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Chemistry 5 Marks Questions

Hydrocarbons · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 14 questions.

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Question 15 Marks
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.
Answer
Addition of HBr to propene is an ionic electrophilic addition reaction in which the electrophile, i.e., $\mathrm{H}^{+}$first adds to give a more stable 2° carbocation. In the $2^{\text {nd }}$ step, the carbocation is rapidly attacked by the nucleophile Br ion to give 2-bromopropane.


In presence of benzoyl peroxide, the reaction is still electrophilic but the electrophile here is a Br free radical which is obtained by the action of benzoyl peroxide on HBr.

In the first step, Br radical adds to propene in such a way so as to generate the more stable 2° free radical. In the second step, the free radical thus obtained rapidly abstracts a hydrogen atom from HBr to give 1-bromopropane.

From the above discussion, it is evident that although both reactions are electrophilic addition reactions but it is due to different order of addition of H and Br atoms which gives different products.
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Question 25 Marks
Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
Answer

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:

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:

Now, 'X' is an addition product of alkene ‘A’ with ozone. Therefore, the possible structure of alkene 'A' is:
$\text{H}_3\text{C}-\text{CH}_2-\text{CH}=\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_2\text{CH}_3$
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Question 35 Marks
An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.
Answer


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:

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:
$\text{H}_3\stackrel{1}{\hbox{C}}-\stackrel{2}{\hbox{C}}\text{H}=\stackrel{3}{\hbox{C}}-\stackrel{4}{\hbox{C}}\text{H}_2-\stackrel{5}{\hbox{C}}\text{H}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_2-\text{CH}_3$
The IUPAC name of 'A' is 3-Ethylpent-2-ene.
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Question 45 Marks
How do you account for the formation of ethane during chlorination of methane?
Answer
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:
$\text{Cl}-\text{Cl}\xrightarrow{\ \ \ \text{hv}\ \ \ \ \ }\dot{\text{Cl}}+\dot{\text{C}}\\ \ \text{Chlorine free radicals}$
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:
$\text{CH}_4+\dot{\text{Cl}}\xrightarrow{\ \ \ \ \text{hv}\ \ \ \ }\dot{\text{C}}\text{H}_3+\text{H}-\text{Cl}\\\text{Methane}$
These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.
$\dot{\text{C}}\text{H}_3+\text{Cl}-\text{Cl}\xrightarrow{\ \ \ \ }\text{CH}_3-\text{Cl}+\dot{\text{Cl}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methyl chloride}$
Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and $\mathrm{CH}_3 \mathrm{Cl}$ are the major products formed, other higher halogenated compounds are also formed as:
$\text{CH}_3\text{Cl}+\dot{\text{Cl}}\xrightarrow{\ \ \ \ }\dot{\text{C}}\text{H}_2\text{Cl}+\text{HCl}$
$\dot{\text{C}}\text{H}_2\text{Cl}+\text{Cl}-\text{Cl}\xrightarrow{\ \ \ \ \ }\text{CH}_2\text{Cl}_2+\dot{\text{Cl}}$
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:
$\dot{\text{Cl}}+\dot{\text{Cl}}\xrightarrow{\ \ \ \ }\text{Cl}-\text{Cl}$
$\text{H}_3\dot{\text{C}}+\dot{\text{CH}_3}\xrightarrow{\ \ \ \ \ }\text{H}_3\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Ethane)}$
Hence, by this process, ethane is obtained as a by-product of chlorination of methane.
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Question 55 Marks
6 An alkene 'A' contains three C-C, eight C–H $\sigma$ bonds 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'.
Answer
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 $\sigma$ 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: $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}=\text{C}-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \ \text{H}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(A)}$ 'A' has 3 C–C bonds, 8 C–H $ \sigma$ bonds, and one C–C π bond. Hence, the IUPAC name of 'A' is But-2-ene. Ozonolysis of 'A' takes place as: https://api.studentbro.in/web/HaqQhKV4.png" style="height:30%; width:30%"> The final product is ethanal with molecular mass = [2 × 12] + (4 × 1) + (1 × 16)] = 44 u.
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Question 65 Marks
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Answer
Hex-2-ene is represented as:
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3$
Geometrical isomers of hex-2-ene are:

The dipole moment of cis-compound is a sum of the dipole moments of $\mathrm{C}-\mathrm{CH}_3$ and $\mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ bonds acting in the same direction.
The dipole moment of trans-compound is the resultant of the dipole moments of $\mathrm{C}-\mathrm{CH}_3$ and $\mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ bonds acting in opposite directions.
Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipoledipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.
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Question 75 Marks
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Answer

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:

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.


The boiling points of alkanes (obtained in the mixture) are very close. Hence, it becomes difficult to separate them.
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Question 85 Marks
Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
Answer

o-xylene has two resonance structures:



All three products, i.e., methyl glyoxal, 1,2-dimethylglyoxal, 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).
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Question 95 Marks
4 In the alkane $\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{C}\left(\mathrm{CH}_3\right)_2-\mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)_2$, identify $1^{\circ}, 2^{\circ}, 3^{\circ}$ carbon atoms and give the number of H atoms bonded to each one of these.
Answer


1° carbon atoms are those which are bonded to only one carbon atom, i.e., they have only one carbon atom as their neighbor. The given structure has five 1° carbon atoms and fifteen hydrogen atoms are attached to it.
2° carbon atoms are those which are bonded to two carbon atoms, i.e., they have two carbon atoms as their neighbors. The given structure has two 2° carbon atoms and four hydrogen atoms are attached to it.
3° carbon atoms are those which are bonded to three carbon atoms, i.e., they have three carbon atoms as their neighbors. The given structure has one 3° carbon atom and only one hydrogen atom is attached to it.
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Question 105 Marks
Why is benzene extra ordinarily stable though it contains three double bonds?
Answer
Benzene is a hybrid of resonating structures given as:

All six carbon atoms in benzene are $\mathrm{sp}^2$ hybridized. The two $\mathrm{sp}^2$ hybrid orbitals of each carbon atom overlap with the $\mathrm{sp}^2$ hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining $\mathrm{sp}^2$ 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 $\pi$ bonds by the lateral overlap of $C_1-C_2, C_3-C_4, C_5-C_6$, or $C_2-C_3, C_4-C_5, C_6-C_1$.


The six $\pi'\text{s}$ are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized $\pi-\text{electrons}$ stabilize benzene.
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Question 115 Marks
Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
Answer
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 $\left(\mathrm{NO}_2-\right)$.
Now, $\mathrm{CH}_3$ - group is electron donating and $\mathrm{NO}_2$ - 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:
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Question 125 Marks
Write chemical equations for combustion reaction of the following hydrocarbons:
  1. Butane.
  2. Pentene.
  3. Hexyne.
  4. Toluene.
Answer
Upon combutions any hydrocarbon produces $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ for the givren hydrocarbons the reaction may be represented as:
  1. Butane: $\text{C}_4\text{H}_{10}+\frac{13}{2}\text{O}_2\xrightarrow{\ \ \ \Delta\ \ \ }4\text{CO}_2+5\text{H}_2\text{O}$
  2. Pentene: $\text{C}_5\text{H}_{10}+\frac{15}{2}\text{O}_2\xrightarrow{\ \ \ \Delta\ \ \ }5\text{CO}_2+5\text{H}_2\text{O}$
  3. Hexyne: $\text{C}_6\text{H}_{10}+\frac{17}{2}\text{O}_2\xrightarrow{\ \ \ \Delta\ \ \ }6\text{CO}_2+5\text{H}_2\text{O}$
  4. Toluene: $\text{C}_7\text{H}_{8}+9\text{O}_2\xrightarrow{\ \ \ \Delta\ \ \ }7\text{CO}_2+4\text{H}_2\text{O}$
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Question 135 Marks
What are the necessary conditions for any system to be aromatic?
Answer
The necessary and sufficient condition for any system to be aromatic is given by Huckel's rule. As per Huckel's rule, any system is said to be aromatic if it satisfies the following 3 conditions:
  1. Contains (4n + 2)71 electrons, where n is any positive integer or 0,
  2. Shows complete delocalisation of $\pi$ electrons and
  3. The molecule must be planar.
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Question 145 Marks
8 Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Answer
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:

As the s-character increases, the electronegativity of carbon increases and the electrons of $\mathrm{C}-\mathrm{H}$ bond pair lie closer to the carbon atom. As a result, partial positive charge of H -atom increases and $\mathrm{H}^{+}$ions are set free.
The s-character increases in the order:
$\mathrm{sp}^3<\mathrm{sp}^2<\mathrm{sp}$
Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.
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