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.}} $

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.

Statement (I) explains a characteristic of $\mathrm{S}_{\mathrm{N}}2$ (Substitution Nucleophilic Bimolecular) reactions. In $\mathrm{S}_{\mathrm{N}}2$ reactions, the attack of the nucleophile and the departure of the leaving group occur simultaneously in a single step. This reaction has a back-side attack mechanism, where the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group. This back-side attack leads to the inversion of configuration at the carbon atom undergoing substitution, a phenomenon often referred to as 'Walden inversion'. Thus, $\mathrm{S}_{\mathrm{N}}2$ reactions are stereospecific, leading to the formation of a product with a specific configuration (either inversion or retention of configuration, but typically inversion in the case of $\mathrm{S}_{\mathrm{N}}2$). Therefore, Statement I is true.

Statement (II) concerns $\mathrm{S}_{\mathrm{N}}1$ (Substitution Nucleophilic Unimolecular) reactions, which proceed in two steps: first, the leaving group departs, forming a carbocation intermediate, and then the nucleophile attacks. The attack of the nucleophile can occur from either side of the planar carbocation, resulting in the formation of both enantiomers of the product. When the reactant is chiral, the product is usually a racemic mixture, consisting of equal amounts of both enantiomers, assuming that there is no steric or electronic preference for the attack on the carbocation. This makes $\mathrm{S}_{\mathrm{N}}1$ reactions non-stereospecific. Therefore, Statement II is true as well.

Based on the explanations provided, the correct answer is Option D: Both Statement I and Statement II are true.

$\text { Number of } 2^{\circ} \text { carbon atom }=1$

In aqueous medium the reaction follows $\mathrm{S_N 1}$ mechanism majorly

Due to Lewis base formation electrophile stops/greatly slowed. Hence statement 2 is false while statement 1 is true.

Hence correct Ans. (4).

$\mathrm{PhCH}=\mathrm{CH}_2 \xrightarrow{\mathrm{B}_2 \mathrm{H}_6 / \mathrm{H}_2 \mathrm{O}_2 \mathrm{OH}^{-}} \mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{OH}$