Class 11 | Chemistry | Chapter 13 | Hydrocarbons 60 Most important MCQs NEET/JEE

Question. From NCERT

The dihedral angle between the H-atoms in staggered conformation of ethane is

1. 60 $°$

2. 0 $°$

3. 120 $°$

4. 180 $°$

Answer

Question. From NCERT NEET - 2021

The dihedral angle of the least stable conformer of ethane is-

1. 60 $°$

2. 0 $°$

3. 120 $°$

4. 180 $°$

Answer

Question. From NCERT NEET - 2021

The major product of the following chemical reaction is:

Answer

solution

Question.

The major product Z obtained in the following reaction scheme is:

Shift 1 Jan 9 JEE Main 2020 Chemistry Paper With Solutions

Shift 1 Jan 9 JEE Main 2020 Chemistry Paper With Solution

Answer

Solution:

Shift 1 Jan 9 JEE Main 2020 Chemistry Papers With Solutions

Hence, major product formed is that of option c.

Answer: (c)

Question.From NCERT NEET - 2020

Elimination reaction of 2-Bromopentane to form pent-2-ene is-

(a) $β$ -Elimination reaction

(b) Follows Zaitsev rule

(d) Dehydration reaction

1. (a), (c), (d)

2. (b), (c), (d)

3. (a), (b), (d)

4. (a), (b), (c)

Answer

Correct option is 4 (a), (b), (c)

This reaction is an example of β-elimination.Hydrogen is removed from β-carbon and halogen from α-carbon, hence dehydrohalogenation reaction.

Pent-2-ene is major product known as Saytzeff's product and it is more stable alkene than Pent-1-ene.

Generally, in E 2 reaction Zaitsev alkene is formed as a major product (more stable alkene).

solution

Question.Easy

When chloro-benzene reacts with sodium in presence of ether than the major product formed is.

Answer

solution

Question.Easy

Molecular mass of Benzene is _______.

$72$

$78$

$79$

$77$

Answer

Correct option is B -78

Benzene is C 6 H 6 . So molecular mass= 12×6+1×6=78

Question.

Which of the following alkane cannot be made in good yield by Wurtz reaction?

1. 2,3-Dimethylbutane

2. n-Heptane

3. n-Butane

4. n-Hexane

Answer

Correct option is

n-Heptane

Wurtz reaction is a reaction in which alkyl halide is converted to alkane by Na/ether. It is limited to the synthesis of symmetrical alkanes.

As we know that n-Heptane involves the odd number of carbon hence it won't be produced by this method. Rest molecules can be formed by the Wurtz reaction.

Option 2 is correct.

Question.

Which of the following alkane cannot be prepared by Wurtz's reaction in good yield?

Answer

Correct option is

Wurtz reaction is a reaction in which alkyl halide is converted to alkane by Na/ether. It is limited to the synthesis of symmetrical alkanes.
(Image ) As this molecule involves ring strain the reaction will be of poor yield.

Question.

Which if the following alkane is not produced when the mixture methyl chloride and ethyl chloride is subjected to wurtz reaction?

Ethane

Propane

n-butane

n-pentane

Answer

solution

Question.NEET - 2020

An alkene on ozonolysis gives methanal as one of the products. Its structure is:

1. 2.

3. 4.

Answer

Correct option is

The structure given in option B gives methanal on ozonolysis as shown in the above reaction.

Question.

An alkene, on ozonolysis gives formaldehyde and acetaldehyde. The alkene is

Ethene

Propene

Butene -1

Butene-2

Answer

Correct option is

Propene

Propene on ozonolysis gives formaldehyde and acetaldehyde.

solution

Question.

Which alkene on ozonolysis gives $C H_{3} C H_{2} C H O$ and

$C H_{3} C H_{2} C H = C H C H_{2} C H_{3}$

$C H_{3} C H_{2} C H = C H C H_{3}$

Answer

Correct option is

option

Question.

One mole of alkene on ozonolysis gives $2$ moles of butanone. The alkene is:

3, 4-dimethylhex-2-ene

2, 3-dimethylhex-3-ene

3, 4-dimethylhex-3-ene

2, 3-dimethylhex-2-ene

2, 5-dimethylhex-3-ene

Answer

Correct option is

3, 4-dimethylhex-3-ene

solution

Question.NEET - 2020

Which of the following is a free radical substitution reaction ?

(1) Benzene with Br2/AlCl3

(2) Acetylene with HBr

(3) Methane with Br2/hv

(4) Propene with HBr/(C6H5COO)2

Answer

Question.

Which one of the following is a free radical substitution reaction ?

$C H_{3} C H O + H C N \to C H_{3} - O H ∣ C H - C N$

Answer

Correct option is

Chlorination of toluene in presence of sunlight follows free radical mechanism. Benzyl radical ( $C_{6} H_{5} C H_{2}^{+})$ is most stable so benzyl chloride is major product formed.

All other reactions follows ion formation mechanism.

Option D is correct.

Question.

Which of the following species is least stable ?

Answer

Correct option is

All compounds have delocalised electrons in their ring.
The compound which is anti-aromatic will be least stable.
Hence, option C is correct

Question.

Which of the following species is expected to yield maximum percentage of meta substitution product ?

$A r C H_{3}$

$A r C H_{2} C l$

$A r C H C l_{2}$

$A r C C l_{3}$

Answer

Correct option is

$A r C C l_{3}$

Among all given reagents $- C C l_{3}$ is strongest electron withdrawing group so it will reduce electron density at ortho and para position and it will direct reacting group at meta position.

So, $- C C l_{3}$ will yield maximum meta product.

Option D is correct.

Question.

Which of the following groups are meta-directing?

$- C H O$

$- O H$

$- O C O C H_{3}$

$- C l$

Answer

Correct option is

$- C H O$

$- C H O is a meta directing group.$

Meta directing groups are the electron withdrawing groups. The attack of the incoming group is at the meta position.

solution

Question.Medium

Which of the following is the strongest ortho, para directing group :

$- O H$

$- C l$

$- O C H_{3}$

$- C H_{3}$

Answer

Correct option is

−OH

OH− is strongest ortho, para directing group.

solution

Question.

A group which deactivates the benzene ring towards electrophilic substitution but directs the reaction at ortho and para position:

$- N H_{2}$

$- C l$

$- N O_{2}$

$- C_{2} H_{5}$

Answer

Correct option is

$- C l$

$- N H_{2}$ is activating group due to $+ M$ effect.
$- N O_{2}$ is a deactivating group due to $- M$ effect.
$- C H_{3}$ is activating group due to $+ I$ effect.
$- C l$ is deactivating group due to $- I$ effect but it is ortho, para directing due to $+ M$ effect.

Question.Hard

The correct order of activation of benzene ring towards an electrophile is :

(IV)>(III)>(II)>(I)

(III)>(I)>(IV)>(II)

(III)>(IV)>(I)>(II)

(IV)>(III)>(II)>(I)

Answer

Correct option is

(III)>(IV)>(I)>(II)

$(I I I) > (I V) > (I) > (I I)$
The trend is on the basis of resonance of functional groups with benzene ring. $O C H_{3}, O = O, - O C O C H_{3}$ are electron donating groups and fluorine is an electron withdrawing group.

Question.NEET - 2019

An alkene "A" on reaction with $O_{3}$ and $Zn - H_{2} O$ gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:

1. $H_{3} C - \underset{Cl}{\underset{|}{CH}} - {CH}_{|}^{|}_{{CH}_{3}}^{{CH}_{3}}$

2. $Cl - {CH}_{2} - {CH}_{2} - {CH}_{|}^{|}_{{CH}_{3}}^{{CH}_{3}}$

3. $H_{3} C - {CH}_{2} - \overset{{CH}_{2} Cl}{\overset{|}{CH}} - {CH}_{3}$

4. $H_{3} C - {CH}_{2} - {C_{|}^{|}}_{Cl}^{{CH}_{3}} - {CH}_{3}$

Answer

solution

Question.

X is

Hard

Answer

option

Question.

The major product of the following reaction is

Hard

Answer

solution

Question.

$3 - P h e n y l$ propene on reaction with $H B r$ gives (major product).

$C_{6} H_{5} C H_{2} C H (B r) C H_{3}$

$C_{6} H_{5} C H (B r) C H_{2} C H_{3}$

$C_{6} H_{5} C H_{2} C H_{2} C H_{2} B r$

$C_{6} H_{5} C H (B r) C H = C H_{2}$

Answer

According to Markownikoff's rule, the negative part of the attacking reagent adds to less hydrogenated carbon atom of the double bond

solution

Question.From NCERT NEET - 2019

The most suitable reagent for the following conversion is-

1. ${Hg}^{2 +} / H^{+}, H_{2} O$

2. Na/liquid NH3

3. H2, Pd/C, quinoline

4. Zn/HCl

Answer

Question.From NCERT NEET - 2019

The reaction among the following that proceeds through an electrophilic substitution, is:

Answer

Question.NEET - 2019

The alkane that gives only one monochloro product on chlorination with $C l_{2}$ in presence of diffused sunlight is -

1. 2,2,-dimethylbutane

2. neopentane

3. n-pentane

4. Isopentane

Answer

solution

Question.From NCERT NEET - 2019

In the following reaction

$H_{3} CC \equiv CH \to_{873 K}^{Red Hot Iron Tube} A$

The number of ( $σ$ ) bonds present in the product (A) is-

1. 21

2. 9

3. 24

4. 18

Answer

Solution Image

Question.From NCERT NEET - 2018

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide B,B undergoes Wurtz reaction to give a gaseous hydrocarbon containing less than four carbon atoms.

formula of (A) is-

CH \equiv CH

{CH}_{2} = {CH}_{2}

{CH}_{3} - {CH}_{3}

{CH}_{4}

Answer

Hydrocarbon (A) is methane $(C H_{4})$ . It reacts with bromine by substitution to form methyl bromide $C H_{3} - B r$
$C H_{4} + B r_{2} \to C H_{3} - B r + H B r$

Methyl bromide by Wurtz reaction is converted to a gaseous hydrocarbon containing less than four carbon atoms.
$2 C H_{3} B r + 2 N a \to C H_{3} - C H_{3} + 2 N a B r$

In Wurtz reaction, when same alkyl halide is used, the product alkane has even number of C atoms.

The hydrocarbon containing less than four carbon atoms with even number of C atoms is ethane $(C H_{3} - C H_{3})$ .

Question.From NCERT NEET - 2018

The compound

C_{7} H_{8}

undergoes the following reactions :

C_{7} H_{8} \overset{3 {Cl}_{2} / hv}{\to} A \overset{{Br}_{2} / Fe}{\to} B \overset{Zn / HCl}{\to}

The product 'C' is-

1. m–Bromotoluene

2. o–Bromotoluene

3. 3–Bromo–2,4,6–trichlorotoluene

4. p–Bromotoluene

Answer

solution

Question.From NCERT NEET - 2018

P, Q and R in the above mentioned sequence of reactions are respectively -

Answer

solution

Question.From NCERT NEET - 2018

The molecules that has hybridisation

{sp}^{2}, {sp}^{2}, sp, sp

from left to right atoms is-

HC \equiv C - C \equiv H

{CH}_{2} = CH - C \equiv CH

{CH}_{2} = CH - CH = {CH}_{2}

{CH}_{2} - CH = CH - {CH}_{3}

Answer

Question.From NCERT NEET - 2017

Predict the correct intermediate and product in the following reaction.

H3C-C $\equiv$ CH $\to_{{HgSO}_{4}}^{H_{2} O, H_{2} {SO}_{4}}$ Intermediate $\to$ Product

(A) (B)

Answer

Question.From NCERT NEET - 2017

The correct order of acidity is-

CH \equiv CH > {CH}_{3} - C \equiv CH > {CH}_{2} = {CH}_{2} > {CH}_{3} - {CH}_{3}

CH \equiv CH > {CH}_{2} = {CH}_{2} > {CH}_{3} - C \equiv CH > {CH}_{3} - {CH}_{3}

{CH}_{3} - {CH}_{3} > {CH}_{2} = {CH}_{2} > {CH}_{3} - C \equiv CH > CH \equiv CH

{CH}_{2} = {CH}_{2} > {CH}_{3} - {CH}_{3} > {CH}_{3} - C \equiv CH > CH \equiv CH

Answer

(Acidic strength order).

More is the s-character, more is the acidic strength.

The s-character is maximum in sp hybrid carbon atom followed by sp 2

and sp 3

Question.

The correct order for acid strength of compounds $C H \equiv C H, C H_{3} - C \equiv C H$ and $C H_{2} = C H_{2}$ is as follows:

$C H \equiv C H > C H_{2} = C H_{2} > C H_{3} - C \equiv C H$

$H C \equiv C H > C H_{3} - C \equiv C H > C H_{2} = C H_{2}$

$C H_{3} - C \equiv C H > C H_{2} = C H_{2} > H C \equiv C H$

$C H_{3} - C \equiv C H > C H \equiv C H > C H_{2} = C H_{2}$

Answer

Correct option is

$H C \equiv C H > C H_{3} - C \equiv C H > C H_{2} = C H_{2}$

$C H \equiv C H > C H_{3} - C \equiv C H > C H_{2} = C H_{2}$
(Acidic strength order).
More is the s-character, more is the acidic strength.
The s-character is maximum in $s p$ hybrid carbon atom followed by $s p^{2}$ and $s p^{3}$ .

Question.

$H C \equiv C H E x c e s s N a A E x c e s s C H_{3} C l B$ .
Product $B$ is :

2-butyne

1-butyne

propyne

monosodium acetylide

Hard

Answer

Correct option is

2-butyne

$H C \equiv C H \to E x c e s s N a N a - C \equiv C - N a$
$N a - C \equiv C - N a E x c e s s C H_{3} C l C H_{3} - C \equiv C - C H_{3}$
Compound 'B' is 2-Butyne

Question.From NCERT NEET - 2016

Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?

Answer

solution

Question.

Compound $(D)$ is :

Hard

Answer

solution

Question.

Compound $(A)$ on reaction with $N H_{2} N H_{2} + O H$ gives:

Hard

Answer

solution

Question.

Compound $(F)$ is:

Answer

solution

Question.

Answer

Question.NEET - 2016

In the given reaction

The product P is

Answer

Question.From NCERT NEET - 2016

The compound that will react most readily with gaseous bromine has the formula is-

1. $C_{3} H_{6}$

2. $C_{2} H_{2}$

3. $C_{4} H_{10}$

4. $C_{2} H_{4}$

Answer

The compound that will react most readily with gaseous bromine has the formula C3H6

Unsymmetrical alkenes generally are more reactive than the symmetrical alkenes, alkynes and alkanes. That is why, propene is more reactive than the other given compounds.

Question.

Bond energy is highest for :

$S n - S n$

$G e - G e$

$C - C$

$S i - S i$

Easy

Answer

Correct option is

$C - C$

Bond energies of these compounds are:
$C - C = 346 k J / m o l, S i - S i = 249 k J / m o l, G e - G e = 188 k J / m o l, S n - S n = 146 k J / m o l$ .

Question.

Which of the following has the highest first ionization energy?

Lithium

Beryllium

Boron

Carbon

Answer

Carbon has the highest first ionization energy.

The order of the ionization energies is Li<Be<B<C

On moving from left t right in a period, the ionization energy increases as the effective nuclear charge increases and the size decreases. The removal of electron becomes more and more difficult.

Question.

Which of the following have highest melting points?

$p -$ block elements

$s -$ block elements

$d -$ block elements

None of these

Answer

d− block elements

d− block elements have highest melting points as they can form covalent bonds due to incompletely filled d orbitals.

They melt above 1000 degree C

Question.

Which of the following has the highest calorific value?

Coal gas

Water gas

Producer gas

Carbon dixode gas

Medium

Answer

Calorific value: The amount of energy produced by the complete combustion of a material or fuel.
Coal gas $H_{2} + C O + C H_{4}$ has the highest calorific value.

Question.From NCERT NEET - 2016

The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is-

1. The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain.

2. The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has a torsional strain.

3. The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain.

4. The staggered conformation of ethane is less stable than eclipsed conformation because staggered conformation has a torsional strain.

Answer

We know that the different spatial arrangements of atoms or groups of atoms which can be converted into one another by the rotation around single bonds are known as conformations or conformational isomers. There are two types of conformations: staggered conformation and eclipsed conformation.

Complete step by step answer:
The molecular formula for ethane is

C_{2} H_{6}

. In the structure of ethane, two carbon atoms are bonded to each other by a single bond. Each carbon atom is bonded to three hydrogen atoms. The structure of ethane is as follows:

Now we can imagine that one carbon atom of the ethane undergoes rotation about the central bond while the other carbon atom is fixed. This rotation gives us an infinite number of conformations depending on the angle of rotation.
The conformations differ in spatial arrangements of atoms but the bond angles and the bond lengths remain the same.
The two conformations of ethane are: staggered conformation and eclipsed conformation.
> Staggered conformation: In staggered conformation, the hydrogen atoms of both the carbon atoms are placed widely apart. The carbon-hydrogen bond pairs of the two carbon atoms are far away from each other.
> Eclipsed conformation: In eclipsed conformation, the hydrogen atoms of both the carbon atoms are facing each other. The carbon-hydrogen bond pairs of the two carbon atoms are very close to each other.
The staggered and eclipsed conformations of ethane are as follows:

The potential energy of the molecule increases as the rotation about the central bond increases. The increase in potential energy is due to the repulsion between electrons in the bond. This increase in the potential energy is known as the torsional strain. The torsional strain thus depends on the rotation about the central bond.
From the structures of staggered and eclipsed conformations, we can conclude that the torsional strain in the staggered conformation is less than that of the eclipsed conformation.
Thus, the staggered conformation is more stable than the eclipsed conformation because staggered conformation has no torsional strain.
Thus, the correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ‘the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain’.

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Important MCQs* Hydrocabons

The dihedral angle between the H-atoms in staggered conformation of ethane is

When chloro-benzene reacts with sodium in presence of ether than the major product formed is.

Molecular mass of Benzene is _______.

72

78

79

77

n-Heptane

Which of the following alkane cannot be prepared by Wurtz's reaction in good yield?

Wurtz reaction is a reaction in which alkyl halide is converted to alkane by Na/ether. It is limited to the synthesis of symmetrical alkanes.(Image ) As this molecule involves ring strain the reaction will be of poor yield.

Which if the following alkane is not produced when the mixture methyl chloride and ethyl chloride is subjected to wurtz reaction?

Ethane

Propane

n-butane

n-pentane

The structure given in option B gives methanal on ozonolysis as shown in the above reaction.

An alkene, on ozonolysis gives formaldehyde and acetaldehyde. The alkene is

Ethene

Propene

Butene -1

Butene-2

Which alkene on ozonolysis gives CH3​CH2​CHO and

CH3​CH2​CH=CHCH2​CH3​

CH3​CH2​CH=CHCH3​

One mole of alkene on ozonolysis gives 2 moles of butanone. The alkene is:

3, 4-dimethylhex-2-ene

2, 3-dimethylhex-3-ene

3, 4-dimethylhex-3-ene

2, 3-dimethylhex-2-ene

2, 5-dimethylhex-3-ene

Which one of the following is a free radical substitution reaction ?

CH3​CHO+HCN→CH3​−OH∣C​H​−CN

Chlorination of toluene in presence of sunlight follows free radical mechanism. Benzyl radical (C6​H5​CH2+​)is most stable so benzyl chloride is major product formed.All other reactions follows ion formation mechanism.Option D is correct.

Which of the following species is least stable ?

All compounds have delocalised electrons in their ring.The compound which is anti-aromatic will be least stable.Hence, option C is correct

Which of the following species is expected to yield maximum percentage of meta substitution product ?

ArCH3​

ArCH2​Cl

ArCHCl2​

ArCCl3​

ArCCl3​

Among all given reagents −CCl3​ is strongest electron withdrawing group so it will reduce electron density at ortho and para position and it will direct reacting group at meta position.So, −CCl3​ will yield maximum meta product.Option D is correct.

Which of the following groups are meta-directing?

−CHO

−OH

−OCOCH3​

−Cl

−CHO

- CHO is a meta directing group. Meta directing groups are the electron withdrawing groups. The attack of the incoming group is at the meta position.

Which of the following is the strongest ortho, para directing group :

−OH

−Cl

−OCH3​

−CH3​

A group which deactivates the benzene ring towards electrophilic substitution but directs the reaction at ortho and para position:

−NH2​

−Cl

−NO2​

−C2​H5​

−Cl

−NH2​ is activating group due to +M effect.−NO2​ is a deactivating group due to −M effect.−CH3​ is activating group due to +I effect.−Cl is deactivating group due to −I effect but it is ortho, para directing due to +M effect.

The correct order of activation of benzene ring towards an electrophile is :

(IV)>(III)>(II)>(I)

(III)>(I)>(IV)>(II)

(III)>(IV)>(I)>(II)

(IV)>(III)>(II)>(I)

(III)>(IV)>(I)>(II)

(III)>(IV)>(I)>(II) The trend is on the basis of resonance of functional groups with benzene ring. OCH3​,O=O,−OCOCH3​ are electron donating groups and fluorine is an electron withdrawing group.

X is

The major product of the following reaction is

3−Phenyl propene on reaction with HBr gives (major product).

C6​H5​CH2​CH(Br)CH3​

C6​H5​CH(Br)CH2​CH3​

C6​H5​CH2​CH2​CH2​Br

C6​H5​CH(Br)CH=CH2​

The correct order for acid strength of compounds CH≡CH,CH3​−C≡CH and CH2​=CH2​ is as follows:

CH≡CH>CH2​=CH2​>CH3​−C≡CH

HC≡CH>CH3​−C≡CH>CH2​=CH2​

$72$

$78$

$79$

$77$

Wurtz reaction is a reaction in which alkyl halide is converted to alkane by Na/ether. It is limited to the synthesis of symmetrical alkanes.
(Image ) As this molecule involves ring strain the reaction will be of poor yield.

Which alkene on ozonolysis gives $C H_{3} C H_{2} C H O$ and

$C H_{3} C H_{2} C H = C H C H_{2} C H_{3}$

$C H_{3} C H_{2} C H = C H C H_{3}$

One mole of alkene on ozonolysis gives $2$ moles of butanone. The alkene is:

$C H_{3} C H O + H C N \to C H_{3} - O H ∣ C H - C N$

Chlorination of toluene in presence of sunlight follows free radical mechanism. Benzyl radical ( $C_{6} H_{5} C H_{2}^{+})$ is most stable so benzyl chloride is major product formed.

All other reactions follows ion formation mechanism.

Option D is correct.

All compounds have delocalised electrons in their ring.
The compound which is anti-aromatic will be least stable.
Hence, option C is correct

$A r C H_{3}$

$A r C H_{2} C l$

$A r C H C l_{2}$

$A r C C l_{3}$

$A r C C l_{3}$

Among all given reagents $- C C l_{3}$ is strongest electron withdrawing group so it will reduce electron density at ortho and para position and it will direct reacting group at meta position.

So, $- C C l_{3}$ will yield maximum meta product.

Option D is correct.

$- C H O$

$- O H$

$- O C O C H_{3}$

$- C l$

$- C H O$

$- C H O is a meta directing group.$

Meta directing groups are the electron withdrawing groups. The attack of the incoming group is at the meta position.

$- O H$

$- C l$

$- O C H_{3}$

$- C H_{3}$

$- N H_{2}$

$- C l$

$- N O_{2}$

$- C_{2} H_{5}$

$- C l$

$- N H_{2}$ is activating group due to $+ M$ effect.
$- N O_{2}$ is a deactivating group due to $- M$ effect.
$- C H_{3}$ is activating group due to $+ I$ effect.
$- C l$ is deactivating group due to $- I$ effect but it is ortho, para directing due to $+ M$ effect.

$(I I I) > (I V) > (I) > (I I)$
The trend is on the basis of resonance of functional groups with benzene ring. $O C H_{3}, O = O, - O C O C H_{3}$ are electron donating groups and fluorine is an electron withdrawing group.

$3 - P h e n y l$ propene on reaction with $H B r$ gives (major product).

$C_{6} H_{5} C H_{2} C H (B r) C H_{3}$

$C_{6} H_{5} C H (B r) C H_{2} C H_{3}$

$C_{6} H_{5} C H_{2} C H_{2} C H_{2} B r$

$C_{6} H_{5} C H (B r) C H = C H_{2}$

The correct order for acid strength of compounds $C H \equiv C H, C H_{3} - C \equiv C H$ and $C H_{2} = C H_{2}$ is as follows:

$C H \equiv C H > C H_{2} = C H_{2} > C H_{3} - C \equiv C H$

$H C \equiv C H > C H_{3} - C \equiv C H > C H_{2} = C H_{2}$

$C H_{3} - C \equiv C H > C H_{2} = C H_{2} > H C \equiv C H$

$C H_{3} - C \equiv C H > C H \equiv C H > C H_{2} = C H_{2}$

$H C \equiv C H > C H_{3} - C \equiv C H > C H_{2} = C H_{2}$

$C H \equiv C H > C H_{3} - C \equiv C H > C H_{2} = C H_{2}$
(Acidic strength order).
More is the s-character, more is the acidic strength.
The s-character is maximum in $s p$ hybrid carbon atom followed by $s p^{2}$ and $s p^{3}$ .

$H C \equiv C H E x c e s s N a A E x c e s s C H_{3} C l B$ .
Product $B$ is :

$H C \equiv C H \to E x c e s s N a N a - C \equiv C - N a$
$N a - C \equiv C - N a E x c e s s C H_{3} C l C H_{3} - C \equiv C - C H_{3}$
Compound 'B' is 2-Butyne

Compound $(D)$ is :

Compound $(A)$ on reaction with $N H_{2} N H_{2} + O H$ gives:

Compound $(F)$ is:

$S n - S n$

$G e - G e$

$C - C$

$S i - S i$

$C - C$

Bond energies of these compounds are:
$C - C = 346 k J / m o l, S i - S i = 249 k J / m o l, G e - G e = 188 k J / m o l, S n - S n = 146 k J / m o l$ .

$p -$ block elements

$s -$ block elements

$d -$ block elements

Calorific value: The amount of energy produced by the complete combustion of a material or fuel.
Coal gas $H_{2} + C O + C H_{4}$ has the highest calorific value.