Chapter 13 Hydrocarbons

Multiple Choice Questions (MCQs)

1. Arrange the following in decreasing order of their boiling points.

A. $n$ butane

B. 2-methylbutane

C. $n$-pentane

D. 2, 2-dimethylpropane

(a) $A>B>C>D$

(b) $B>C>D>A$

(c) $D>C>B>A$

(d) $C>B>D>A$

Show Answer

Answer

(d) As the number of carbon atom increases, boiling point increases.

Boiling point decreases with branching

(4 carbon atoms with no branching)

2. Arrange the halogens $F_{2}, Cl_{2}, Br_{2}, I_{2}$, in order of their increasing reactivity with alkanes.

(a) $I_{2}<Br_{2}<Cl_{2}<F_{2}$

(b) $Br_{2}<Cl_{2}<F_{2}<I_{2}$

(c) $F_{2}<Cl_{2}<Br_{2}<I_{2}$

(d) $Br_{2}<I_{2}<Cl_{2}<F_{2}$

Show Answer

Answer

(a) Rate of reaction of alkanes with halogens is $F_2>Cl_2>Br_2>I_2$

Alkane react with $F_{2}$ is vigorously and with $I_{2}$ the reaction is too slow that it requires a catalyst. It is because of high electronegativity of fluorine. Reactivity decreases with decrease in electronegativity and electronegativity decreases down the group

3. The increasing order of reduction of alkyl halides with zinc and dilute $\mathrm{HCl}$ is

(a) $\mathrm{R}-\mathrm{Cl}<\mathrm{R}-\mathrm{I}<\mathrm{R}-\mathrm{Br}$

(b) $R \quad \mathrm{Cl}<R-\mathrm{Br}<R-\mathrm{I}$

(c) $\mathrm{R}-\mathrm{I}<\mathrm{R}-\mathrm{Br}<\mathrm{R}-\mathrm{Cl}$

(d) $\mathrm{R}-\mathrm{Br}<\mathrm{R}-\mathrm{I}<\mathrm{R}-\mathrm{Cl}$

Show Answer

Answer

(b) The reactivity of halogens with alkane is $F_{2}>Cl_{2}>Br_{2}>I_{2}$ Hence, reduction of alkyl halide with $Zn$ and dilute $HCl$ follows reverse order i.e., $R-I>R-Br>R-Cl$. Further, the reactivity of this reduction increases as the strength of $C-X$ bond decreases

4. The correct IUPAC name of the following alkane is

(a) 3,6-diethyl-2-methyloctane

(b) 5-isopropyl -3-ethyloctane

(c) 3-ethyl-5-isopropyloctane

(d) 3-isopropyl-6-ethyloctane

Show Answer

Answer

(a) The correct IUPAC name is

Longest chain $-8 \mathrm{C}$ atom alkane $=$ octane

Branch on 2, 3,6 follows lowest sum rule.

Branch of 2-C-methyl; 3, 6, C atom-ethyl.

Ethyl comes alphabetically before methyl.

Hence, 3,6-diethyl 2-methyl octane.

5. The addition of $\mathrm{HBr}$ to 1-butene gives a mixture of products $A, B$ and $C$.

The mixture consists of

(a) $A$ and $B$ as major and $C$ as minor products

(b) $B$ as major, $A$ and $C$ as minor products

(c) $B$ as minor, $A$ and $C$ as major products

(d) $A$ and $B$ as minor and $C$ as major products

Show Answer

Answer

(a) The alkene is unsymmetrical, hence will follow Markownikoff’s rule to give major product.

Since, I contains, a chiral carbon, it exists in two enantiomers ( $A$ and $B$ ) which are mirror images of each other.

$(A)$

$(B)$

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

Show Answer

Thinking Process

This question is based upon geometrical isomerism. For geomterical isomerism, it is essential that each carbon atom of the double bond must have different substituents.

Answer

(d) In option (d), a carbon with double bond has two same functional groups $\left(\mathrm{CH}_{3}\right)$ attached. The rotation around carbon will not produce a new compound. Hence, geometrical isomerism is not possible.

7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene.

(a) $\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$

(b) $\mathrm{HBr}>\mathrm{HI}>\mathrm{HCl}$

(c) $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

(d) $\mathrm{HCl}>\mathrm{HI}>\mathrm{HBr}$

Show Answer

Answer

(c) Bond energy of $\mathrm{HI}$ is $296.8 \mathrm{~kJ} / \mathrm{mol}, \mathrm{HBr}$ is $36.7 \mathrm{~kJ} / \mathrm{mol}$ and $\mathrm{HCl}$ is $430.5 \mathrm{~kJ} / \mathrm{mol}$. Hence, $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$ is the order of reactivity with propene.

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

A. $\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}^{-}$

B. $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}^{-}$

C. $H_{3} C-CH_{2}$

(a) $A>B>C$

(b) $B>A>C$

(c) $\mathrm{C}>\mathrm{B}>\mathrm{A}$

(d) $\mathrm{C}>\mathrm{A}>\mathrm{B}$

Show Answer

Answer

(b) $+I$-effect decreases the stability of carbon anion. Since, $\left(CH_{3}\right)$ group has $+I$-effect, therefore, it intensifies the negative charge and hence destabilises $(A)$ relative to $(B)$.

$s p$ hybridised carbanion is more stabilised than $s p^{3}$

$CH \equiv \underset{(B)^{s p}}{C^{-}}>CH_{3}-C \equiv \underset{(A)^{s p}}{C^{-}}>\underset{(C)}{CH_{3}}-\underset{\text { sp } ^{3}}{CH_{2}}$

Hence, $B>A>C$

9. Arrange the following alkyl halides in decreasing order of the rate of $\beta$-elimination reaction with alcoholic $\mathrm{KOH}$.

C. $CH_{3}-CH_{2}-CH_{2}-Br$

(a) $\mathrm{A}>\mathrm{B}>\mathrm{C}$

(b) $\mathrm{C}>\mathrm{B}>\mathrm{A}$

(c) $B>C>A$

(d) $A>C>B$

Show Answer

Answer

(d) Alkyl halides on heating with alcoholic potash eliminates one molecule of halogen acid to form alkene. Hydrogen is eliminated from $\beta$-carbon atom. Nature of alkyl group determines rate of reaction i.e., $3^{\circ}>2^{\circ}>1^{\circ}$ or $A>\underset{2^{\circ} \beta \text {-Carbon }}{C}>B$

$CH_{3}{ }^{1^{\circ} \beta \text {-carbon }} CH_{2}-Br$ $CH_{3}-CH_{2}-CH_{2}-Br$

$(A)$

$(B)$

$(C)$

10. Which of the following reactions of methane is incomplete combustion?

(a) $2 CH_{4}+O_{2} Cu / 523 ~K / 100 ~atm \rightarrow 2 CH_{3} OH$

(b) $CH_{4}+O_{2} \quad Mo_{2} O_{3} \rightarrow HCHO+H_{2} O$

(c) $CH_{4}+O_{2} \rightarrow C(s)+2 H_{2} O(l)$

(d) $CH_{2}+2 O_{2} \rightarrow CO_{2}(g)+2 H_{2} O(l)$

Show Answer

Answer

(c) During incomplete combustion of alkanes with insufficient amount of air or dioxygen carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. Thus,

Multiple Choice Questions (More Than One Options)

11. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions?

Show Answer

Answer

$(c, d)$

Alkanes on heating with a regulated supply of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products.

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

Show Answer

Answer

$(c, d)$

Alkenes which have two substituents on each carbon atom of the double bond, give mixture of ketones on ozonolysis. Thus, option (c) and (d) give mixture of ketones.

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

(a) 5-Butyl-4-isopropyldecane

(b) 5-Ethyl-4-propyldecane

(c) 5-sec-Butyl -4-iso-propyldecane

(d) 4-(1-methylethyl) - 5 - (1-methylpropyl)-decane

Show Answer

Answer

$(c, d)$

5- sec-Butyl -4 iso -propyldecane

4-(1-methylethyl)-5-(1-methylpropyl)- decane

Although IUPAC name for sec- butyl and isopropyl groups are 1methyl propyl and 1-methylethyl respectively yet both these names, are also recommended for IUPAC nomenclatiro

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

(a) 5-(2’, 2’ - Dimethylpropyl)-decane

(b) 4-Butyl-2,2-dimethylnonane

(c) 2,2- Dimethyl-4- pentyloctane

(d) 5-neo-Pentyldecane

Show Answer

Answer

$(a, d)$

5- (2’,2’- Dimethylpropyl)- decane 5-neo- pentyldecane

The IUPAC name for neopentyl groups is 2, 2 dimethyl propyl.

15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring

(a) deactivates the ring by inductive effect

(b) deactivates the ring by reasonance

(c) increases the charge density at ortho and para position relative to meta position by resonance.

(d) directs the incoming electrophile to meta position by increasing the charge density relative to ortho and para position.

Show Answer

Answer

(a, c)

For an electrophilic substitution reaction, the presence of halogen atom in the benzene ring deactivates the ring by inductive effect and increases the charge density at ortho and para position relative to meta position by resonance.

When chlorine is attached to benzene ring, chlorine being more electronegative pulls the electron i.e., $-I$-effect. The electron cloud of benzene is less dense. Chlorine makes aryl halide, moderately deactivating group. But due to resonance the electron density on ortho and para position is greater than in meta position.

The last structure contributes more to the orientation and hence halogen are $o$-and $p$-directors.

16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group.

(a) deactivates the ring by inductive effect

(b) activates the ring by inductive effect

(c) decreases the charge density at ortho and para position of the ring relative to meta position by resonance

(d) increases the charge density at meta position relative to the ortho and para positions of the ring by resonance

Show Answer

Answer

$(a, c)$

Nitro group by virtue of $-I$-effect withdraw electrons from the ring and increase the charge and destabilises carbocation.

In ortho, para-attack of electrophile on nitrobenzene, we are getting two structures (A) and $(B)$ in which positive charge is appearing on the carbon atom directly attached to the nitro group.

As nitro group is electron withdrawing by nature, it decreases the stability of such product and hence meta attack is more feasible when electron withdrawing substituents are attached.

17. Which of the following are correct?

(a) $\mathrm{CH} _{3}-\mathrm{O}-\mathrm{CH} _{2}^{\oplus}$ is more stable than $\mathrm{CH} _{3}-\mathrm{CH} _{2}^{\oplus}$

(b) $\left(\mathrm{CH} _{3}\right) _{2} \mathrm{CH}^{\oplus}$ is less stable than $\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CH} _{2}^{\oplus}$

(c) $\mathrm{CH} _{2}=\mathrm{CH}-\mathrm{CH} _{2}^{\oplus}$ is more stable than $\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CH} _{2}^{\oplus}$

(d) $\mathrm{CH} _{2}=\mathrm{CH}^{\oplus}$ is more stable than $\mathrm{CH} _{3}-\mathrm{CH} _{2}^{\oplus}$

Show Answer

Answer

$(a, c)$

(i) $+I$-effect increases the stability of carbocation $+I$-effect of i.e., $\mathrm{CH} _{3}-\mathrm{O}>\mathrm{CH} _{3}$.

Thus, $\mathrm{CH} _{3}-\mathrm{O}-\stackrel{+}{\mathrm{C}} \mathrm{H} _{2}$ is more stable than $\mathrm{CH} _{3}-\stackrel{+}{\mathrm{C}} \mathrm{H} _{2}$.

(ii) $(\mathrm{CH} _{3})_2 -\mathrm{CH}^+ $ is more stable than $\mathrm{CH} _{3}-{CH} _{3}-\stackrel{+}{\mathrm{C}} \mathrm{H} _{2}$ because former has stabilised by + I-effect of two $-CH_3$ groups. $I$-effect of two $-\mathrm{CH} _{3}$ groups.

(iii) $\mathrm{CH} _{2}=\mathrm{CH} \quad \mathrm{CH} _{2} \leftrightarrow \mathrm{CH} _{2}^{+}-\mathrm{CH}=\mathrm{CH} _{2}$ is stabilised by strong resonance effect while $\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CH} _{2}$ is stabilised by weak $+I$-effect of the $\mathrm{CH} _{3} \mathrm{CH} _{2}$ group.

(iv) In $\mathrm{CH} _{2}=\stackrel{+}{\mathrm{C}} \mathrm{H}$, +ve charge is present, on the more electronegatiue, sp-hybridised carbon while in $\mathrm{CH} _{3}-\mathrm{CH} _{2}$, +ve charge is present on the less electronegative $s p^{2}$-hybridised carbon therefore, $\mathrm{CH} _{2}=\stackrel{+}{\mathrm{C}} \mathrm{H}$ is less stable than $\mathrm{CH} _{3}-\stackrel{+}{\mathrm{C}} \mathrm{H} _{2}$.

$$ \mathrm{CH} _{2}=\underset{s p^{2}}{\mathrm{CH}}-\mathrm{CH}^{+} $$

$$ \mathrm{CH} _{3}-\underset{s p^{3}}{\mathrm{CH} _{2}}-\mathrm{CH} _{2}^{+} $$

18. Four structures are given in options (a) to (d). Examine them and select the aromatic structures.

(a)

(b)

(c)

(d)

Show Answer

Answer

(a, c) Aromaticity requires following condition

(i) planarity

(ii) complete delocalisation of $\pi$ electrons in the ring.

(iii) presence of $(4 n+2) \pi$ electrons in the ring.

(a)

planar $\pi$ electrons $=2$ $n=0$

(b)

(c)

$(4 n+2)=6$ in each ring $n=1$

(d)

$\pi$ planar $n=$ not integer

19. The molecules having dipole moment are ……..

(a) 2,2-Dimethylpropane

(b) trans-Pent-2-ene

(c) cis-Hex-3-ene

(d) 2, 2, 3, 3 - Tetramethylbutane

Show Answer

Answer

$(b, c)$

Thus, trans-pent-2-ene show net diple moment because different group attached and cis- Hex -3- ene show dipole moment because both groups $\left(\mathrm{C} _{2} \mathrm{H} _{5}\right)$ are inclined to each other at angle of $60^{\circ}$ therefore have a finite resultant.

Short Answer Type Questions

20. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.

Show Answer

Answer

Alkenes are rich source of loosely held pi $(\pi)$ electrons, due to which they show electrophilic addition reaction. Electrophilic addition reaction of alkenes are accompanied by large energy changes so these are energetically favourable than that of electrophilic substitution reactions. In special conditions alkenes also undergo free radical substitution reactions.

In arenes during electrophilic addition reactions, aromatic character of benzene ring is destroyed while during electrophilic substitution reaction it remains intact. Electrophilic substitution reactions of arenes are energetically more favourable than that of electrophilic addition reaction.

That’s why alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reaction.

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?

Show Answer

Thinking Process

In geometrical isomerism, when same groups are on the same side it is cis and if same groups are on the opposite side it is trans isomer.

Answer

Trans-2-butene formed by the reduction of 2-butyne is capable of showing geometrical isomerism.

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

Show Answer

Thinking Process

The infinite number of momentary arrangements of the atoms in space which result through rotation about a single bond are called conformations.

In ethane, if one carbon atom is kept stationary and other rotated around $\mathrm{C}-\mathrm{C}$ axis, we have eclipsed, skew and staggered conformation.

Answer

Alkanes can have infinite number of conformations by rotation around $\mathrm{C}-\mathrm{C}$ single bonds. This rotation around a C-C single bond is hindered by a small energy barrier of 1-20 kJ $\mathrm{mol}^{-}$due to weak repulsive interaction between the adjacent bonds. such a type of repulsive interaction is called torsional strain. In staggered form of ethane, the electron cloud of carbon hydrogen bonds are far apart.

Hence, minimum repulsive force. In eclipsed electron cloud of carbon-hydrogen become close resulting in increase in electron cloud repulsion. This repulsion affects stablity of a conformer.

In all the conformations of ethane the staggered form has least torsional strain and the eclipsed form has the maximum torsional strain. Hence, rotation around C-C bond in ethtane is not completely free.

Eclipsed

Staggered

Newman’s projection of ethane

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

Show Answer

Answer

Staggered form of ethane is more stable than the eclipsed conformation, by about $12.55 \mathrm{~kJ} / \mathrm{mol}$. This is because any two hydrogen atoms on adjacent carbon atoms of staggered conformation are maximum apart while in eclipsed conformation, they cover or eclipse each other in space. Thus, in staggered form, there is minimum repulsive forces, minimum energy and maximum stability of the molecule.

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

Show Answer

Answer

Addition of halogen acids to an alkene is an electrophilic addition reaction.

First step is slow so, it is rate determining step. The rate of this step depends on the availability of proton. This in turn depends upon the bond dissociation enthalpy of the $\mathrm{H}-X$ molecule.

Lower the bond dissociation enthalpy of $\mathrm{H}-\mathrm{X}$ molecule, greater the reactivity of halogen halide. Since the bond dissociation energy decreases in the order;

$$ \mathrm{HI}\left(296.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)<\mathrm{HBr}\left(363.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)<\mathrm{HCl}\left(430.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) $$

Therefore, the reactivity of the halogen acids decreases from $\mathrm{HI}$ to $\mathrm{HCl}$. i.e., $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

25. What will be the product obtained as a result of the following reaction and why?

Show Answer

Answer

When Friedel-Craft alkylation is carried out with higher alkyl halide, e.g., $n$-propyl chloride, then the electrophile $n$-propyl carbocation ( $1^{\circ}$ carbocation) formed which rearranges to form more stable iso-propyl carbocation ( $2^{\circ}$ carbocation). Afterward the main product iso-propyl benzene will be formed.

26. How will you convert benzene into (a) p-nitrobromobenzene

(b) m-nitrobromobenzene

Show Answer

Answer

Halogens attached to benzene ring is ortho and para directing where as nitro group is meta directing.

(a)

$p$-bromo nitro benzene (b)

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

Show Answer

Answer

The methoxy group $\left(-\mathrm{OCH} _{3}\right)$ is electron releasing group. It increases the electron density in benzene nucleus due to resonance effect ( $+R$-effect). Hence, it makes anisole more reactive than benzene towards electrophile.

$$ Chlorobenzene $$

$$(-I-effect)$$

In case of alkyl halides, halogens are moderately deactivating because of their strong $-I$ effect. Thus, overall electron density on benzene ring decreases. It makes further substitution difficult.

$-\mathrm{NO} _{2}$ group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong - $R$ - effect and strong $-I$-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order

28. Despite their $-I$ effect, halogens are $o$ - and $p$ - directing in haloarenes. Explain.

Show Answer

Answer

Halogens have $(-I)$ and $(+R)$ effect, these groups are deactivating due to their $(-I)$ effect and they are ortho, para directing due to $(+R)$ effect.

$$ \text{ortho, para-directing influence} $$

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

Show Answer

Answer

The meta - directing substituents (like - $\mathrm{NO} _{2}$ group) withdraw electrons from the benzene ring and thus, deactivate the benzene ring for further substitution and make the benzene ring less reactive in comparison to the unsubstituted benzene ring.

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

Show Answer

Answer

Acetylene when passed through red hot iron tube at $873 \mathrm{~K}$, undergoes cyclic polymerisation benzene which upon subsequent nitration gives nitrobenzene.

Note In nitration of benzene ring conc $\mathrm{H} _{2} \mathrm{SO} _{4}$ acts as an catalyst to produce an electrophile $+\mathrm{NO} _{2} \cdot\left(\right.$ from $\mathrm{HNO} _{3}$ )

31. Predict the major product(s) of the following reactions and explain their formation.

Show Answer

Answer

In presence of organic peroxides, the addition of $\mathrm{HBr}$ to propene follows anti Markowinkov’s rule (or peroxide effect) to form 1-bromopropane (n-propyl bromide)

However, in absence of peroxides, addition of $\mathrm{HBr}$ to propene follows Markownikoff’s rule and gives 2- bromopropane as major product.

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) $\mathrm{H} _{3} \mathrm{CO}^{-}$

(ii) $ H_3C-\stackrel{\stackrel{\large O}{||}}{C}-O^- $

(iii) $\dot{\mathrm{Cl}}$

(iv) $\mathrm{Cl} _{2} \mathrm{C}$ :

(v) $\left(\mathrm{H} _{3} \mathrm{C}\right) _{3} \mathrm{C}^{+}$

(vi) $\mathrm{Br}^{-}$

(vii) $\mathrm{H} _{3} \mathrm{COH}$

(viii) $R$-NH- $R$

Show Answer

Answer

Electrophiles are electron deficient species. They may be natural or positively charged e.g., (iii) $\dot{\mathrm{Cl}}$, (iv) $\mathrm{Cl} _{2} \mathrm{C}$ : , (v) $\left(\mathrm{H} _{3} \mathrm{C}\right) _{3} \mathrm{C}^{+}$

Nucleophiles are electron rich species. They may be neutral or negatively charged e.g.,

(i) $\mathrm{H} _{3} \mathrm{CO}^{-}$,

(ii) $ H_3C-\stackrel{\stackrel{\large O}{||}}{C}-O^- $

(vi) $\mathrm{Br}^{-}$,

(vii) $\mathrm{H} _{3} \mathrm{C}-\ddot{O}-\mathrm{H}$,

(viii) $R \ddot{\mathrm{N}} \mathrm{HR}$

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

Show Answer

Answer

The given organic compound is

$$ 2-methyl \quad butane $$

This compound has 9 primary hydrogen, 2 secondary and one tertiary hydrogen atoms. The relative reactivity of $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ hydrogen atoms towards chlorination is $1: 3.8: 5$. Relative amount of product after chlorination $=$ Number of hydrogen $\times$ relative reactivity

Total amount of mono chloro product $=9+7.6+5=21.6$

Percentage of $1^{\circ}$ mono chloro product $=\frac{9}{21.6} \times 100=41.7 $ %

Percentage of $2^{\circ}$ mono chloro product $=\frac{7.6}{21.6} \times 100=35.2 $ %

Percentage of $3^{\circ}$ mono chloro product $=\frac{5}{21.6} \times 100=23.1 $.%

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.

Show Answer

Thinking Process

This question is based upon Wurtz reaction. Wurtz reaction represent that two alkyl groups can be coupled by reacting alkyl halide with

$ 2RX+2Na \xrightarrow[\text{dry ether}]{\Delta} R-R + 2NaX $

Answer

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

Show Answer

Answer

2-methylpropane gives two types of radicals.

Radical (I) is more stable because it is $3^{\circ}$ free radical and stabilised by nine hyperconjugative structures (as it has $9 \alpha$-hydrogens)

Radical (II) is less stable because it is $1^{\circ}$ free radical and stabilised by only one hyperconjugative structure (as it has only $1 \alpha$ - hydrogen)

36. An alkane $\mathrm{C} _{8} \mathrm{H} _{18}$ 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.

Show Answer

Answer

From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction of a primary alkyl halide gives an alkane $\left(\mathrm{C} _{8} \mathrm{H} _{18}\right)$, therefore, the alkyl halide must contain four carbon atoms. Now the two possible primary alkyl halides having four corbon atoms each are I and II.

Since, alkane $\mathrm{C} _{8} \mathrm{H} _{18}$ on monobromination yields a single isomer of tertiary alkyl halide, therefore, the alkane must contain tertiary hydrogen. This is possible, only if primary alkyl halide (which undergoes Wurtz reaction) has a tertiary hydrogen.

37. The ring systems having following characteristics are aromatic.

(i) Planar ring containing conjugated $\pi$ bonds.

(ii) Complete delocalisation of the $\pi$-electrons in ring system i.e., each atom in the ring has unhybridised $p$-orbital, and

(iii) Presence of $(4 n+2) \pi$ - electrons in the ring where $n$ is an integer $(n=0,1,2, \ldots . . . .$.$) \text{[Huckel rule]}.$

Using this information classify the following compounds as aromatic/non-aromatic.

(A)

(B)

(C)

(D)

(E)

$$ (G) $$

Show Answer

Answer

Compound Planar
ring
Complete
delocalisation of
$\pi$-electron
Huckel rule
$(4 n+2) \pi$ electron
Aromatic or
non-aromatic
A. $P$ P $6 \pi e^{-}$
Huckel rule obeyed
Aromatic
B. í Í Incomplete (sp ${ }^{3}$
hybrid carbon)
$6 \pi \mathrm{e}^{-}$ Non-aromatic
C. P P $6 \pi e^{-}(4 n+2+$ lone
pair $\left.e^{-}\right)$Huckel rule
verified
Aromatic
D. P í $4 \pi e^{-}$ Anti-aromatic
E. P P Huckel rule obeyed Aromatic
F. P P $2 \pi e^{-}$Huckel rule
verified $n=0$
Aromatic
G. P í $8 \pi e^{-}$Huckel rule
not verified
Non-aromatic
38. Which of the following compounds are aromatic according to Huckel’s rule?

(A)

(B)

(C)

(D)

(E)

(F)

Show Answer

Answer

A. The compound has $8 \pi$ electrons. It will be non-aromatic. Both rings are non-benzenoid.

B. The compound is aromatic. It has $6 \pi e^{-}$delocalised electron $\left(4 \pi e^{-}+2\right.$ lone pair electrons), all the four carbon atoms and the $\mathrm{N}$ atom are $s p^{2}$ hybridised.

C. The compound contains $6 \pi$ electrons but not in the ring hence it is non-aromatic. D. $10 \pi e^{-}$obeying Huckel rule and the ring is planar. It is aromatic.

E. In this compound one six membered planar ring has $6 \pi e^{-}$although it has $8 \pi$ electrons in two rings. It is therefore aromatic.

F. It has $14 \pi$ electrons in conjugation and in the planar ring, Huckel rule is verified. It will be aromatic.

39. Suggest a route to prepare ethyl hydrogensulphate $\left(\begin{array}{llll}\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{OSO} _{2}-\mathrm{OH}) \text { starting from ethanol }\left(\mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OH}\right) \text {. }\end{array}\right.$

Show Answer

Answer

For preparation of ethyl hydrogensulphate $\left(\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{OSO} _{2}-\mathrm{OH}\right)$ starting from ethanol $\left(\mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OH}\right)$, it is the two steps mechanism.

Step I Protonation of alcohol

$$\quad \quad \quad \quad \quad \quad \quad \text { protonated ethanol } $$

Step II Attack of nucleophile

$$ \mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{O}-\mathrm{SO} _{2} \mathrm{OH}+\mathrm{H} _{2} \mathrm{O} $$

$$ \text{ethyl hydrogen surphate} $$

Temperature should not be allowed to rise above $383 \mathrm{~K}$, otherwise diethyl ether will be produced at $413 \mathrm{~K}$ or ethene at $433 \mathrm{~K}$.

Matching The Columns

40. Match the reagent from Column I which on reaction with $\mathrm{CH} _{3}-\mathrm{CH}=\mathrm{CH} _{2}$ gives some product given in Column II as per the codes given below

Column I Column II
A. $\mathrm{O} _{3} / \mathrm{Zn}+\mathrm{H} _{2} \mathrm{O}$ 1. Acetic acid and $\mathrm{CO} _{2}$
B. $\mathrm{KMnO} _{4} / \mathrm{H}^{+}$ 2. Propan-1-ol
C. $\mathrm{KMnO} _{4} / \mathrm{OH}^{-}$ 3. Propan-2-ol
D. $\mathrm{H} _{2} \mathrm{O} / \mathrm{H}^{+}$ 4. Acetaldehyde and formaldehyde
E. $\mathrm{B} _{2} \mathrm{H} _{6} / \mathrm{NaOH}^{+}$and $\mathrm{H} _{2} \mathrm{O} _{2}$ 5. Propane-1, 2-diol

Show Answer

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(5)$

D. $\rightarrow(3)$

E. $\rightarrow(2)$

Reagent Recation with propene
A. $\mathrm{O} _{3} / \mathrm{Zn}+\mathrm{H} _{2} \mathrm{O}$
B. $\mathrm{KMnO} _{4} / \mathrm{H}^{+}$ $\mathrm{CH} _{3} \mathrm{CH}=\mathrm{CH} _{2} \underset{\mathrm{H}^{+}}{\stackrel{\mathrm{KMnO} _{4}}{\rightarrow}} \underset{\text { Acetic acid }}{\mathrm{CH} _{3} \mathrm{COOH}}+\mathrm{CO} _{2}$
C. $\mathrm{KMnO} _{4} / \mathrm{OH}^{-}$
D. $\mathrm{H} _{2} \mathrm{O} / \mathrm{H}^{+}$
E. $\mathrm{B} _{2} \mathrm{H} _{6} / \mathrm{NaOH}^{+}$and $\mathrm{H} _{2} \mathrm{O} _{2}$
41. Match the hydrocarbons in Column I with the boiling points given in Column II.

Column I Column II
A. n-pentane 1. $282.5 \mathrm{~K}$
B. iso-pentane 2. $309 \mathrm{~K}$
C. neo-pentane 3. $301 \mathrm{~K}$

Show Answer

Thinking Process

To solve this question, it keep in mind that branching of hydrocarbons decreases boiling point of the compound

Answer

A. $\rightarrow(2)$

B. $\rightarrow$ (3) $\quad$

C. $\rightarrow$ (1)

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

Column I Column II
A. Benzene $+\mathrm{Cl} _{2} \xrightarrow{\mathrm{AlCl} _{3}} $ 1. Benzoic acid
B. Benzene $+\mathrm{CH} _{3} \mathrm{Cl} \xrightarrow{\mathrm{AlCl} _{3}} $ 2. Methyl phenyl ketone
C. Benzene $+\mathrm{CH} _{3} \mathrm{COCl} \xrightarrow{\mathrm{AlCl} _{3}} $ 3. Toluene
D. Toluene $\stackrel{\mathrm{KMnO} _{4} / \mathrm{NaOH}}{\rightarrow}$ 4. Chlorobenzene
5. Benzene hexachloride

Show Answer

Answer

A. $\rightarrow(4) \quad$

B. $\rightarrow$ (3) $\quad$

C. $\rightarrow$ (2) $\quad$

D. $\rightarrow$ (1)

reactants Products
A. Benzene $+\mathrm{Cl} _{2} \xrightarrow{\mathrm{AlCl} _{3}} $
B. Benzene $+\mathrm{CH} _{3} \mathrm{Cl} \xrightarrow{\mathrm{AlCl} _{3}} $
C. Benzene $+\mathrm{CH} _{3} \mathrm{COCl} \xrightarrow{\mathrm{AlCl} _{3}} $
D. Toluene $\stackrel{\mathrm{KMnO} _{4} / \mathrm{NaOH}}{\rightarrow}$
43. Match the reactions given in Column I with the reaction types in Column II.

Column I Column II
A. $\mathrm{CH} _{2}=\mathrm{CH} _{2}+\mathrm{H} _{2} \mathrm{O} \xrightarrow{\mathrm{H}^{+} } \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}$ 1. Hydrogenation
B. $\mathrm{CH} _{2}=\mathrm{CH} _{2}+\mathrm{H} _{2} \xrightarrow{\mathrm{Pd}} \mathrm{CH} _{3}-\mathrm{CH} _{3}$ 2. Halogenation
C. $\mathrm{CH} _{2}=\mathrm{CH} _{2}+\mathrm{Cl} _{2} \rightarrow \mathrm{Cl}-\mathrm{CH} _{2}-\mathrm{CH} _{2}-\mathrm{Cl}$ 3. Polymerisation
4. Hydration
5. Condensation

Show Answer

Answer

A. $\rightarrow(4)$

B. $\rightarrow$ (1)

C. $\rightarrow(2)$

D. $\rightarrow(3)$

Assertion and Reason

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 in each question.

44. Assertion (A) The compound tetraene has the following structural formula.

It is cyclic and has conjugated 8 8 -electron system but it is not an aromatic compound.

Reason (R) $(4 n+2) \pi$ electrons rule does not hold good and ring is not planar.

(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$

(b) Both $\mathrm{A}$ and $\mathrm{R}$ are correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$

(c) Both $\mathrm{A}$ and $\mathrm{R}$ are not correct

(d) $\mathrm{A}$ is not correct but $\mathrm{R}$ is correct

Show Answer

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

According to Huckel rule Aromaticity is shown by compounds possessing following characteristics

(i) Compound must be planar and cyclic

(ii) Complete delocalisation of $\pi$ electrons in the ring

(iii) Presence of conjugated $(4 n+2) \pi$ electrons in the ring where $n$ is an integer $(n=0,1,2, \ldots)$ cyclo octatetraene (given) has a tub like structure. It loses planarity. No. of $\pi e^{-}$delocalised $=8$. and $n$ is not integer. Hence, cycloctatetraene is a non-aromatic compound.

45. Assertion (A) Toluene on Friedal Crafts methylation gives $o$ - and $p$ xylene.

Reason (R) $\mathrm{CH} _{3}$-group bonded to benzene ring increases electron density at $o$ - and $p$ - position.

(a) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(b) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(c) Both $\mathrm{A}$ and $\mathrm{R}$ are not correct

(d) $\mathrm{A}$ is not correct but $\mathrm{R}$ is correct

Show Answer

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

Toluene has $-\mathrm{CH} _{3}$ group attached to benzene. $-\mathrm{CH} _{3}$ group activates the benzene ring for the attack of an electrophile.

In resonating structure of toluene, electronic density is more on ortho and para position. Hence, substitution takes place mainly at these positions.

46. 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, $\mathrm{NO} _{2}^{+}$.

(a) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(b) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(c) Both $\mathrm{A}$ and $\mathrm{R}$ are not correct

(d) $\mathrm{A}$ is not correct but $\mathrm{R}$ is correct

Show Answer

Answer

(a) Both assertion and reason are correct and reason is the correct explanation of assertion. In nitration of benzene with nitric acid sulphuric acid acts as a calatyst. It helps in the formation of electrophile i.e., nitronium ion $\mathrm{NO} _{2}^{+}$.

$$ HNO_3+H_2 SO_4 \rightarrow NO_2^+ +2 HSO_4^- +H_3 O^+ $$

47. Assertion (A) Among isomeric pentanes, 2, 2-dimethylpentane has highest boiling point.

Reason (R) Branching does not affect the boiling point.

(a) Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

(b) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

(c) Both $\mathrm{A}$ and $\mathrm{R}$ are not correct

(d) $\mathrm{A}$ is not correct but $\mathrm{R}$ is correct

Show Answer

Answer

(c) Both assertion and reason are correct

Correct assertion Among isomeric pentanes, 2, 2 - dimethylpentane has the lowest boiling point.

Correct reason Branching decrease the boiling point.

Long Answer Type Questions

48. An alkyl halide $\mathrm{C} _{5} \mathrm{H} _{11}$ (A) reacts with ethanolic $\mathrm{KOH}$ to give an alkene ’ $B$ ‘, which reacts with $\mathrm{Br} _{2}$ to give a compound ’ $C$ ‘, which on dehydrobromination gives an alkyne ’ $D$ ‘. On treatment with sodium metal in liquid ammonia, one mole of ’ $D$ ’ gives one mole of the sodium salt of ’ $D$ ’ and half a mole of hydrogen gas. Complete hydrogenation of ’ $D$ ’ yields a straight chain alkane. Identify $A, B, C$ and $D$. Give the reactions involved.

Show Answer

Answer

The reaction scheme involved in the problem is

Hydrogenation of alkyne (D) gives straight chain alkane hence all the compounds (A), (B), $(C)$ and $(D)$ must be straight chain compounds. Alkyne (D) form sodium salt which proves that it is terminal alkyne. Involved reactions are as follows

It is important point that alkyl halide (A) can not be 2-bromopentane because dehydrobromination of $(A)$ would have given 2-pentene as the major product in accordance with Markownikoff’s rule.

49. $896 \mathrm{~mL}$ vapour of a hydrocarbon ’ $A$ ’ having carbon $87.80 $ % and hydrogen $12.19 $ % weighs $3.28 \mathrm{~g}$ at STP. Hydrogenation of ’ $A$ ’ gives 2-methylpentane. Also ’ $A$ ’ on hydration in the presence of $\mathrm{H} _{2} \mathrm{SO} _{4}$ and $\mathrm{HgSO} _{4}$ gives a ketone ’ $B$ ’ having molecular formula $\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O}$. The ketone ’ $B$ ’ gives a positive iodoform test. Find the structure of ’ $A$ ’ and give the reactions involved.

Show Answer

Answer

To determine the molecular mass of hydrocarbon (A) $896 \mathrm{~mL}$ vapour of $\mathrm{C} _{x} \mathrm{H} _{y}(A)$ weighs $3.28 \mathrm{~g}$ at $\mathrm{STP}$

$22700 \mathrm{~mL}$ vapour of $\mathrm{C} _{x} \mathrm{H} _{y}(A)$ weighs $\frac{3.28 \times 22700}{896} \mathrm{~g} / \mathrm{mol}$ at STP

$$ =83.1 \mathrm{~g} / \mathrm{mol} $$

Hence, molecular mass of $\mathrm{C} _{x} \mathrm{H} _{y}(A)=83.1 \mathrm{~g} \mathrm{~mol}^{-1}$. To determine the empirical formula of hydrocarbon $(A)$.

Element % Atomic mass Relative ratio Relative no.
of atoms
Simplest ratio
$\mathrm{C}$ 87.8 12 7.31 1 3
$\mathrm{H}$ 12.19 1 12.19 1.66 $4.98 \approx 5$

Thus, Empirical formula of $A$ is $\mathrm{C} _{3} \mathrm{H} _{5}$.

$\therefore$ Empirical formula mass $=36+5=41$.

$$ n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{83.1}{41}=2.02 \approx 2 $$

Molecular mass is double of empirical formula mass.

$\therefore$ Molecular formula is $\mathrm{C} _{6} \mathrm{H} _{10}$

To determine the structure of compounds $(A)$ and $(B)$

Hence, hydrogenation of hydrocarbon (A) requires 2 moles of hydrogen to form 2-methylpentane. Therefore, hydrocarbon(A) is an alkyne having five carbon atoms in a staight chain and a methyl substituent at position 2 . Thus, the possible structures for the alkyne $(A)$ are I and II.

Since, addition of $\mathrm{H} _{2} \mathrm{O}$ to alkyne $(A)$ in presence of $\mathrm{Hg}^{2+}$, give a ketone which gives positive iodoform test, therefore, ketone $(B)$ must be a methyl ketone, i.e., it must contain a $\mathrm{COCH} _{3}$ group.

Now addition of $\mathrm{H} _{2} \mathrm{O}$ to alkyne (II) should give a mixture of two ketones in which 2- methyl pentan -3 one (minor) and 4-methylpentan -2-one ketone (B) (which shows + ve iodoform test) predominates.

In contrast, addition of $\mathrm{H} _{2} \mathrm{O}$ to alkyne (I) will give only one ketone, i.e., 4- methylpentan-2one which gives iodoform test.

Thus, hydrocabon $\mathrm{C} _{x} \mathrm{H} _{y}(A)$ is 4-methylpent-1-yne. 4- methylpentan -2 one (gives + ve iodoform test)

50. An unsaturated hydrocarbon ’ $A$ ’ adds two molecules of $\mathrm{H} _{2}$ 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.

Show Answer

Answer

The scheme of reaction is

Compound $(A)$

Thus, structure of $A$ may be given as

The reactions involved in the question

51. In the presence of peroxide addition of $\mathrm{HBr}$ to propene takes place according to anti Markownikoff’s rule but peroxide effect is not seen in the case of $\mathrm{HCl}$ and $\mathrm{HI}$. Explain.

Show Answer

Answer

$\mathrm{CH} _{3}-\mathrm{CH}=\mathrm{CH} _{2}+\mathrm{HBr} \xrightarrow{\text { Peroxide }} \mathrm{CH} _{3}-\mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{Br}$

$$ \text { propene } \quad \quad \quad \quad \quad \quad \quad n \text {-propyl bromide } $$

The mechanism of the reaction is

Step 1

Step II

Peroxide effect is effective only in the case of $\mathrm{HBr}$ and not seen in the case of $\mathrm{HCl}$ and $\mathrm{HI}$. This is due to the following reasons.

(i) $\mathrm{H} \quad \mathrm{Cl}$ bond $(103 \mathrm{kcal} / \mathrm{mol})$ is stronger than $\mathrm{H} \quad \mathrm{Br}$ bond $(87 \mathrm{kcal} / \mathrm{mol})$

$\mathrm{H} \quad \mathrm{Cl}$ bond is not decomposed by the peroxide free radical whereas the $\mathrm{H} \quad \mathrm{I}$ bond is weaker $(71 \mathrm{kcal} / \mathrm{mol})$ form iodine free radicals.

(ii) lodine free radical $\left(\mathrm{I}^{\circ}\right)$ formed as $\mathrm{H} \quad \mathrm{I}$ bond is weaker but iodine free radicals readily combine with each other to form iodine molecules rather attacking the double bond.



Table of Contents