Text Book Back Questions and Answers - Chapter 13 Hydrocarbons 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Hydrocarbons
Textual Evaluation Solved
Multiple Choice Questions.
Question 1.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is 
(a) the eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
Question 2.
$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+2 \mathrm{Na} \stackrel{\text { Dry ether }}{\longrightarrow} \mathrm{C}_4 \mathrm{H}_{10}+2 \mathrm{NaBr}
$
The above reaction is an example of which of the following?
(a) Reirner Tiemann reaction
(b) Wurtz reaction
(c) Aldol condensation
(d) Hoffmann reaction
Answer:
(b) Wurtz reaction
Question 3.
An alkyl bromide (A) reacts with sodium in ether to form 4, 5-diethyloctane, the compound (A) is
(a) $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{Br}$
(b) $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_5 \mathrm{Br}$
(c) $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$

Answer:

Question 4.
The $\mathrm{C}-\mathrm{H}$ bond and $\mathrm{C}-\mathrm{C}$ bond in ethane are formed by which of the following types of overlap
(a) $\mathrm{sp}^3-\mathrm{s}$ and $\mathrm{sp}^3-\mathrm{sp}^3$
(b) $\mathrm{sp}^2-\mathrm{s}$ and $\mathrm{sp}^3-\mathrm{sp}^3$
(c) $\mathrm{sp}-\mathrm{sp}$ and $\mathrm{sp}-\mathrm{sp}$
(d) $p-s$ and $p-p$
Answer:
(a) $s p^3-s$ and $s p^3-s p^3$
Question 5.

In the following Reaction

the major product obtained is ....................

Answer:

Question 6 .
Which of the following is optically active?
(a) 2 -Methylpentane
(b) Citric acid
(c) Glycerol
(d) none of these
Answer:
(a) 2 -Methylpentane
Question 7.
The compounds formed at anode in the electrolysis of an aqueous solution of potassium acetate are
(a) $\mathrm{CH}_4$ and $\mathrm{H}_2$
(b) $\mathrm{CH}_4$ and $\mathrm{CO}_2$
(c) $\mathrm{C}_2 \mathrm{H}_6$ and $\mathrm{CO}_2$
(d) $\mathrm{C}_2 \mathrm{H}_6$ and $\mathrm{Cl}_2$
Answer:
(c) $\mathrm{C}_2 \mathrm{H}_6$ and $\mathrm{CO}_2$
Question 8 .
The general formula for cycloalkanes is
(a) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{\mathrm{n}}$
(b) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}$
(c) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}$
(d) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}$
Answer:
(b) $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}$
Question 9.
The compound that will react most readily with gaseous bromine has the formula
(a) $\mathrm{C}_3 \mathrm{H}_6$
(b) $\mathrm{C}_2 \mathrm{H}_2$
(c) $\mathrm{C}_4 \mathrm{H}_{10}$
(d) $\mathrm{C}_2 \mathrm{H}_4$
Answer:
(a) $\mathrm{C}_3 \mathrm{H}_6$
Question 10 .
Which of the following compounds shall not produce propene by reaction with $\mathrm{HBr}$ followed by elimination (or) only direct elimination reaction? 
(a) 
(b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$
(c) $\mathrm{H}_2 \mathrm{C}-\mathrm{C}=\mathrm{O}$
(d) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
Answer:
(c) $\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{O}$
Question 11.
Which among the following alkenes on reductive ozonolysis produces only propanone?
(a) 2 -Methylpropene
(b) 2-Methylbut-2-ene
(c) 2,3-Dimethylbut-1-ene
(d) 2,3-Dimethylbut-2-ene
Answer:
(d) 2,3-Dimethylbut-2-ene
Question 12.
The major product formed when 2 bromo - 2 - methylbutane is refluxed with ethanolic $\mathrm{KOH}$ is
(a) 2 - methylbut - 2 - ene
(b) 2 - methylbutan - 1-ol
(c) 2 - methyl but - 1 - ene
(d) 2-methylbutan-2-ol
Answer:
(a) 2-methylbut-2-ene
Question 13.
Major product of the below mentioned reaction is
$
\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2 \stackrel{\mathrm{ICl}}{\longrightarrow}
$
(a) 2-chloro-1-iodo-2-methylpropane
(b) 1 - chloro - 2 - iodo - 2 - methylpropane
(c) 1 ,2-dichioro-2-methylpropane

(d) 1,2 diiodo 2 -methylpropane
Answer:
(a) 2-chioro-1-lodo-2-methylpropane
Question 14.
The IUPAC name of the following compound is .............................

(a) trans -2 - chloro- 3 iodo-2-pentane
(b) cis -3 - iodo $-4-$ chloro -3 - pentane
(c) trans -3 - iodo - $4-$ chloro-3-pentene
(d) cis -2 chloro -3 - lodo -2 - pdntene
Answer:
(a) trans -2 - chloro -3 - iodo-2-pentane
Question 15.
cis -2 - butene and trans -2 - butene are ........
(a) conformational isomers
(b) structural isomers
(c) configurational isomers
(d) optical isomers
Answer:
(c) configurational isomers
Question 16.
Identify the compound $(\mathrm{A})$ in the following reaction.

Answer:

Question 17.

where A is ..................

(a) $\mathrm{Zn}$
(b) $\mathrm{Conc} . \mathrm{H}_2 \mathrm{SO}_4$
(c) Alc. $\mathrm{KOH}$
(d) $\mathrm{Dil} . \mathrm{H}_2 \mathrm{SO}_4$
Answer:
(c) Alc. $\mathrm{KOH}$
Question 18.
Consider the nitration of benzene using mixed conc. $\mathrm{FeSO}_4$ and $\mathrm{HNO}_3$, if a large quantity of $\mathrm{KHSO}_4$ is added to the mixture, the rate of nitration will be
(a) unchanged
(b) doubled
(c) faster
(d) slower
Answer:
(d) slower
Question 19.
In which of the following molecules, all atoms are co-planar?

Answer:
(d) both (a) and (b)
Question 20.
Propyne on passing through red hot iron tube gives ..........

Answer:

Question 21

Answer:

Question 22.
Which one of the following is non-aromatic?

Answer:

Question 23.

Which of the following compounds will not undergo Friedal - crafts reaction easily? [NEET]
(a) Nitrobenzene
(b) Toluene
(c) Cumene
(d) Xyiene
Answer:
(a) Nitrobenzene
Question 24 .
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) $-\mathrm{COOH}$
(b) $-\mathrm{NO}_2$
(c) $-\mathrm{CN}$
(d) $-\mathrm{SO}_3 \mathrm{H}$
Answer:
(b) $-\mathrm{NO}_2$
Question 25.
Which of the following can be used as the halide component for friedal - crafts reaction?
(a) Chiorobenzene
(b) Bromobenzene
(c) Chloroethene
(d) Isopropyl chloride
Answer:
(d) Isopropyl chloride
Question 26.
An alkane is obtained by decarboxylation of sodium propionate. Same alkane can be prepared by
(a) Catalytic hydrogenation of propene
(b) action of sodium metal on iodomethane
(c) reduction of 1 - chloropropane
(d) reduction of bromomethane
Answer:
(b) action of sodium metal on iodomethane
Question 27.
Which of the following is aliphatic saturated hydrocarbon?
(a) $\mathrm{C}_8 \mathrm{H}_{18}$
(b) $\mathrm{C}_9 \mathrm{H}_{18}$
(c) $\mathrm{C}_8 \mathrm{H}_{14}$

(d) All of these
Answer:
(a) $\mathrm{C}_8 \mathrm{H}_{18}$
Question 28.
Identify the compound ' $Z$ ' in the following reaction.

(a) Formaldehyde
(b) Acetaldehyde
(c) Formic acid
(d) None of these
Answer:
(a) Formaldehyde
Question 29.
Peroxide effect (Kharasch effect) can be studied in case of
(a) Oct-4-ene
(b) Hex-3-ene
(c) Pent-1-ene
(d) But-2- ene
Answer:
(a) Pent-1-ene
Question 30.
2 - butyne on chlorination gives
(a) 1 - chlorobutane
(b) 1,2-dichlorobutane
(c) 1, 1,2,2-tetrachlorobutane
(d) 2, 2,3,3-tetrachlorobutane
Answer:
(d) 2, 2, 3, 3 tetra chiorobutane
Short Answer Questions
Question 31.
Give IUPAC names for the fllowing compounds
(i) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_3$

(ii)

(iii) $\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}\left(\mathrm{CH}_3\right)_2$
(iv) Ethyl - isopropyl - acetytene
(v) $\mathrm{CH} \equiv \mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{C} \equiv \mathrm{CH}$
Answer:

(i)

(ii)

(iii)

(iv)

(v)

Question 32 .
Identify the compound $\mathrm{A} . \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ in the following series of reactions.

Answer:

Question 33.
Write a short note on ortho - para directors aromat\&c elropiJic substitution reactions.
Answer:
The group which increases the eleiron deity at oio and para positions of the ring are known as ortho-para directors.
Example:
$-\mathrm{OH},-\mathrm{NH}_2-\mathrm{NHR}-\mathrm{CH}_3,-\mathrm{OCH}_3$ etc.
Let us consider the directive influences of phenolic $(-\mathrm{OH})$ group. Phenol is the resonace hybrid of following structure.

In these resonance structures the negative charge residue is present on ortho and para posrtions of the ring structure. Therefore the electron density at ortho and para positions increases as compared to the metci
position, thus phenolic group activities the benzene ring for electrophilic attack at ortho and para positions and hetice $-\mathrm{OH}$ group is an ortho-para direction or and an activator.
Question 34.
How is propyne prepared from an alkyene dihalide?
Answer:

Question 35.
An alkyl halide with molecular formula $\mathrm{C}_6 \mathrm{H}_{13} \mathrm{Br}$ on dehydrohalogenation gave two isomeric alkenes $\mathrm{X}$ and $\mathrm{Y}$ with molecular formula $\mathrm{C}_6 \mathrm{H}_{12}$. On reductive ozonolysis, $\mathrm{X}$ and $\mathrm{Y}$ gave four compounds $\mathrm{CH}_3 \mathrm{COCH}_3$. $\mathrm{CH}_3 \mathrm{CHO}, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$ and $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCHO}$. Find the alkyl halide.
Answer:
1. $\mathrm{C}_6 \mathrm{H}_{13} \mathrm{Br}$ is $3-$ Bromo -4 methylpentanc.

$\text { 2. } 3 \text { - Bromo }-4 \text { methylpentane on dehydrogenation give two isomers } \mathrm{X} \text { and } \mathrm{Y} \text { as follows: }$

There fore $\mathrm{C}_6 \mathrm{H}_{13} \mathrm{Br}$ is $3-$ Bromo-4-methy ipentane.
Question 36.
Describe the mechanism of Nitration of benzene.
Answer:
Step-1 :

Step-2:
Attack of the electrophile on benzene ring to form arenium ion.

Step-3:
Rearomatisation of arenium ion.

Overall Reaction: 

Question 37.
How does Huckel rule help to decide the aromatic character of a compound?
Answer:
A compound is said to be aromatic, if it obeys the following rules:
- The molecule must be cyclic.
- The molecule must be co-planar.
- Complete delocalisation of it-electrons in the ring.
- Presence of $(4 n+2) \pi$ electrons in the ring where $n$ is an integer $(n=0,1,2 \ldots)$
This is known as Huckel's rule.
Example:

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised ir electrons.
4. $4 n+2=6$
$4 \mathrm{n}=6-2$
$4 n=4$
$\mathrm{n}=1$
It obey Huckel's rule, with $\mathrm{n}=1$, hence benzene is aromatic in nature.
Question 38 .
Suggest the route for the preparation of the following from benzene.
1.3-chioro - nitrobenzene
2. 4- chlorotoluene
3. Bromobenzene
4. in-dinitrobenzene
Answer:
1. Preparation of 3 - chloronitro - benzene from benzene:
Benzene undergoes nitration and followed by chlorination and it leads to the formation of 3chloronitrobenzene.

2. Preparation 4-chiorotoluene from benzene:
Benzene undergoes Fnedel crafi's alkylation followed by chlorination and it leads to the formation of 4- chiorotoluene.

3. Preparation of Bromobenzene from benzene: Bezene undergo bromination to give bromobenzene.

4. Preparation of m-dinitrobenzene from benzene:
Benzene undergo twice the time nitration to give $\mathrm{m}$-dinitrobenzene.

Question 39.
Suggest a simple chemical test to distinguish propane and propene.
Answer:
Chemical test to distinguish between propane and propene:
1. Bromine water test:
Propene contains double bond, therefore when we pour the bromine water to propene sample, it decolounses the bromine water whereas propane which is a saturated hydrocarbon does not decolourise the bromine water.
2. Baeyer's test:
When propene reacts with Bayer's reagent it gives 1,2 dihydroxypropene. Propane does not react with Baeyer's reagent.

Question 40 .
What happens when isobutylene is treated with acidified potassium permanganate?
Answer:
Isobutylene is treated with acidified $\mathrm{KMnO}_4$ to give acetone.

Question 41 .
how will you convert ethyl chloride in to -
1. ethane
2. $\mathrm{n}$ - butane
Answer:
1. Conversion of ethyl chloride into ethane:

2. Conversion of ethyl chloride into n-butane:
Wurtz reaction:

Question 42 .
Describe the conformers of n-butane.
Answer:
n-butane may be considered as a derivative of ethane as one hydrogen on each carbon atom is replaced by a methyl group.
Edipsed conformation:
In this conformation, the distance between the two methyl groups is minimum so there is maximum repulsion between them and it is the least stable conformer.
Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. It is the most stable conformer. The following potentially energy diagram shows the relative stability of various conformers of $n$-butane.

Question 43.
Write the chemical equations for combustion of propane.

Answer:
Chemical equations for combustion of propane: The general combustion reaction for any alkane is:

Question 44.
Explain Markornikoffs rule with suitable example.
Answer:
Markornikoff's rule: When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon atom that has more number of hydrogen and halogen adds to the carbon atom having fewer hydrogen atoms.
Example:
Addition of $\mathrm{HBr}$ to Propene:

Question 45.
What happens when ethylene is passed through cold dilute alkaline potassium permanganate.
Answer:
Ethylene reacts with cold dilute alkaline $\mathrm{KMnO}_4$ solution to give ethylene glycol:

Question 46.
Write the structures of following alkanes.
1. 2,3-Dimethyl-6-(2-methylpropyl) decane
2. 5-(2-Ethylbutyl) -3,3-dimethyldecane
3. 5 (1,2-Dimethyipropyl) - 2-methylnonane
Answer:
1. 2,3-Dimethyl-6-(2-rnethylpropyl) decane

2. 5-(2-Ethylbutyl)-3,3-dimethyldecane

3. 5-(1,2-Dimethylpropyl) -2-methylnonane 

Question 47.
How will you prepare propane from a sodium salt of fatty acid?
Answer:

Sodium salt of butyric acid on heating with sodalime gives propane.
Question 48 .

Answer:

Question 49.
How will you distinguish 1 - butyne and 2 - butyne?
Answer:

(i)

(ii)

(iii)

(iv)

Answer:

(i)

(ii)

(iii)

(iv)

Question 50 .
How will you distinguish 1 - butyene and 2 - butyne?
Answer:

In 1-butyne, terminal carbon atom contains atom one acidic hydrogen, therefore it will react with silver nitrate in the presence of ammonium hydroxide to give silver butynide. Whereas 2-butyne does not undergo such type of the reaction, because of the absence of acidic hydrogen.

Evaluate Your self
Question 1.
Write the structural formula and carbon skeleton formula for all possible chain isomers of $\mathrm{C}_6 \mathrm{H}_{14}$ ( $\mathrm{Hexane}$ ).

Answer:
$\mathrm{C}_6 \mathrm{H}_{14}$ has five possible isomeric structures:

Question 2.
Give the IUPAC name for the following alkane.

(a)

(b)

Answer:

(a)

(b)

Question 3.
Draw the structural formula for 4,5 -diethyl -3,4,5-trimethyloiane

Answer:

Question 4.
Water destroys Grignard reagents. Why?
Answer:
$
\mathrm{CH}_3 \mathrm{MgX}+\mathrm{HOH} \rightarrow \mathrm{CH}_4-\mathrm{Mg}(\mathrm{OH}) \mathrm{X}
$
Water would protonate the grignard reagent and destroy the gngnard reagent, because the grignard carbon atom is highly nucleophilic. This would form a hydrocarbon. Therefore to make a grignard solution, only ether is the best solvent and water or alcohol are not used for that purpose.
Question 5.
Is it possible to prepare methane by Kolbe's electrolytic method.
Answer:
Kolbe's electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe's electrolytic method.
Question 6.
Write down the combustion reaction of propane whose $\mathrm{AH}^{\circ}=-2220 \mathrm{~kJ}$

Answer:

Question 7.
Why ethane is produced in chlorination of methane?
Answer:
Chlorination of methane involves free radical mechanism. During the propagation step methyl free radical is produced, which is involved in the termination step, the two methyl free radical then combines to form
ethane.

Question 8 .
How toluene can be prepared by this method?
(i) From n-heptane,
(ii) From 2-methyihexane
Answer:

(i)

(ii)

Question 9.
Write the IUPAC names for the following alkenes.

(i)

(ii)

(iii)

(iv)

Answer:

(i)

(ii)

(iii)

(iv)

Question 10 .
Draw the structures for the following alkenes.
1. 6-Bromo-2,3-dimethyl 2 - hexene
2. 5 -Bromo-4-chloro 1 -heptene
3. 2,5-Dimethyl 4 - octene
4. 4-Methyl-2 pentene
Answer:
1. 6-Bromo-2,3-dirnethyl-2-hexene:

$\text { 2. } 5 \text { Bromo -4-Chloro-1-heptene: }$

$\text { 3. 2.5-dimethyl - } 4 \text { - Octene: }$

$\text { 4. } 4 \text {-methyl } 2 \text { pentene: }$

Question 11.
Draw the structure and write down the IUPAC name for the isomerism exhibited by the molecular formulae:
1. $\mathrm{C}_5 \mathrm{H}_{10}-$ Pentene ( 3 isomers)
2. $\mathrm{C}_6 \mathrm{H}_{12}-$ Hexene (5 isomers)
Answer:
1. $\mathrm{C}_5 \mathrm{H}_{10}-$ Pentene ( 3 isomers)
(a) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2 \rightarrow 1$ - penlene (or) pent- 1-ene
(b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \rightarrow 2$ - pentene (or) pent-2-ene

(c)

2. $\mathrm{C}_6 \mathrm{H}_{12}-\mathrm{Hexene}(5$ isomers)
(a) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2 \rightarrow$ Hex-I-ene
(b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \rightarrow \mathrm{Hex}-2$ - ene
(c) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_3 \rightarrow \mathrm{Hex}-3-$ ene

These two compounds exhibits constitutional isomerism.
Question 12 .
Determine whether each of the following alkenes can exist as cis-trans isomers?
(a) 1 -Chloropropene
(b) 2 -Chloropropene
Answer:
(a) 1 -Chloropropene:
$
\mathrm{CH}_3-\mathrm{CH}=\mathrm{CHCl}
$

Therefore $-1-$ Chloropropene has cis-trans isomers.
(b) 2-Chloropropene:

In 2-Chloropropene, it deviates from the rule, that is "At least one same group is present on the two doubly bonded carbon atom". Therefore-2-chloropropenc cannot exist as cis - trans isomers.
Question 13.
Draw cis-trans isomers for the following compounds
(a) 2-chloro-2-butene
(b) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_3$
Answer:
(a) 2-Chloro-2-butene:

$\text { (b) } \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}-\mathrm{CH}_3$

Question 14.
How propene is prepared form 1,2-dichloropropane?
Answer:

Question 15.
How ozone reacts with 2-methyl propene?
Answer:

Question 16.
An organic compound (A) on ozonolysis gives only acetaldehyde. (A) reacts with $\mathrm{Br}_2 / \mathrm{CCl}_4$ to give compound (B). Identify the compounds (A) and (B). Write the IUPAC name of (A) and (B). Give the geometrical isomers of $(\mathrm{A})$.
Answer:
2-Rutene undergo ozonolysis to give acetaldehyde only.

$\text { Geometrical isomers of } 2-\text { Butene (A): }$

Question 17.
An organic compund (A) $\mathrm{C}_2 \mathrm{H}_4$ decolourises bromine water. (A) on reaction with chlorine gives (Br). A reacts with $\mathrm{HBr}$ to give (C). Identify (A),(B),(C). Explain the reactions.
Answer:
(i) $\mathrm{C}_2 \mathrm{H}_4(\mathrm{~A})$, decolourises the bromine water. Therefore it contains double bond. Hence (A) is ethylene.

Question 18 .
Prepare propyne from its corresponding alkene.
Answer:
Preparation of propyne from propene:

Question 19.
Write the products A \& B for the following reaction.

Answer:

Question 20.

Answer:

Question 21.
Calculate the number of rings present in $\mathrm{C}_{18} \mathrm{H}_{12}$.
Answer:
Double bond equivalent formula $=\left(\mathrm{C}-\frac{1}{2}+\frac{1}{2}\right)$
Where $\mathrm{C}=$ no. of carbon atoms. $\mathrm{H}=$ no. of hydrogen and halogen atoms and $\mathrm{N}=$ no. of nitrogen atoms.
So. in $\mathrm{C}_{18} \mathrm{H}_{12}$ Double bond equkalent $=18-\frac{12}{2}+0+1=18-6+1=13$

One ring is equal to one double bond equivalent.
$\therefore$ here four rings are there, four double bond equivalent arc used. So remaining, $13-4=9$. nine double bonds are present in the ring. Hence, $\mathrm{C}_{18} \mathrm{H}_{12}$ contain four aromatic rings.
Question 22

write all possible isomers for an aromatic benzenoid compound having the molecular formula $\mathrm{C}_8 \mathrm{H}_{10}$.

Answer:
Possible isomers for $\mathrm{C}_8 \mathrm{H}_{10}$ :

(i)

(ii)

(iii)

(iv)

Question 23.
Write all possible isomers Iòr a monosubstituted aromatic benzenoid compound having the molecular formula $\mathrm{C}_9 \mathrm{H}_{12}$
Answer:
Possible isomers for $\mathrm{C}_9 \mathrm{H}_{12}$ :

Question 24.
how benzene can be prepared by Grignard Reagent?
Answer:
Phenyl magnesium bromde reagent reacts with water molecule to give benzene:

Question 25 .
Why benzene undergoes electrophilic substitution reaction whereas alkenes undergoes addition reaction?

Answer:
- Benzene possess an unhybridised p-orbital containing one electron. The lateral overlap of theirporbitals produces 3 it bond.
- The six electron of the p-orbitais cover all the six carbon atoms and arc said to be delocalised.
- Due to delocalisation, strong it-bond is formed which makes the molecule stable. Therefore benzene undergoes electrophilic substitution reaction, whereas alkenes undergoes addition reaction.
Question 26.
Convert Ethyne to Benzene and name the process.
Answer:
Conversion of Ethyne into Benzene:

This process is one of the cyclic polymerisation process.
Question 27.
Toluene undergoes nitration easily than benzene. Why?
Answer:
1. Toluene has a methyl group on the benzene ring which is electron releasing group and hence activate the benzene ring by pushing the electrons on the benzene ring.
2. $\mathrm{CH}_3$ group is ortho - para director and ring activator. Therefore in toluene, ortho and para positions are the most reactive towards an electrophile, thus promoting electrophilic substitution reaction.
3. The methyl group hence makes it around 25 times more reactive than benzene. Therefore it undergoes nitration easily than benzene.