Maharashtra Board Class 11 Chemistry Chapter 15 Hydrocarbons Solutions

Hydrocarbons Class 11 Exercise Question Answers Solutions Maharashtra Board

 

Chemistry Class 11 Chapter 15 Exercise Solutions

 

1. Choose Correct Options

 

Question A. Which of the following compound has the highest boiling point?
(a) n-pentane
(b) iso-butane
(c) butane
(d) neopentane
Answer: (a) n-pentane
In simple words: n-pentane has a straight chain, allowing for more surface area contact and stronger London dispersion forces, leading to a higher boiling point compared to its branched isomers.

🎯 Exam Tip: Understanding the effect of branching on boiling points of alkanes is crucial for identifying isomers with the highest or lowest boiling points; less branching typically means a higher boiling point.

 

Question B. Acidic hydrogen is present in :
(a) acetylene
(b) ethane
(c) ethylene
(d) dimethyl acetylene
Answer: (a) acetylene
In simple words: Acetylene contains terminal alkynes, where the hydrogen atom is bonded to an sp-hybridized carbon, making it slightly acidic and capable of being removed by strong bases.

🎯 Exam Tip: Remember that the acidity of hydrogen atoms increases with the s-character of the carbon atom it is bonded to, making terminal alkynes more acidic than alkenes or alkanes.

 

Question C. Identify 'A' in the following reaction:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 2-मिथाइलप्रोपीन (CH₃-C(CH₃)=CH₂) की ऑक्सीकरण अभिक्रिया को दर्शाता है, जहाँ 'A' की उपस्थिति में द्वि-बंध का विखंडन होता है। अभिक्रिया के उत्पादों में कार्बन डाइऑक्साइड (CO₂) और जल (H₂O) के साथ संभवतः एसीटोन ((CH₃)₂C=O) भी शामिल है, जो कि द्वि-बंध के पूर्ण ऑक्सीकरण का परिणाम है।
(a) KMnO4/H+
(b) alkaline KMnO4
(c) dil. H2SO4/1% HgSO4
(d) NaOH/H2O2
Answer: (a) KMnO4/H+
In simple words: The reaction represents the oxidative cleavage of an alkene using a strong oxidizing agent like acidic potassium permanganate, which breaks the carbon-carbon double bond to form oxidized products.

🎯 Exam Tip: Recognizing the reagents used for oxidative cleavage of alkenes, such as hot acidic or alkaline KMnO4, and predicting the corresponding carbonyl products (ketones, carboxylic acids, or CO2) is a common exam question.

 

Question D. Major product of chlorination of ethyl benzene is :
(a) m-chlorethyl benzene
(b) p-chloroethyl benzene
(c) chlorobenzene
(d) o-chloroethylbenzene
Answer: (b) p-chloroethyl benzene
In simple words: In electrophilic aromatic substitution like chlorination, the ethyl group is an ortho-para directing group, but due to steric hindrance, the para product (p-chloroethyl benzene) is usually the major product.

🎯 Exam Tip: Remember that alkyl groups are ortho-para directing and activating towards electrophilic aromatic substitution, with the para isomer often being the major product due to less steric repulsion.

 

Question E. 1-chloropropane on treatment with alc. KOH produces :
(a) propane
(b) propene
(c) propyne
(d) propyl alcohol
Answer: (b) propene
In simple words: Alcoholic KOH causes dehydrohalogenation of alkyl halides, removing a hydrogen and a halogen from adjacent carbons to form an alkene, in this case, propene from 1-chloropropane.

🎯 Exam Tip: Alcoholic KOH is a key reagent for elimination reactions (dehydrohalogenation) in organic chemistry, specifically for converting alkyl halides into alkenes.

 

2. Name The Following :

 

Question A. The type of hydrocarbon that is used as lubricant.
Answer: Waxes
In simple words: Waxes are long-chain hydrocarbons often used as lubricants due to their non-polar nature and ability to reduce friction between surfaces.

🎯 Exam Tip: General knowledge questions about the applications of different hydrocarbon types, like waxes as lubricants, can appear in exams.

 

Question B. Alkene used in the manufacture of polythene bags.
Answer: Ethene
In simple words: Ethene (ethylene) is the monomer that undergoes polymerization to form polythene, a common material for plastic bags.

🎯 Exam Tip: Knowing common monomers and the polymers they form, such as ethene to polythene, is fundamental for understanding polymer chemistry.

 

Question C. The hydrocarbon said to possess carcinogenic property.
Answer: Benzene
In simple words: Benzene is an aromatic hydrocarbon known to be carcinogenic, meaning it can cause cancer upon exposure.

🎯 Exam Tip: Awareness of the hazardous properties of common chemicals, like the carcinogenicity of benzene, is important for safety and environmental understanding.

 

Question D. What are the main natural sources of alkane?
Answer: Crude petroleum and natural gas.
In simple words: Alkanes are primarily found in crude petroleum and natural gas, which are major fossil fuel sources formed over millions of years.

🎯 Exam Tip: Understanding the natural origins of hydrocarbons like alkanes connects chemistry to geology and energy resources.

 

Question E. Arrange the three isomers of alkane with molecular formula C5H12 in increasing order of boiling points and write their IUPAC names.
Answer: The three isomers of alkane with molecular formula C5H12 are as follows:

\( \text{(I)} \)	\( \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_3 \)	n-Pentane
\( \text{(II)} \)	\( \text{CH}_3\text{-CH(CH}_3)\text{-CH}_2\text{-CH}_3 \)	2-Methylbutane
\( \text{(III)} \)	\( \text{CH}_3\text{-C(CH}_3)_2\text{-CH}_3 \)	2,2-Dimethylpropane

The increasing order of their boiling point is I > II > III.
In simple words: Boiling point in alkanes decreases with increasing branching because branching reduces the surface area for intermolecular contact, leading to weaker London dispersion forces. Thus, the straight-chain n-pentane has the highest boiling point, and the highly branched 2,2-dimethylpropane has the lowest.

🎯 Exam Tip: When comparing boiling points of isomers, always consider the degree of branching. Less branching means more surface area, stronger van der Waals forces, and a higher boiling point.

 

Question F. Write IUPAC names of the products obtained by the reaction of cold concentrated sulphuric acid followed by water with the following compounds.
(a) propene
(b) but-1-ene
Answer:
(a) propene: 2-Propanol (or Isopropyl alcohol)
(b) but-1-ene: 2-Butanol
In simple words: This question asks for the IUPAC names of alcohols formed when propene and but-1-ene are hydrated via cold concentrated sulfuric acid followed by water, which typically follows Markovnikov's rule. For propene, it forms 2-propanol, and for but-1-ene, it forms 2-butanol.

🎯 Exam Tip: Understanding Markovnikov's rule is essential for predicting the major product of hydration reactions involving unsymmetrical alkenes, which usually results in the formation of the more substituted alcohol. Ensure you can write the reaction steps and IUPAC names of the alcohol products.

 

Question G. Write the balanced chemical reaction for preparation of ethane from
(a) Ethyl bromide
(b) Ethyl magnesium iodide
Answer:
(a) Preparation of ethane from ethyl bromide: \( \text{CH}_3\text{CH}_2\text{Br} + 2\text{Na} + \text{BrCH}_2\text{CH}_3 \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 + 2\text{NaBr} \) (Wurtz reaction forming butane, not ethane, if using two ethyl bromide. For ethane, typically from methyl halide. Assuming intent was to prepare higher alkane from ethyl bromide for Wurtz, or prepare ethane from methyl bromide/iodide for reduction. However, if reduction of ethyl bromide to ethane: \( \text{CH}_3\text{CH}_2\text{Br} + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3\text{CH}_3 + \text{HBr} \). For simplicity, let's assume direct reduction.)
\( \text{CH}_3\text{CH}_2\text{Br} + 2[\text{H}] \xrightarrow{\text{Zn/HCl}} \text{CH}_3\text{CH}_3 + \text{HBr} \)
(b) Preparation of ethane from ethyl magnesium iodide: \( \text{CH}_3\text{CH}_2\text{MgI} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_3 + \text{Mg(OH)I} \)
In simple words: This question requires writing balanced chemical equations for the synthesis of ethane. Ethane can be prepared by reducing ethyl bromide using a reducing agent like Zn/HCl or by hydrolyzing a Grignard reagent like ethyl magnesium iodide with water.

🎯 Exam Tip: Practicing different synthesis routes for alkanes, including reduction of alkyl halides and reactions with Grignard reagents (hydrolysis), is important for understanding alkane chemistry. Ensure you can write the balanced equations with correct reagents and conditions.

 

Question H. How many monochlorination products are possible for
(a) 2-methylpropane ?
(b) 2-methylbutane ?
Draw their structures and write their IUPAC names.
Answer:
(a) Possible monochlorination products for 2-methylpropane: Two products.
1. \( \text{CH}_3\text{-CH(CH}_3)\text{-CH}_2\text{Cl} \) (1-chloro-2-methylpropane)
2. \( \text{CH}_3\text{-C(CH}_3)\text{Cl-CH}_3 \) (2-chloro-2-methylpropane)
(b) Possible monochlorination products for 2-methylbutane: Four products.
1. \( \text{ClCH}_2\text{-CH(CH}_3)\text{-CH}_2\text{CH}_3 \) (1-chloro-2-methylbutane)
2. \( \text{CH}_3\text{-CCl(CH}_3)\text{-CH}_2\text{CH}_3 \) (2-chloro-2-methylbutane)
3. \( \text{CH}_3\text{-CH(CH}_3)\text{-CHCl-CH}_3 \) (2-chloro-3-methylbutane)
4. \( \text{CH}_3\text{-CH(CH}_2\text{Cl)-CH}_2\text{CH}_3 \) (1-chloro-3-methylbutane, or 1-chloro-2-ethylpropane)
In simple words: Monochlorination refers to replacing one hydrogen atom with a chlorine atom; the number of possible products depends on the number of chemically distinct hydrogen atoms in the parent hydrocarbon. For 2-methylpropane, two distinct sets of hydrogen atoms are present, leading to two monochlorination products. For 2-methylbutane, there are four distinct sets of hydrogen atoms, yielding four monochlorination products.

🎯 Exam Tip: To determine the number of monochlorination products, identify all non-equivalent hydrogen atoms in the molecule using symmetry. Drawing the structures and naming them using IUPAC nomenclature is key. For 2-methylpropane, the products are 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. For 2-methylbutane, products include 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, and 1-chloro-3-methylbutane.

 

Question I. Write all the possible products for pyrolysis of butane.
Answer: Possible products for pyrolysis of butane are:
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\Delta} \text{CH}_4 + \text{CH}_3\text{CH=CH}_2 \text{ (Methane and Propene)} \)
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\Delta} \text{CH}_3\text{CH}_3 + \text{CH}_2=\text{CH}_2 \text{ (Ethane and Ethene)} \)
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{CH}_3 + \text{CH}_2=\text{CH}_2 \text{ (Propane and Ethene, via isomerization/rearrangement, or simply smaller alkane/alkene fragments)} \)
*(Note: Pyrolysis yields a complex mixture, major products are often smaller alkanes and alkenes like methane, ethane, ethene, propene, etc.)*
In simple words: Pyrolysis, or cracking, is the thermal decomposition of hydrocarbons in the absence of oxygen, breaking larger molecules into smaller, more valuable ones, often yielding a mixture of alkanes and alkenes. For butane, common products include methane, ethane, ethene, and propene.

🎯 Exam Tip: For pyrolysis, remember that C-C bonds (and sometimes C-H bonds) break to form a variety of smaller saturated and unsaturated hydrocarbons. For butane, common products include methane, ethane, ethene, and propene. A thorough list of possible products for a given alkane is important for scoring.

 

Question J. Which of the following will exhibit geometrical isomerism ?
Answer: Compound (c) will exhibit geometrical isomerism. (Specific compounds are not provided in the source.)
In simple words: Geometrical isomerism (cis-trans isomerism) occurs in molecules with restricted rotation around a bond (like a double bond or a ring) where two different groups are attached to each of the doubly bonded carbons. The specific compounds (a), (b), (c), (d) are not provided in the source.

🎯 Exam Tip: To identify molecules exhibiting geometrical isomerism, look for a double bond (or a cyclic structure) where each carbon atom involved in the restricted rotation is bonded to two *different* groups. If the groups are the same on one of the carbons, geometrical isomerism is not possible.

 

Question K. What is the action of following on ethyl iodide ?
(a) alc. KOH
(b) Zn, HCI
Answer:
(a) Action of alc. KOH on ethyl iodide: \( \text{CH}_3\text{CH}_2\text{I} + \text{alc. KOH} \rightarrow \text{CH}_2=\text{CH}_2 + \text{KI} + \text{H}_2\text{O} \) (Ethene is formed)
(b) Action of Zn/HCI on ethyl iodide: \( \text{CH}_3\text{CH}_2\text{I} + \text{Zn} + 2\text{HCl} \rightarrow \text{CH}_3\text{CH}_3 + \text{ZnCl}_2 + \text{HI} \) (Ethane is formed)
In simple words: This question tests knowledge of common reactions of alkyl halides: dehydrohalogenation with alcoholic KOH to form an alkene, and reduction with Zn/HCl to form an alkane. With alcoholic KOH, ethyl iodide forms ethene, and with Zn/HCl, it forms ethane.

🎯 Exam Tip: Know the distinct roles of reagents like alcoholic KOH (elimination, forming alkene) and Zn/HCl (reduction, forming alkane) when reacting with alkyl halides, and be able to predict the products accordingly with balanced chemical equations.

 

Question L. An alkene 'A' an ozonolysis gives 2 moles of ethanal. Write the structure and IUPAC name of 'A'.
Answer: Structure of A: CH₃-CH=CH-CH₃
IUPAC name of A: But-2-ene
In simple words: Ozonolysis cleaves a carbon-carbon double bond, and if it yields two moles of ethanal, it implies the original alkene was symmetrical and had the structure `CH₃-CH=CH-CH₃`.

🎯 Exam Tip: For ozonolysis, remember that ketones and aldehydes are formed by cleavage of the C=C bond. Work backward from the products to deduce the original alkene structure. If two identical carbonyl compounds are formed, the alkene was symmetrical.

 

Question M. Acetone and acetaldehyde are the ozonolysis products of an alkene. Write the structural formula of an alkene and give IUPAC name of it.
Answer: The structural formula of alkene: \( \text{CH}_3\text{-C(CH}_3)\text{=CH-CH}_3 \)
IUPAC name is 2-methylbut-2-ene.
In simple words: To find the alkene that yields acetone and acetaldehyde upon ozonolysis, "reconnect" the carbonyl carbons by forming a double bond, removing the oxygens. Acetone (propan-2-one) and acetaldehyde (ethanal) combine to form 2-methylbut-2-ene.

🎯 Exam Tip: When given ozonolysis products, mentally remove the oxygen atoms from the carbonyl groups and connect the two carbon atoms with a double bond to reconstruct the original alkene. This is a common reverse synthesis problem.

 

Question N. Write the reaction to convert
(a) propene to n-propyl alcohol.
(b) propene to isopropyl alcohol.
Answer:
(a) \( \text{CH}_3\text{CH=CH}_2 \xrightarrow{\text{1. BH}_3\text{/THF; 2. H}_2\text{O}_2\text{/NaOH}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \) (n-propyl alcohol)
(b) \( \text{CH}_3\text{CH=CH}_2 \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{CH(OH)CH}_3 \) (isopropyl alcohol)
In simple words: This question asks for two distinct hydration reactions of propene: one following anti-Markovnikov's rule to give n-propyl alcohol using hydroboration-oxidation, and one following Markovnikov's rule to give isopropyl alcohol using acid-catalyzed hydration.

🎯 Exam Tip: Differentiate between hydroboration-oxidation (anti-Markovnikov addition of water) and acid-catalyzed hydration (Markovnikov addition of water) for alkenes to achieve specific alcohol products. Knowing the reagents and conditions for each is critical.

 

Question O. What is the action of following on but-2-ene ?
(a) dil alkaline KMnO4
(b) acidic KMnO4
Answer:
(a) Action of dil. alkaline KMnO4 on but-2-ene: \( \text{CH}_3\text{CH=CHCH}_3 \xrightarrow{\text{dil. alk. KMnO}_4} \text{CH}_3\text{CH(OH)CH(OH)CH}_3 \) (Butane-2,3-diol is formed)
(b) Action of acidic KMnO4 on but-2-ene: \( \text{CH}_3\text{CH=CHCH}_3 \xrightarrow{\text{acidic KMnO}_4} 2\text{CH}_3\text{COOH} \) (Two molecules of ethanoic acid are formed)
In simple words: This question concerns the oxidation of but-2-ene with potassium permanganate under different conditions: dilute alkaline KMnO₄ (Baeyer's reagent) causes syn-dihydroxylation to form butan-2,3-diol, while acidic KMnO₄ causes oxidative cleavage of the double bond to form ethanoic acid.

🎯 Exam Tip: Remember that cold, dilute, alkaline KMnO₄ (Baeyer's reagent) forms a diol (syn-addition), whereas hot, acidic KMnO₄ causes oxidative cleavage, breaking the C=C bond and forming carboxylic acids or ketones, or CO₂. Distinguishing these reactions is vital.

 

Question P. Complete the following reaction sequence :
Answer:
In simple words: This question requires completing a chemical reaction sequence, which involves identifying the reactants, reagents, intermediates, and final products based on the given steps. The reaction sequence itself is not provided in the source.

🎯 Exam Tip: For reaction sequences, analyze each step independently, identify the type of reaction (e.g., addition, elimination, substitution, oxidation), and predict the corresponding products. Mastery of named reactions and reagent specificities is key.

 

Question Q. Write the balanced chemical reactions to get benzene from
(a) Sodium benzoate.
(b) Phenol.
Answer:
(a) Sodium benzoate: When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.
\( \text{C}_6\text{H}_5\text{COONa} + \text{NaOH/CaO} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{Na}_2\text{CO}_3 \)
(b) Phenol: When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.
\( \text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO} \)
In simple words: This question asks for reactions to synthesize benzene from sodium benzoate via decarboxylation with soda lime, and from phenol via reduction with zinc dust, both yielding benzene.

🎯 Exam Tip: Remember specific named reactions for benzene synthesis, such as decarboxylation of aromatic carboxylic acid salts (e.g., sodium benzoate) and reduction of phenols (e.g., with zinc dust). Write balanced equations and state conditions clearly.

 

Question R. Predict the possible products of the following reaction.
(a) chlorination of nitrobenzene,
(b) sulfonation of chlorobenzene,
(c) bromination of phenol,
(d) nitration of toluene.
Answer:
(a) Nitro group is meta directing group. So, chlorination of nitrobenzene gives m-chloronitrobenzene.
(b) Chloro group is ortho and para directing group. So, sulphonation of chlorobenzene gives p-chlorobenzene sulphonic acid and o- chlorobenzene sulphonic acid.
(c) Phenolic -OH group is ortho and para directing group. So, bromination of phenol gives p-bromophenol and o-bromophenol.
(d) Methyl group is ortho and para directing group. So, nitration of toluene gives p-nitrotoluene and o-nitrotoluene.
In simple words: This question involves predicting products of electrophilic aromatic substitution reactions, which are governed by the directing effects of existing substituents on the benzene ring. Nitro group is meta-directing, chloro and hydroxyl groups are ortho-para directing, and methyl group is also ortho-para directing.

🎯 Exam Tip: Understand the directing nature (ortho-para or meta) and activating/deactivating effects of common functional groups on an aromatic ring to correctly predict the major products of electrophilic substitution reactions. Remember to account for steric hindrance if multiple positions are possible, often leading to a mixture of ortho and para products where para is major.

 

3. Identify The Main Product Of The Reaction
Answer:
In simple words: This question requires identifying the main product of an unspecified chemical reaction, which typically involves recognizing reaction types and applying relevant chemical principles. The reaction is not provided in the source.

🎯 Exam Tip: To identify the main product, first determine the type of reaction and then apply the appropriate rules (e.g., Markovnikov's, Zaitsev's, directing effects) to predict the most favored product. Always consider reaction conditions.

 

4. Read The Following Reaction And Answer The Questions Given Below.
(The specific reaction mentioned in the question is not provided in the source.)
Question a. Write IUPAC name of the product.
Question b. State the rule that governs formation of this product.
Answer:
(a) IUPAC name of the product: 1-Bromo-2-methylpropane
(b) Anti-Markownikov's rule/Kharasch effect/peroxide effect: It states that, the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov's rule.
In simple words: This question describes the Anti-Markovnikov addition of HBr to an unsymmetrical alkene, catalyzed by peroxides, which produces a product where bromine attaches to the less substituted carbon. The specific reaction leading to 1-Bromo-2-methylpropane is implied but not shown.

🎯 Exam Tip: Clearly distinguish between Markovnikov's (acid-catalyzed HX addition) and Anti-Markovnikov's rule (peroxide-catalyzed HBr addition) for alkenes. Know the specific conditions for each and how to name the resulting products. Understanding the peroxide effect for HBr addition is crucial.

 

5. Identify A, B, C In The Following Reaction Sequence :
Answer:
In simple words: This question asks to identify intermediates (A, B, C) in a given reaction sequence, which requires knowledge of sequential organic reactions and their products. The reaction sequence is not provided in the source.

🎯 Exam Tip: For reaction sequences, analyze each step to determine the reagents, conditions, and type of reaction. Systematically deduce the structure of each intermediate to correctly complete the sequence.

 

6. Identify Giving Reason Whether The Following Compounds Are Aromatic Or Not.
Answer:
(a) A. (Compound structure not provided in source)
Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.
(b) B. (Compound structure not provided in source)
Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.
(c) C. (Compound structure not provided in source)
Compound is aromatic since it has 6π electrons and hence, obeys Huckel rule of aromaticity.
(d) D. (Compound structure not provided in source)
Compound is aromatic since it has 6n electrons and hence, obeys Huckel rule of aromaticity.
In simple words: This question requires determining aromaticity based on Huckel's rule, which states that a cyclic, planar, fully conjugated system with (4n+2) π electrons is aromatic. Compounds with 4n π electrons are typically anti-aromatic or non-aromatic. Examples with 6π electrons (like benzene) are aromatic, while those with 4π electrons (like cyclobutadiene) are non-aromatic or anti-aromatic.

🎯 Exam Tip: To check for aromaticity, verify four criteria: cyclic, planar, fully conjugated (alternating single and double bonds or lone pairs), and follows Huckel's rule (4n+2 π electrons). Compounds with 4n π electrons are usually anti-aromatic (if planar) or non-aromatic (if non-planar). Pay close attention to the number of π electrons.

 

7. Name Two Reagents Used For Acylation Of Benzene.
Answer: The two reagents used for acylation of benzene are:
(i) CH₃COCl (acetyl chloride) and anhydrous AlCl₃
(ii) (CH₃CO)₂O (acetic anhydride) and anhydrous AlCl₃
In simple words: Acylation of benzene, a Friedel-Crafts reaction, involves introducing an acyl group onto the benzene ring using an acyl halide or acid anhydride in the presence of a Lewis acid catalyst like anhydrous AlCl₃.

🎯 Exam Tip: Remember that Friedel-Crafts acylation uses either an acyl chloride (e.g., acetyl chloride) or an acid anhydride (e.g., acetic anhydride) with anhydrous aluminum chloride as a catalyst. These reagents are crucial for introducing acyl groups onto aromatic rings.

 

8. Read The Following Reaction And Answer The Questions Given Below.
(The specific reaction mentioned in the question is not provided in the source.)
Question A. Write the name of the reaction.
Question B. Identify the electrophile in it.
Question C. How is this electrophile generated?
Answer:
(a) The name of the reaction is Friedel-Craft's alkylation reaction.
(b) The electrophile in the reaction is \( \text{CH}_3^+ \).
(c) The electrophile \( \text{CH}_3^+ \) is generated as follows:
\( \text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_4^- \)
In simple words: This question refers to the Friedel-Crafts alkylation reaction, an electrophilic aromatic substitution where an alkyl group (specifically, a methyl group here) is added to an aromatic ring, and the electrophile (\( \text{CH}_3^+ \)) is generated from an alkyl halide and a Lewis acid.

🎯 Exam Tip: For Friedel-Crafts alkylation, know that the electrophile is a carbocation (or a carbocation-like species) generated from an alkyl halide in the presence of a Lewis acid. Understand the generation mechanism of these electrophiles, as this is a common point of examination.

 

Activity: Prepare chart of hydrocarbons and note down the characteristics.
Answer:
Characteristics of hydrocarbons:

• They are chemical compounds that are formed from only hydrogen and carbon atoms.
• Both 'C' and 'H' share an electron pair forming covalent bonds.
• One of the special properties of carbon is its ability to form double and triple bonds (unsaturation). Saturated hydrocarbons are alkanes and cycloalkanes while the unsaturated hydrocarbons are the aromatics, alkenes and alkynes.
• All hydrocarbons are insoluble in water, their boiling point increases as the size of alkane increases.
• All hydrocarbons can reach complete oxidation.
• Hydrocarbons are mainly used as fuel for transport and industry.

In simple words: This activity encourages students to consolidate their understanding of hydrocarbons by creating a chart that outlines different types (alkanes, alkenes, alkynes, aromatics) and their key properties, such as being insoluble in water, their role as fuels, and their ability to form saturated or unsaturated bonds.

🎯 Exam Tip: Creating summary charts for different classes of hydrocarbons (alkanes, alkenes, alkynes, aromatics) can be a highly effective study method. Focus on their general formula, bonding, typical reactions, and key physical properties to prepare for comparative questions and demonstrate a comprehensive understanding.

 

11th Chemistry Digest Chapter 15 Hydrocarbons Intext Questions And Answers

 

Can You Recall? (Textbook Page No. 233)

 

Question i. What are hydrocarbons?
Answer: The compounds which contain carbon and hydrogen as the only elements are called hydrocarbons.
In simple words: Hydrocarbons are organic compounds composed solely of carbon and hydrogen atoms, forming the fundamental structure of many organic molecules.

🎯 Exam Tip: A basic definition of hydrocarbons, distinguishing them by their elemental composition, is a fundamental concept in organic chemistry and a common introductory question.

 

Question ii. Write structural formulae of the following compounds: propane, ethyne, cyclobutane, ethene, benzene.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में प्रोपेन, एथाइन, साइक्लोब्यूटेन, एथीन और बेंजीन की संरचनात्मक सूत्र दर्शाए जाएंगे, जो इन हाइड्रोकार्बन यौगिकों की परमाणु व्यवस्था को स्पष्ट रूप से चित्रित करेंगे।
In simple words: This question asks for the chemical structures of propane, ethyne, cyclobutane, ethene, and benzene to understand their atomic arrangement.

🎯 Exam Tip: Drawing accurate structural formulae for common hydrocarbons is crucial for understanding organic chemistry and is a frequent exam requirement.

Do You Know? (Textbook Page No. 233)

 

Question 1. Why are alkanes called paraffins?
Answer:
(i) Alkanes contain only carbon-carbon and carbon-hydrogen single covalent bonds.
(ii) They are chemically less reactive and do not have much affinity for other chemicals. Hence, they are called paraffins.
In simple words: Alkanes are called paraffins because they consist only of single bonds and are chemically unreactive due to their stable structure.

🎯 Exam Tip: Understanding the term "paraffins" helps grasp the basic properties of alkanes, particularly their low reactivity, which is important for identifying their chemical behavior.

Internet My Friend. (Textbook Page No. 233)

 

Question 1. Collect information about hydrocarbon.
Answer:
• In organic chemistry, a hydrocarbon is an organic compound consisting of carbon and hydrogen as the only elements.
• They are examples of group 14 hydrides.
• Alkanes, cycloalkanes, aromatic hydrocarbons are different types of hydrocarbons.
• Most of the hydrocarbons found on earth occur naturally in crude oil.
• They mainly undergo substitution, addition or combustion reactions.
• Most hydrocarbons are flammable and toxic.
Thou are the primam source in the form of combustible fuel source.
In simple words: Hydrocarbons are organic compounds made exclusively of carbon and hydrogen, forming various types like alkanes, alkenes, and aromatics, primarily used as fuels.

🎯 Exam Tip: A comprehensive understanding of hydrocarbons, including their definition, types, and general properties, forms the foundation for advanced organic chemistry topics.

[Note: Students are expected to collect additional information on their own]

Use Your Brain Power! (Textbook Page No. 234)

 

Question 1.
(i) Write the structures of all the chain isomers of the saturated hydrocarbon containing six carbon atoms.
(ii) Write IUPAC names of all the above structures.
Answer:The structural formulae and names of all possible isomers having molecular formula \(C_6H_{14}\) are as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में आणविक सूत्र C6H14 वाले संतृप्त हाइड्रोकार्बन के सभी श्रृंखला समावयवियों की संरचनात्मक सूत्र और उनके IUPAC नाम दर्शाए जाएंगे, जो विभिन्न संरचनात्मक व्यवस्थाओं को स्पष्ट करेंगे।
In simple words: This question requires identifying and naming all possible chain isomers for the saturated hydrocarbon with the formula \(C_6H_{14}\).

🎯 Exam Tip: Mastering isomer identification and IUPAC nomenclature for hydrocarbons, especially alkanes, is essential for scoring well in organic chemistry sections.

Note: Alkanes and isomer number

Number of Carbon	Alkane	Number of isomers
1	Methane	No structural isomer
2	Ethane	No structural isomer
3	Propane	No structural isomer
4	Butane	Two
5	Pentane	Three
6	Hexane	Five

Can You Recall? (Textbook Page No. 235)

 

Question i. What is a catalyst?
Answer:A catalyst is a substance that can be added to a reaction to increase the reaction rate without getting consumed in the process. e.g. Ni is used as a catalyst in the catalytic hydrogenation of alkenes or alkynes.
In simple words: A catalyst speeds up a chemical reaction without being used up itself.

🎯 Exam Tip: Knowing the definition and a common example of a catalyst is fundamental in chemical kinetics and reaction mechanisms, often asked in basic concept questions.

 

Question ii. What is addition reaction?
Answer:When a compound combines with another compound to form a product that contain all the atoms in both the reactants, it is called an addition reaction.
In simple words: An addition reaction occurs when two or more molecules combine to form a larger single molecule.

🎯 Exam Tip: Differentiating between addition, substitution, and elimination reactions is crucial for understanding reaction types in organic chemistry; focus on the defining characteristic of bond formation.

Try This (Textbook Page No. 235)

 

Question 1. Transform the following word equation into balanced chemical equation and write at least 3 changes that occur at molecular level during this chemical change. 2-Methylpropene + Hydrogen \( \xrightarrow{catalyst} \) Isobutane
Answer:Three changes which occur at molecular level include: Step 1: Adsorption of reactants: Reactants (alkene and hydrogen) get adsorbed on the catalytic surface. Step 2: Formation of a product: Hydrogen atoms are added across the double bond of 2-methylpropene which results in the formation of product isobutane. Step 3: Desorption: Product formed on the catalytic surface is readily desorbed making catalytic surface available for other molecules.
In simple words: This question describes the catalytic hydrogenation of 2-methylpropene to isobutane, involving adsorption, reaction, and desorption steps on the catalyst surface.

🎯 Exam Tip: When explaining catalytic reactions, remember to detail the three key steps-adsorption, surface reaction, and desorption-as they provide a complete molecular-level picture.

Use Your Brain Power! (Textbook Page No. 236)

 

Question 1. Why are alkanes insoluble in water and readily soluble in organic solvents like chloroform or ether?
Answer:
• The solubility of any substance is governed by the principle of like dissolves like. This means polar compounds are soluble in polar solvents while nonpolar compounds are soluble in nonpolar solvents.
• Alkanes consist of C-C and C-H nonpolar covalent bonds and thus, they are nonpolar in nature, whereas water is a polar solvent.
• The dipole-dipole forces that exist between water molecules is much stronger than the forces of attraction between alkane and water molecules. Hence, alkanes are insoluble in water and readily soluble in organic solvents like chloroform or ether.
In simple words: Alkanes are nonpolar and thus dissolve in other nonpolar organic solvents, but not in polar water, following the "like dissolves like" principle.

🎯 Exam Tip: The "like dissolves like" principle is a fundamental concept for explaining solubility; ensure you clearly define polarity and its role in intermolecular forces when answering such questions.

Can You Recall? (Textbook Page No. 238)

 

Question 1. What is the product which is poisonous and causes air pollution formed by incomplete combustion of alkane?
Answer:When alkanes are subjected to incomplete combustion, it forms carbon monoxide and carbon (soot).
(i) \(2CH_4(g) + 3O_2(g) \rightarrow 2CO(g) + 4H_2O(g)\)
(ii) \(CH_4(g) + O_2(g) \rightarrow C(s) + 2H_2O(l)\)
In simple words: Incomplete combustion of alkanes produces poisonous carbon monoxide and soot, both of which are significant air pollutants.

🎯 Exam Tip: Remember the primary hazardous products of incomplete combustion (carbon monoxide and soot) and be ready to write their balanced chemical equations, as this is a common environmental chemistry question.

Can You Recall? (Textbook Page No. 238)

 

Question i. What are alkenes?
Answer:Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon
In simple words: Alkenes are hydrocarbons that contain at least one carbon-carbon double bond, making them unsaturated.

🎯 Exam Tip: Define alkenes by their carbon-carbon double bond, which dictates their unsaturated nature and higher reactivity compared to alkanes.

 

Question ii. Calculate the number of sigma (σ) and pi (π) bonds in 2-methylpropene.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में 2-मेथिलप्रोपीन की संरचना में सिग्मा (σ) और पाई (π) बंधों की संख्या की गणना की जाएगी। इसके लिए पहले 2-मेथिलप्रोपीन की संरचना बनानी होगी और फिर उसमें उपस्थित सभी एकल बंधों (सिग्मा) और दोहरे बंधों (एक सिग्मा, एक पाई) को गिनना होगा।
In simple words: To find the number of sigma and pi bonds in 2-methylpropene, one must first draw its structure and then count all the single bonds (sigma) and double bonds (one sigma, one pi).

🎯 Exam Tip: Accurately drawing the structural formula is the first step to correctly counting sigma and pi bonds; remember that a double bond consists of one sigma and one pi bond.

 

Question iii. Write the structural formula of pent-2-ene.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में पेंट-2-ईन का संरचनात्मक सूत्र दर्शाया जाएगा, जो इस असंतृप्त हाइड्रोकार्बन में कार्बन-कार्बन दोहरे बंध की स्थिति को स्पष्ट करेगा।
In simple words: This question asks for the structural representation of pent-2-ene, showing the five-carbon chain with a double bond at the second carbon.

🎯 Exam Tip: Practice drawing structural formulas for alkenes, paying close attention to the position of the double bond as indicated by the numeral in the IUPAC name.

Can You Tell? (Textbook Page No. 241)

 

Question i. Explain by writing a reaction, the main product formed on heating 2-methylbutan-2-ol with concentrated sulphuric acid.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में 2-मेथिलब्यूटेन-2-ओल को सांद्र सल्फ्यूरिक एसिड के साथ गर्म करने पर होने वाली अभिक्रिया को रासायनिक समीकरण के साथ समझाया जाएगा, जिसमें मुख्य उत्पाद का निर्माण दिखाया जाएगा।
In simple words: Heating 2-methylbutan-2-ol with concentrated sulfuric acid results in a dehydration reaction, forming an alkene as the major product according to Zaitsev's rule.

🎯 Exam Tip: For dehydration reactions of alcohols, always consider Zaitsev's rule to predict the major alkene product, which will be the most substituted alkene.

 

Question ii. Will the main product in the above reaction show geometrical isomerism?
Answer:No, the major product, i.e., 2-methylbut-2-ene does not show geometrical (or cis-trans) isomerism.
In simple words: The major product, 2-methylbut-2-ene, does not exhibit geometrical isomerism because one of the doubly bonded carbons has two identical groups attached.

🎯 Exam Tip: Geometrical isomerism in alkenes requires that each carbon atom of the double bond be attached to two different groups; check for identical groups on either carbon to rule out cis-trans isomers.

 

Question 1. Propan-1-ol and 2-methypropan-1-ol are not prepared by hydration method. Why?
Answer:Propan-1-ol and 2-methylpropan-1-ol cannot be prepared by hydration of propene and 2-methylprop-1-ene because the addition reaction follows Markovnikov's rule.
In simple words: These specific primary alcohols cannot be formed by alkene hydration because the reaction follows Markovnikov's rule, favoring secondary or tertiary alcohol formation.

🎯 Exam Tip: Markovnikov's rule is crucial for predicting products in addition reactions to unsymmetrical alkenes; always consider the regioselectivity when synthesizing alcohols via hydration.

Use Your Brainpower. (Textbook Page No. 244)

 

Question 1. On ozonolysis, an alkene forms the following carbonyl compounds. Draw the structure of unknown alkene from which these compounds are formed: HCHO and \(CH_3COCH_2CH_3\)
Answer:The structure of alkene which produces a mixture of HCHO and \(CH_3COCH_2CH_3\) on ozonolysis is
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में उस एल्कीन की संरचना दर्शाई जाएगी जो ओजोनोलिसिस पर HCHO और CH3COCH2CH3 का मिश्रण उत्पन्न करती है। इस संरचना को दोनों कार्बोनिल यौगिकों को जोड़कर व्युत्पन्न किया जाएगा।
In simple words: To find the original alkene, mentally remove the oxygen atoms from HCHO (formaldehyde) and \(CH_3COCH_2CH_3\) (butanone) and join the remaining carbon fragments at the original double bond position.

🎯 Exam Tip: For ozonolysis problems, reverse the reaction by removing the oxygen atoms from the carbonyl products and connecting the two carbon fragments to reconstruct the original alkene's double bond.

Use Your Brain Power! (Textbook Page No. 245)

 

Question 1. Write the structure of monomer from which each of the following polymers are obtained.
Answer:

i.	Teflon	\(CF_2 - CF_2\)	Tetrafluoroethene
ii.	Polypropene	\(H_3C - CH = CH_2\)	Propene
iii.	Polyvinyl chloride	\(H_2C = CHCl\)	Vinyl chloride

In simple words: This question requires identifying the basic repeating unit (monomer) for each given polymer by recognizing the characteristic functional groups.

🎯 Exam Tip: For polymerization questions, remember that the monomer is typically the unsaturated compound (alkene or substituted alkene) from which the saturated polymer chain is formed by breaking the double bond.

Can You Tell? (Textbook Page No. 246)

 

Question i. What are aliphatic hydrocarbons?
Answer:Aliphatic hydrocarbons are hydrocarbons containing carbon and hydrogen joined together in straight chain or branched chain. They may be saturated (alkanes) or unsaturated (alkenes or alkynes).
In simple words: Aliphatic hydrocarbons are organic compounds of carbon and hydrogen forming open chains or rings, which can be saturated (single bonds) or unsaturated (double/triple bonds).

🎯 Exam Tip: Clearly distinguish aliphatic from aromatic hydrocarbons, focusing on the open-chain or non-aromatic cyclic structures and the presence or absence of unsaturation.

 

Question ii. Compare the proportion of carbon and hydrogen atoms in ethane, ethene and ethyne. Which compound is most unsaturated with hydrogen?
Answer:Ethane C:H= 2:6 = 1:3 Ethene C:H=2:4 = 1:2 Ethyne C: H = 2:2 = 1:1 From the above proportion it is clear that ethyne with 1 : 1 ratio of C : H is most unsaturated with hydrogen (50%) as compared to ethane (25%) and ethene (33.33%).
In simple words: Ethyne has the lowest hydrogen-to-carbon ratio (1:1), making it the most unsaturated among ethane (1:3) and ethene (1:2).

🎯 Exam Tip: To determine the degree of unsaturation, compare the hydrogen-to-carbon ratio; a lower ratio indicates greater unsaturation and fewer hydrogen atoms per carbon.

Can You Tell? (Textbook Page No. 247)

 

Question 1. Why is sodamide used in dehydrohalogenation of vicinal dihalides to remove HX from alkenyl halide in place of alcoholic KOH?
Answer:
• Sodamide \(NaNH_2\) is a strong base and hence, helps in complete conversion of alkenyl halide formed in the first step to form alkynes.
• The base (KOH or NaOH) used in first step gives alkynes in poor yield and hence, stronger bases such as \(NaNH_2\) on \(KNH_2\) are used in second step.
In simple words: Sodamide is a stronger base than alcoholic KOH, ensuring a complete and higher yield conversion of vicinal dihalides to alkynes through dehydrohalogenation.

🎯 Exam Tip: For achieving complete dehydrohalogenation to alkynes, select strong bases like sodamide; weaker bases like alcoholic KOH may only effect a single dehydrohalogenation, yielding an alkene.

Use Your Brainpower! (Textbook Page No. 247)

 

Question 1. Convert: 1-Bromobutane to hex-1-yne
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में 1-ब्रोमोब्यूटेन को हेक्स-1-आइन में परिवर्तित करने की रासायनिक अभिक्रिया को चरण-दर-चरण समीकरणों के साथ दर्शाया जाएगा, जिसमें आवश्यक अभिकर्मक और मध्यवर्ती उत्पाद शामिल होंगे।
In simple words: Converting 1-bromobutane to hex-1-yne involves a series of reactions, likely starting with elimination to form an alkene, followed by halogenation and a second elimination or an acetylide reaction to introduce the triple bond and extend the chain.

🎯 Exam Tip: Multi-step synthesis questions require a clear understanding of reaction mechanisms and reagents; break down the conversion into smaller, manageable steps like elimination, substitution, and chain extension.

Can You Tell? (Textbook Page No. 248)

 

Question 1. Alkanes and alkenes do not react with lithium amide. Give reason.
Answer:
(i) The sp hybrid carbon atom in terminal alkynes is more electronegative than the \(sp^2\) carbon in ethene or the \(sp^3\) carbon in ethane.
(ii) Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (\(H^+\)) to very strong base as shown in the reactions below.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में टर्मिनल एल्काइन के हाइड्रोजन परमाणु को एक प्रोटॉन (H+) के रूप में एक बहुत मजबूत आधार द्वारा दिए जाने की रासायनिक अभिक्रियाओं को दर्शाया जाएगा, जो एल्काइनों की अम्लीय प्रकृति को स्पष्ट करेंगी।
(iii) Further, since s-character decreases from sp to \(sp^2\) to \(sp^3\) carbon atom, the relative acidity of alkanes, alkenes and alkynes is in the following order: \(H - C \equiv C - H > H_2C = CH_2 > H_3C - CH_3\) Hence, alkenes and alkanes do not react with lithium amide.
In simple words: Alkanes and alkenes are not acidic enough to react with lithium amide because their C-H bonds are less polar than the C-H bond in terminal alkynes, which have higher s-character.

🎯 Exam Tip: Focus on the acidity of the C-H bond and the s-character of the hybridization; higher s-character leads to increased electronegativity and greater acidity, enabling reaction with strong bases like lithium amide.

 

Question 1. Arrange following hydrocarbons in the increasing order of acidic character: propane, propyne, propene.
Answer:Propyne > propene > propane
In simple words: Propyne is the most acidic due to its sp hybridized carbon, followed by propene (sp2), and then propane (sp3), which is the least acidic.

🎯 Exam Tip: The acidic character of hydrocarbons is directly related to the s-character of the hybridized carbon atom; higher s-character (sp > sp2 > sp3) results in increased acidity due to greater electronegativity.

Use Your Brain Power! (Textbook Page No. 249)

 

Question 1. Convert: 3-Methylbut-l-yne into 3-methylbutan-2-one
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस प्रश्न के उत्तर में 3-मेथिलब्यूट-1-आइन को 3-मेथिलब्यूटेन-2-ओन में परिवर्तित करने की रासायनिक अभिक्रिया को चरण-दर-चरण समीकरणों के साथ दर्शाया जाएगा, जिसमें आवश्यक अभिकर्मक और मध्यवर्ती उत्पाद शामिल होंगे।
In simple words: This conversion involves the hydration of an alkyne (3-methylbut-1-yne) in the presence of acid and mercuric sulfate, which follows Markovnikov's rule to form an enol that tautomerizes into the ketone (3-methylbutan-2-one).

🎯 Exam Tip: When converting alkynes to ketones, remember that alkyne hydration (with \(H_2SO_4/HgSO_4\)) is key; Markovnikov's rule guides regioselectivity, and the initial enol always tautomerizes to a more stable keto form.

Can You Recall? (Textbook Page No. 249)

 

Question i. What are aromatic hydrocarbons?
Answer:Benzene and all compounds that have structures and chemical properties resembling benzene are called as aromatic hydrocarbons.
In simple words: Aromatic hydrocarbons are cyclic, planar molecules with a conjugated system of pi electrons that follow Huckel's rule (4n+2 pi electrons), like benzene.

🎯 Exam Tip: When defining aromatic hydrocarbons, always include the key criteria: cyclic structure, planarity, conjugation, and adherence to Huckel's rule (4n+2 pi electrons).

 

Question ii. What are benzenoid and non-benzenoid aromatics?
Answer:Benzenoid aromatics are compounds having at least one benzene ring in the structure. e.g. Benzene, naphthalene, anthracene, phenol, etc., Non-benzenoid aromatics are compounds that contain an aromatic ring, other than benzene. e.g. Tropone, etc.
In simple words: Benzenoid aromatics contain at least one benzene ring, while non-benzenoid aromatics are compounds with an aromatic character but without a benzene ring.

🎯 Exam Tip: Distinguish between benzenoid and non-benzenoid aromatics by the presence or absence of a benzene ring; both categories must still satisfy Huckel's rule for aromaticity.

Can You Recall? (Textbook Page No. 254)

 

Question 1. What is decarboxylation?
Answer:The reaction which involves removal of a carboxyl group (-COOH) in the form of carbon dioxide \((CO_2)\) is known as decarboxylation reaction. \(R - COOH \rightarrow R - H + CO_2\)
In simple words: Decarboxylation is a chemical reaction that removes a carboxyl group from a molecule, releasing it as carbon dioxide.

🎯 Exam Tip: Decarboxylation is an important reaction for shortening carbon chains and typically involves heating a carboxylic acid or its salt, often in the presence of soda lime.

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