Hydrocarbons

Stability order :

Trans is more stable than cis.

Dipole moment $\mathrm{x} \neq 0$

When benzene is nitrated, it forms nitrobenzene (benzene with a $-\mathrm{NO}_2$ group).

Now, Friedel–Crafts acylation ($\mathrm{CH}_3\mathrm{COCl}/\mathrm{AlCl}_3$) needs an aromatic ring that is sufficiently reactive to attack the electrophile.

The $-\mathrm{NO}_2$ group is a strongly deactivating group. It pulls electron density away from the benzene ring, so the ring becomes much less reactive.

Because of this strong deactivation, nitrobenzene does not undergo Friedel–Crafts acylation with $\mathrm{CH}_3\mathrm{COCl}/\mathrm{AlCl}_3$.

So, nitrobenzene will not give the shown acylated product under these conditions.

$\begin{aligned} & 264-80\left(x+\frac{y}{4}\right)+80 x=224 \\ & 264-\frac{80 y}{4}=224 \\ & 40=\frac{80 y}{4} \Rightarrow y=2 \\ & 264-80\left(x+\frac{y}{4}\right)=64 \\ & 264-80\left(x+\frac{1}{2}\right)=64 \\ & 264-80 x-40=64 \\ & x=2\end{aligned}$

Total dichloro products (excluding stereoisomers) = 4

Analysis of Statement I

Statement I says: "Methane can be prepared by decarboxylation of sodium ethanoate, Kolbe's electrolysis of sodium acetate and reaction of $\mathrm{CH}_3 \mathrm{MgBr}$ with water."

1. Decarboxylation of sodium ethanoate:

   This reaction involves heating a sodium salt of a carboxylic acid with soda-lime (a mixture of NaOH and CaO). The general reaction is the removal of a carboxyl group.

   For sodium ethanoate ($CH_3COONa$), the reaction is:

   $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$

   As you can see, methane ($CH_4$) is indeed the product. So, this part is correct.

2. Kolbe's electrolysis of sodium acetate:

   Kolbe's electrolytic method involves the electrolysis of an aqueous solution of a sodium or potassium salt of a carboxylic acid.

   For sodium acetate ($CH_3COONa$), at the anode, the acetate ions lose electrons and form methyl free radicals, which then combine to form ethane.

   At Anode:

   $2CH_3COO^- \rightarrow 2\dot{C}H_3 + 2CO_2 + 2e^-$

   $\dot{C}H_3 + \dot{C}H_3 \rightarrow CH_3-CH_3 \text{ (Ethane)}$

   The product is ethane, not methane. As per our NCERT textbook, this method is used to prepare symmetrical alkanes with an even number of carbon atoms. Methane cannot be prepared by this method.

3. Reaction of $\mathrm{CH}_3 \mathrm{MgBr}$ with water:

   $\mathrm{CH}_3 \mathrm{MgBr}$ is a Grignard reagent (methylmagnesium bromide). Grignard reagents are highly reactive and react with any source of a proton, such as water, to form hydrocarbons.

   The reaction is:

   $CH_3MgBr + H_2O \rightarrow CH_4 + Mg(OH)Br$

   Here, the methyl group from the Grignard reagent takes a hydrogen atom from water to form methane ($CH_4$). So, this part is also correct.

Conclusion for Statement I:

The statement claims that methane can be prepared by all three methods. Since Kolbe's electrolysis of sodium acetate produces ethane and not methane, the entire statement becomes false.

Analysis of Statement II

Statement II says: "Methane cannot be prepared from unsaturated hydrocarbons and by Wurtz reaction."

Let's check both parts.

1. Preparation from unsaturated hydrocarbons:

   Alkanes are prepared from unsaturated hydrocarbons (alkenes and alkynes) by hydrogenation, which is the addition of hydrogen ($H_2$) across the double or triple bond.

   For example:

   $CH_2=CH_2 \text{ (Ethene)} + H_2 \xrightarrow{Ni/Pt/Pd} CH_3-CH_3 \text{ (Ethane)}$

   To prepare methane ($CH_4$), which has only one carbon atom, we would need an unsaturated hydrocarbon with one carbon atom. The smallest alkene (ethene) and alkyne (ethyne) have two carbon atoms. A one-carbon alkene or alkyne is not possible. Therefore, methane cannot be prepared by the hydrogenation of unsaturated hydrocarbons. This part is correct.

2. Preparation by Wurtz reaction:

   The Wurtz reaction involves the treatment of alkyl halides with sodium metal in dry ether to produce higher alkanes. It essentially joins two alkyl groups together.

   The general reaction is:

   $2R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$

   This reaction always results in an alkane with an even number of carbon atoms, and it doubles the carbon chain. To prepare methane ($CH_4$), which has only one carbon atom, we would need to start with an alkyl halide having half a carbon atom, which is impossible.

   If we use the smallest alkyl halide, methyl chloride ($CH_3Cl$), the Wurtz reaction would give ethane:

   $2CH_3Cl + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 \text{ (Ethane)} + 2NaCl$

   So, methane cannot be prepared by the Wurtz reaction. This part is also correct.

Conclusion for Statement II:

Both parts of the statement are correct. Methane indeed cannot be prepared from unsaturated hydrocarbons or by the Wurtz reaction. Therefore, Statement II is true.

Final Decision

• Statement I is false.

• Statement II is true.

Looking at the options, this corresponds to the choice where Statement I is false and Statement II is true.

The correct answer is Option D.

To compare the stability of alkenes, we mainly look at how many alkyl groups are attached to the double bond. More alkyl groups means more hyperconjugation, so the alkene becomes more stable.

(D) $\mathrm{H}_3 \mathrm{C}-\mathrm{HC}=\mathrm{CH}_2 \quad(3 \propto-\mathrm{H})$

More substituted alkenes are more stable because they have more $\alpha-H$ atoms. These $\alpha-H$ atoms take part in hyperconjugation, which stabilizes the double bond.

Now compare each alkene:

C. $2,3$-Dimethylbut-$2$-ene is the most substituted (tetra-substituted), so it is the most stable.

A. $2$-Methylbut-$2$-ene is tri-substituted, so it is next in stability.

B. cis-But-$2$-ene is di-substituted, so it is less stable than A.

D. Prop-$1$-ene is mono-substituted, so it is the least stable.

Order: $\quad C>A>B>D$.

Option (B) and (D) reaction are not able to form alkene as a product.

St-I : Correct statement

St-II : In correct statement because product has chiral centre.

Both Statement I and Statement II are correct.

The compound given is hex - 1 - en - 4 - yne

Structure is

a double bond has one $\pi$ bond and one sigma bond.

A triple bond has one $\sigma$ bond and two $\pi$ bonds.

So, the number of $\pi$ bonds $=1+2=3$

For sigma bonds, count all carbon $-$ carbon bonds and carbon $-$ hydrogen bonds except the $\pi$ bonds in double bond and triple bond.

The number of sigma bonds = 13

So, total number of bonds $\Rightarrow \sigma=13\quad \pi=3$

Answer : option (C) 13 and 3

Ozonolysis product

Formation of more stable free radical intermediate.

4-Ethyl-3,4-dimethyloctane

Statement I

“One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_2$ gas.”

Reaction and Stoichiometry

Propyne (terminal alkyne): $\mathrm{CH_3C \equiv CH}$.

Reaction with sodium:

$ 2\,\mathrm{CH_3C \equiv CH} \;+\; 2\,\mathrm{Na} \;\longrightarrow\; 2\,(\mathrm{CH_3C \equiv C^-Na^+}) \;+\; \mathrm{H_2}. $

From this balanced equation, 2 moles of propyne produce 1 mole of $\mathrm{H_2}$.

Hence, 1 mole of propyne will produce $\tfrac{1}{2}$ mole of $\mathrm{H_2}$.

$ \boxed{\text{Statement I is correct.}} $

Statement II

“Four grams of propyne reacts with $\mathrm{NaNH_2}$ to liberate $\mathrm{NH_3}$ gas which occupies 224 mL at STP.”

Analysis

Moles of propyne

Molecular mass of propyne ($\mathrm{C_3H_4}$):

$ 3 \times 12 + 4 \times 1 = 36 + 4 = 40\,\mathrm{g/mol}. $

Four grams of propyne is:

$ \frac{4\,\mathrm{g}}{40\,\mathrm{g/mol}} = 0.1\,\mathrm{mol}. $

Reaction with $\mathrm{NaNH_2}$

For a terminal alkyne:

$ \mathrm{CH_3C \equiv CH} \;+\; \mathrm{NaNH_2} \;\longrightarrow\; \mathrm{CH_3C \equiv C^-Na^+} \;+\; \mathrm{NH_3}. $

1 mole of propyne produces 1 mole of $\mathrm{NH_3}$.

Moles of $\mathrm{NH_3}$ produced

With $0.1\,\mathrm{mol}$ of propyne, we get $0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$.

Volume of $\mathrm{NH_3}$ at STP

1 mole of any ideal gas at STP $\approx 22.4\,\mathrm{L} = 22400\,\mathrm{mL}.$

$0.1\,\mathrm{mol}$ of $\mathrm{NH_3}$ occupies $0.1 \times 22.4\,\mathrm{L} = 2.24\,\mathrm{L} = 2240\,\mathrm{mL}.$

However, Statement II says the liberated $\mathrm{NH_3}$ occupies only 224 mL at STP, which corresponds to $0.01\,\mathrm{mol}$ of $\mathrm{NH_3}$, not $0.1\,\mathrm{mol}$. Therefore, the statement’s volume is off by a factor of 10 and is thus incorrect if the reaction goes to completion in a typical way.

$ \boxed{\text{Statement II is incorrect.}} $

Conclusion

Statement I is correct.

Statement II is incorrect.

Hence, the best choice is:

$ \boxed{\text{Option D: Statement I is correct but Statement II is incorrect.}} $

$X$ 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)

Among the given sets, only III is correct i.e.

cis-but-2-ene > trans-but-2-ene (dipole moment) while I and II are incorrect. Their correct form is (I) torsional strain is more in eclipsed ethane than in staggered ethane. It is because in eclipsed form hydrogen atoms attached to two carbons are as closed together as possible while in staggered form hydrogen are as far apart as possible. 2-dimethyl butane < 2-methyl pentane (B.P)

The complete reaction sequence is as follows

So, the molecular formula is $\mathrm{C}_6 \mathrm{H}_6 \mathrm{Cl}$ and its empirical formula is CHCl . Thus, the empirical formula weight of $Z$ is 48.5

$t$-butyl benzene does not give benzoic acid when treated with alkaline $\mathrm{KMnO}_4$ because it lack the necessary benzylic hydrogen required for oxidation to form benzoic acid, since the tertiary carbon in $t$-butylbenzene has no hydrogen attached directly to it, the oxidation cannot proceeds to form carboxylic acid.

Statement I is wrong, but Statement II is right.

Statement I should say: In Kolbe's electrolysis of sodium propionate, the product is butane, not n-hexane.

The chemical reaction is:

$ \begin{aligned} & 2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COONa} + 2 \mathrm{H}_2 \mathrm{O} \longrightarrow \\ & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 + 2 \mathrm{CO}_2 + 2 \mathrm{NaOH} + \mathrm{H}_2 \uparrow \\ & \text { Butane } \end{aligned} $

In Statement II, it is correct that in Kolbe's process, carbon dioxide ($\mathrm{CO}_2$) is released at the anode and hydrogen gas ($\mathrm{H}_2$) is released at the cathode.