Hydrocarbons

$X$ is an isomer of $\mathrm{C}_6 \mathrm{H}_{14}$. It has 4 primary and 2 tertiary C .

The compound is 2,2-dimethyl butane. It can be prepared by Wurtz reaction.

The complete reaction sequence is as follows

The correct match is A-II, B-IV, C-I, D-III.

Fittig Reaction When chlorobenzene react with sodium metal in dry ether it gives biphenyl.

The number of $\sigma$ and $\pi$ bonds in both phenyl ring are 23 and 6 respectively. Therefore, their sum is $23+6=29$.

Cumene on oxidation in air gives cumene hydroperoxide via radical chain mechanism.

Phenol $(Y)$ reacts with sodium metal ( Na ) form sodium phenoxide and hydrogen gas.

Acetone does not reacts with sodium metal. So $Z$ is acetone (propanone).

The boiling point of alkanes is primarily determined by molecular mass and branching.

(a) Molecular mass : Hexane $\left(\mathrm{C}_6 \mathrm{H}_{14}\right)$ has a higher molecular mass than the other compounds $\left(\mathrm{C}_5 \mathrm{H}_{12}\right)$. Therefore, it has the highest boiling point.

• Order so far : $D>(A, B, C)$

(b) Branching : For the isomers of pentane $\left(\mathrm{C}_5 \mathrm{H}_{12}\right)$, boiling point decreases with increasing branching due to a reduction in surface area for intermolecular forces.

• Pentane $(C)$ is a straight chain.

• 2-Methylbutane $(A)$ has one branch.

• 2,2-Dimethylpropane ( $B$ ) has two branches and is the most compact.

• Order for isomers : $C>A>B$. Combining both factors, the final decreasing order of boiling points is $D>C>A>B$.

The number of 1-alkyne isomer possible for the compound with the molecular formula $\mathrm{C}_6 \mathrm{H}_{10}$ is 4 . This are as follows,

• $\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$

(1-hexyne)

• $\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{CH}_2-\mathrm{CH}_3$

(3-methyl-1-pentyne)

• $\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{CH}_3$

(4-methyl-1-pentyne)

• $\mathrm{CH} \equiv \mathrm{C}-\mathrm{C}\left(\mathrm{CH}_3\right)_2-\mathrm{CH}_3$

(3,3-dimethyl-1-butyne)

Thus, 4 isomers are possible.

The reaction involved is,

Thus, the hybridisation of $Y$ is $s p$ and its nature is neutral.

$ \text { The complete reaction is as follows, } $

$ \mathrm{C}_2 \mathrm{H}_6+\frac{3}{2} \mathrm{O}_2 \xrightarrow{\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Mn}, \Delta} \underset{\text { (Product } X \text { ) }}{2 \mathrm{CH}_3 \mathrm{COOH}} \xrightarrow{\mathrm{Na}} \underset{\text { (Product } Y \text { ) }}{\mathrm{CH}_3 \mathrm{COO}^{-}} $

$ \begin{aligned} &\text { Now Kolbe's electrolysis, }\\ &2 \mathrm{CH}_3 \mathrm{COO}^{-} \xrightarrow{\text { Electrolysis }} \underset{\substack{(P) \\ \text { Ethane }}}{\mathrm{C}_2 \mathrm{H}_6}+\underset{\text { (Q gas) }}{2 \mathrm{CO}_2}+2 e^{-} \end{aligned} $

Among the given options (c) is an example of electrophilic substitution reaction.

$ \mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl} \xrightarrow{\mathrm{AlCl}_3} \mathrm{C}_6 \mathrm{H}_5\left(\mathrm{COCH}_3\right)+\mathrm{HCl} $

The given reaction is Friedal-Crafts acylation reaction, here the reaction of benzene with an arylhalide or acid anhydride is the presence of Lewis acid $\left(\mathrm{AlCl}_3\right)$ yields acyl benzene. Here, the attacking reagent is as electrophile.

$ \text { The complete reaction is } $

$ \text { The IUPAC name of } X=\text { 3-ethyl-2-methylpent-2-ene } $

The complete reaction mechanism is as follows

So, in product $Z$, number of $s p^2$ carbon $(b)=4$, number of $s p^3$ carbon $(a)=4$

So, $a+b=8$

Given that molecular formula is $\mathrm{C}_6 \mathrm{H}_{10}$. First of all let us calculate the degree of unsaturation for $\mathrm{C}_6 \mathrm{H}_{10}$.

$ \text { } \begin{aligned}So,\,\, \mathrm{DBE} & =\mathrm{C}+1 \frac{-\mathrm{H}}{2} \\ \mathrm{DBE} & =6+1 \frac{-10}{2}=2 \end{aligned} $

Here, DBE of 2 indicates the presence of one triple bond or two double bonds or one ring and one double bond. So, for alkynes let us see the structure with one triple bond.

Now, the possible number of alkyne monomers of $\mathrm{C}_6 \mathrm{H}_{10}$ are

Hex-1-yne (Terminal alkyne)

Hex-2-yne (Internal alkyne)

Hex-3-yne (Internal alkyne)

3-methyl pent-1-yne (Terminal alkyne)

4-methylpent-1-yne (Terminal alkyne)

4-methylpent-2-yne (Internal alkyne)

3, 3-Dimethyl but-1-yne (Terminal alkyne)

So, terminal alkynes contain acidic hydrogen while internal alkynes contains no acidic hydrogen. Thus $x$ and $y$ are 4,3 respectively.

The complete reaction is as follows,

Number of $s p^3$ carbon $=2$ (methyl group)

Number of $s p^2$ carbon $=2$(double bond carbon)

The complete reaction is as follows

respectively.

Next alkane to methane is ethane and it is prepared by reaction given in statement (I) and statement (IV). The reactions are

$ 2 \mathrm{CH}_{3} \mathrm{Br} \xrightarrow{2 \mathrm{Na}} \mathrm{CH}_{3}-\mathrm{CH}_{3}+2 \mathrm{NaBr} $

$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+2[\mathrm{H}] \xrightarrow[\mathrm{H}^{+}]{\mathrm{Zn}} \mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{HBr} $

When methane is heated with catalyst $ X $ (which is copper at 573 K ) it gives methanol on methyl alcohol. When methane is reacted with $ \mathrm{O}_{2} $ in presence of $ \mathrm{MO}_{2} \mathrm{O}_{3} $ then methanal (formaldehyde) is formed.

The complete reaction are as follows,

$ \begin{aligned} & \underset{\text { Methane }}{2 \mathrm{CH}_4}+\mathrm{O}_2 \xrightarrow[573 \mathrm{~K} / 100 \text { atm }]{\mathrm{Cu}} \underset{\text { Methanol }}{2 \mathrm{CH}_3 \mathrm{OH}} \\\\ & \underset{\text { Methane }}{\mathrm{CH}_4}+\mathrm{O}_2 \xrightarrow[\Delta]{\mathrm{Mo}_2 \mathrm{O}_3}\underset{\text { Methanal }}{\mathrm{HCHO}}+\mathrm{H}_2 \mathrm{O} \end{aligned} $

The complete reaction sequence is as follows

The compound $ A $ is sodium butanoate.

When sodium salt of butanoic acid is heated with soda lime, it readily undergoes decarboxylation. The reaction is as follows,

$\underset{\text { Sodium butanoate }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COONa}}+\underset{\text { Sodalime }}{(\mathrm{NaOH}+\mathrm{CaO})} \longrightarrow \underset{\text { Propane }}{\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3}$

$\begin{gathered}\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \\ {\left[X: \mathrm{C}_4 \mathrm{H}_8\right]}\end{gathered}\xrightarrow[{\mathrm{[O]}}]{\mathrm{KMnO}_4 / \mathrm{H}^{+}} 2 \mathrm{CH}_3 \mathrm{COOH} \underset{(Y)}{ }$

$\Rightarrow \underset{(\text { Sodium salt of } Y)}{\mathrm{CH}_3 \mathrm{COONa}} \xrightarrow[\substack{\text { (Sodalime) } \\-\mathrm{Na}_2 \mathrm{CO}_3}]{(\mathrm{NaOH}+\mathrm{CaO}) / \Delta} \underset{(Z)}{\mathrm{CH}_4}$

Note: But-2-ene $\left(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3\right)$ exhibits geometrical isomerism which undergoes oxidative cleavage of $\mathrm{C}=\mathrm{C}$ with $\mathrm{KMnO}_4 / \mathrm{H}^{+}$to give $Y$.

The complete reaction is as follows,

$ \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 $

$ \left(\mathrm{C}_3 \mathrm{H}_6\right) \xrightarrow[(X)]{\mathrm{HBr} / R \mathrm{OOR}, \text { Kharasch affect }} $

$ \underset{(Y)}{\mathrm{C}_3 \mathrm{H}_7 \mathrm{Br}} \underset{\mathrm{AlCl}_3}{\mathrm{C}_6 \mathrm{H}_6} $

The acidic strength can be determined on the basis of hybridisation. More is the $s$-character higher will be the acidity. Triple bond has highest

$s$-character while double bond has a less $s$-character than triple bond but higher than a single bond.

So, correct order of acidic strength is IV $<$ I $<$ III $<$ II.

Propyne undergoes hydration in presence of $\mathrm{Hg}^{2+} / \mathrm{H}^{+}$to give unsaturated alcohol ( $X$ ). This alcohol undergoes tautomerism to form ketone $(Y)$.

HBr is added to propene in presence of peroxide such as $\mathrm{H}_2 \mathrm{O}_2$ follows anti-Markownikoff's addition. This mechanism involves formation of radical intermediate.

The rate of attack of electrophile is least when $X$ is $\mathrm{NO}_2$. It is because $\mathrm{NO}_2$ is an electron withdrawing group which makes the benzene ring electrophilic in nature. Therefore, benzene ring with $\mathrm{NO}_2$ groups are attacked by nucleophiles rather than electrophiles.

Thus, among all the compounds phenanthrene has maximum number of $\pi$-electrons.

$ \text { Alkene = 2-methylpropene } $

To decide the structure of alkene from ozonolysis bring the products together in such a way that their O -atoms are face to face, replace O by $\mathrm{C}=\mathrm{C}$ bond.

The alkene $(X)$ with formula $\mathrm{C}_4 \mathrm{H}_8$ does not exhibit geometrical isomerism. Thus, $(X)$ is $\mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{CH}_3$.

By using the reagents given in option (c) ketone is formed which gives iodoform test.

From sequence of reagents given in option (a),

From sequence of reagents given in option (b)

From sequence of reagents given in option (d),

$ \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{CH}_3 $

The alkene will be pent-2-ene. pent-2-ene on ozonolysis gives ethanal and propanal as products.

The correct match is A-IV, B-III, C-II, D-I.

Benzene reacts with $\mathrm{Cl}_2$ in presence of anhy. $\mathrm{AlCl}_3$ to form chlorobenzene.

Benzene reacts with methylchloride in presence of anhy. $\mathrm{AlCl}_3$ to give toluene . This is Friedel-Craft's alkylation reaction.

Benzene reacts with acetylchloride in presence of anhy. $\mathrm{AlCl}_3$ to give alkyl phenyl ketone.This reaction is known as Friedel-Craft's acylation reaction.

Toluene reacts with $\mathrm{KMnO}_4 / \mathrm{KOH}$ followed by acidic hydrolysis to give benzoic acid.

2 molecules of bromopropane when treated with sodium ( Na ) in presence of ether to give $n$-hexane $(X)$. $(X)$ reacts with $\mathrm{Mo}_2 \mathrm{O}_3$ at high temperature 773 K and pressure $10-20$ atm to give benzene $(Y)$ as product. Further $(Y)$ reacts with methyl chloride in presence of anhy. $\mathrm{AlCl}_3$ to give methylbenzene or toluene ( Z ).

The reaction is as follows:

$ \text { In this product, } 5-\mathrm{OH} \text { are present. } $

The reaction proceeds via free radical mechanism and hence forms gammexane $\left(\mathrm{C}_6 \mathrm{H}_6 \mathrm{Cl}_6\right)$.

The compound which is most reactive towards an electrophilic attack will be electron-rich in nature. So, out of benzene, toluene, chlorobenzene and phenol. Phenol is most electron-rich compound followed by toluene and then benzene, the chlorobenzene is least electron-rich compound since chlorine is electron-withdrawing group which makes benzene electron-deficient in nature.

Hence correct order is

(i) When glucose is heated with Hi, it completely reduces to yield $n$-hexane as HI is a reducing agent.

(ii) In next step, $n$-hexane reacts with $\mathrm{Mo}_2 \mathrm{O}_3$ at 773 K and $10-20 \mathrm{~atm}$ to form benzene. This reaction is known as aromatisation reaction.

Alkynes are reduced to cis-alkenes on reacting with $\mathrm{H}_2$ in presence of Lindlar's catalyst. So, $P$ is cis-pent-2-ene. This addition of $\mathrm{H}_2$ is syn-addition.

Alkynes are reduced to trans-alkenes on treating with Na in liquid $\mathrm{NH}_3$. Here, anti-addition of hydrogen takes place. So, Q is trans-2-pentene.

This reaction occurs in three steps

Mechanism

Step-I Alcoholic KOH eliminates HBr from the dihalide to form alkene.

Step-II In second step, $\mathrm{NaNH}_2$ is a strong base. So, it can further eliminate one more HBr molecule to form alkyne.

Step-III Alkynes on heating with red hot iron tube polymerise to form aromatic benzene ring.

Dry ether is the most sultable solvent for Wurtz reaction because

Attack of the alkene on bromine gives the bromonium ion, which is positive ion. Alkyl groups are electron donating and carbocation-stabilising groups.

$D$ has 2 alkyl groups, $C$ has one and $A$ has none, but the $B$ has a electron

withdrawing nitro group. So, higher the stability of bromonium ion formed higher will be the rate of bromination reaction.

So, the correct order of rate of reaction is $D>C>A>B$.

$ \text { 6-oxoheptanal heptane is } $

$ \text { It is formed by reductive ozonolysis of } $

$ \text { Hence, option (a) is the correct answer. } $

In the reaction(s),

$ \begin{aligned} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}}(P) \\ & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{H}_2 \mathrm{O}_2 \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}]{\left(\mathrm{BH}_3\right)_2}(Q) \end{aligned} $

The products are,

$ \text { because, } $

Br add according to, Markownikoff's rule, while in

$ \mathrm{C}_8 \mathrm{H}_8 \text { is } $

It is a derivative of benzene and decolorised the cold alkaline $\mathrm{KMnO}_4$ solution.

The reaction occurs as follows :

$ \mathrm{C}_8 \mathrm{H}_8+2 \mathrm{KMnO}_4 \longrightarrow \mathrm{KC}_8 \mathrm{H}_5 \mathrm{O}_2+2 \mathrm{MnO}_2+\mathrm{KOH}+\mathrm{H}_2 \mathrm{O} $

Addition of bromine can be used as the test for unsaturation with regard to colour change of reaction. e.g.

Unsaturated compound $+\mathrm{Br}_2 \rightarrow$ colour of $\mathrm{Br}_2$ rapidly disappears.

i.e. Bromine water is decolourised.

Lindlar catalyst is partially deactivated catalyst composed of $\mathrm{BaSO}_4$ coated with Pd poisoned with quinoline.

It reduce ( $\mathrm{C} \equiv \mathrm{C}$ ) to ( $\mathrm{C}=\mathrm{C}$ ) bond, occurs via syn-addition and gives cis-alkene.