Answer

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

Boiling point decreases with branching

(4 carbon atoms with no branching)

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

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

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.

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.

$(A)$

$(B)$

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.

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.

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, $B>A>C$

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 $A>\underset{2^{\circ} \beta \text {-Carbon }}{C}>B$

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

$(A)$

$(B)$

$(C)$

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,

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.

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.

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

Answer

$(a, d)$

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

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

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.

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.

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 $\mathrm{CH} _{3}-\stackrel{+}{\mathrm{C}} \mathrm{H} _{2}$.

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

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

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

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.

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.

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.

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

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.

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., $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

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.

Answer

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

(a)

$p$-bromo nitro benzene (b)

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

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.

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

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.

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 $\mathrm{HNO} _{3}$ )

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.

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

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 $.%

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

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)

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.

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

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.

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 $433 \mathrm{~K}$.

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 $\mathrm{H} _{2} \mathrm{O} _{2}$

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)

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 $\stackrel{\mathrm{KMnO} _{4} / \mathrm{NaOH}}{\rightarrow}$

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.

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.

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.

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

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

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.

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.

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 $(A)$.

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)

Answer

The scheme of reaction is

Compound $(A)$

Thus, structure of $A$ may be given as

The reactions involved in the question

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.