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.

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

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.

The primary catalyst used for the isomerisation of $n$-alkanes to branch

chain alkanes are acid catalyst, particularly zeolites and sulphated zirconia.

Anhydrous $\mathrm{AlCl}_3$ or HCl can be also used.

The complete reaction is as follows

This is Wurtz-fittig reaction

$ \text { This is Friedel-Craft alkylation. } $

The complete reaction is as follows,

The final product is benzene, which is also formed by aromatisation of $n$-hexane.

The reaction is a Wurtz reaction, In the given case, three alkanes are formed

$ 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+2 \mathrm{Na} \xrightarrow[\text { Ether }]{\text { Dry }} \underset{\text { Butane }}{\mathrm{C}_4 \mathrm{H}_{10}}+2 \mathrm{NaBr} $

$ 2 \mathrm{C}_3 \mathrm{H}_7 \mathrm{Br}+2 \mathrm{Na} \xrightarrow[\text { Ether }]{\text { Dry }} \underset{\text { Hexane }}{\mathrm{C}_6 \mathrm{H}_{14}}+2 \mathrm{NaBr} $

$ \begin{aligned} \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+\mathrm{C}_3 \mathrm{H}_7 \mathrm{Br}+2 \mathrm{Na} \longrightarrow & \underset{\text { Pentene }}{\mathrm{C}_5 \mathrm{H}_{12}} +2 \mathrm{NaBr} \end{aligned} $

The complete reaction sequence is as follows

Thus, statement given in option $(C)$ is incorrect $z$ on reaction with $\mathrm{CH}_3 \mathrm{MgBr}$ followed by hydrolysis will give $3^{\circ}$ alcohol.

The complete reaction is as follows,

$ \begin{aligned} & x=\mathrm{H}_2 \mathrm{O} / \mathrm{H}_2 \mathrm{SO}_4, \mathrm{Hg}^{2+} \\ & y=\mathrm{PCC} \end{aligned} $

To evaluate which statement regarding ethyne is incorrect, it's necessary to review the characteristics of the chemical bonds in ethyne (acetylene, $\mathrm{C_2H_2}$) compared to ethene (ethylene, $\mathrm{C_2H_4}$).

Option A: The statement that the carbon-carbon bonds in ethyne are weaker than that in ethene is incorrect. In ethyne, the carbon-carbon triple bond is composed of one sigma ($\sigma$) bond and two pi ($\pi$) bonds. This configuration makes the triple bond in ethyne much stronger and shorter compared to the double bond in ethene, which consists of one $\sigma$ bond and one $\pi$ bond. Thus, the carbon-carbon bond in ethyne is actually stronger than that in ethene.

Option B: The statement that both carbons are sp hybridised in ethyne is correct. In ethyne, each carbon atom is bonded to the other carbon atom and a single hydrogen atom, necessitating the sp hybridisation to form a linear molecule with $180^\circ$ angles between the bonds.

Option C: The statement that the $\mathrm{C-C}$ bonds in ethyne are shorter than that in ethene is correct. The presence of a triple bond between the carbon atoms in ethyne results in a shorter bond length compared to the double-bonded carbons in ethene. This is because the additional pi bonds in a triple bond pull the carbon atoms closer together.

Option D: The statement that ethyne is linear is correct. Due to the sp hybridisation of carbon atoms in ethyne, the molecule adopts a linear geometry with bond angles of $180^\circ$.

Therefore, the incorrect statement regarding ethyne is Option A: The carbon-carbon bonds in ethyne are weaker than that in ethene.