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Page No 311: - Chapter 10 Haloalkanes & Halo Arenes Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 10.4:

Which one of the following has the highest dipole moment?

(i) CH2Cl2

(ii) CHCl3

(iii) CCl4

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6824/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_56c2ee55.jpg

Dichlormethane (CH2Cl2)

μ = 1.60D

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6824/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_4e6e8267.jpg

Chloroform (CHCl3)

μ = 1.08D

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6824/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m1dc322ab.jpg

Carbon tetrachloride (CCl4)

μ = 0D

CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds cancel each other. Hence, its resultant dipole moment is zero.

As shown in the above figure, in CHCl3, the resultant of dipole moments of two C−Cl bonds is opposed by the resultant of dipole moments of one C−H bond and one C−Cl bond. Since the resultant of one C−H bond and one C−Cl bond dipole moments is smaller than two C−Cl bonds, the opposition is to a small extent. As a result, CHCl3 has a small dipole moment of 1.08 D.

On the other hand, in case of CH2Cl2, the resultant of the dipole moments of two C−Cl bonds is strengthened by the resultant of the dipole moments of two C−H bonds. As a result, CH2Cl2 has a higher dipole moment of 1.60 D than CHCl3 i.e., CH2Cl2 has the highest dipole moment.

Hence, the given compounds can be arranged in the increasing order of their dipole moments as:

CCl4 < CHCl3 < CH2Cl2

Question 10.5:

A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Answer:

A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane.

Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane.

Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6825/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m19bc237b.jpg

Cyclopentane (C5H10)

The reactions involved in the Question are:

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/1(535).png

Question 10.6:

Write the isomers of the compound having formula C4H9Br.

Answer:

There are four isomers of the compound having the formula C4H9Br. These isomers are given below.

(a)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6827/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m1c6b608d.jpg

1−Bromobutane

(b)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6827/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_f3a25b5.jpg

2−Bromobutane

(c)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6827/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_2dd2052a.jpg

1−Bromo−2−methylpropane

(d)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6827/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m360ca553.jpg

2−Bromo−2−methylpropane

Question 10.7:

Write the equations for the preparation of 1−iodobutane from

(i) 1-butanol

(ii) 1-chlorobutane

(iii) but-1-ene.

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6828/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m7396140b.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6828/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m6bf33829.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6828/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_74978295.jpg

Question 10.8:

What are ambident nucleophiles? Explain with an example.

Answer:

Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.

For example, nitrite ion is an ambident nucleophile.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6829/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_5454f67e.jpg

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6829/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_me81e426.jpg

Question 10.9:

Which compound in each of the following pairs will react faster in SN2 reaction with OH?

(i) CH3Br or CH3I

(ii) (CH3)3CCl or CH3Cl

Answer:

(i) In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group.

R−F << R−Cl < R−Br < R−I

Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH.

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6830/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m6006a9e4.jpg

The SN2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH3Cl. Hence, CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH.

Question 10.10:

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

(i) 1-Bromo-1-methylcyclohexane

(ii) 2-Chloro-2-methylbutane

(iii) 2,2,3-Trimethyl-3-bromopentane.

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/1%20(6)(2).png

In the given compound, there are two types of β-hydrogen atoms are present. Thus, dehydrohalogenation of this compound gives two alkenes.

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/2%20(5).png ​

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6832/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_64ac5af8.jpg

In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6832/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m2dfb93ab.jpg

Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.

Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6832/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m6fece1.jpg 2,2,3-Trimethyl-3-bromopentane

In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6832/NCERT_5-12-08_Utpal_12_Chemistry_10_10_GSX_html_m7057b929.jpg

According to Saytzeff’s rule, in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to the doubly bonded carbon atom is preferably formed.

Hence, alkene (I) i.e., 3,4,4-trimethylpent-2-ene is the major product in this reaction.

Question 10.11:

How will you bring about the following conversions?

(i) Ethanol to but-1-yne

(ii) Ethane to bromoethene

(iii) Propene to 1-nitropropane

(iv) Toluene to benzyl alcohol

(v) Propene to propyne

(vi) Ethanol to ethyl fluoride

(vii) Bromomethane to propanone

(viii) But-1-ene to but-2-ene

(ix) 1-Chlorobutane to n-octane

(x) Benzene to biphenyl.

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_6be2560c.gif

(ii)

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/fig%201(278).png

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m2e7a36dc.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m45edb8ba.jpg

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m1ce0a2d9.jpg

(vi)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m2116cafd.jpg

(vii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m49c6cd96.jpg

(viii)

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/fig%202(246).png

(ix)

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/fig%203(220).png

(x)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6834/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m6ba9b15.jpg

Question 10.12:

Explain why

(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

(ii) alkyl halides, though polar, are immiscible with water?

(iii) Grignard reagents should be prepared under anhydrous conditions?

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6836/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_511465de.jpg

In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of C−Cl bond near the Cl-atom is less in chlorobenzene than in cydohexyl chloride.

Moreover, the −R effect of the benzene ring of chlorobenzene decreases the electron density of the C−Cl bond near the Cl-atom. As a result, the polarity of the C−Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held together by dipole-dipole interactions. Similarly, strong H-bonds exist between the water molecules. The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water.

(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6836/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_4b0a1e38.gif

Therefore, Grignard reagents should be prepared under anhydrous conditions.

Question 10.13:

Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Answer:

Uses of Freon − 12

Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.

Uses of DDT

DDT (p, p−dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973.

Uses of carbontetrachloride (CCl4)

(i) It is used for manufacturing refrigerants and propellants for aerosol cans.

(ii) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.

(iii) It is used as a solvent in the manufacture of pharmaceutical products.

(iv) Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

Uses of iodoform (CHI3)

Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

Question 10.14:

Write the structure of the major organic product in each of the following reactions:

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_57783b52.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_62901c13.gif

(iii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m48eb75d0.gif

(iv) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m48d6a2e6.gif

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m21cd431a.gif

(vi) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m5fdee271.gif

(vii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_40c50296.gif

(viii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_70dd5ee5.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m182eb64a.gif

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_b012ab3.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m530683f1.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m71dc6617.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m19becf9a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_21ad1223.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_400a60b6.gif

(viii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6838/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_4932b73f.jpg

Question 10.15:

Write the mechanism of the following reaction:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6840/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m3919f156.gif

Answer:

The given reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6840/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m3919f156.gif

The given reaction is an SN2 reaction. In this reaction, CN acts as the nucleophile and attacks the carbon atom to which Br is attached. CN ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6840/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_309f0430.jpg

Question 10.16:

Arrange the compounds of each set in order of reactivity towards SN2 displacement:

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6841/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m483cc4f0.jpg

An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards SN2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.

1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane

Hence, the increasing order of reactivity towards SN2 displacement is:

2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6841/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_78278e60.jpg

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is

3° < 2° < 1°.

Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2 displacement as:

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

[2-Bromo-3-methylbutane is incorrectly given in NCERT]

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/269/6841/NCERT_8-12-08_Utpal_Chemistry_12-10-8_GSX_html_m4a85dcee.jpg

The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents. Therefore, the increasing order of steric hindrances in the given compounds is as below:

1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane

< 1-Bromo-2, 2-dimethylpropane

Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane

Also Read : Page-No-312:-Chapter-10-Haloalkanes-&-Halo-Arenes-Exercise-Solutions-class-12-ncert-solutions-Chemistry

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