Synthesis of Secondary and Tertiary Amines

Q.

Both lithium$\left(\mathrm{Li}\right)$ and magnesium$\left(\mathrm{Mg}\right)$ display several similar properties due to the diagonal relationship; However, the incorrect one is:

1. Both form nitrides.

2. Nitrates of both $\mathrm{Li}$ and $\mathrm{Mg}$ yield ${\mathrm{NO}}_2$and ${\mathrm{O}}_2$ on heating.

3. Both form basic carbonates.

4. Both form soluble bicarbonates.

Q. Based on the reaction given above, choose the correct options:

1. The products are stereoisomers
2. The products are constitutional isomers
3. The reaction proceeds via electrophilic aromatic substitution
4. The mechanism involves a benzyne intermediate

Q. Based on the reaction given above, choose the correct options:

1. The products are stereoisomers
2. The products are constitutional isomers
3. The reaction proceeds via electrophilic aromatic substitution
4. The mechanism involves a benzyne intermediate

Q.

When methyl iodide is heated with ammonia, the product obtained is

1. Methylamine

2. Dimethylamine

3. Trimethylamine

4. All of the above.

Q. A hydrocarbon $A\left(C_4{H}_8\right)$ on reaction with$HCl$ gives a compound $B\left(C_4{H}_{9}Cl\right)$ which on reaction with $1$ mole of $N{H}_3$ gives a compound $C\left(C_4{H}_{11}N\right)$. On reacting with $NaN{O}_2$ and $HCl$ (low temperature) followed by treatment with water, compound $C$ yields optically active alcohol(s). Ozonolysis of $A\left(1\text{mole}\right)$ gives $2$ moles of acetaldehyde.
The compound C is:

1. Butan-2-amine
2. N, N-dimethylethanamine
3. Diethylamine
4. Butylamine

Q. Choose the correct answer form the alternatives given.
$N{H}_{4}Cl+\left(A\right)\to Microcosmicsalt$
$\downarrow Heat$
$\left(B\right)\underset{+MnO}{\overset{Heat}{\to }}\left(C\right)$
Violet bead
(A), (B) and (C) respectively are :

1. $N{a}_{3}P{O}_4,NaP{O}_3,\left(Mn{)}_3\right(P{O}_4{)}_2$
2. $N{a}_{2}HP{O}_4,N{a}_{3}P{O}_4,M{n}_3(P{O}_4{)}_2$
3. $N{a}_{2}HP{O}_4,NaP{O}_3,Mn(P{O}_3{)}_2$
4. $N{a}_{2}HP{O}_4,NaP{O}_3,NaMnP{O}_4$

Q. $\text{Butanedial}+\text{Ammonium carbonate}\stackrel{△}{\to }P$
Then:

1. P is aromatic
2. P is antiaromatic
3. P on reaction with chloroform and sodium hydroxide gives:
   
4. P on reaction with chloroform and sodium hydroxide gives:
   

Q. What happens when ethylamine is made to react with excess formic acid and formaldehyde?

1. The intermediate is a secondary amine
2. The final product is a tertiary amine
3. This reaction is called Eschweiler-Clarke methylation
4. The mechanism involves reductive amination