Class 11 - Chemistry - Hydrocarbons
How do you account for the formation of ethane during chlorination of methane?
Chlorination of methane occurs by the following mechanism. It is a free radical reaction mechanism.
Write IUPAC names of the following compounds:
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)
Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
An alkene ‘A’ on ozonolysis gives a mixture of ethanol and pentan-3-one. Write structure and IUPAC name of ‘A’.
An alkene (A) when subjected to ozonloysis cleaves at the double bond and give rise to carbonyl compounds.
Since, given compound is ethanal and pentan-3-one, alkene can be
An alkene ‘A’ contains three C-C, eight C-H sigma bond and one C-C pi bond. ‘A’ on ozonolysis gives
two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
Hence, there are 3 C-C, 8 C-H a-bonds and one C-C pi bond.
Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
Write chemical equations for combustion reaction of the following hydrocarbons:
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Structures of cis- and trans-isomer of hex-2-ene are as below.
Cis-Hex-2-ene have higher boiling point and higher dipole moment.
Trans-Hex-2-ene has lower dipole moment and lower boiling point.
Boiling point depends on dipole-dipole interactions.
Why is benzene extra ordinarily stable though it contains three double bonds?
Extra stability of benzene:
What are the necessary conditions for any system to be aromatic?
Where, n = number of electrons (1, 2, 3, etc.)
Explain why the following systems are not aromatic?
How will you convert benzene into
In the alkane H3C-CH2-C(CH3)2-CH2-CH(CH3)2, identify 1o,2o,3o carbon atoms and give the number of H atoms bonded to each one of these.
Expanded formula is given below:
What effect does branching of an alkane chain has on its boiling point?
Branching of carbon atom chain, decreases boiling point of alkane. Reduced surface area results in
decreased Van der Waal’s interaction and finally leads to lower boiling point.
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.
Addition of HBr to propene is an ionic electrophilic addition reaction in which H+ is an electrophile which adds to gives a more stable 2o carbocation.
During the next step, carbocation is rapidly attached by the nucleophile Br-ion to give 2-bromopropane.
In the presence of benzoyl peroxide, reaction is electropilic. Br is free radical which is obtained by the action of benzoyl peroxide on HBr.
Br radical adds to propene in such a way to generate the more stable 2o free radical.
In the next step, free radical obtained rapidly abstract a hydrogen atom from HBr to give 1-Bromopropane.
Both these reactions are electrophilic addition reactions, but gives different products due to different order of addition of H and Br atoms.
Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?
o-xylene is regarded as a resonance hybrid of the following two Kekule structures.
Ozonolysis of these products is given below:
All the three products formed cannot be obtained from anyone of two Kekule structures.
This shows that o-xylene is the resonance hybrid of two kekule structures.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behavior. Also give reason for this behavior.
s-electrons are closer to the nucleus. As the s-character of the orbital making C-H bond increases,
the electrons of C-H bond lie closer to the C-atom.
So the partial positive charge on H-atom and acidic character increases as s-character of the orbital increases.
Hybridization state of C in three carbon compounds are
Decreasing order of acidic character is
Ethyne > Benzene > Hexane
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Due to the presence of electron cloud containing 6-electrons below and above the plane of benzene ring,
it is rich a source of electrons.
It attracts electrophiles reagents towards it and repels nucleophiles.
Benzene does not undergoes electrophilic substitution reactions and nucleophilic substitutions easily.
How would you convert the following compounds into benzene?
Write structures of all alkenes which in hydrogenation give 2-methylbutane.
Structures which produce 2-methyl butane on hydrogenation are
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+.
Higher electron density in benzene ring, more reactive is the compound towards these reactions.
NO2 is more powerful electron-withdrawing group than Cl. Hence, more the number of nitro groups, less reactive is the compound.
Methyl group is electron donating but NO2 group is electron withdrawing.
Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
Methyl group is electron donating and nitro group is electron withdrawing.
Maximum electron density will be in toluene followed by benzene and least in m-dinitrobenzene.
Toulene > Benzene > m-dinitrobenzene
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Anhydrous FeCl3 (Ferric Chloride) can be used.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms?
Illustrate your answer by taking one example.
Wurtz reaction is used for preparation of alkanes.
For preparing alkanes having odd number of C-atoms, a mixture of alkyl halides have to be used.