Chemistry Test - 22

In autocatalysis, a reaction product is itself a catalyst for that reaction leading to positive feedback.

A single chemical reaction is said to have undergone autocatalysis, or be autocatalytic, if one of the reaction products is also a reactant and therefore a catalyst in the same or a coupled reaction. The reaction is called an autocatalytic reaction. A catalyst is a substance that accelerates the rate of a chemical reaction but remains chemically unchanged afterwards.

\(\mathrm{P C l_{3}}\) or phosphorous trichloride is a chemical compound which readily reacts with water to produce \(\mathrm{H C l}\) gas, and with \(\mathrm{H C l}\) gas the by-product formed is \(\mathrm{H_{3} P O_{3}}\).

The reaction is as follows:

\(\mathrm{PCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}+3 \mathrm{HCl}\)

Therefore, the products formed by the hydrolysis of \(\mathrm{P C l_{3}}\) are \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{HCl}\).

Number of donor atoms in \(N \left( CH _{2} CH _{2} NH _{2}\right)_{3}\) is four so it is a tetradentate ligand. Each nitrogen atom is a donor atom.

Carbylamine reaction mechanism includes the addition of amine to the intermediate created from the dehydrohalogenation of chloroform. This intermediate is called dichlorocarbene. The carbylamine reaction is also known as Hofmann isocyanide synthesis. It is the reaction of a primary amine, chloroform and a base to synthesize isocyanides. The dichlorocarbene intermediate is very important for this conversion. The carbylamine reaction cannot be used to synthesize isocyanides from secondary or tertiary amines. In general, the carbylamine reaction can be written as -

The electronic configurations are:

\(T i^{+}-[A r] 3 d^{1} 4 s^{1}: T i^{3+}-[A r] 3 d^{1} 4 s^{0}\)

\(C u^{+}-[A r] 3 d^{10} 4 s^{0} : C u^{2+}-[A r] 3 d^{9} 4 s^{0}\)

\(C r^{+}-[A r] 3 d^{5} 4 s^{0} : C r^{3+}-[A r] 3 d^{2} 4 s^{0}\)

In solutions, hydration enthalpy of ions also affect their stability.

So, based upon both fact i.e., electronic configuration and hydration enthalpy \(T i^{+}\) is more stable than \(T i^{3+}\) in solution while in other two cases higher. oxidation state ions are more stable.

All the alkylated benzenes will oxidize to benzoic acid. t-butyl benzene is inert to \(KMnO _{4}\). tert. Butylbenzene has no benzylic hydrogens and hence does not give benzoic acid on oxidation. But as the substituted tert. Butyl group get seperated from the benzene ring due to steric effect and forms tert.-butanoic acid.

Depression in freezing point is a colligative property which depends upon the amount of the solute.

\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}\)

Thus, for a given solvent and given concentration, \(\Delta T_{f}\) is directly porportional to \(i\) (Van't Hoff factor) i.e. maximum \(T_{f}\) (and hence lowest freezing point) will correspond to maximum value of \(i \).

(A) \(A l_{2}\left(S O_{4}\right)_{3} \stackrel{H_{2} O}{\longrightarrow} 2 A l^{3+}+3 S O_{4}^{2-}\)

Here, Van't Hoff factor, i = 5

(B) \(C_{5} H_{10} O_{8} \stackrel{H_{2} O}{\longrightarrow}\) No ionization

Here, Van't Hoff factor, i = 1

(C) \(K I \stackrel{H_{2} O}{\longrightarrow}K^{+}+I^{-}\)

Here, Van't Hoff factor, i = 2

(D) \(C_{12} H_{22} O_{11} \stackrel{H_{2} O}{\longrightarrow}\) No ionization

Here, Van't Hoff factor, i = 2

Therefore, \(0.10 M(\approx 0.10 \mathrm{~m})\) aqueous solution will have the lowest feezing point.

We know that, electronegativity of halogen group decreases on moving down the group because size of atom increases. Decreasing order of electronegativity of halogens is F, Cl, Br, I.

In case of \(\mathrm{s p^{3}}\) hybridization:

• The axial position of metal has less s-character and the equatorial position of metal has more s-character.
• The ligand which are s-phile (s-lover) will tend to occupy equatorial position, and the ligands which are s-nonphile will tend to occupy axial position.
• The s-phile ligand have less electronegativity or we can say s-nonphile (having more electronegativity) atoms have tendency to occupy axial position is called apicophilicity.
• Thus, \(\mathrm{C l^{-}}\)is more electronegative than \(\mathrm{B r^{-}}\).

Therefore, it is s-nonphile and will occupy axial position. Thus, the correct structure of \(\mathrm{P C l_{3} B r_{2}}\) will be as shown:

The two monosaccharides are held together by a glycosidic linkage between \(\mathrm{C}-1\) of \(\alpha\)-D-glucopyranose and \(\mathrm{C}-2\) of \(\beta\)-D-fructofuranose.

So, the correct set is\(\alpha{-D}\)-glucopyranose and \(\beta{-D}\)-fructofuranose.

Ethyl alcohol is industrially prepared from ethylene by Absorbing in H₂SO₄ followed by hydrolysis.

Acid-catalyzed hydration of Alkene:

Alkenes react with water (hydrolysis) in the presence of acid as a catalyst to form alcohols.

In the case of unsymmetrical alkenes, the addition reaction occurs in accordance with Markovnikov’s rule.

Mechanism:
Step 1: Protonation of an alkene to form carbocation by an electrophilic attack of hydronium ion (H₃O⁺).

Step 2: Nucleophilic attack of water on carbocation.

Step 3: Deprotonation to form an alcohol.

Industrial preparation of Ethyl alcohol:

• Ethanol reacts with sulphuric acid to form an intermediate product.
• This is followed by hydrolysis to give ethanol and H₂SO₄ as byproducts.
• Here, the H₂SO₄ formed as a byproduct is reused for further reaction with ethyl alcohol.

The reaction is as follows:

\(CH _{2}= CH _{2} \stackrel{ H _{2} SO _{4}}{\longrightarrow} CH _{3}- CH _{2}- HSO _{4} \stackrel{\text { Hydrolysis }}{\longrightarrow} CH _{3} CH _{2}- OH + H _{2} SO _{4}\)