Chemistry Test-5

Transamination catalyzed by aminotransferase occurs in two stages. In the first step, the α amino group of an amino acid is transferred to the enzyme, producing the corresponding α-keto acid and the aminated enzyme. During the second stage, the amino group is transferred to the keto acid acceptor, forming the amino acid product while regenerating the enzyme. The chirality of an amino acid is determined during transamination.

The binding energy of a particle can be defined as the minimum energy required to remove nucleons (Proton or neutron) to an infinite distance from the nucleus.

It can be expressed as,

\(\Delta E =\left( Zm _{ p }+( A - Z ) m _{ n }- M \right) c ^{2} \times 931.5 MeV\)

This is derived by using Einstein's mass-energy relation and we can consider binding energy per nucleons as an average energy per nucleon needed to separate a nucleus into its individual nucleons.

The binding energy of \(Li\) is,

\(\Rightarrow_{3}^{7} Li =7 \times 5.60=39.20 MeV\)

The binding energy of \(He\) is,

\(\Rightarrow{ }_{2}^{4} He =2 \times 4 \times 7.06=56.48 MeV\)

The reaction between \(Li\) and \(He\) is given as \(p+{ }_{3}^{7} Li \rightarrow 2_{2}^{4} He\)

The energy of the proton is,

\(\Rightarrow p=2{ }_{2}^{4} He -{ }_{3}^{7} Li\)

\(\Rightarrow p =(56.48-39.20) MeV =17.28 MeV\)

The properties of elements are periodic function of thier atomic number.

The modern periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Modern periodic table is based on atomic number. Thus, the properties of elements are related to their atomic number or electronic configuration.  Elements are arranged according to increasing order of atomic number. Elements with similar properties occur at regular intervals in the periodic table.

Potassium \(\mathrm{K}\)metal reacts with water \(\mathrm{ H _{2} O}\) to give potassium hydroxide \((\mathrm{KOH} )\) and hydrogen gas \(\mathrm{( H _{2}}).\) The reaction for this process is:

\( \mathrm{K + H _{2} O \rightarrow KOH + H _{2} \uparrow}\) (gas)

A balanced chemical equation represents a chemical equation in which the number of atoms of each element on both reactant and product sides are the same. The balanced chemical equation for the given reaction is:

\(2 \mathrm{K ( s )+2 H _{2} O ( I ) \rightarrow 2 KOH ( aq )+ H _{2} \uparrow( g )}+\) heat energy.

Decreasing order of the \(\mathrm{pK}_{\mathrm{b}}\) values:

\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\)

\(\mathrm{pK}_{\mathrm{b}}=-\log \mathrm{K}_{\mathrm{b}}\)

If \(\mathrm{pK}_{\mathrm{b}}\) value is less, then more the value of \(\mathrm{K}_{\mathrm{b}}\) and the basic character.

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}\) are aromatic amine while \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) and \(\mathrm{CH}_{3}-\mathrm{NH}_{2}\) are aliphatic amines.

\(-\mathrm{C}_{2} \mathrm{H}_{5}\) group is showing \(+\mathrm{I}\) effect.

In \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), only one \(-\mathrm{C}_{2} \mathrm{H}_{5}\) group is present while in \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\), two \(-\mathrm{C}_{2} \mathrm{H}_{5}\) groups are present.

Thus, the \(+\mathrm{I}\) effect more in \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}\) than in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\).

Therefore, the electron density over the \(\mathrm{N}-\) atom is more in \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) than in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\).

So, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) is more basic than \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\).

Both \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) are less basic than \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) due to the delocalization of the lone pair in the former two.

Further, among \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), the former will be more basic due to the \(+\mathrm{I}\) effect of \(-\mathrm{CH}_{3}\) group.

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}\) (basic character)

The basic character of aromatic amine is less than aliphatic amine because lone pair of nitrogen in aromatic amine is involved in delocalisation while in aliphatic amine it is not involved in delocalisation hence overall basic character is \(\left(\mathrm{K}_{\mathrm{b}}\right)\)

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NHCH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)

Thus the order of \(\mathrm{pK}_{\mathrm{b}}\) is:

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \)

\(\because \mathrm{pK}_{\mathrm{b}}=-\log \mathrm{K}_{\mathrm{b}}\)

Buna-S is an example of co-polymer.

An antacid is a substance which neutralizes stomach acidity and is used to relieve heartburn, indigestion or an upset stomach.

Example- \(\mathrm{Mg ( OH )_2, Na _2 CO _3, NaHCO _3}\).

Aluminium phosphate is not used as an antacid.

Iron \((\mathrm{Fe})\) does not react with hot or cold water. It only reacts with steam to give hydrogen gas.

Magnesium \((\mathrm{Mg})\) does not react with cold water but reacts with hot water to give hydrogen gas.

Potassium \((\mathrm{K})\) reacts violently with cold water to give hydrogen gas.

Copper \((\mathrm{Cu})\) does not react with hot/cold water or steam.

We know that fluorine is the highly electronegative atom in the entire periodic table. So it has always an oxidation number of -1 in all its compounds.

Annihilation is the complete destruction of an object. In Physics, when a particle and its antiparticle collide and disappears and release energy, It's called Annihilation.

The amount of energy (E) produced by annihilation is equal to the mass \((m)\) that disappears multiplied by the square of the speed of light in a vacuum (c).

\(\mathrm{E}=\mathrm{mc}^{2}\)

The rest mass of an electron and positron is \(9.109 \times 10^{-31} \mathrm{~kg}\) The energy released when both collide:

\(E=2 \times \mathrm{mc}^{2}\)

\(\Rightarrow E=2 \times 9.109 \times 10^{-31} \times\left(3 \times 10^{8}\right)^{2}\)

\(\Rightarrow E=163.962 \times 10^{-15} \mathrm{~J}\)

Now, we know that \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\)

\(\therefore\) The energy released in Electron-volt (eV) will be:

\(=\frac{163.962 \times 10^{-15}}{1.6 \times 10^{-19}} \mathrm{eV} \)

\(=102.47 \times 10^{4} \mathrm{eV} \)

\(=1.02 \mathrm{MeV}\)