Chemistry Test - 24

Primary and secondary amines react with Grignard reagents to form alkanes.

For the reaction to occur, there have to be hydrogen atoms attached to the nitrogen atom. The hydrogen is extracted by the negatively charged alkyl group of the Grignard reagents leading to the formation of alkanes.

As there are no hydrogens attached to the Nitrogen atom of tertiary amines, they do not undergo the reaction with Grignard reagents. The reaction of ethylamine with methyl magnesium bromide, Methane is formed from the alkyl part of the Grignard Reagent.

The reaction can be written as:

When ethyl amine will be treated with ethyl magnesium bromide, it will give ethane. The reaction will be:

\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{MgBr} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}\)

Hence, when ethylamine is treated with methyl magnesium bromide, the product formed is methane.

Element with electronic configuration 2,8,2 belongs to the 3rd period and 2nd group. The valence electrons give us the idea of group numbers. As the electrons are divided into 3different shells therefore the period number is 3.

Tertiary butyl alcohol is most reactive with Lucas Reagent.

Ethanol is a primary alcohol and does not give instant turbidity with Lucas Reagent. Methanol is also primary alcohol that does not give turbidity with Lucas Reagent. Isopropyl alcohol is secondary alcohol that gives turbidity with Lucas Reagent after 5 to 10 minutes.

Tertiary butyl alcohol is tertiary alcohol that gives turbidity instantaneously on reaction with Lucas Reagent. The reaction can be given as follows:

Hydrogen gas subjected to alpha particles shrank in volume and diatomic hydrogen was converted to triatomic.It is named "Hyzone" in analogy to ozone and its extra reactivity over normal hydrogen.In other words, H₃is also called Hyzone.

We use the expression \(q=\frac{\mu}{\mathrm r}\) to solve for \(\mathrm q\). The debye unit, \(\mathrm D\), equals \(3.34 \times 10^{-30} \mathrm{C} \mathrm{m}\), so the dipole moment of \(\mathrm{HF}\) is \(\mu=1.83 \times\left(3.34 \times 10^{-30} \mathrm{C} \mathrm{m}\right)=6.11 \times 10^{-30} \mathrm{C} \mathrm{m}\)

The S.I. prefix p (pico) means \(\times 10^{-12}\), so the bond length \(r=\) \(91.7 \times 10^{-12} \mathrm{~m}\). Substituting in the equation above gives \(\mathrm q=\frac{6.11 \times 10^{-30} \mathrm{C} \mathrm{m}}{91.7 \times 10^{-12} \mathrm{~m}}=6.66 \times 10^{-20} \mathrm{C}\)

The amount of charge on an electron (i.e., an electronic charge unit) equals \(1.602 \times 10^{-19} \mathrm{C}\), which we can express as \(1\mathrm e=1.602 \times 10^{-19} \mathrm{C}\)

The value of \(q\) in electronic charge units is therefore \(q=6.66 \times 10^{-20}\left(\frac{1 e}{1.602 \times 10^{-19}}\right)\)\(=0.416 \mathrm e\)

The given structure is Peroxymonosulphuric acid. The image is shown below:

Highly reactive metals with lower reduction potentials than hydrogen are not obtained by electrolysis of aqueous solutions of their salts since their cations cannot be reduced at cathode in presence of water. Instead of reduction of these metal cations, water undergoes reduction. Mg are reactive metals with lower reduction potentials.

Usually d-block elements are less reactive and hence can be prepared by electrolysis of their aqueous solutions.

Given,

Concentration \(=1.2 \times 10^{3} \mathrm{M}\)

\(1 \mathrm{~mm}^{3}\) will contain \(\frac{1.2 \times 10^{3} \mathrm{~mol} / \mathrm{dm}^{3}}{10^{6} \mathrm{~mm}^{3} / \mathrm{dm}^{3}}\)

\(=1.2 \times 10^{-3}\) moles of soap

1 mole of soap \(=6.023 \times 10^{23}\) molecules

\(1.2 \times 10^{-3}\) moles of soap \(=1.2 \times 10^{-3} \times 6.023 \times 10^{23}\)

\(=7.227 \times10^{20}\) stearate ions

The average number of stearate ions in one colloidal particle \( (\text { micelle })=\frac{7.227 \times 10^{20}}{2.4 \times 10^{13}}=3 \times 10^{7}\)

(1) In compound A, positive charge on O atom is not stable that is why OH2⊖ get's removed.

(2) In Carbocation (B), two α−H present and after 1, 2-methyl shift number of α−H becomes six so it becomes more stable as far carbocation when number of α-H increases stability increases.

(3) Among product D, E and F, D is major product as in case of alkene most stable, alkene are those which have more α−H. And every reactant produce that product which is more stable.

Ostwald's process is :

\(4 \mathrm{NH}_3+5 \mathrm{O}_2 \longrightarrow \underset{\text{(A)}}{4 \mathrm{NO}}+6 \mathrm{H}_2 \mathrm{O}\)

\(\underset{\text{(A)}}{2 \mathrm{NO}}+\mathrm{O}_2 \longrightarrow \underset{\text{(A)}}{2 \mathrm{NO}_2}\)

\(4 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \longrightarrow 4 \mathrm{HNO}_3\)

\(\therefore\) A and B are NO and NO2 respectively