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NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons


Ques 13.1: How do you account for the formation of ethane during chlorination of methane? 

Ans: Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps.

Step 1: Initiation:

The reaction begins with the homolytic cleavage of Cl – Cl bond as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Step 2: Propagation:

In the second step, chlorine free radicals attack methane molecules and break down the C–H bond to generate methyl radicals as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and CH3Cl are the major products formed, other higher halogenated compounds are also formed as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Step 3: Termination:

Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

Ques 13.2: Write IUPAC names of the following compounds:

a. NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                                            

b.  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

c. NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                                                                 

d.  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons 

 e. NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                                                               

f.  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons         

 

g.  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 

Ans: (a)

  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons  
IUPAC name: 2-Methylbut-2-ene

(b)   NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons          
IUPAC name: Pen-1-ene-3-yne

(c) NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons  can be written as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons
IUPAC name: 1, 3-Butadiene or Buta-1,3-diene

(d) NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons               
IUPAC name: 4-Phenyl but-1-ene

(e)   NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                     
IUPAC name: 2-Methyl phenol

(f)

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

IUPAC name: 5-(2-Methylpropyl)-decane

(g)

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

IUPAC name: 4-Ethyldeca-1, 5, 8-triene

Ques 13.3: For the following compounds, write structural formulas and IUPAC names for  all possible isomers having the number of double or triple bond as indicated:

(a) C4H8 (one double bond)                                               

(b) C5H8 (one triple bond)

Ans: (a) The following structural isomers are possible for C4H8 with one double bond:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

   NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                                      

The IUPAC name of

Compound (I) is But-1-ene,

Compound (II) is But-2-ene, and

Compound (III) is 2-Methylprop-1-ene.

(b) The following structural isomers are possible for C5C8 with one triple bond:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons          
 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons       
NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                                        

The IUPAC name of

Compound (I) is Pent-1-yne,

Compound (II) is Pent-2-yne, and

Compound (III) is 3-Methylbut-1-ene.

Ques 13.4: Write IUPAC names of the products obtained by the ozonolysis of the following compounds:

(i) Pent-2-ene                                                

(ii) 3,4-Dimethyl-hept-3-ene

(iii) 2-Ethylbut-1-ene                                  

(iv) 1-Phenylbut-1-ene

Ans: (i) Pent-2-ene undergoes ozonolysis as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The IUPAC name of Product (I) is ethanal and Product (II)is propanal.

(ii) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The IUPAC name of Product (I)is butan-2-one and Product (II)is Pentan-2-one.

(iii) 2-Ethylbut-1-ene undergoes ozonolysis as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The IUPAC name of Product (I)is pentan-3-one and Product (II)is methanal.

(iv) 1-Phenylbut-1-ene undergoes ozonolysis as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The IUPAC name of Product (I)is benzaldehyde and Product (II)is propanal.

Ques 13.5: An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.

Ans:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

During ozonolysis, an ozonide having a cyclic structure is formed as an intermediate which undergoes cleavage to give the final products. Ethanal and pentan-3-one are obtained from the intermediate ozonide. Hence, the expected structure of the ozonide is:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

This ozonide is formed as an addition of ozone to ‘A’. The desired structure of ‘A’ can be obtained by the removal of ozone from the ozonide. Hence, the structural formula of ‘A’ is:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The IUPAC name of ‘A’ is 3-Ethylpent-2-ene.

Ques 13.6: An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Ans: As per the given information, ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Hence, the structure of ‘A’ can be represented as:  XC = CX

There are eight C–H σ bonds. Hence, there are 8 hydrogen atoms in ‘A’. Also, there are three C–C bonds. Hence, there are four carbon atoms present in the structure of ‘A’.

Combining the inferences, the structure of ‘A’ can be represented as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

‘A’ has 3 C–C bonds, 8 C–H σ bonds, and one C–C π bond.

Hence, the IUPAC name of ‘A’ is But-2-ene.

Ozonolysis of ‘A’ takes place as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The final product is ethanal with molecular mass

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Ques 13.7: Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Ans: As per the given information, propanal and pentan-3-one are the ozonolysis products of an alkene. Let the given alkene be ‘A’. Writing the reverse of the ozonolysis reaction, we get:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The products are obtained on the cleavage of ozonide ‘X’. Hence, ‘X’ contains both products in the cyclic form. The possible structure of ozonide can be represented as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Now, ‘X’ is an addition product of alkene ‘A’ with ozone. Therefore, the possible structure of alkene ‘A’ is:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Ques 13.8: Write chemical equations for combustion reaction of the following hydrocarbons:

(i) Butane 
 (ii) Pentene
 (iii) Hexyne
 (iv) Toluene

Ans: Combustion can be defined as a reaction of a compound with oxygen.

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

(iv)     NCERT Solutions Class 11 Chemistry Chapter 9 - HydrocarbonsNCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons                   

         Toluene

Ques 13.9: Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Ans: Hex-2-ene is represented as:

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

trans isomer

The dipole moment of cis-compound is a sum of the dipole moments of C–CH3 and 

C–CH2CH2CH3 bonds acting in the same direction.

The dipole moment of trans-compound is the resultant of the dipole moments of C–CH3 and C–CH2CH2CH3 bonds acting in opposite directions.

Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.

Ques 13.10: Why is benzene extra ordinarily stable though it contains three double bonds?

Ans: Benzene is a hybrid of resonating structures given as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

All six carbon atoms in benzene are sp2 hybridized. The two sp2 hybrid orbitals of each carbon atom overlap with the sp2 hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining sp2 hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six sigma C–H bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three π bonds by the lateral overlap of NCERT Solutions Class 11 Chemistry Chapter 9 - HydrocarbonsNCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons  .

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The six π ’s are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized π -electrons stabilize benzene.

Ques 13.11: What are the necessary conditions for any system to be aromatic?

Ans: A compound is said to be aromatic if it satisfies the following three conditions:

(i) It should have a planar structure.

(ii) The π –electrons of the compound are completely delocalized in the ring.

(iii) The total number of π –electrons present in the ring should be equal to (4n + 2), where n = 0, 1, 2 … etc. This is known as Huckel’s rule.

Ques 13.12: Explain why the following systems are not aromatic?

(i)       NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons   
 (ii)     NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons  
 (iii) NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons   

 

Ans: (i) 

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

For the given compound, the number of π-electrons is 6.But only four π-electrons are present within the ring. Also there is no conjugation of π-electrons within the ring and the compound is not planar in shape. Hence, the given compound is not aromatic in nature.

(ii) 

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

For the given compound, the number of π -electrons is 4.

By Huckel’s rule,

4n + 2 = 4

4n = 2

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…), which is not true for the given compound. Hence, it is not aromatic in nature.

(iii)  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

For the given compound, the number of π-electrons is 8.

By Huckel’s rule,

4n + 2 = 8

4n = 6

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2…). Since the value of n is not an integer, the given compound is not aromatic in nature.

Ques 13.13: How will you convert benzene into

(i) p-nitrobromobenzene
 (ii) m-nitrochlorobenzene
 (iii) p -nitrotoluene
 (iv) acetophenone

Ans: (i) Benzene can be converted into p-nitrobromobenzene as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

(ii) Benzene can be converted into m-nitrochlorobenzene as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

(iii) Benzene can be converted into p-nitrotoulene as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

(iv) Benzene can be converted into acetophenone as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Ques 13.14: In the alkane H3C–CH2–C(CH3)2–CH2–CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.

Ans:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

1° carbon atoms are those which are bonded to only one carbon atom i.e., they have only one carbon atom as their neighbour. The given structure has five 1° carbon atoms and fifteen hydrogen atoms attached to it.

2° carbon atoms are those which are bonded to two carbon atoms i.e., they have two carbon atoms as their neighbours. The given structure has two 2° carbon atoms and four hydrogen atoms attached to it.

3° carbon atoms are those which are bonded to three carbon atoms i.e., they have three carbon atoms as their neighbours. The given structure has one 3° carbon atom and only one hydrogen atom is attached to it.

Ques 13.15: What effect does branching of an alkane chain has on its boiling point?

Ans: Alkanes experience inter-molecular Van der Waals forces. The stronger the force, the greater will be the boiling point of the alkane.

As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.

Ques 13.16: Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Ans: Addition of HBr to propene is an example of an electrophilic substitution reaction.

Hydrogen bromide provides an electrophile, H . This electrophile attacks the double bond to form 1° and 2° carbocations as shown:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br attacks the carbocation to form 2 – bromopropane as the major product.

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

This reaction follows Markovnikov’s rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms.

In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov’s rule. The reaction follows a free radical chain mechanism as:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 

Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1 – bromopropane is obtained as the major product.

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide. 

Ques 13.17: Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Ans: o-xylene has two resonance structures:

  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

All three products, i.e., methyl glyoxal, 1, 2-demethylglyoxal, and glyoxal are obtained from two Kekule structures. Since all three products cannot be obtained from any one of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule structures (I and II).

Ques 13.18: Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.

Ans: Acidic character of a species is defined on the basis of ease with which it can lose its H–atoms.

The hybridization state of carbon in the given compound is:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

As the s–character increases, the electronegativity of carbon increases and the electrons of C–H bond pair lie closer to the carbon atom. As a result, partial positive charge of H atom increases and H+  ions are set free.

The s–character increases in the order:

sp3sp2sp

Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

Ques 13.19: Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

Ans: Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles.

Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.

Ques 13.20: How would you convert the following compounds into benzene?

(i) Ethyne
 (ii) Ethene
 (iii) Hexane

Ans: (i) Benzene from Ethyne:

  NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

(ii) Benzene from Ethene:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

(iii) Hexane to Benzene

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 

Ques 13.21: Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Ans: The basic skeleton of 2-methylbutane is shown below:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

On the basis of this structure, various alkenes that will give 2-methylbutane on hydrogenation are:

(a)

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons           

(b)

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons            

 

(c)

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 

Ques 13.22: Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+

(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene

(b) Toluene, p-H3C–C6H4–NO2p-O2N–C6H4–NO2.

Ans: Electrophiles are reagents that participate in a reaction by accepting an electron pair in order to bond to nucleophiles. The higher the electron density on a benzene ring, the more reactive is the compound towards an electrophile, E+  (Electrophilic reaction).

(a) The presence of an electron withdrawing group (i.e., NO2and Cl) deactivates the aromatic ring by decreasing the electron density.

Since NO2–group is more electron withdrawing (due to resonance effect) than the Cl– group (due to inductive effect), the decreasing order of reactivity is as follows:

Chlorobenzene > p – nitrochlorobenzene > 2, 4 – dinitrochlorobenzene

(b) While CH3– is an electron donating group, NO2– group is electron withdrawing. Hence, toluene will have the maximum electron density and is most easily attacked by E+. NO2– is an electron withdrawing group. Hence, when the number of NO2– substituents is greater, the order is as follows:

Toluene > p–CH3–C6H4–NO2p –O2 N–C6H4–NO2

Ques 13.23: Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?

Ans: The ease of nitration depends on the presence of electron density on the compound to form nitrates. Nitration reactions are examples of electrophilic substitution reactions where an electron-rich species is attacked by a nitronium ion (NO2–).

Now, CH3– group is electron donating and NO2– is electron withdrawing. Therefore, toluene will have the maximum electron density among the three compounds followed by benzene. On the other hand, m– Dinitrobenzene will have the least electron density. Hence, it will undergo nitration with difficulty. Hence, the increasing order of nitration is as follows:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

 

Ques 13.24: Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

Ans: The ethylation reaction of benzene involves the addition of an ethyl group on the benzene ring. Such a reaction is called a Friedel-Craft alkylation reaction. This reaction takes place in the presence of a Lewis acid.

Any Lewis acid like anhydrous FeCl3, SnCl4, BF3 etc. can be used during the ethylation of benzene.

Ques 13.25: Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

Ans: Wurtz reaction is limited for the synthesis of symmetrical alkanes (alkanes with an even number of carbon atoms) In the reaction, two similar alkyl halides are taken as reactants and an alkane, containing double the number of carbon atoms, are formed. Example:

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

Wurtz reaction cannot be used for the preparation of unsymmetrical alkanes because if two dissimilar alkyl halides are taken as the reactants, then a mixture of alkanes is obtained as the products. Since the reaction involves free radical species, a side reaction also occurs to produce an alkene. For example, the reaction of bromomethane and iodoethane gives a mixture of alkanes.

 NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

The boiling points of alkanes (obtained in the mixture) are very close. Hence, it becomes difficult to separate them.

The document NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons is a part of the NEET Course Chemistry Class 11.
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FAQs on NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

1. What are hydrocarbons and their significance?
Ans. Hydrocarbons are organic compounds consisting of hydrogen and carbon atoms. They are significant because they form the basis of fossil fuels such as coal, petroleum, and natural gas, which are vital for energy production and various industrial processes.
2. How are hydrocarbons classified?
Ans. Hydrocarbons are classified into two main types: aliphatic and aromatic hydrocarbons. Aliphatic hydrocarbons have straight or branched chains, while aromatic hydrocarbons have a ring structure. Both types can further be categorized based on the presence of single, double, or triple bonds between carbon atoms.
3. What is the difference between saturated and unsaturated hydrocarbons?
Ans. Saturated hydrocarbons only contain single bonds between carbon atoms, while unsaturated hydrocarbons have double or triple bonds. Saturated hydrocarbons are typically more stable and less reactive, while unsaturated hydrocarbons are more reactive and prone to undergo addition reactions.
4. How do hydrocarbons contribute to air pollution?
Ans. Hydrocarbons, especially volatile organic compounds (VOCs), can contribute to air pollution. When hydrocarbons react with nitrogen oxides and sunlight, they form ground-level ozone, which is a major component of smog. Ozone can cause respiratory problems and damage to plants and materials.
5. What are the environmental impacts of hydrocarbon spills?
Ans. Hydrocarbon spills can have severe environmental impacts. They can contaminate soil, water bodies, and groundwater, leading to the death of aquatic life and long-term damage to ecosystems. Hydrocarbons can also volatilize and contribute to air pollution, posing health risks to humans and animals. Cleanup and remediation of hydrocarbon spills are challenging and costly processes.
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