matter.
Human sources: Burning of coal and oil, particularly in power plants and industrial processes like refining.

Q. 6 What is acid rain and how does it affect our environment?
Acid rain is rainfall that contains elevated levels of hydrogen ions (low pH) due to the presence of sulphuric and nitric acids, formed from sulfur dioxide (SO₂) and nitrogen oxides (NOx) emissions.
Effects on the environment:

Aquatic life: Acid rain lowers the pH of water bodies, harming fish and other aquatic organisms.
Soil: It leaches essential nutrients like calcium and magnesium from the soil, reducing soil fertility.
Vegetation: It damages the leaves of plants, stunts growth, and weakens trees, making them more vulnerable to diseases.
Built structures: Acid rain corrodes buildings, statues, and infrastructure, especially those made of limestone and marble.

Q. 7 What is smog? Explain the pollutants which are the main cause of photochemical smog.
Smog is a type of air pollution caused by the interaction of sunlight with pollutants such as nitrogen oxides (NOx) and volatile organic compounds (VOCs).

Photochemical smog: Formed when sunlight reacts with nitrogen oxides and hydrocarbons from vehicle emissions, creating harmful ozone (O₃) at ground level.
Main pollutants:
Nitrogen oxides (NOx): Emitted by vehicles and industrial processes.
Volatile organic compounds (VOCs): Released from vehicle exhaust, industrial emissions, and solvents.
Ozone (O₃): A secondary pollutant formed through the reaction of NOx and VOCs in sunlight, contributing to respiratory issues and environmental damage.

Q. 8 Why is the ozone layer depleting? What will happen when the concentration of ozone will be decreased?
The ozone layer is depleting due to the release of chlorofluorocarbons (CFCs) and other ozone-depleting substances (ODS), which break down ozone molecules (O₃) in the stratosphere.

Consequences of ozone depletion:
Increased exposure to harmful ultraviolet (UV) radiation, leading to higher rates of skin cancer, cataracts, and immune system suppression in humans.
Disruption of marine ecosystems, particularly affecting plankton, which form the base of the aquatic food chain.
Harm to plant life, which could reduce agricultural productivity.

Q. 9 How is oil spillage affecting the marine life?
Oil spills cause significant harm to marine ecosystems:

Coating of animals: Oil coats the fur and feathers of marine animals such as birds and otters, reducing their insulation and buoyancy, leading to hypothermia or drowning.
Toxicity: The chemicals in oil are toxic to marine organisms, damaging their organs and reproductive systems.
Contamination of food chains: Oil contaminants enter the food chain, affecting not only marine organisms but also humans who consume seafood.

Q. 10 How detergents are a threat to aquatic animal life?
Detergents contain surfactants, phosphates, and other chemicals that can harm aquatic life:

Surfactants: Disrupt the membranes of aquatic organisms, leading to cell damage and death.
Phosphates: Promote eutrophication, leading to algal blooms that deplete oxygen in water, creating dead zones where aquatic life cannot survive.
Bioaccumulation: Some detergent chemicals accumulate in aquatic organisms, affecting their health and the health of those higher up the food chain.

Q. 12 Explain how pesticides are dangerous to human beings.
Pesticides are hazardous to humans through:

Acute toxicity: Exposure to high doses can cause immediate health issues such as nausea, dizziness, and respiratory problems.
Chronic exposure: Long-term exposure can lead to serious health effects, including cancer, neurological disorders, reproductive issues, and developmental problems in children.
Bioaccumulation: Pesticides can accumulate in the food chain, leading to higher concentrations in humans through consumption of contaminated food and water.

Q. 13 Discuss industrial waste effluents.
Industrial waste effluents are liquid waste products discharged by industries. They may contain:

Toxic chemicals: Such as heavy metals (lead, mercury), volatile organic compounds (VOCs), and acids, which can contaminate water sources and harm both human and environmental health.
Nutrients: Excess nitrogen and phosphorus from agricultural and food industries can cause eutrophication in water bodies.
Thermal pollution: Waste heat discharged into water bodies can disrupt ecosystems by altering temperature conditions and reducing oxygen levels.

Q. 14 How water is purified i.e., made potable. Discuss in detail.
Water purification involves several steps to remove contaminants and make it safe for drinking:

Coagulation and flocculation: Chemicals (coagulants) are added to water to bind small particles into larger clumps (flocs).
Sedimentation: The flocs settle at the bottom of a sedimentation tank.
Filtration: The water passes through sand, gravel, or charcoal filters to remove smaller particles.
Disinfection: Chlorine, ozone, or ultraviolet light is used to kill harmful bacteria, viruses, and parasites.
Additional treatment: Fluoridation and pH adjustments may be done before distribution to consumers.

Q. 15 What are leachates?
Leachates are liquids that have percolated through solid waste (in landfills or contaminated soil) and extracted dissolved or suspended materials. Leachates often contain harmful chemicals and pollutants, which can contaminate soil and groundwater if not properly managed.

Q. 16 Explain the process of incineration of industrial waste.
Incineration is a waste treatment process that involves burning industrial waste at high temperatures (800°C to 1000°C). The waste is converted into ash, flue gases, and heat.
Steps involved in incineration:

Waste preparation: Sorting and pre-treating waste materials to ensure safe combustion.
Combustion: The waste is burned in a controlled environment, reducing the volume of solid waste by up to 90%.
Energy recovery: The heat generated during incineration can be used to produce electricity or steam.
Emission control: Pollutants in flue gases are treated using filters and scrubbers to minimize environmental harm.

Common Chemical Industries Solved Exercise PTB

December 1, 2024October 9, 2024 by samreen

Prepare for second-year exams with solved exercises on Common Chemical Industries, based on the PTB curriculum. Cover essential topics such as industrial processes, applications, and key concepts to boost your understanding.

Q. 5 (a) What are fertilizers? Why are they needed?
Fertilizers are chemical or natural substances added to soil to supply essential nutrients to plants and promote growth. They are needed to replenish the soil with nutrients that are depleted due to continuous crop cultivation, ensuring high yields and healthy plant growth.

(b) Discuss the classification of fertilizers and their uses.
Fertilizers are classified into two main types:

Organic fertilizers: These come from natural sources such as manure, compost, and bone meal. They improve soil structure and water retention while providing nutrients.
Inorganic (chemical) fertilizers: These are synthesized to provide specific nutrients like nitrogen, phosphorus, and potassium (NPK).
Nitrogen fertilizers (e.g., urea): Promote leaf and stem growth.
Phosphorus fertilizers (e.g., superphosphate): Enhance root development and flowering.
Potassium fertilizers (e.g., potassium chloride): Improve overall plant health and disease resistance.

(c) How is urea manufactured in Pakistan? Describe in detail the process used.
Urea is synthesized from ammonia (NH₃) and carbon dioxide (CO₂) using the Bosch-Meiser process:

Ammonia synthesis: Nitrogen is obtained from air, and hydrogen is derived from natural gas. These two gases react at high pressure and temperature in the presence of a catalyst to form ammonia.
N₂ + 3H₂ → 2NH₃
Urea synthesis: Ammonia reacts with carbon dioxide under high pressure to form ammonium carbamate, which is then dehydrated to form urea.
2NH₃ + CO₂ → NH₂CONH₂ + H₂O
The resulting urea is then prilled or granulated for use as fertilizer.

Q. 6 (a) What are the prospects of fertilizer industry in Pakistan?
The fertilizer industry in Pakistan has significant growth potential due to the country’s agrarian economy. Increasing demand for agricultural productivity, government subsidies, and rising export opportunities can further boost the industry. However, challenges include the high cost of raw materials and energy shortages.

(b) What are essential nutrient elements and why are these needed for plant growth?
Essential nutrient elements are chemical elements required for plants to grow and complete their life cycle. They are divided into macronutrients and micronutrients:

Macronutrients: Nitrogen (N), Phosphorus (P), Potassium (K), Calcium (Ca), Magnesium (Mg), and Sulfur (S).
Micronutrients: Iron (Fe), Zinc (Zn), Manganese (Mn), Boron (B), Copper (Cu), Molybdenum (Mo), and Chlorine (Cl).
These nutrients play vital roles in photosynthesis, respiration, enzyme activation, and overall plant health.

Contain the right proportion of essential nutrients.
Be easily absorbed by plants.
Be cost-effective and easy to apply.
Have minimal environmental impact (low toxicity).
Improve soil fertility over time without causing nutrient imbalance.

Q. 7 (a) Describe the composition of a good Portland cement.
Portland cement is composed of the following main ingredients:

Calcium oxide (CaO): 60-65%
Silica (SiO₂): 20-25%
Alumina (Al₂O₃): 5-10%
Iron oxide (Fe₂O₃): 2-4%
Magnesia (MgO): 1-3%
Sulfur trioxide (SO₃): 1-2%
Additionally, small amounts of gypsum are added to control the setting time.

(b) Discuss the wet process for the manufacturing of cement with the help of flow sheet diagram.
In the wet process, the raw materials (limestone and clay) are ground with water to form a slurry. The steps include:

Raw material preparation: Limestone and clay are crushed and mixed with water to form slurry.
Mixing: The slurry is blended to achieve uniform composition.
Heating in a kiln: The slurry is fed into a rotary kiln and heated to high temperatures (up to 1400-1500°C), resulting in clinker formation.
Clinker cooling and grinding: The clinker is cooled, mixed with gypsum, and ground to produce cement.

(c) What do you understand by the term “setting of cement”? Also discuss the reactions taking place in the first 24 hours.
The “setting of cement” refers to the process by which the cement paste transforms from a fluid to a solid state upon hydration.

Initial setting: The hydration of tricalcium silicate (C₃S) and dicalcium silicate (C₂S) produces calcium silicate hydrate (C-S-H) and calcium hydroxide (Ca(OH)₂), resulting in the hardening of the cement.
Within 24 hours: The initial set occurs as C₃S reacts rapidly, contributing to early strength development, while C₂S reacts slowly, providing long-term strength.

Q. 8 What are the essential non-woody raw materials used in the production of pulp and paper in Pakistan?
The essential non-woody raw materials used in pulp and paper production in Pakistan include:

Wheat straw
Bagasse (sugarcane waste)
Rice straw
Kenaf (a fibrous plant)
These materials are chosen for their availability and lower environmental impact compared to wood-based raw materials.

Q. 9 (a) What are the principal methods of chemical pulping used for the production of paper?
The principal methods of chemical pulping are:

Kraft process (sulfate process): Wood chips are cooked in sodium hydroxide (NaOH) and sodium sulfide (Na₂S) to break down lignin and release cellulose fibers.
Sulfite process: Uses sulfurous acid (H₂SO₃) and bisulfites to dissolve lignin.
Soda process: Uses sodium hydroxide (NaOH) alone to separate fibers.

(b) Describe the neutral sulphite semi-chemical (NSSC) process for the manufacturing of pulp and paper.
In the NSSC process, wood chips are treated with sodium sulfite (Na₂SO₃) and sodium carbonate (Na₂CO₃) to partially break down lignin, followed by mechanical refining. This process produces a pulp with a balance between strength and quality, suitable for making corrugated boards and packaging material.

Q. 10 (a) What are the common bleaching agents used in the paper industry in Pakistan? Briefly describe the bleaching process.
Common bleaching agents used in the paper industry include:

Chlorine dioxide (ClO₂)
Hydrogen peroxide (H₂O₂)
Oxygen (O₂)
Ozone (O₃)

Bleaching process:
The bleaching process removes the residual lignin and brightens the pulp. It typically involves multiple stages:

Lignin removal: Chlorine dioxide or oxygen is used to break down lignin.
Brightening: Hydrogen peroxide or ozone is used to further whiten the pulp by oxidizing remaining chromophores.

(b) What are the prospects of the paper industry in Pakistan?
The paper industry in Pakistan has significant potential for growth due to the increasing demand for packaging and paper products. However, the industry faces challenges such as dependency on imported raw materials, lack of advanced technology, and environmental concerns related to deforestation. Development of non-wood raw materials and modernizing production processes can improve the industry’s prospects.

Aldehydes and Ketones Exercise PTB

November 30, 2024October 5, 2024 by samreen

Prepare for PTB exams with solved exercises on Aldehydes and Ketones. Focus on key topics like structure, properties, reactions, and applications to deepen your understanding and improve exam performance.

Q. 4: Give one laboratory and one industrial method for the preparation of formaldehyde.

Laboratory method:
Formaldehyde can be prepared in the lab by the oxidation of methanol. This is done by passing methanol vapor mixed with air over a heated copper or silver catalyst at around 400°C. The reaction is as follows:

[ \text{CH}_3\text{OH} + \frac{1}{2}\text{O}_2 \xrightarrow{Cu/Ag} \text{HCHO} + \text{H}_2\text{O} ]

Industrial method:
On an industrial scale, formaldehyde is produced by the catalytic oxidation of methanol. Methanol vapor is passed over a catalyst of iron oxide-molybdenum oxide (Fe-Mo) or silver at 300-400°C:

[ \text{CH}_3\text{OH} + \frac{1}{2}\text{O}_2 \rightarrow \text{HCHO} + \text{H}_2\text{O} ]

This method is highly efficient and commonly used in formaldehyde production.

Q. 5: How does formaldehyde react with the following reagents?

(i) CH₃MgI (Grignard reagent):
Formaldehyde reacts with methyl magnesium iodide to give a primary alcohol upon hydrolysis.

[ \text{HCHO} + \text{CH}_3\text{MgI} \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{OH} ]

(ii) HCN:
Formaldehyde reacts with hydrogen cyanide (HCN) to form cyanohydrin.

[ \text{HCHO} + \text{HCN} \rightarrow \text{HOCH}_2\text{CN} ]

(iii) NaHSO₃ (Sodium bisulfite):
Formaldehyde reacts with sodium bisulfite to form a bisulfite addition compound.

[ \text{HCHO} + \text{NaHSO}_3 \rightarrow \text{HOCH}_2\text{SO}_3\text{Na} ]

(iv) Conc. NaOH (Cannizzaro Reaction):
In the presence of concentrated NaOH, formaldehyde undergoes the Cannizzaro reaction, where one molecule is reduced to methanol and another is oxidized to formate.

[ 2 \text{HCHO} + \text{NaOH} \rightarrow \text{HCOONa} + \text{CH}_3\text{OH} ]

(v) NaBH₄/H₂O (Sodium borohydride):
Formaldehyde is reduced to methanol when treated with sodium borohydride.

[ \text{HCHO} + \text{NaBH}_4 \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{OH} ]

(vi) Tollens’ reagent:
Formaldehyde is oxidized by Tollens’ reagent (ammoniacal silver nitrate solution) to form formic acid.

[ \text{HCHO} + 2 \text{Ag(NH}_3\text{)}_2\text{OH} \rightarrow \text{HCOOH} + 2 \text{Ag} + \text{H}_2\text{O} + 2 \text{NH}_3 ]

(vii) Fehling’s reagent:
Formaldehyde reduces Fehling’s reagent, precipitating copper(I) oxide (Cu₂O).

[ \text{HCHO} + 2 \text{Cu}^{2+} + 5 \text{OH}^- \rightarrow \text{HCOO}^- + \text{Cu}_2\text{O} + 3 \text{H}_2\text{O} ]

Q. 6: Give one laboratory and one industrial method for the preparation of acetaldehyde.

Laboratory method:
Acetaldehyde can be prepared in the lab by the oxidation of ethanol using potassium dichromate and dilute sulfuric acid.

[ \text{C}_2\text{H}_5\text{OH} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{CHO} + \text{Cr}_2\text{(SO}_4\text{)}_3 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O} ]

Industrial method:
Acetaldehyde is produced industrially by the Wacker process, which involves the oxidation of ethylene using a palladium(II) chloride catalyst.

[ \text{C}_2\text{H}_4 + \frac{1}{2}\text{O}_2 \xrightarrow{PdCl_2, CuCl_2} \text{CH}_3\text{CHO} ]

Q. 7: How does acetaldehyde react with the following reagents?

(i) C₂H₅MgI (Ethyl magnesium iodide):
Acetaldehyde reacts with ethyl magnesium iodide to give a secondary alcohol after hydrolysis.

[ \text{CH}_3\text{CHO} + \text{C}_2\text{H}_5\text{MgI} \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} ]

(ii) HCN:
Acetaldehyde reacts with HCN to form α-hydroxypropionitrile (cyanohydrin).

[ \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN} ]

(iii) NaHSO₃:
Acetaldehyde reacts with sodium bisulfite to form a bisulfite addition compound.

[ \text{CH}_3\text{CHO} + \text{NaHSO}_3 \rightarrow \text{CH}_3\text{CH(OH)SO}_3\text{Na} ]

(iv) Dilute NaOH (Aldol condensation):
In dilute NaOH, acetaldehyde undergoes aldol condensation, forming 3-hydroxybutanal.

[ 2 \text{CH}_3\text{CHO} \xrightarrow{\text{NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} ]

(v) LiAlH₄:
Acetaldehyde is reduced to ethanol by lithium aluminum hydride.

[ \text{CH}_3\text{CHO} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} ]

(vi) NaBH₄/H₂O:
Acetaldehyde is reduced to ethanol by sodium borohydride.

[ \text{CH}_3\text{CHO} + \text{NaBH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} ]

(vii) NH₂OH (Hydroxylamine):
Acetaldehyde reacts with hydroxylamine to form acetaldehyde oxime.

[ \text{CH}_3\text{CHO} + \text{NH}_2\text{OH} \rightarrow \text{CH}_3\text{CH(OH)NOH} ]

(viii) K₂Cr₂O₇/H₂SO₄:
Acetaldehyde is oxidized by potassium dichromate to form acetic acid.

[ \text{CH}_3\text{CHO} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{COOH} + \text{Cr}_2\text{(SO}_4\text{)}_3 ]

Q. 8: Describe briefly the mechanism of nucleophilic addition to a carbonyl compound.

Nucleophilic addition to a carbonyl compound typically involves the following steps:

Attack of the nucleophile: The nucleophile attacks the electrophilic carbon of the carbonyl group (C=O), breaking the π-bond and pushing electrons onto oxygen. [ \text{Nu}^- + \text{R}_2\text{C=O} \rightarrow \text{R}_2\text{C(OH)Nu} ]
Formation of a tetrahedral intermediate: The nucleophile forms a bond with the carbon, and the oxygen temporarily carries a negative charge as part of the intermediate.
Protonation: In the final step, a proton is typically transferred to the oxygen to stabilize the compound.

The reaction proceeds faster if the carbonyl group is more electrophilic

(i.e., aldehydes are more reactive than ketones).

Q. 9: Explain with mechanism the addition of ethylmagnesium bromide to acetaldehyde. What is the importance of this reaction?

Mechanism:
Ethylmagnesium bromide (Grignard reagent) reacts with acetaldehyde as follows:

The ethyl group (acting as a nucleophile) from ethylmagnesium bromide attacks the carbonyl carbon in acetaldehyde. [ \text{CH}_3\text{CHO} + \text{C}_2\text{H}_5\text{MgBr} \rightarrow \text{CH}_3\text{CH(C}_2\text{H}_5)\text{OMgBr} ]
The resulting alkoxide ion is then hydrolyzed with water to give butanol. [ \text{CH}_3\text{CH(C}_2\text{H}_5)\text{OMgBr} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} ]

Importance:
This reaction is a key step in organic synthesis for forming new carbon-carbon bonds, allowing the formation of alcohols from aldehydes.

Q. 10: Explain with mechanism the addition of sodium bisulphite to acetone. What is the utility of this reaction?

Mechanism:
Sodium bisulfite reacts with acetone by nucleophilic attack on the carbonyl carbon.

The sulfur atom in bisulfite (HSO₃⁻) acts as the nucleophile and attacks the carbonyl carbon. [ \text{CH}_3\text{COCH}_3 + \text{HSO}_3^- \rightarrow \text{CH}_3\text{C(OH)(SO}_3\text{H)}\text{CH}_3} ]
A bisulfite addition product (also called a hydrosulfite compound) is formed.

Utility:
The formation of bisulfite addition compounds is used for the purification of aldehydes and ketones, as they can easily be crystallized and later regenerated by acid treatment.

Q. 11: Describe with mechanism aldol condensation. Why formaldehyde does not give this reaction?

Aldol condensation mechanism:

Formation of enolate ion: Under basic conditions, an aldehyde like acetaldehyde forms an enolate ion. [ \text{CH}_3\text{CHO} \xrightarrow{\text{OH}^-} \text{CH}_2\text{CHO}^- ]
Nucleophilic attack: The enolate ion attacks the carbonyl carbon of another acetaldehyde molecule, forming 3-hydroxybutanal (aldol). [ \text{CH}_3\text{CH=O} + \text{CH}_2\text{CHO}^- \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CHO} ]
Dehydration: Under higher temperatures, the aldol product can lose water to form an α,β-unsaturated aldehyde. [ \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \rightarrow \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O} ]

Why formaldehyde does not undergo aldol condensation:
Formaldehyde lacks an α-hydrogen, which is necessary to form the enolate ion in the first step of the aldol condensation. Hence, it cannot participate in the reaction.

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Q. 12: What types of aldehydes give Cannizzaro’s reaction? Give its mechanism.

Aldehydes without α-hydrogen atoms undergo Cannizzaro’s reaction. Examples include formaldehyde and benzaldehyde.

Mechanism:

Nucleophilic attack: In the presence of a strong base (like NaOH), a hydroxide ion attacks the carbonyl carbon of one aldehyde molecule, forming an alkoxide ion. [ \text{RCHO} + \text{OH}^- \rightarrow \text{RCH(OH)}^- ]
Hydride transfer: The alkoxide ion then transfers a hydride ion to another aldehyde molecule, reducing it to an alcohol and oxidizing the original molecule to a carboxylate ion. [ \text{RCH(OH)}^- + \text{RCHO} \rightarrow \text{RCH}_2\text{OH} + \text{RCOO}^- ]
Final products: The products are a primary alcohol and a carboxylate ion. [ 2 \text{RCHO} \rightarrow \text{RCH}_2\text{OH} + \text{RCOO}^- ]

Q. 13: Explain the mechanism of the reaction of phenylhydrazine with acetone.

Mechanism:

Formation of hydrazone: Phenylhydrazine (C₆H₅NHNH₂) reacts with acetone, a carbonyl compound, in a condensation reaction to form a hydrazone. The lone pair on nitrogen of phenylhydrazine attacks the carbonyl carbon of acetone. [ \text{CH}_3\text{COCH}_3 + \text{C}_6\text{H}_5\text{NHNH}_2 \rightarrow \text{CH}_3\text{C=NNHC}_6\text{H}_5 + \text{H}_2\text{O} ]
Removal of water: Water is eliminated, and a double bond is formed between the carbon and nitrogen.

The product is acetone phenylhydrazone.

Q. 14: Using ethyne as a starting material, how would you get acetaldehyde, acetone, and ethyl alcohol?

Acetaldehyde:

Ethyne (acetylene) undergoes hydration (Kucherov’s reaction) in the presence of mercury(II) sulfate and sulfuric acid, forming acetaldehyde. [ \text{CH≡CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}, \text{H}_2\text{SO}_4} \text{CH}_3\text{CHO} ]

Acetone:

Convert acetaldehyde to acetone by the oxidation of isopropanol, which can be formed from ethylene through Markovnikov hydration. [ \text{CH}_2\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CHOHCH}_3 ]

Ethyl alcohol:

Hydrate ethyne using sulfuric acid to form acetaldehyde, then reduce acetaldehyde using a reducing agent such as sodium borohydride (NaBH₄) to form ethyl alcohol. [ \text{CH}_3\text{CHO} + \text{NaBH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} ]

Q. 15: Give the mechanism of addition of HCN to acetone.

Mechanism:

Nucleophilic attack: The cyanide ion (CN⁻) attacks the electrophilic carbonyl carbon in acetone. [ \text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow \text{CH}_3\text{C(CN)OHCH}_3^- ]
Protonation: The negatively charged oxygen in the intermediate is protonated by water, forming acetone cyanohydrin. [ \text{CH}_3\text{C(CN)OHCH}_3^- + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{C(CN)OHCH}_3 ]

Q. 16: How would you bring about the following conversions?

(i) Acetone into t-butyl alcohol:

React acetone with methylmagnesium bromide (Grignard reagent) followed by hydrolysis to form t-butyl alcohol. [ \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgBr} \rightarrow \text{(CH}_3)_3\text{COH} ]

(ii) Propanal into 1-propanol:

Reduce propanal with sodium borohydride or lithium aluminum hydride to form 1-propanol. [ \text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} ]

(iii) Propanone into 2-propanol:

Reduce propanone (acetone) using sodium borohydride or lithium aluminum hydride to form 2-propanol. [ \text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CHOHCH}_3 ]

(iv) Ethanol into propanone:

Oxidize ethanol to acetaldehyde, followed by a Grignard reaction with methylmagnesium bromide to form isopropyl alcohol. Oxidation of isopropyl alcohol gives propanone. [ \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CHO} \xrightarrow{\text{CH}_3\text{MgBr}} \text{CH}_3\text{CHOHCH}_3 ]

(v) Ethyne into ethanol:

Hydration of ethyne in the presence of sulfuric acid and mercury sulfate produces acetaldehyde, which can be reduced to ethanol. [ \text{CH≡CH} \xrightarrow{\text{Hg}^{2+}} \text{CH}_3\text{CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH}_2\text{OH} ]

(vi) Ethanol into ethene:

Dehydrate ethanol by heating it with concentrated sulfuric acid. [ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_2=CH}_2 ]

(vii) Ethanol into ethanoic acid:

Oxidize ethanol using potassium permanganate or potassium dichromate to form ethanoic acid. [ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}_2\text{SO}_4} \text{CH}_3\text{COOH} ]

(viii) Methanol into ethanol:

Convert methanol to formaldehyde, followed by the reaction with a Grignard reagent (methylmagnesium bromide), and then hydrolysis. [ \text{CH}_3\text{OH} \rightarrow \text{CH}_3\text{CHO} \xrightarrow{\text{CH}_3\text{MgBr}} \text{CH}_3\text{CH}_2\text{OH} ]

Q. 17: How will you distinguish between:

(i) Methanal and Ethanal:
Methanal gives Cannizzaro’s reaction, whereas ethanal gives aldol condensation under basic conditions.

(ii) Ethanol and Propanone:
Ethanol reacts with iodoform reagent (NaOH/I₂) to give a yellow precipitate, whereas propanone does not.

(iii) Ethanol and Propanal:
Ethanol does not react with Tollens’ reagent, whereas propanal gives a silver mirror.

(iv) Acetone and Ethyl Alcohol:
Acetone gives a positive iodoform test, whereas ethyl alcohol does not.

(v) Butanone and 3-Pentanone:
Butanone gives a positive iodoform test, whereas 3-pentanone does not.

(vi) Acetaldehyde and Benzaldehyde:
Acetaldehyde gives a positive Fehling’s test, but benzaldehyde does not.

Q. 18: Discuss the oxidation of:

(a) Aldehydes:

With K₂Cr₂O₇/H₂SO₄: Aldehydes are oxidized to carboxylic acids

With Tollens’ reagent: Aldehydes give a silver mirror.
With Fehling’s solution: Aldehydes reduce Fehling’s solution to a red precipitate of Cu₂O.

(b) Ketones:

Ketones resist oxidation under mild conditions but can be cleaved into carboxylic acids by strong oxidizing agents.

Q. 19: Discuss reduction of:

(a) Aldehydes:
Aldehydes are reduced to primary alcohols by NaBH₄ or LiAlH₄.

(b) Ketones:
Ketones are reduced to secondary alcohols by NaBH₄ or LiAlH₄.

Q. 20: Give three uses for each of formaldehyde and acetaldehyde.

Formaldehyde:

Used in the production of polymers such as bakelite.
Used as a preservative in laboratories.
Used in the manufacture of disinfectants.

Acetaldehyde:

Used as an intermediate in the production of acetic acid.
Used in the manufacture of perfumes.
Used in the synthesis of drugs.

Alcohol, Phenol and Ether Solved Exercise

November 30, 2024October 4, 2024 by samreen

Prepare for exams with solved exercises on Alcohol, Phenol, and Ether. Cover key concepts like structure, properties, reactions, and uses to strengthen your understanding and improve exam performance.

Q.4. What are alcohols? How are they classified? How will you distinguish between primary, secondary, and tertiary alcohols?

Alcohols are organic compounds containing the hydroxyl (-OH) functional group attached to a carbon atom.
Classification:
Primary (1°): The -OH group is attached to a carbon atom that is connected to only one other alkyl group (e.g., ethanol).
Secondary (2°): The -OH group is attached to a carbon atom bonded to two other alkyl groups (e.g., isopropanol).
Tertiary (3°): The -OH group is attached to a carbon atom bonded to three alkyl groups (e.g., tert-butanol).
Distinguishing:
Lucas test: Tertiary alcohol reacts rapidly, secondary reacts slowly, and primary alcohol shows no immediate reaction with Lucas reagent.

Q.5. How is methyl alcohol obtained on a large scale? How can it be distinguished from ethyl alcohol?

Methyl alcohol (methanol) is obtained on a large scale by the catalytic hydrogenation of carbon monoxide (CO) using a catalyst like zinc oxide at high temperature and pressure.
Distinguishing: Methanol is poisonous, and when treated with iodine and sodium hydroxide, methanol forms a yellow precipitate (iodoform test), which does not happen with ethanol.

Q.6. What is fermentation? Which compound may be obtained on an industrial scale by fermentation?

Fermentation is a biochemical process where sugars (glucose) are converted into alcohol and carbon dioxide by the action of enzymes from yeast.
Compound obtained: Ethanol is obtained on an industrial scale by fermentation of sugars.

Q.7. Explain the following terms: Absolute alcohol, Methylated spirit, Rectified spirit, Denaturing of alcohols.

Absolute alcohol: Pure ethanol that is 100% alcohol, free from water.
Methylated spirit: Ethanol mixed with a small quantity of methanol or other chemicals to make it unfit for consumption.
Rectified spirit: A mixture of ethanol and water, typically containing about 95% ethanol.
Denaturing of alcohols: The process of adding substances (like methanol) to ethanol to make it unsuitable for drinking.

Q.8. How does ethyl alcohol react with the following reagents?

i) Conc. H₂SO₄: Dehydration to form ethene at 170°C.
ii) Na: Produces sodium ethoxide and hydrogen gas.
iii) PCl₅: Produces ethyl chloride and phosphorus oxychloride (POCl₃).
iv) CH₃COOH: Esterification to form ethyl acetate in the presence of an acid catalyst.
v) SOCl₂: Produces ethyl chloride and sulfur dioxide (SO₂) and HCl.

Q.9. How will you obtain primary, secondary, and tertiary alcohols by reacting Grignard reagent with suitable carbonyl compounds?

Primary alcohol: React a Grignard reagent (RMgX) with formaldehyde (HCHO) followed by hydrolysis.
Secondary alcohol: React a Grignard reagent with an aldehyde (R-CHO) followed by hydrolysis.
Tertiary alcohol: React a Grignard reagent with a ketone (R-CO-R) followed by hydrolysis.

Q.10. How will you distinguish between the following?

i) An alcohol and a phenol: Phenols react with neutral FeCl₃ to give a violet color, while alcohols do not.
ii) An alcohol and an ether: Alcohols react with sodium metal to release hydrogen, while ethers do not.
iii) Methanol and ethanol: Methanol gives a positive iodoform test, while ethanol does not.
iv) A tertiary alcohol and a primary alcohol: Tertiary alcohols react immediately with Lucas reagent, but primary alcohols do not.
v) 1-propanol and 2-propanol: 2-Propanol gives a positive iodoform test, whereas 1-propanol does not.

Q.11. Give reasons for the followings:

i) Ethyl alcohol is a liquid while methyl chloride is a gas:

Ethyl alcohol has hydrogen bonding due to the -OH group, which leads to higher intermolecular forces, making it a liquid. Methyl chloride, lacking hydrogen bonds, is a gas at room temperature.

ii) Ethanol has a higher boiling point than diethyl ether:

Ethanol has hydrogen bonding, which increases its boiling point, whereas diethyl ether has weak Van der Waals forces leading to a lower boiling point.

iii) Absolute alcohol cannot be prepared by fermentation process:

Fermentation always produces ethanol along with water and some impurities. It is difficult to obtain 100% pure ethanol by fermentation alone.

iv) Ethanol gives different products with conc. H₂SO₄ under different conditions:

At 170°C, ethanol dehydrates to ethene. At 140°C, it forms diethyl ether, depending on the temperature and reaction conditions.

v) Water has a higher boiling point than ethanol:

Water has stronger hydrogen bonding due to its two hydrogen atoms bonded to oxygen, leading to a higher boiling point compared to ethanol.

Q.12. How will you convert:

i) Methanol into ethanol:

Treat methanol with methyl iodide (CH₃I) in the presence of NaOH to obtain ethanol.

ii) Ethanol into methanol:

First oxidize ethanol to acetaldehyde (CH₃CHO) using K₂Cr₂O₇/H₂SO₄, then reduce it to methanol using LiAlH₄.

iii) Ethanol into isopropyl alcohol:

First oxidize ethanol to acetone using K₂Cr₂O₇, then reduce acetone with H₂/Pd to isopropyl alcohol.

iv) Formaldehyde into ethyl alcohol:

Treat formaldehyde with methyl magnesium bromide (CH₃MgBr) followed by hydrolysis to obtain ethanol.

v) Acetone into ethyl alcohol:

First reduce acetone to isopropanol using NaBH₄, then dehydrogenate isopropanol to obtain ethyl alcohol.

Q.13. Explain the following terms using ethyl alcohol as an example:

i) Oxidation:

Ethanol can be oxidized to acetaldehyde (CH₃CHO) using a mild oxidizing agent like PCC. Further oxidation leads to acetic acid.

ii) Dehydration:

When ethanol is heated with concentrated H₂SO₄ at 170°C, it undergoes dehydration to form ethene (CH₂=CH₂).

iii) Esterification:

Ethanol reacts with acetic acid in the presence of concentrated H₂SO₄ to form ethyl acetate (CH₃COOCH₂CH₃) and water.

iv) Ether formation:

Ethanol, when treated with H₂SO₄ at 140°C, forms diethyl ether (CH₃CH₂OCH₂CH₃).

Q.14. Compare the reactions of phenol with those of ethanol. Discuss the difference if any.

Phenol is more acidic than ethanol due to the resonance stabilization of the phenoxide ion. Phenol reacts with NaOH to form sodium phenoxide, while ethanol does not react with NaOH. Phenol undergoes electrophilic substitution reactions (like nitration) more easily due to the activating effect of the -OH group, while ethanol shows typical reactions of alcohols like oxidation and esterification.

Q.15. Arrange the following compounds in order of their increasing acid strength and give reasons:

H₂O, C₂H₅OH, C₆H₅OH, C₆H₅COOH

Order: C₂H₅OH < H₂O < C₆H₅OH < C₆H₅COOH
C₂H₅OH (ethanol) is the least acidic as it lacks resonance or strong electronegative groups.
H₂O (water) is slightly more acidic than ethanol.
C₆H₅OH (phenol) is more acidic due to resonance stabilization of the phenoxide ion.
C₆H₅COOH (benzoic acid) is the most acidic due to the strong electron-withdrawing effect of the carboxyl group.

Q.16. Write down two methods for preparing phenol. What is the action of the following on phenol:

i) HNO₃: Nitration of phenol yields nitrophenol (o- and p-nitrophenol).
ii) NaOH: Phenol reacts with NaOH to form sodium phenoxide.
iii) Zn: Reduction of phenol with Zn forms benzene.
iv) Bromine water: Bromination of phenol gives 2,4,6-tribromophenol.

Q.17. Give the uses of phenols. How bakelite is prepared from it?

Uses of phenols:
Antiseptic (e.g., in disinfectants).
Used in the manufacture of plastics (e.g., Bakelite).
Used in the production of drugs and dyes.
Bakelite preparation:
Phenol reacts with formaldehyde under heat and pressure in the presence of a catalyst (like HCl or ZnCl₂) to form Bakelite, a hard plastic.

Q.18. (a) Write IUPAC names of the following compounds:

i) (CH₃)₃CH – OH: 2-Methylpropan-2-ol
ii) (CH₃)₂CHCH₂CH₂OH: 3-Methylbutan-1-ol
iii) CH₃ – CH – OH | CH₃: Propan-2-ol
iv) C₂H₅ – CH – OH | I: 1-Iodo-2-propanol

Q. 18.(b) Write structure formulas for the following compounds:

Glycol (Ethylene glycol)

Structural formula: HO-CH₂-CH₂-OH

Glycerol

Structural formula: HO-CH₂-CH(OH)-CH₂OH

Carbolic Acid (Phenol)

Structural formula: C₆H₅OH

Acetophenone

Structural formula: C₆H₅COCH₃

Picric Acid

Structural formula: C₆H₂(NO₂)₃OH

Q.19. (a) Name the following compounds:

i) CH₃ — CH₂ — CH₂ — O — CH₃

Name: 1-Methoxypropane

ii) (CH₃)₂CH₂ — O — CH(CH₃)₂

Name: 2-Isopropoxypropane

iii) CH₃ — CH₂ — CH₂ — O — CH₂ — CH₃

Name: 1-Ethoxypropane

iv) C₂H₅ — O — C₆H₅

Name: Ethoxybenzene

v) CH₃ — O — C₆H₅

Name: Methoxybenzene (Anisole)

Q.19. (b) Write down structural formulas of the following compounds:

Methoxyethane

Structural formula: CH₃-O-CH₂CH₃

Ethoxybenzene

Structural formula: C₂H₅-O-C₆H₅

Sodium Ethoxide

Structural formula: CH₃CH₂ONa

Sodium Phenoxide

Structural formula: C₆H₅ONa

Propoxypropane

Structural formula: CH₃CH₂CH₂-O-CH₂CH₂CH₃

Alkyl Halide Solved Exercise PTB

November 30, 2024September 29, 2024 by samreen

Prepare for PTB exams with solved exercises on Alkyl Halides. Focus on key topics like structure, reactions, and applications to enhance your understanding and perform well in exams.

Q.4. Define alkyl halide. Which is the best method of preparing alkyl halides?

Alkyl Halide:
Alkyl halides (also known as haloalkanes) are organic compounds in which one or more halogen atoms (fluorine, chlorine, bromine, or iodine) are covalently bonded to an alkyl group (an aliphatic carbon chain). They have the general formula ( R-X ), where ( R ) represents the alkyl group, and ( X ) represents the halogen atom.

Best Method of Preparing Alkyl Halides:
The best method to prepare alkyl halides is by the free radical halogenation of alkanes. This process involves substituting a hydrogen atom in an alkane with a halogen atom under the influence of heat or light (UV radiation).

General Reaction:
R-H + X2 → (Heat or UV) R-X + H-X

Where ( R-H ) is an alkane, and ( X2 ) is a halogen molecule such as ( Cl2 ) or ( Br2 ). This process produces alkyl halides and a hydrogen halide as a byproduct.

Q.5. Write down a method for the preparation of ethyl magnesium bromide in the laboratory.

Preparation of Ethyl Magnesium Bromide:

Ethyl magnesium bromide is a Grignard reagent, which can be prepared by reacting ethyl bromide with magnesium in the presence of dry ether.

Reaction:
C2H5-Br + Mg → (Dry ether) C2H5MgBr

Procedure:

Place magnesium turnings in a dry flask.
Add dry ether to the flask to keep the reaction environment anhydrous.
Slowly add ethyl bromide (C2H5Br) to the flask while stirring the mixture.
The reaction begins with the formation of ethyl magnesium bromide (C2H5MgBr).
The reaction is exothermic, so it is crucial to maintain the temperature by adding the reactants slowly.

Grignard reagents are highly reactive and must be prepared in an anhydrous environment because they react with water.

Q.6. Give IUPAC names to the following compounds.

i.
Structure: CH3-CH2-CH2-CH2-CH2-Cl
IUPAC Name: 1-Chloropentane

ii.
Structure: CH3CH2CH2CH3
IUPAC Name: Butane

iii.
Structure: CH3-CH(CH3)-CH2-CH2-CH2-CH3
IUPAC Name: 2-Methylhexane

iv.
Structure: CH2(Cl)-CH(CH3)-CH3
IUPAC Name: 1-Chloro-2-methylpropane

v.
Structure: ( \text{CH}_3\text{CBr}_3 )
IUPAC Name: Bromoform

vi.
Structure: ( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} )
IUPAC Name: 1-Chlorobutane

vii.
Structure: ( \text{CH}_3\text{CH}_2\text{CHClCH}_2\text{CH}_2\text{CH}_2\text{Cl} )
IUPAC Name: 1,6-Dichlorohexane

viii.
Structure:
( \text{Br-CH}_2\text{CH}_3 )
IUPAC Name: Ethyl bromide or Bromoethane

ix.
Structure: CH2(Cl)-CH2-CH3
IUPAC Name: 1-Chloropropane

x.
Structure: CH2(Br)-C(CH3)2-CH3
IUPAC Name: 1-Bromo-2,2-dimethylpropane

xi.
Structure: CH2(Cl)-CH(CH3)-CH2-CH3
IUPAC Name: 1-Chloro-2-methylbutane

xii.
Structure: CH3-C(CH3)2-CH2-CH3
IUPAC Name: 2,2-Dimethylbutane

xiii.
Structure: CH3-C(Br)(CH3)-CH3
IUPAC Name: Bromotert-butane or 2-Bromo-2-methylpropane

Q.7. Draw all the possible structures that have the molecular formula ( C4H9Cl ). Classify each as primary, secondary, or tertiary chloride. Give their names according to the IUPAC system.

1-Chlorobutane (Primary chloride):
CH2(Cl)-CH2-CH2-CH3
2-Chlorobutane (Secondary chloride):
CH3-CH(Cl)-CH2-CH3
2-Chloro-2-methylpropane (Tertiary chloride):
CH3-C(Cl)(CH3)-CH3
1-Chloro-2-methylpropane (Primary chloride):
CH2(Cl)-CH(CH3)-CH3

Q.8. Using ethyl bromide as a starting material, how would you prepare the following compounds?

(a) n-Butane:
Ethyl bromide + sodium metal → n-Butane (Wurtz reaction).
Reagent: Na, Condition: Dry ether

(b) Ethyl alcohol:
Ethyl bromide + aqueous KOH → Ethyl alcohol
Reagent: Aqueous KOH, Condition: Reflux

(d) Ethane:
Ethyl bromide + Zn/HCl → Ethane (Reduction reaction)
Reagent: Zn/HCl

(e) Ethene:
Ethyl bromide + alcoholic KOH → Ethene (Dehydrohalogenation)
Reagent: Alcoholic KOH, Condition: Heat

(f) Propanoic acid:
Ethyl bromide + KCN → Ethyl cyanide → Hydrolysis → Propanoic acid
Reagent: KCN and H3O+

(g) Propane:
Ethyl bromide + Mg → Grignard reagent → Addition of water → Propane
Reagent: Mg, Condition: Dry ether, followed by H2O

Q.9. Write a detailed note on the mechanism of nucleophilic substitution reactions.

Nucleophilic substitution reactions involve the replacement of a leaving group (usually a halide ion) by a nucleophile. There are two main types of nucleophilic substitution mechanisms:

SN1 Mechanism (Unimolecular Nucleophilic Substitution):

Occurs in two steps.
Step 1: Formation of a carbocation after the leaving group departs.
Step 2: Nucleophile attacks the carbocation to form the product.
Occurs in tertiary alkyl halides due to the stability of the carbocation.

SN2 Mechanism (Bimolecular Nucleophilic Substitution):

Occurs in one concerted step.
The nucleophile attacks from the opposite side of the leaving group, leading to inversion of configuration.
Occurs primarily in primary and secondary alkyl halides.

Q.10. What do you understand by the term β-elimination reaction? Explain briefly the two possible mechanisms of 3-elimination reactions.

β-Elimination refers to a reaction where a hydrogen atom from the β-carbon (the carbon adjacent to the carbon bonded to the leaving group) and the leaving group are removed, resulting in the formation of a double bond.

Two possible mechanisms for β-elimination are:

E1 Mechanism (Unimolecular Elimination):

Occurs in two steps: first, the leaving group departs to form a carbocation, and then the base removes a proton from the β-carbon, resulting in an alkene.
Typically seen in tertiary halides.

E2 Mechanism (Bimolecular Elimination):

Occurs in a single step: the base simultaneously abstracts a proton from the β-carbon as the leaving group leaves, resulting in the formation of an alkene.
Occurs in primary and secondary halides.

Q.11. What products are formed when the following compounds are treated with ethyl magnesium bromide, followed by hydrolysis in the presence of an acid?

i) HCHO (Formaldehyde) → 1-Propanol
ii) CH3CHO (Acetaldehyde) → 2-Propanol
iii) CO2 → Propanoic acid
iv) (CH3)2CO (Acetone) → Tert-Butyl alcohol
v) CH3—CH2—CHO (Propanal) → 2-Butanol
vi) ClCN → Propanoic acid (after hydrolysis of nitrile)

Q.12. How will you carry out the following conversions?

i) CH4 → CH3COOH
Method: First, convert methane to methyl chloride (CH3Cl) by chlorination. Then, convert CH3Cl to acetic acid (CH3COOH) via carboxylation using a Grignard reagent and carbon dioxide.

ii) CH3—CH3 → (CH3—CH2)2N+ Br
Method: Ethane can be halogenated to form ethyl bromide (CH3CH2Br), which can then be reacted with ammonia to form a quaternary ammonium salt.

iii) CH2 = CH2 → CH3—CH2—CH2—CH2—OH
Method: Ethene can be reacted with HBr to form ethyl bromide, which can then be converted to 1-butanol through Wurtz coupling and hydration.

iv) CH3—CH2Cl → CH3—CH=CH2
Method: Ethyl chloride can undergo dehydrohalogenation with alcoholic KOH to form propene.

v) CH3COOH → CH3—CH=CH2
Method: Decarboxylation of acetic acid leads to the formation of propene.

Aromatic Hydrocarbons Solved Exercise

November 30, 2024September 28, 2024 by samreen

Prepare for exams with solved exercises on Aromatic Hydrocarbons. Cover key topics like structure, properties, reactions, and applications to reinforce your understanding and improve exam performance.

Q. 4
What are aromatic hydrocarbons? How are they classified?

Aromatic hydrocarbons are organic compounds that contain one or more benzene rings or similar structures with alternating double bonds. These hydrocarbons are highly unsaturated, and their electrons are delocalized across the ring, giving them extra stability. Classification:

Benzenoid Aromatic Hydrocarbons: Contain one or more benzene rings (e.g., benzene, toluene, xylene).
Non-benzenoid Aromatic Hydrocarbons: Do not contain a benzene ring but still have delocalized π-electrons that give them aromatic character (e.g., tropolone, azulene).

Q. 5
What happens when:

(a) Benzene is heated with conc. H₂SO₄ at 250°C?

When benzene is heated with concentrated sulfuric acid at a high temperature, sulfonation occurs, resulting in benzene sulfonic acid (C₆H₅SO₃H). This is an electrophilic substitution reaction.

(b) Chlorine is passed through benzene in sunlight?

In the presence of sunlight, chlorine reacts with benzene to form hexachlorocyclohexane (C₆H₆Cl₆) via a free radical substitution reaction.

This leads to the formation of maleic anhydride (C₄H₂O₃) through oxidation of benzene.

(d) Benzene is burnt in a free supply of air?

When benzene is burnt in an excess of air, it combusts completely to form carbon dioxide (CO₂) and water (H₂O), releasing a large amount of heat.

Q. 6
What is meant by the terms:

i) Aromatic: Refers to organic compounds that contain a conjugated π-electron system within a closed loop, typically a benzene ring, that follows Huckel’s rule (4n + 2 π-electrons).

ii) Oxidation: A chemical reaction that involves the loss of electrons or the addition of oxygen or removal of hydrogen from a molecule.

iii) Sulphonation: The introduction of a sulfonic acid group (-SO₃H) into an organic compound, typically via reaction with sulfuric acid.

iv) Nitration: The introduction of a nitro group (-NO₂) into an organic molecule, usually via reaction with nitric acid in the presence of sulfuric acid.

v) Halogenation: The addition or substitution of a halogen atom (Cl, Br, I) into an organic molecule, typically through an electrophilic substitution or free radical mechanism.

Q. 7
(a) Draw structural formulas for the following compounds:

m-Chlorobenzoic acid:
A benzene ring with a carboxylic acid group (-COOH) at position 1 and a chlorine atom at position 3.
p-Hydroxybenzoic acid:
A benzene ring with a hydroxyl group (-OH) at position 4 and a carboxylic acid group (-COOH) at position 1.
m-Bromonitrobenzene:
A benzene ring with a nitro group (-NO₂) at position 1 and a bromine atom at position 3.
o-Ethyltoluene:
A benzene ring with a methyl group (-CH₃) at position 1 and an ethyl group (-C₂H₅) at position 2.
p-Nitroaniline:
A benzene ring with an amino group (-NH₂) at position 1 and a nitro group (-NO₂) at position 4.
2,4,6-Trinitrotoluene (TNT):
A benzene ring with a methyl group (-CH₃) at position 1 and nitro groups (-NO₂) at positions 2, 4, and 6.
m-Nitrophenol:
A benzene ring with a hydroxyl group (-OH) at position 1 and a nitro group (-NO₂) at position 3.
p-Dibenzylbenzene:
A benzene ring attached to two benzyl groups (-CH₂Ph) at positions 1 and 4.
2-Amino-5-bromo-3-nitrobenzenesulphonic acid:
A benzene ring with an amino group (-NH₂) at position 2, a bromine atom at position 5, a nitro group (-NO₂) at position 3, and a sulfonic acid group (-SO₃H) at position 1.

(b) Give names and possible isomeric structures of:

Xylenes:
Ortho-xylene, Meta-xylene, Para-xylene (isomers of dimethylbenzene).
Trimethylbenzene:
1,2,3-Trimethylbenzene, 1,2,4-Trimethylbenzene, 1,3,5-Trimethylbenzene.
Bromonitrotoluene:
Isomers where bromine, nitro, and methyl groups are positioned differently on the benzene ring.

Q. 8
Write IUPAC names of the following molecules:

4-Chlorobenzaldehyde
3-Bromo-2-methylpropanoic acid
3,4-Dibromophenol

Q. 9
General mechanism of electrophilic aromatic substitution reactions:

Formation of the electrophile: The reaction begins with the generation of an electrophile (e.g., Br⁺, NO₂⁺) via a catalyst like FeBr₃ or H₂SO₄.
Electrophilic attack: The π-electrons of the aromatic ring attack the electrophile, forming a non-aromatic carbocation (an arenium ion).
Deprotonation: A proton (H⁺) is removed from the carbocation, restoring the aromaticity of the ring, resulting in the substitution of a hydrogen atom with the electrophile.

Q. 10
(a) Describe the structure of benzene on the basis of:

i) Atomic orbital treatment:

Benzene’s carbon atoms are sp² hybridized, and each carbon forms three sigma bonds: two with neighboring carbon atoms and one with a hydrogen atom. The unhybridized p orbitals overlap sideways to form a delocalized π-electron cloud above and below the plane of the carbon atoms.

ii) Resonance method:

Benzene is represented as a resonance hybrid of two structures where the double bonds alternate between the carbon atoms. This delocalization of π-electrons provides extra stability to the benzene molecule.

(b) Prove that benzene has a cyclic structure:

Experimental evidence shows that all six C–C bond lengths in benzene are identical and shorter than a typical single bond but longer than a double bond, which is consistent with a resonance-stabilized cyclic structure. Additionally, the molecular formula (C₆H₆) and aromatic behavior (like undergoing substitution reactions rather than addition reactions) further confirm its cyclic nature.

Q. 11

Predict the major products of bromination of the following compounds:

(a) Toluene

Major product: o-Bromotoluene and p-Bromotoluene
Toluene (methylbenzene) undergoes electrophilic substitution in the ortho- and para-positions relative to the methyl group due to the electron-donating nature of the methyl group.

(b) Nitrobenzene

Major product: m-Bromonitrobenzene
The nitro group is electron-withdrawing and directs incoming electrophiles to the meta-position, leading to the formation of m-bromonitrobenzene.

Major product: o-Dibromobenzene and p-Dibromobenzene
Bromine is an ortho/para-directing group, so bromination will occur in the ortho and para positions relative to the already present bromine atom.

(d) Benzoic acid

Major product: m-Bromobenzoic acid
The carboxyl group (-COOH) is an electron-withdrawing group, so bromination occurs at the meta-position.

(e) Benzaldehyde

Major product: m-Bromobenzaldehyde
The formyl group (-CHO) is an electron-withdrawing group, directing the bromine to the meta-position.

(f) Phenol

Major product: o-Bromophenol and p-Bromophenol
The hydroxyl group (-OH) is an electron-donating group, leading to substitution in the ortho and para positions.

Q. 12

How will you prepare the following compounds from benzene in two steps:

(a) m-Chloronitrobenzene

Nitration of benzene:
Benzene reacts with concentrated nitric acid and sulfuric acid to form nitrobenzene (C₆H₅NO₂).
Chlorination of nitrobenzene:
Nitrobenzene is treated with chlorine in the presence of a Lewis acid catalyst (FeCl₃), directing substitution at the meta position to give m-chloronitrobenzene (C₆H₄ClNO₂).

(b) p-Chloronitrobenzene

Chlorination of benzene:
Benzene is chlorinated using chlorine in the presence of FeCl₃, yielding chlorobenzene (C₆H₅Cl).
Nitration of chlorobenzene:
Chlorobenzene undergoes nitration with HNO₃ and H₂SO₄, giving p-chloronitrobenzene (C₆H₄ClNO₂) as the major product.

Q. 13

Complete the following reactions. Also mention the conditions needed to carry out these reactions:

Benzene + H₂ → Cyclohexane

Condition: High pressure and nickel (Ni) catalyst.
This is the hydrogenation of benzene, where three equivalents of hydrogen add to the benzene ring to form cyclohexane.

Benzene + O₂ → Carbon dioxide (CO₂) and water (H₂O)

Condition: Combustion in the presence of oxygen.
Complete combustion of benzene yields carbon dioxide and water.

Phenol + Zn → Benzene

Condition: Heated with zinc dust.
Zinc reduces phenol (C₆H₅OH) to benzene (C₆H₆).

Benzene + SO₃ → Benzene sulfonic acid (C₆H₅SO₃H)

Condition: Sulfonation with concentrated sulfuric acid or SO₃.
Sulfonation of benzene introduces a sulfonic acid group (-SO₃H) onto the ring.

Benzene + HOH (hydration) → No reaction

Condition: Hydration requires more drastic conditions, and benzene does not react with water directly under normal conditions.

Benzene + H₂SO₄ → Benzene sulfonic acid

Condition: Sulfonation occurs when benzene is treated with concentrated sulfuric acid or oleum.
This introduces a sulfonic acid group (-SO₃H) into the benzene ring.

Q. 14

Detail out three reactions in which benzene behaves as if it is a saturated hydrocarbon and three reactions in which it behaves as if it is unsaturated:

As a saturated hydrocarbon:

Hydrogenation:
Benzene can undergo hydrogenation in the presence of a nickel catalyst under high pressure to form cyclohexane (C₆H₁₂), similar to a saturated hydrocarbon.
Halogenation (under UV light):
Benzene reacts with halogens (e.g., chlorine) under UV light to form hexachlorocyclohexane (C₆H₆Cl₆), indicating a saturated character.
Combustion:
Like saturated hydrocarbons, benzene combusts completely in air to form carbon dioxide and water.

As an unsaturated hydrocarbon:

Bromination (in the presence of FeBr₃):
Benzene undergoes electrophilic substitution rather than addition with bromine, acting like an unsaturated hydrocarbon.
Nitration:
In the presence of concentrated H₂SO₄ and HNO₃, benzene undergoes nitration, an electrophilic substitution reaction, indicative of unsaturation.
Friedel-Crafts alkylation/acylation:
Benzene reacts with alkyl halides or acyl chlorides in the presence of AlCl₃, undergoing electrophilic substitution, which is a feature of unsaturation.

Q. 15

What are Friedel-Crafts reactions? Give the mechanism with an example of the following reactions:

(i) Friedel-Crafts alkylation reaction:

Definition: It is the alkylation of an aromatic ring using an alkyl halide (e.g., CH₃Cl) and a Lewis acid catalyst (e.g., AlCl₃). Mechanism:

Formation of the electrophile: The alkyl halide reacts with AlCl₃, forming a carbocation (CH₃⁺).
Electrophilic attack: The benzene ring donates its π-electrons to the carbocation, forming an arenium ion (carbocation intermediate).
Deprotonation: A proton is lost from the arenium ion, restoring the aromaticity and forming the alkylated product. Example:
Benzene + CH₃Cl (in the presence of AlCl₃) → Toluene (methylbenzene)

(ii) Friedel-Crafts acylation reaction:

Definition: This involves the acylation of benzene using an acyl chloride (e.g., CH₃COCl) in the presence of a Lewis acid catalyst (e.g., AlCl₃). Mechanism:

Formation of the acylium ion: The acyl chloride reacts with AlCl₃, forming an acylium ion (CH₃CO⁺).
Electrophilic attack: The acylium ion is attacked by the π-electrons of benzene, forming an arenium ion.
Deprotonation: A proton is lost, and the aromaticity is restored, giving the acylated benzene. Example:
Benzene + CH₃COCl (in the presence of AlCl₃) → Acetophenone (C₆H₅COCH₃)

Aliphatic hydrocarbons Solved Exercise

December 28, 2024September 26, 2024 by samreen

Unlock the key to mastering aliphatic hydrocarbons with this comprehensive guide to solved exercises. Covering topics such as alkanes, alkenes, and alkynes, this resource is aligned with the latest syllabus for Lahore Board, Federal Board, and other educational boards. Understand the structure, nomenclature, reactions, and properties of aliphatic hydrocarbons through step-by-step solutions to textbook exercises, conceptual questions, MCQs, and past paper problems. Designed to simplify learning, this content is ideal for students preparing for their chemistry exams

Q4. Write the structural formula for each of the following compounds:

i) 2-Methylpropane:
Structure:
CH3-CH(CH3)-CH3

ii) 3-Ethylpentane:
Structure:
CH₃-CH₂-CH(CH₂CH₃)-CH₂-CH₃

iii) 2,2,3,4-Tetramethylpentane:
Structure:
CH₃-C(CH₃)-C(CH₃)(CH₃)-CH(CH₃)-CH₃

iv) Neopentane (2,2-Dimethylpropane):
Structure:
(CH₃)₄C

v) 4-Ethyl-3,4-Dimethylheptane:
Structure:
CH₃-CH₂-CH(CH₃)-CH(CH₂CH₃)-CH₂-CH₃

vi) 4-iso-Propylheptane:
Structure:
CH₃-CH₂-CH(CH₂CH(CH₃)₂)-CH₂-CH₂-CH₃

vii) 2,2-Dimethylbutane:
Structure:
(CH₃)₃C-CH₂CH₃

Q5. Write down the names of the following compounds according to the IUPAC system:

i)
Structure: CH₃-CH₂-CH₂-CH₂-CH(CH₃)-CH₃
Name: 3-Methylhexane

ii)
Structure: (CH₃)₃C-CH₂-C(CH₃)₃
Name: 2,2,3,3-Tetramethylbutane

iii)
Structure: CH₃-C(CH₃)-CH(CH₃)-CH₂CH₃
Name: 2,3-Dimethylpentane

iv)
Structure: (CH₃)₃-C-CH₂CH₃
Name: 2,2-Dimethylbutane

v)
Structure: CH₃CH₂C(CH₂CH₃)-CH(CH₂CH₃)-CH₂CH₃
Name: 3-Ethyl-4-propylhexane

vi)
Structure: (C₆H₅)₃CH
Name: Triphenylmethane

vii)
Structure: CH₃-CH(CH₃)-CH₂-CH(CH₃)-CH₂-CH₃
Name: 2,4-Dimethylhexane

viii)
Structure: (CH₃)₂CH-CH₂-CH(CH₃)-CH₃
Name: 3-Methylhexane

Q6. What are the rules for naming alkanes? Explain with suitable examples.

The IUPAC rules for naming alkanes are:

Identify the longest continuous chain of carbon atoms. This chain determines the base name of the alkane (e.g., “pentane” for five carbon atoms).

Example: In CH₃-CH₂-CH₂-CH₃, the longest chain has four carbon atoms, so it’s called “butane.”

Number the chain from the end closest to the first substituent (branching group).

Example: In 2-Methylbutane, the chain is numbered from the side closest to the methyl group.

Name the substituents (side chains) attached to the main chain, giving them the lowest possible numbers.

Example: In CH₃CH₂CH(CH₃)-CH₃, the substituent (CH₃) is on the second carbon, so it is named “2-methylbutane.”

If there are multiple identical substituents, use prefixes like “di-“, “tri-“, etc., and number them.

Example: In 2,3-Dimethylbutane, two methyl groups are attached at positions 2 and 3.

Arrange the substituents alphabetically when writing the complete name.

Example: In 3-Ethyl-2,3-Dimethylpentane, “ethyl” comes before “methyl” alphabetically.

Q7. (a) Write down the structural formulas for all the isomeric hexanes and name them according to the IUPAC system.

n-Hexane:
CH₃-(CH₂)₄-CH₃
2-Methylpentane:
CH₃-CH(CH₃)-CH₂-CH₂-CH₃
3-Methylpentane:
CH₃-CH₂-CH(CH₃)-CH₂-CH₃
2,2-Dimethylbutane:
(CH₃)₂C-CH₂-CH₃
2,3-Dimethylbutane:
CH₃-CH(CH₃)-CH(CH₃)-CH₃
3-Ethylpentane:
CH₃-CH₂-CH(CH₂CH₃)-CH₂-CH₃

(b) The following names are incorrect. Give the correct IUPAC names:

i) 4-Methylpentane
Correct name: 2-Methylpentane

ii) 2-Methyl-3-Ethylbutane
Correct name: 3-Methylpentane

iii) 3,5,5-Trimethylhexane
Correct name: 2,2,4-Trimethylpentane

Q8. (a) Explain why alkenes are less reactive than alkynes? What is the effect of branching on the melting point of alkanes?

Reactivity of Alkenes vs. Alkynes:

Alkenes have a double bond (C=C) consisting of one sigma (σ) bond and one pi (π) bond.
Alkynes have a triple bond (C≡C), consisting of one sigma (σ) bond and two pi (π) bonds. The presence of two pi bonds in alkynes makes them more reactive than alkenes because pi bonds are weaker and more easily broken in chemical reactions.

Effect of Branching on Melting Point of Alkanes:

Branching decreases the surface area available for van der Waals forces, leading to a lower melting point. Linear alkanes can pack more tightly, while branched alkanes are less dense, resulting in a lower melting point.

(b) Three different alkenes yield 2-methylbutane when they are hydrogenated in the presence of a metal catalyst. Give their structures and write equations for the reactions involved.

2-Methyl-1-butene:
Structure: CH₃-CH=CH-CH₃
Equation: CH₃-CH=CH-CH₃ + H₂ → CH₃CH₂CH₂CH₃
2-Methyl-2-butene:
Structure: CH₂=C(CH₃)-CH₃
Equation: CH₂=C(CH₃)-CH₃ + H₂ → CH₃CH₂CH₂CH₃
3-Methyl-1-butene:
Structure: CH₂=CH-CH(CH₃)-CH₃
Equation: CH₂=CH-CH(CH₃)-CH₃ + H₂ → CH₃CH₂CH₂CH₃

Q9. (a) Outline the methods available for the preparation of alkanes.

Hydrogenation of Alkenes and Alkynes:

Alkanes can be prepared by the catalytic hydrogenation of alkenes and alkynes using a metal catalyst such as nickel (Ni), palladium (Pd), or platinum (Pt).
Example:
CH2=CH2 + H2 → Ni CH3-CH3

Wurtz Reaction:

Alkanes can be formed by the coupling of alkyl halides in the presence of sodium metal in dry ether.
Example:
2CH3Cl + 2Na → CH3-CH3 + 2NaCl

Decarboxylation of Carboxylic Acids (Kolbe’s Electrolysis):

Alkanes are produced when sodium salts of carboxylic acids undergo electrolysis.
Example:
CH3COONa + NaOH → CaO, heat CH4 + Na2CO3

Reduction of Alkyl Halides:

Alkyl halides can be reduced to alkanes by treatment with zinc and hydrochloric acid.
Example:
CH3Cl + H2 → Zn, HCl CH4 + HCl

(b) How will you bring about the following conversions?

i) Methane to Ethane:

Wurtz reaction:
2CH3Cl + 2Na → dry ether C2H6 + 2NaCl

ii) Ethane to Methane:

Cracking of Ethane
C2H6 → high temp CH4 + C

iii) Acetic Acid to Ethane:

Decarboxylation of acetic acid:
CH3COONa + NaOH → CaO, heat C2H6 + Na2CO3

iv) Methane to Nitromethane:

Nitration of Methane (Halogenation):
CH4 + HNO3 → CH3NO2 + H2O

Q10. (a) What is meant by octane number? Why does a high-octane fuel have a less tendency to knock in an automobile engine?

Octane Number:

The octane number of a fuel is a measure of its ability to resist knocking during combustion. It is based on a scale where iso-octane (2,2,4-trimethylpentane) is given a rating of 100 (least knocking), and n-heptane is given a rating of 0 (most knocking). Fuels with higher octane numbers burn more smoothly in engines.

Why High-Octane Fuels Have Less Knocking:

High-octane fuels are composed of branched-chain hydrocarbons, which burn more evenly. This reduces the chances of pre-ignition or “knocking” (where the air-fuel mixture combusts prematurely). This smooth combustion is essential for modern engines to operate efficiently without damage.

(b) Explain the free radical mechanism for the reaction of chlorine with methane in the presence of sunlight.

The chlorination of methane proceeds via a free radical chain reaction with the following steps:

Initiation Step:

UV light causes the dissociation of chlorine molecules into free radicals.
Cl2 → UV light 2Cl .

Propagation Step:

The chlorine radical abstracts a hydrogen atom from methane, forming hydrogen chloride and a methyl radical.
CH4 + Cl → CH3^. + HCl ]
The methyl radical reacts with another chlorine molecule to form chloromethane and regenerate a chlorine radical.
CH3. + Cl2 →CH3Cl + Cl.

Termination Step:

The chain reaction is terminated when two radicals combine to form a stable molecule.
Cl + CH3. →CH3Cl ]
Cl. + Cl. →Cl2 ]

Q11. (a) Write structural formulas for each of the following compounds:

i) Isobutylene (2-methylpropene):
Structure:
CH2=C(CH3)-CH3

ii) 2,3,4,4-Tetramethyl-2-pentene:
Structure:
CH2=C(CH3)-CH(CH3)-CH2-CH3

iii) 2,5-Heptadiene:
Structure:
CH2=CH-CH2-CH2-CH=CH2

iv) 4,5-Dimethyl-2-hexene:
Structure:
CH3-CH=CH-CH(CH3)-CH(CH3)-CH3

v) Vinylacetylene (1-buten-3-yne):
Structure:
CH2=CH-C ≡ CH

vi) 1,3-Pentadiene:
Structure:
CH2=CH-CH=CH-CH3

vii) 1-Butyne:
Structure:
CH≡ CH-CH2-CH3

viii) 3-n-Propyl-1,4-pentadiene:
Structure:
CH2=CH-CH2-CH2-CH=CH-CH2-CH2-CH3

ix) Vinyl bromide:
Structure:
CH2=CH-Br

x) But-1-en-3-yne:
Structure:
CH2=CH-C≡CH

xi) 4-Methyl-2-pentyne:
Structure:
CH3-CH2-C≡CH-CH3

xii) Isopentane (2-methylbutane):
Structure:
CH3-CH(CH3)-CH2-CH3

(b) Name the following compounds by IUPAC system:

i) 3-Methylhex-2-ene:
Structure:
CH3-CH=CH-CH2-CH(CH3)-CH3

ii) 2-Methyl-1-propene:
Structure:
CH2=C(CH3)-CH3

iii) 1-Heptyne:
Structure:
C≡CH-(CH2)4-CH3

Q. 12
(a) Methods for Preparation of Alkenes:

Dehydration of alcohols: Alcohols are heated with a strong acid like sulfuric acid (H₂SO₄) to remove water and form an alkene.
Dehydrohalogenation of alkyl halides: Alkyl halides react with a strong base such as KOH in ethanol, which removes hydrogen and halogen atoms to form an alkene.
Catalytic cracking: Larger hydrocarbons are broken down into smaller alkenes through thermal or catalytic cracking.

To establish that ethylene contains a double bond, use Bromine Water Test: Ethylene decolorizes bromine water, indicating the presence of a double bond.

(b) Structure formulas of alkenes formed by dehydrohalogenation:

1-Chloropentane:
CH2=CH-CH2-CH2-CH3 (Pent-1-ene)
2-Chloro-3-methylbutane:
CH2=CH-CH3-CH3 (3-methylbut-1-ene)
1-Chloro-2,2-dimethylpropane: No reaction, as no hydrogen is available on adjacent carbon for elimination.

Q. 13
(a) Preparation of Propene:

From 1-propanol (CH₃–CH₂–CH₂–OH):
From propyne (CH≡C–CH₃):
From isopropyl chloride (CH₃–CHCl–CH₃):

(b) Skeletal formula of all possible alkenes (C₄H₈):

But-1-ene: CH₂=CH-CH₂-CH₃
But-2-ene: CH₃-CH=CH-CH₃
2-methylprop-1-ene: CH₂=C-CH₃-CH₃

Q. 14
(a) Conversion of ethene to ethyl alcohol:

(Ethene reacts with water in the presence of an acid catalyst to form ethanol.)

(b) Reactions for preparation of:

1,2-Dibromoethane:
CH2=CH2 + Br2 —–>CH2Br-CH2Br
]
Ethyne:
CHBr=CHBr —–>HC≡CH + ZnBr2
]
Ethane:
CH2=CH2 + H2 ——>Ni CH3-CH3
]
Ethylene glycol:

1-Butene to 1-Butyne:
1-Propanol to propene:

Q. 15
Reaction scheme:

(Ethane is first cracked to ethene, which then undergoes further dehydrogenation to form ethyne.)

Q. 16
Products from 1-butene (CH₂=CH–CH₂–CH₃):

H₂ (Hydrogenation):
Br₂ (Bromination):
HBr (Hydrobromination):

Here are the answers to the new set of questions:

Q. 17
Identify each lettered product in the given reactions:

(i) Ethyl alcohol:

A: Ethene (CH₂=CH₂)
(Dehydration of ethyl alcohol using concentrated H₂SO₄)
B: 1,2-Dibromoethane (CH₂Br–CH₂Br)
(Addition of Br₂ in CCl₄ to ethene)
C: Ethylene glycol (CH₂OH–CH₂OH)
(Oxidation with cold dilute KMnO₄)

(ii) Propene:

D: 1-Butene (CH₂=CH–CH₂–CH₃)
(Dehydrohalogenation with alcoholic KOH)
E: 2-Butanone (CH₃CO–CH₂–CH₃)
(Oxidation with alkaline KMnO₄)
F: Butanenitrile (CH₃CH₂CH₂CN)
(Reaction with HCN)

Q. 18
After ozonolysis of the compound, acetaldehyde (CH₃CHO) was obtained, which suggests the structural formula of the compound was ethene (CH₂=CH₂).

Q. 19
(a) Markovnikov’s Rule Alcohol Products:

Propene: Propan-2-ol (CH₃CH(OH)CH₃)
1-Butene: Butan-2-ol (CH₃CH₂CH(OH)CH₃)
2-Butene: Butan-2-ol (same as for 1-butene)

(b) The most likely product from the addition of hydrogen iodide (HI) to 2-methyl-2-butene is 2-iodo-2-methylbutane. This follows Markovnikov’s rule, where the iodine attaches to the more substituted carbon.

Q. 20
Hydrocarbons are classified as:

Saturated hydrocarbons (alkanes): Contain only single bonds (e.g., ethane, propane).
Unsaturated hydrocarbons: Contain double or triple bonds (e.g., ethene, ethyne).

Characteristic reactions:

Saturated hydrocarbons: Undergo substitution reactions.
Unsaturated hydrocarbons: Undergo addition reactions.

Q. 21
(a) Preparation of Ethyne:

Calcium carbide (CaC₂) reacts with water to produce ethyne.

(b) Ethyne Reactions:

With Hydrogen: Ethyne (HC≡CH) reacts with hydrogen (H₂) to form ethane (CH₃CH₃).
With Halogen acid (e.g., HCl): Produces chloroethene.
With alkaline KMnO₄: Ethyne is oxidized to oxalic acid.
With 10% H₂SO₄ (in presence of HgSO₄): Forms acetaldehyde (CH₃CHO).
With ammoniacal cuprous chloride: Forms copper acetylide (Cu₂C₂).

Ethene: Used in the production of polyethylene (plastics) and ethanol.
Ethane: Important fuel and feedstock for ethylene production.
Ethyne: Used in welding (as acetylene) and for organic synthesis.

Q. 22
To distinguish ethane, ethene, and ethyne:

Ethane: Does not react with bromine water.
Ethene: Decolorizes bromine water and reacts with KMnO₄ to give diols.
Ethyne: Forms a red precipitate with ammoniacal cuprous chloride and also decolorizes bromine water.

Q. 23
(a) Synthesis of compounds from ethyne:

Ethene: Hydrogenate ethyne using a Lindlar’s catalyst.
Ethanol: React ethene with water in the presence of an acid catalyst.

Q. 24
(a) Comparison of the Reactivity of Ethane, Ethene, and Ethyne:

Ethane (C₂H₆): It is a saturated hydrocarbon, so it undergoes substitution reactions (e.g., with halogens) but is less reactive than unsaturated hydrocarbons.
Ethene (C₂H₄): Ethene has a double bond, making it more reactive than ethane. It readily undergoes addition reactions, such as halogenation, hydration, and hydrogenation.
Ethyne (C₂H₂): Ethyne is even more reactive than ethene due to its triple bond. It can undergo addition reactions similar to ethene but can also react with certain metals (e.g., forming acetylides).

(b) Comparison of Physical Properties of Alkanes, Alkenes, and Alkynes:

Alkanes (e.g., ethane): Non-polar, low boiling and melting points, relatively inert, less dense than water, and generally insoluble in water but soluble in organic solvents.
Alkenes (e.g., ethene): Also non-polar but with slightly higher boiling points than corresponding alkanes due to the presence of a double bond, which introduces some polarity in the molecule.
Alkynes (e.g., ethyne): Non-polar and exhibit higher boiling points than both alkanes and alkenes. Due to the linear structure of the triple bond, alkynes have stronger intermolecular forces than alkanes and alkenes, leading to slightly higher melting and boiling points.

Q. 25
Reactions of Propyne with the following reagents:

(a) AgNO₃/NH₄OH (Tollens’ reagent):
Propyne (HC≡C–CH₃) reacts with Tollens’ reagent to form a white precipitate of silver acetylide (AgC≡C–CH₃). This test is specific for terminal alkynes due to the acidic hydrogen attached to the terminal carbon.

(b) Cu₂Cl₂/NH₄OH:
Propyne reacts with ammoniacal cuprous chloride to form a red precipitate of copper acetylide (CuC≡C–CH₃). This is also a test for terminal alkynes.

(c) H₂O/H₂SO₄/HgSO₄ (Hydration):
Propyne undergoes hydration in the presence of H₂SO₄ and HgSO₄ to form a ketone, specifically propanone (acetone), via the addition of water across the triple bond.

Q. 26
A compound with the molecular formula C₄H₆ suggests an alkyne. When treated with hydrogen and a Ni catalyst, a new compound C₄H₁₀ is formed, indicating the complete hydrogenation of an alkyne to an alkane.

The reaction of C₄H₆ with ammoniacal silver nitrate forms a white precipitate, indicating a terminal alkyne. This points to but-1-yne (CH≡C–CH₂–CH₃) as the structure of the original compound. Upon hydrogenation, butane (C₄H₁₀) is formed.

Q. 21
(a) Identification of A and B:

Starting compound: 1-Propanol (CH₃CH₂CH₂OH)
Reaction with PCl₅ forms A: 1-chloropropane (CH₃CH₂CH₂Cl).
Reaction with sodium in ether leads to the formation of B: Hexane (CH₃(CH₂)₄CH₃) (via the Wurtz reaction).

(b) General Mechanism of Electrophilic Addition Reactions of Alkenes:

Formation of carbocation (Electrophilic attack): The double bond in an alkene is electron-rich and attacks an electrophile (e.g., H⁺ from HCl), leading to the formation of a carbocation intermediate.
Nucleophilic attack: The nucleophile (e.g., Cl⁻) then attacks the positively charged carbon atom (carbocation) to complete the addition, forming a saturated product.

Short Answer Questions:

i. Define cognitive bias.

ii. What is a false cause fallacy?

iii. Describe the straw man fallacy.

iv. What does the fallacy of exclusion involve?

v. Give an example of a faulty analogy.

vi. List one pro and one con of chemical substances.

vii. Responsibility of scientists/companies in chemical production.

viii. Importance of regulations in the chemical industry.

ix. What is a claim in a scientific argument?

x. Example of an assumption in renewable energy debates.

Summary Cheat Sheet

i. Explain confirmation bias and its potential impact on scientific research.

ii. Discuss ethical considerations in chemical production/use, balancing benefits and risks.

iii. Deconstruct a scientific argument using EVs and air pollution.

iv. Analyze regulations in the chemical industry with examples.

v. Evaluate pros/cons of pesticides in agriculture.

Summary Cheat Sheet

Question i:

Question ii:

Question iii:

Question iv:

Summary of Correct Answers:

Question iv:

Question v:

Question vi:

Question vii:

Question viii:

Question ix:

Question x:

Answer Summary:

Main Reaction

Mechanism: Free Radical Halogenation

1. Initiation

2. Propagation

3. Termination

Tips and Tricks for the Mechanism

Detailed Insights on Selectivity in Halogenation

1. Reactivity and Selectivity of Halogens

2. Factors Influencing Selectivity

Radical Stability

Energy Considerations

4. Controlling Polyhalogenation

5. Selectivity Ratio

Key Takeaways

Q. 4: Give one laboratory and one industrial method for the preparation of formaldehyde.

Q. 5: How does formaldehyde react with the following reagents?

Q. 6: Give one laboratory and one industrial method for the preparation of acetaldehyde.

Q. 7: How does acetaldehyde react with the following reagents?

Q. 8: Describe briefly the mechanism of nucleophilic addition to a carbonyl compound.

Q. 9: Explain with mechanism the addition of ethylmagnesium bromide to acetaldehyde. What is the importance of this reaction?

Q. 10: Explain with mechanism the addition of sodium bisulphite to acetone. What is the utility of this reaction?

Q. 11: Describe with mechanism aldol condensation. Why formaldehyde does not give this reaction?

Q. 12: What types of aldehydes give Cannizzaro’s reaction? Give its mechanism.

Q. 13: Explain the mechanism of the reaction of phenylhydrazine with acetone.

Q. 14: Using ethyne as a starting material, how would you get acetaldehyde, acetone, and ethyl alcohol?

Q. 15: Give the mechanism of addition of HCN to acetone.

Q. 16: How would you bring about the following conversions?

Q. 17: How will you distinguish between:

Q. 18: Discuss the oxidation of:

Q. 19: Discuss reduction of:

Q. 20: Give three uses for each of formaldehyde and acetaldehyde.

Q.4. Define alkyl halide. Which is the best method of preparing alkyl halides?

Q.5. Write down a method for the preparation of ethyl magnesium bromide in the laboratory.

Q.6. Give IUPAC names to the following compounds.

Q.7. Draw all the possible structures that have the molecular formula ( C4H9Cl ). Classify each as primary, secondary, or tertiary chloride. Give their names according to the IUPAC system.

Q.8. Using ethyl bromide as a starting material, how would you prepare the following compounds?

Q.9. Write a detailed note on the mechanism of nucleophilic substitution reactions.

Q.10. What do you understand by the term β-elimination reaction? Explain briefly the two possible mechanisms of 3-elimination reactions.

Q.11. What products are formed when the following compounds are treated with ethyl magnesium bromide, followed by hydrolysis in the presence of an acid?

Q.12. How will you carry out the following conversions?

Q. 11

Q. 12

Q. 13

Q. 14

Q. 15

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