NCERT Exemplar Class 11 Chemistry Solutions Chapter 13 Hydrocarbons

Question 2. Arrange the halogens $F_{2}, Cl_{2}, Br_{2}, I_{2}$, in order of their increasing reactivity with alkanes.
$(i) I_{2}< Br_{2} < Cl_{2} < F_{2}$
$(ii) Br_{2} < Cl_{2} < F_{2} < I_{2}$
$(iii) F_{2} < Cl_{2} < Br_{2} < I_{2}$
$(iv) Br_{2} < I_{2} < Cl_{2} < F_{2}$
Answer:

The answer is the option $(i) I_{2}< Br_{2} < Cl_{2} < F_{2}$
Explanation: Since electronegativity of halogens decreases down the group, Fluorine is the most electronegative. As electronegativity of Fluorine decreases their reactivity with alkanes also decreases. Thus, $F_{2}$ is highly reactive, whereas $I_{2}$ is the least reactive.

Question 3. The increasing order of reduction of alkyl halides with zinc and dilute HCl is
(i) R–Cl < R–I < R–Br
(ii) R–Cl < R–Br < R–I
(iii) R–I < R–Br < R–Cl
(iv) R–Br < R–I < R–Cl
Answer:

The answer is the option (ii) R-Cl < R-Br < R-I
Explanation: We know that reactivity of halogen decreases down the group; therefore, reduction of alkyl halides with Zn/HCl follow reverse order.
Reactivity of reduction $\alpha$ 1bond strength of C-X
$Reactivity of reduction \propto \frac{1}{bond strength C-X}$
It also depends on the size of the halogen. Therefore, we can say that to increase the reactivity; the bond strength must be reduced.

Question 4. The correct IUPAC name of the following alkane is

(i) 3,6 – Diethyl – 2 – methyloctane
(ii) 5 – Isopropyl – 3 – ethyloctane
(iii) 3 – Ethyl – 5 – isopropyloctane
(iv) 3 – Isopropyl – 6 – ethyloctane

Answer:

The answer is the option (i) 3,6-Diethyl-2-methyloctane
Explanation: Since the alkane has 8 carbon atoms, it is octane, viz., the longest chain. There is a methyl group on carbon 2, ethyl groups on carbon 3 & 6. Since 2 ethyl groups are present, it will be diethyl, and alphabetically it comes before methyl. The lowest sum rule is followed by side chains present on the carbon atoms 2, 3 & 6. Hence, the option (i).

Question 5. The addition of HBr to 1-butene gives a mixture of products A, B and C
$CH_{3}-CH_{2}-CH_{2}-CH_{2}-Br$
(C)
The mixture consists of
(i) A and B as major and C as minor products
(ii) B as major, A and C as minor products
(iii) B as minor, A and C as major products
(iv) A and B as minor and C as major products
Answer:

The answer is option (i) A and B as major and C as minor products.
Explanation: According to Markovnikov’s rule, the major product is 2-Bromobutane, and the minor product is I-Bromobutane. Butane-1 is unsymmetrical. 2-Bromobutane has chiral carbon and hence exists in two enantiomers.
Thus, it is clear that the major products are A & B while the minor product is C.

Question 6. Which of the following will not show geometrical isomerism ?






Answer:

The answer is the option
Explanation: Alkenes show geometrical isomerism where the double-bonded carbons must have different atoms or groups.
Therefore, the structure in option (iv) does not show geometrical isomerism since the double-bonded carbons have 3 same groups and one different group.

Question7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
(i) HCl > HBr > HI
(ii) HBr > HI > HCl
(iii) HI > HBr > HCl
(iv) HCl > HI > HBr

Answer:

The answer is the option (iii) HI > HBr > HCl
Explanation: The factors that affect the reactivity of hydrogen halides are bond strength and bond dissociation energy. We know that, in halogen halides the size of the halogen atom increases, the bond dissociation energy, as well as bond strength, decreases.
Hence, the option (iii) is the correct order.

Question 8. Arrange the following carbanions in order of their decreasing stability.

(i) A > B > C
(ii) B > A > C
(iii) C > B > A
(iv) C > A > B
Answer:

The answer is option (ii) B > A > C
Explanation: CH₃ group has +I effect, which decreases the stability of the carbon anion and in C this effect direct to the negatively charged carbon. +I effect is also present in A, but there is more distant from the negatively charged carbon, and in B there is no +I effect

Question 9. Arrange the following alkyl halides in decreasing order of the rate of $\beta$ – elimination reaction with alcoholic KOH.
(B) $CH_{3}-CH_{2}-Br$ (C) $CH_{3}-CH_{2}-CH_{2}-Br$
(i) A > B > C
(ii) C > B > A
(iii) B > C > A
(iv) A > C > B
Answer:

The answer is the option (iv) A > C > B
Explanation: When alkyl halides are heated with alc. KOH, an alkene is formed by eliminating one molecule of halogen acid. Hydrogen is eliminated from the beta carbon atom. Order of reactivity is $3^{\circ} > 2^{\circ} > 1^{\circ}$, hence option (iv). The rate of reaction is determined by the nature of alkyl groups.

Question10. Which of the following reactions of methane is incomplete combustion:
$(i)2CH_{4}+O_{2}\xrightarrow[Cu/523]{K/100atm}2CH_{3}OH$
$(ii)CH_{4}+O_{2}\overset{Mo_{2}O_{3}}{\rightarrow}HCHO+H_{2}O$
$(iii)CH_{4}+O_{2}\rightarrow C(s)+2H_{2}O(l)$
$(iv)CH_{4}+2O_{2}\rightarrow CO_{2}(g)+2H_{2}O(l)$

Answer:

The answer is the option $(iii)CH_{4}+O_{2}\rightarrow C(s)+2H_{2}O(l)$
Explanation: Insufficient supply of air or oxygen leads to incomplete combustion and carbon black is formed.
Hence, $CH_{4}+O_{2}\rightarrow C(s)+2H_{2}O(l)$

Question 12. Which of the following alkenes on ozonolysis give a mixture of ketones only?

Answer:

The answer is the option (iii) and (iv)
Explanation: Depending on groups or atoms attached, alkenes give two molecules of carbonyl compounds on ozonolysis. Ketones are formed if the double-bonded carbon atoms have alkyl groups.

Question 13. Which are the correct IUPAC names of the following compound?

(i) 5– Butyl – 4– isopropyldecane
(ii) 5– Ethyl – 4– propyldecane
(iii) 5– sec-Butyl – 4– iso-propyldecane
(iv) 4–(1-methylethyl)– 5 – (1-methylpropyl)-decane
Answer:

The answer is the option (iii) 5-sec-Butyl-4-iso-propyldecane & (iv) 4-(1-methylethyl)-5-(1-mrthylpropyl)-decane
Explanation: The longest carbon chain has 10 atoms, hence ‘decane’. According to the lowest sum rule, side chains are on 4^th(isopropyl group) & 5^th(sec-butyl group) carbon atoms. Isopropyl is called 1-Methylethyl group and sec-butyl is called 1-Methylpropyl group in IUPAC. Hence the correct names are in option (iii) & (iv).

Question14. Which are the correct IUPAC names of the following compound?

(i) 5 – (2′, 2′–Dimethylpropyl)-decane
(ii) 4 – Butyl – 2,2– dimethylnonane
(iii) 2,2– Dimethyl – 4– pentyloctane
(iv) 5 – neo-Pentyl decane
Answer:

The answer is the option (i) 5-(2’,2’-Dimethylpropyl)-decane & (iv) 5-neo-Pentyldecane
Explanation: The longest carbon chain has 10 carbon atoms, hence ‘decane’. Sidechain is present on 5th carbon atom, viz., neo-Pentyl or 2’,2’-Dimethylpropyl according to IUPAC. Hence option (i) & (iv).

Question15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring _______.
(i) deactivates the ring by inductive effect
(ii) deactivates the ring by resonance
(iii) increases the charge density at ortho and para position relative to meta position by resonance
(iv) directs the incoming electrophile to meta position by increasing the charge density relative to ortho and para position.

Answer:

The answer is the option (i) deactivates the ring by inducive effect & (iii) increases the charge density at ortho and para position relative to meta position by resonance.
Explanation: Presence of halogen atom on benzene ring direct electrophile on ortho and para positions and increases electron density on ortho and para positions. It also shows the +R effect. Due to -I effect halogen pulls an electron from benzene since they are more electronegative and decreases electron density. The electron density on ortho and para position is greater than meta-position due to resonance.

Question 16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group ________.
(i) deactivates the ring by inductive effect.
(ii) activates the ring by inductive effect.
(iii) decreases the charge density at ortho and para position of the ring relative to meta position by resonance.
(iv) increases the charge density at meta position relative to the ortho and para positions of the ring by resonance.
Answer:

The answer is the option (i) deactivates the ring by inducive effect & (iii) decreases the charge density at ortho and para position of the ring relative to meta position by resonance.
Explanation: Meta position is attacked by an electrophile. Due to the -I effect nitro group deactivates the benzene ring and decreases electron density at ortho and para positions than meta position.

Question 17. Which of the following are correct?
(i) $CH_{3}-O-CH_{2}^{+}$ is more stable than $CH_{3}-CH_{2}^{+}$
(ii) $\left (CH_{3} \right )_{2}CH^{+}$ less table than $CH_{3}-CH_{2}-CH_{2}^{+}$
(iii) $CH_{2}=CH-CH_{2}^{+}$ is more stable than $CH_{3}-CH_{2}-CH_{2}^{+}$
(iv) $CH_{2}=CH^{+}$ is more stable than $CH_{3}-CH_{2}^{+}$

Answer:

The answer is the option $(i)CH_{3}-O-CH_{2}^{+}$ is more stable than $CH_{3}-CH_{2}^{+}$ & (iii) $CH_{2}=CH-CH_{2}^{+}$ is more stable than $CH_{3}-CH_{2}-CH_{2}^{+}$.
Explanation: $(i)CH_{3}-O-CH_{2}^{+}$is more stable than $CH_{3}-CH_{2}^{+}$ because +R effect of $CH_{3}-O$ is greater than $-CH_{3}$ (stability of carbocation increases due to the +R effect).
(iii) We know that the resonance effect is stronger than +I effect. Here, $CH_{2}=CH-CH_{2}^{+} \leftrightarrow ^+CH_{3}-CH=CH_{2}$ is stabilised by resonance effect; whereas $CH_{3}-CH_{2}-CH_{2}^{+}$ is stabilised by +I effect only . Hence, $CH_{2}=CH-CH_{2}^{+}$ is more stable than $CH_{3}-CH_{2}-CH_{2}^{+}$.

Question 18. Four structures are given in options (i) to (iv). Examine them and select the aromatic structures.

(i)


(ii)

(iii)

(iv)

Answer:

The answer is the option (i) & (iii)

Explanation: The following conditions should be fulfilled by a compound in order to become aromatic-

• $\pi$ electrons in the ring should be completely delocalised.

• Planarity

• Huckel Rule should be followed, viz., presence of $(4n + 2) \pi$ electrons in the ring.

Now, analysing the given options:
(i) + n = 0, planar,
$\pi$ electrons = 2
(ii) $\pi$ electrons = 8, not planar,
n = not an integer
(iii) planar
$\pi$electrons = (4n + 2) = 6 in each ring
(iv) lone pair is not in conjugation, hence it is non aromatic
n = not an integer
Hence, option (i) and (iii) are aromatic structure.

Question 19. The molecules having dipole moment are __________.
(i) 2,2-Dimethylpropane
(ii) trans-Pent-2-ene
(iii) cis-Hex-3-ene
(iv) 2, 2, 3, 3 – Tetramethylbutane
Answer:

The answer is the option (ii) trans-Pent-2-ene & (iii) cis-Hex-3-ene
Explanation: (ii) Because of different groups attached, trans-Pent-2-ene shows net dipole moment.
(iii) Here, both $C_2H_5$ groups are inclined to each other at an angle of $60^{\circ}$ and hence has a resultant dipole moment. Thus, cis-Hex-3-ene shows a dipole moment.

Question 21. Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?
Answer:

Butene-2, where either both the methyl groups are on the same side or opposite side to show geometrical isomers, is formed on the reduction of 2-Butyne.
$H_{3}C-C\equiv C-CH_{3}\overset{Na/liq.NH_{3}}{\rightarrow}$

Question 22. Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
Answer:

(i)Alkanes have a carbon-carbon sigma bond where the distribution of electron of sigma molecular orbit is symmetrical around the internuclear axis of C-C bond, viz., not distributed due to rotation about its axis.
(ii) Hence, C-C single bond is permitted for free rotation which results in different spatial arrangements of atoms in space which can change into one another.
Such spatial arrangements of atoms are called conformations or rotamers or conformers.
(iii) Due to rotation around C-C bonds alkenes have infinite no. of conformations. However, rotation around C-C single bond is not completely free and is hindered by small energy barrier of1-20 kJ mol^-1 because of weak repulsive interaction between adjacent bonds. It is called a torsional strain.

Question 23. Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why?
Answer:

(i) Sawhorse projections of ethane


(ii) Newman projections of ethane


Stability of conformation:
(i) There are minimum repulsive forces, minimum energy and maximum stability in the staggered form of ethane. Here, the electrons clouds of carbon-hydrogen bonds are as far as possible.
(ii) the electron clouds come closer when the staggered form changes to the eclipsed form. It results in increased electron cloud repulsions. Here, molecules will have more energy and lesser stability.
Therefore, the staggered form is more stable.

Question 24. The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol^-1, 363.7 kJ mol^-1 and 296.8 kJ mol^-1 respectively. What will be the order of reactivity of these halogen acids?
Answer:

Reactivity of hydrogen halides depends on the dissociation enthalpy of H-X. The order of reactivity of hydrogen halides is HI > HBr > HCl. They add up to alkanes to form alkyl halides.
Bond enthalpy of HI<HBr<HCl, hence reactivity in reverse order is:
HCl < HBr < HI

Question 25. What will be the product obtained as a result of the following reaction and why?
$+H_{3}C-CH_{2}-CH_{2}Cl\overset{AlCl_{3}}{\rightarrow}$

Answer:

When alkylation of Friedel-Crafts takes place with a higher alkyl halide in which primary carbocation is formed first and then converted into secondary carbocation, which is more stable. It is formed by rearrangement; therefore, the product is isopropyl benzene.
Primary carbocation- $CH_{3}-CH_{2}-CH_{2}Cl+AlCl_{3}\rightarrow AlCl_{4}^{+}+CH_{3}-CH_{2}-CH_{2}^{+}$
Secondary carbocation- $CH_{3}-CH_{2}-CH_{2}^{+}\rightarrow CH_{3}-CH^{+}-CH_{3}$

$+H_{3}C-CH_{2}-CH_{2}Cl\overset{AlCl_{3}}{\rightarrow}$

Question 26. How will you convert benzene into
(i) p – nitrobromobenzene
(ii) m – nitrobromobenzene
Answer:

(i) Converting benzene into p-nitro bromobenzene:
$C_{6}H_{6} + Br_{2}$ (In presence of anhyd. $FeBr_{3}$) → $C_{6}H_{5} Br_{2}+HBr$
$C_{6}H_{5}Br + Conc. HNO_{3} + Conc. H_{2}SO_{4}$ → p-nitrobromobenzene
(ii) Converting benzene into m-nitrobromobenzene:
$C_{6}H_{6} + Conc. HNO_{3} + Conc. H_{2}SO_{4}$ → $C_{6}H_{5}NO_{2}$
$C_{6}H_{5}NO_{2} + Br$ (In presence of anhyd. $FeBr_{3}$) → m-nitrobromobenzene

Question 27. Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give reason.


Answer:

The methoxy group$(-OCH_{3})$makes anisole more reactive than benzene because it is an electron-releasing group and increases the electron density of the benzene ring due to the +R effect.



Cl group is less reactive than methoxybenzene as it gives +R and -I effect. And, Chlorobenzene is more reactive than $-NO_{2}$ group since the $-NO_{2}$ gives -I and -R effect.

Question 28. Despite their – I effect, halogens are o- and p-directing in haloarenes. Explain.
Answer:

Halogens are ortho and para-directing because they have +R and -I effect. Now, halogens present on the benzene ring has these effects in which the -I effect deactivates the ring and +R effect increases the electron density on ortho and para positions.

Question 29. Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.

Answer:

The benzene ring deactivated and the electron density decreases on the ortho and para positions as compared to the meta positions due to the presence of the nitro group which has -I and -R effects.

Question 30. Suggest a route for the preparation of nitrobenzene starting from acetylene?
Answer:

Ethylene undergoes cyclic polymerisation when passed through red hot iron tube at 873 K and gives benzene from which

nitrobenzene can be obtained through nitration.

Question 31. Predict the major product (s) of the following reactions and explain their formation.
$H_{3}C-CH=CH_{2}\xrightarrow[HBr]{(Pb-CO-O)_{2}}$
$H_{3}C-CH=CH_{2}\overset{HBr}{\rightarrow}$
Answer:

(i) $H_{3}C-CH=CH_{2}\xrightarrow[HBr]{(Pb-CO-O)_{2}}$ $H_{3}C-CH_{2}-CH_{2}-Br$
Step 1: Homolysis of peroxide for forming free radicals

Step 2: Formation of Bromine free radical
$C_6H_5 + H-Br \rightarrow ^\cdot C_6H_6 + Br^\cdot$

Step 3: Hydrogen bromide reaction with an alkyl radical

(ii)

Question 32. Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles.
$(i)H_{3}CO^{-}$
(ii)

(iii) $Cl^{.}$
$(iv)Cl_{2}C:$
$(v)(H_{3}C)_{3}C^{+}$
$(vi)Br^{-}$
$(vii)H_{3}COH$
$(viii)R-NH-R$
Answer:

Electrophiles can be defined as the electron seeking species or electron-deficient species. They can be either positively charged or neutral.
Therefore, the following species are electrophiles-
(iii) $Cl^{.}$
$(iv)Cl_{2}C:$
$(v)(H_{3}C)_{3}C^{+}$
electron rich species are called nucleophiles; they can be either negatively charged or neutral.
The following species are nucleophiles-
$(i)H_{3}CO^{-}$
(ii)

$(vi)Br^{-}$
$(vii)H_{3}COH$
$(viii)R-NH-R$

Question 33. The relative reactivity of $1^{\circ}, 2^{\circ}, 3^{\circ}$° hydrogen’s towards chlorination is 1 : 3.8 : 5. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.

Answer:

There are 9 primary, 2 secondary and 1 tertiary hydrogen atoms in 2-methylbutane and the relative reactivity of $1^{\circ}, 2^{\circ}, 3^{\circ}$hydrogen atoms towards chlorination is 1:3:8:5.
The relative amount of product after chlorination = no. of hydrogen atom X relative reactivity.

Therefore, the total amount of monochloro product will be :
9 + 7.6 + 5 = 21.6
Now, $1^{\circ}$ monochloro product% = $9 \times \frac{100}{21.6} =41.7$ %
$2^{\circ}$ monochloro product % = $7.6\times \frac{100}{21.6} = 35.2$%
$3^{\circ}$ monochloro product % = $5\times \frac{100}{21.6}=23.1$%

Question 34. Write the structures and names of products obtained in the reactions of sodium with a mixture of 1-iodo-2-methylpropane and 2-iodopropane.
Answer:

Alkanes having double bonds are obtained when n-alkyl halide is treated with name talin in the presence of ether, products obtained are based on Wurtz reaction.
$RX + 2Na +XR \rightarrow R-R + 2NaX$
$CH_{3}-CH(CH_{3})-CH_{2}I + 2Na + ICH_{2}-CH(CH_{3})-CH_{3}\overset{Dry ether}{\rightarrow} 2NaI + CH_{3}-CH(CH_{3})-CH_{2}-CH_{2}-CH(CH_{3})-CH_{3}$
(2,5-dimethylHexane)
$CH_{3}-CH(CH_{3})I + 2Na + I(CH_{3})CH-CH_{3} \rightarrow 2NaI + CH_{3}-CH(CH_{3})-CH(CH_{3})-CH_{3}$
(2,4-dimethylbutane)
$CH_{3}-CH(CH_{3})-CH_{2}I + 2Na + I-CH(CH_{3})-CH_{3} \rightarrow 2NaICH_{3}-CH(CH_{3})-CH-CH_{3}-CH_{3}$
(2,4-dimethylpentane)

Question 35. Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.
Answer:

Due to hyperconjugation and 9 $\alpha$ hydrogen tertiary $3^{\circ}$ free radical is more stable and sit stabilised; whereas, $1^{\circ}$ free radical has 1 $\alpha$ hydrogen and one hyper conjugative structure and hence is less stable.

Question 36. An alkane $C_{8}H_{18}$ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.

Answer:

Question 37. The ring systems having the following characteristics are aromatic.
(i) Planar ring containing conjugated $\pi$ bonds.
(ii) Complete delocalisation of the $\pi$−electrons in the ring system i.e. each atom in the ring has an unhybridized p-orbital, and
(iii) Presence of $(4n+2) \pi$−electrons in the ring where n is an integer (n = 0, 1, 2,………..) [Huckel rule].
Using this information classify the following compounds as aromatic/nonaromatic.

Answer:

Question 38. Which of the following compounds are aromatic according to Huckel’s rule?

Answer:

Question 39. Suggest a route to prepare ethyl hydrogen sulphate $(CH_{3}-CH_{2}-OSO_{2}-OH)$ starting from ethanol $(C_{2}H_{5}OH)$
Answer:

Preparing ethyl hydrogen sulphate starting from ethanol.
Step I-
Protonation of alcohol and formation of a carbocation.
$H_{2}SO_{4} \rightarrow H^{+} + ^{-}OSO_{2}OH$
$CH_{3}- CH_{2} - O - H + H^{+} \rightarrow CH_{3} - CH_{2} - ^{+}OH2$
$CH_{3} - CH_{2} - ^{+}OH2 \rightarrow CH_{3}- ^{+}CH_2 + H_{2}O$
Step II-
Attack of nucleophile
$HO - SO_{2} - O^{-} + ^{+}CH_{2} - CH_{3} \rightarrow CH_{3} - CH_{2} - O - SO_{2}- OH$

Question 41. Match the hydrocarbons in Column I with the boiling points given in Column II.

Answer:

$(i) \rightarrow (b); (ii) \rightarrow (c); (iii) \rightarrow (a)$
Explanation:
(i) There are more van der Waal’s forces in n-pentane, and its boiling point is also high since there is no branching and surface area.
(ii) The boiling point of iso-pentane is less because the molar mass is the same except there’s one brach which reduces the surface area.
(iii) The boiling point of neo-pentane is the lowest because it has two side chains which have the same molar mass.

Question 42. Match the following reactants in Column I with the corresponding reaction products in Column II.

Answer:

$(i) \rightarrow(d); (ii) \rightarrow (c); (iii) \rightarrow (b); (iv) \rightarrow (a)$
Explanation:

Question 43. Match the reactions given in Column I with the reaction types in Column II.

Answer:

$(i) \rightarrow (d); (ii) \rightarrow (a); (iii) \rightarrow (b); (iv) \rightarrow (c)$
Explanation:
(i) Hydration refers to the addition of water molecule
$CH_{2}=CH_{2}+H_{2}O\overset{H^{+}}{\rightarrow}CH_{3}CH_{2}OH$
(ii) Addition of H₂ in the presence of a catalyst is called as Hydrogenation
$CH_{2}=CH_{2}+H_{2}\overset{Pd}{\rightarrow}CH_{3}-CH_{3}$
(iii) Addition of halogens means halogenation
$CH_{2}=CH_{2}+Cl_{2}\rightarrow Cl-CH_{2}-CH_{2}-Cl$
(iv) When a no. of smaller molecules come together to give one bigger molecule, it is known as polymerisation.

Question 45. In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Toluene on Friedel Crafts methylation gives o– and p–xylene.
Reason (R) : $CH_3$ group bonded to benzene ring increases electron density at o– and p– position.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (i) Both A and R are correct, and R is the correct explanation of A.
Explanation:$CH_3$ group is an electron-withdrawing group. It activates the benzene ring due to hyperconjugation effect. Toluene has -$CH_3$ group attached to the benzene ring. This group also increases the electron density at ortho and para positions for electrophile attacks.

Question 46. In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Nitration of benzene with nitric acid requires the use of concentrated sulphuric acid.
Reason (R) : The mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile, $NO_{2}^{+}$ .
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.

Answer:

(i) Both A and R are correct, and R is the correct explanation of A.
Explanation: During nitration, benzene is treated with the nitrating mixture, viz., Conc. $HNO_{3}$ and $H_{2}SO_{4}$. $H_{2}SO_{4}$ helps in furnishing the electrophile, i.e., $NO_{2}^{+}$.
$H_{2}SO_{4}+HNO_{3}\underset{H_{2}O}{\rightarrow}NO_{2}^{+}+HSO_{3}^{-}$

Question 47. In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Among isomeric pentanes, 2, 2-dimethylpentane has highest boiling point.
Reason (R) : Branching does not affect the boiling point.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.

Answer:

The answer is the option (iii) Both A and R are not correct.
Explanation: 2,2-dimethylpentane has the lowest boiling point among isomeric pentanes, and its boiling point decreases further on branching.

Question 49. 896 mL vapour of a hydrocarbon ‘A’ having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of ‘A’ gives 2-methylpentane.
Also ‘A’ on hydration in the presence of $H_{2}SO_{4}$and $HgSO_{4}$ gives a ketone ‘B’ having molecular formula $C_{6}H_{12}O$. The ketone ‘B’ gives a positive iodoform test. Find the structure of ‘A’ and give the reactions involved.
Answer:

To calculate molar mass of hydrocarbon A it is given that 896 ml of Hydrocarbon AC_xH_y weighs 3.28 g at STP.
22400 ml of has AC_xH_y mass = $\frac{3.28 \times 22400 ml}{896 ml}$
= 83.1 g/mol.
Thus, the molar mass of A = 83.1 g/mol

We know that empirical formula = $C_{3}H_{5}$
Empirical formula with weight = $3 \times12 = 36 + 5 = 41$
Molecular formula = (empirical formula) n, where n = mol mass/empirical mass = $\frac{831}{41}=2.02$
Molecular formula = $[C_{3}H_{5}]_{2} = C_{6}H_{10}$
Now, determining the structure of (A) and (B)-
$C_{6}H_{12}$ is obtained when hydrogenation of $C_{5}H_{10}$ happens with two moles of H₂. The structure is methylpentane.
$C_{6}H_{12}O$ (which gives positive iodoform test) is obtained when hydration of (A) takes place in the presence of dil. H=$H_{2}SO_{4}$ and $HgSO_{4}$.
Thus, the structure of $A = (CH_{3})_{2}CH-CH-C\equiv CH$(4-methyl pent-yne) and (B) is 4-methyl pentanone-2.

Question 50. An unsaturated hydrocarbon ‘A’ adds two molecules of H₂ and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of ‘A’, write its IUPAC name and explain the reactions involved.
Answer:

Two molecules of hydrogen add on ‘A’ this shows that ‘A’ is either an alkadiene or an alkyne.
On ozonolysis compound A gives
$CH_{3}CHO + O = CH - CH_{2}- CH_{2}- CHO + O = C(CH_{3})_{2}$
Hence, the structure of A(Its IUPAC name will be 2-methyl octa 2,6 diene) is,

Question 51. In the presence of peroxide addition of HBr to propene takes place according to anti Markovnikov’s rule but peroxide effect is not seen in the case of HCl and HI. Explain.

Answer:

(i)
(ii) $^.C_{6}H_{5}+H-Br\overset{homolysis}{\rightarrow}C_{6}H_{6}+Br^.$
(iii)
(iv) $CH_{3}+\dot{C}H-CH_{2}Br+H-Br\overset{Homolysis}{\rightarrow}CH_{3}-CH_{2}-CH_{2}Br+\dot{B}r$
Since the H-Cl bond is stronger than H-Br bond, peroxide effect is observed only in the case of HBr. It is not observed in the case of HI as well.
H-Br has lesser bond energy than H-Cl; thus, H-Cl bond is not cleaved by the free radical, whereas the H-I bond is weaker and iodine free radicals combine to form dimer iodine molecules.