General Organic Chemistry

Condensed form of $n$ butane

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \text {. } $

Bond line form of $n$ butane

Complete formulae of $n$ butane

Distillation under reduced pressure (vacuum distillation) is generally used for the purification of high-boiling organic liquids that decompose at or below their normal boiling points.

In this method, the external pressure is reduced so that the liquid boils at a lower temperature than its normal boiling point. This prevents thermal decomposition and allows safe purification of heat-sensitive liquids such as glycerol, aniline, and nitrobenzene.

An aromatic species must be cyclic, planar, have a continuous ring of $p$-orbitals, and contain ( $4 n+2$ ) $\pi$-electrons (Huckel's rule). Let's apply these rules to each option

(c) Cycloheptatrienyl cation : It is cyclic, planar, and has $6 \pi$-electrons. This follows Huckel's rule aromatic.

(b) Cyclopentadienly anion: It is cyclic, planar, and has $6 \pi$-electrons. This follows Huckel's rule aromatic.

(a) Cyclohexadienyl cation: It is not aromatic. It is anti-aromatic as it has $4 n$ electrons, which violates the Huckel's rule for aromaticity.

(d) Naphthalene : It is cyclic, planar, and has $10 \pi$-electrons. This follows Huckel's rule aromatic.

Therefore, the cyclohexadienyl cation is the only species that is not aromatic.

$\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ (Ethyl carbocation) :

This is a primary carbocation. It's stabilised by the $+I$ (inductive) effect of the methyl group and some hyperconjugation from the three alpha-hydrogens. The positive carbon is $s p^2$-hybridised.

(b) $\mathrm{CH}_2=\mathrm{C} \mathrm{H}$ (Vinyl carbocation) : In this carbocation, the positive charge is on an $s p$ hybridised carbon. An $s p$ hybridised carbon is more electronegative than an $s p^2$ or $s p^3$ hybridised carbon, meaning it's less capable of accommodating a positive charge. The double bond itself cannot provide effective resonance stabilisation to this directly attached positive charge, and in fact, the proximity of the electron-withdrawing $s p$ hybridised carbon makes it highly unstable. This is due to the electron-withdrawing nature of the double bond.

(c) $\mathrm{CH}_2=\mathrm{CH}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ (Allyl carbocation) : This carbocation is highly stabilised by resonance. The positive charge is delocalised over two carbon atoms due to conjugation with the double bond.

(d) $\mathrm{C}_6 \mathrm{H}_5-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ (Benzyl carbocation) This carbocation is also highly stabilised by resonance. The positive charge is delocalised into

the benzene ring due to conjugation.

• Comparing these, the vinyl carbocation $\left(\mathrm{CH}_2=\stackrel{+}{\mathrm{C}} \mathrm{H}\right)$ is the least stable.

Smaller the $\mathrm{p} K_a$ value, stronger is the acid. Among the given compounds

$ \text { lowest } \mathrm{p} K_a \text { value. } $

Hyperconjugation involves delocalisation of electrons from CH $\sigma$-bond adjacent to a $\pi$ system or carbocation.

Hence, option (a) shows hyperconjugation.

$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$ molecules shows hyperconjugation as electron from CH bond of methyl group interact with $\pi$ bond of alkene.

The correct increasing order of acidic strength of given compounds is I $<$ II $<$ III $<$ IV.

As acidity increases with electron withdrawing group.

$ -\mathrm{OCH}_3<-\mathrm{CH}_3<-\mathrm{CN}<-\mathrm{NO}_2 $

The stability of carbocation decreases with increase in $S$-character on carbon. So, the correct order will be I $>$ III $>$ IV $>$ II $>$ V.

The molar mass of the compound ' $ Z $ ' i.e aspirin is $ 180 \mathrm{~g} / \mathrm{mol} $. It contain 9 carbon atoms

$ \begin{aligned} C \% & =\frac{9 \times 12}{180} \times 100 \\\\ & =60 \% \end{aligned} $

Less is the acidic strength more will be the $\mathrm{p} K_a$ value. So, the correct order increasing $\mathrm{p} K_a$ is

$ \text { IV }<\text { II }<\mathrm{I}<\text { III. } $

Considering electronic effects for resonance/mesomeric ( $\mathrm{R} / \mathrm{M}$ ) which essentially actives or deactivates an organic molecules towards a reagent. The activating groups are:

and the deactivating groups are:

The delocalisation of $\sigma$ electrons of $\mathrm{C}-\mathrm{H}$ bond of an alkyl group linked directly to an atom of unsaturated system or to an atom having unshared $p$-orbital (as in benzene) is observed in hyperconjugation effect.

The uncharged resonating structure is more stable than charged ones. So, (II) is more stable than I and III. Out of I and III, I is more stable as negative charge is acquired by more electronegative atom i.e., oxygen while in (III) positive charge is more oxygen atom.

So, correct order of stability of resonance structures are : II > I > III.

Hyperconjugation is not possible in compounds in which $\alpha$-hydrogen is not present.

Thuscompound does notshow hyperconjugation.

I, III and V contain both $s p$ and $s p^2$ hybridised carbons.

Acidic strength depends on the stability of conjugate base formed. More stable the conjugate base, stronger is the acid. The order of stability of conjugate base formed will be

$ \begin{aligned} \mathrm{H}-\mathrm{C} \equiv & \mathrm{C}-\mathrm{COO}^{\ominus} \\ & >\mathrm{CH}_2=\mathrm{CH}-\mathrm{COO}^{\ominus} \\ & >\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{COO}^{\ominus} \\ & >\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}^{\ominus} \end{aligned} $

(Increase in \% character increases stability of conjugate base)

∴ Correct order of acidic strength will be-

$ \begin{aligned} \mathrm{H}-\mathrm{C} \equiv & \mathrm{C}-\mathrm{COOH} \\ & >\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH} \\ & >\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} \\ & >\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} . \end{aligned} $

For hyperconjugation, compound/ molecule should possess $\alpha$-hydrogen atoms.

A. $\mathrm{H}_3 \mathrm{C}-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2-3 \alpha \mathrm{H}$-atoms

D. $-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$-zero $\alpha \mathrm{H}$-atom.

Therefore, $\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3$ lacks hyperconjugative stability.

Radicals on carbon atom are stabilised when they are in more substituted position that means tertiary radical is more stable than secondary and secondary is more stable than primary radical.

Therefore, $[B]$ is least stable followed by $[C,[A]$ and $[D]$ are most stable ones.

Out of $[A]$ and $[D],[A]$ is more stable as it has more $\alpha$-hydrogen atom than $[D]$, which lead to more number of hyperconjugated structures.

Correct order of stability is $[B]<[C]<[D]<[A]$

Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a $\sigma$-bond (usually $\mathrm{C}-\mathrm{H}$ ) with an adjacent empty or partially filled p-orbital or a $\pi$-orbital to give an extended molecular orbital that increases the stability of the

system. Consider the molecule propene, where $\pi$ bond is in conjugation with $3 \sigma \mathrm{C}-\mathrm{H}$ bond of adjacent methyl group. So, higher the number of adjacent $\sigma \mathrm{C}-\mathrm{H}$ bond to $\pi$-bond higher will be the hyperconjugation and higher will be the stability.

The adjacent $\sigma \mathrm{C}-\mathrm{H}$ bond to $\pi$ bond increases in the order of IV $>$ III $>$ II $>$ I.

Therefore, stability follows the same order.

Bond dissociation energy or bond energy (BE) of $\mathrm{C}-\mathrm{H}$ bond of a hydrocarbon $(R \mathrm{H})$ can be predicted from stability of free radical ( $R^{\bullet}$ ) formed by homolytic bond fission of $\mathrm{C}-\mathrm{H}$ bond of the hydrocarbon.

$ \text { So, }(\mathrm{BE})_{\mathrm{C}-\mathrm{H}} \propto \frac{1}{\text { Stability of free radical }\left(R^{\bullet}\right)} $

$ \begin{aligned} &\text { Stability order of free radicals }\\ &(C)>(D)>(B)>(A) \end{aligned} $

$ \text { In the reaction, } $

The hybridisation of carbon- 2 in $(P)$ is $s p^3$ and in $(Q)$ in $s p^2$, because of 4 sigma bonds in ( $P$ ) and one pi $(\pi)$ bond in $(Q)$.