In Kolbe’s reaction, 

the product formed will be -

1.  Para-salicylic aldehyde

2.  Ortho-salicylic aldehyde

3. Para-salicyclic acid

4. Ortho-salicyclic acid

In Reimer-Tiemann reaction, 

the product will be -

1. ortho-Salicylaldehyde

2. para-Salicylaldehyde

3. meta-Salicylaldehyde

4. all of the above

In Williamson ether synthesis, the major product would be:

1. 1-Methoxypropane

2. 2-Methoxypropane

3. 3-Methoxypropane

4. All of the above

Unsymmetrical ether will be the product in -

(1) Kolbe’s reaction.

(2) Reimer-Tiemann reaction.

(3) Williamson ether synthesis.

(4) None of the above.

The IUPAC name of the product that formed in the following chemical reaction is:

\(\text{Acetone}\xrightarrow[\bf\text{(ii) H}_2\text{O, H}^+]{\bf\text{(i) C}_2\text H_5\text{MgBr, dry Ether}}\text{Product } \)

Match the following:

	List I		List II
i.	O2N-C6H5-OCH3 (p)	a.	4-Ethoxy-1,1-dimethylcyclohexane
ii.	C2H5OCH2-CH(CH3)2	b.	1-Chloro-2-methoxyethane
iii.	CH3OCH2CH2Cl	c.	1-Ethoxy-2-methylpropane
iv.		d.	1-Methoxy-4-nitrobenzene

Dehydration of ethanol in acidic medium yields  -

1. Ethyne

2. Ethanol

3. Ethane

4. Ethene

In the following Williamson’s synthesis reaction,

CH₃CH₂ONa + CH₃-Br → ?

The product will be:

In Williamson’s synthesis reaction major product will be

$$\begin{gathered} C{H}_{3 } \\ | \\ C{H}_3—C—ONa +C{H}_{3}Br\to ? \\ | \\ C{H}_3 \end{gathered}$$

1. 1-methoxy- 2-methylpropane

2. 2-methoxy- 2-methylpropane

3. 2-methoxy- 1-methylpropane

4. 1-methoxy- 1-methylpropane

The major product of Williamson’s ether synthesis reaction would be:


1. Benzene
2. Methoxy benzene
3. Ethoxy benzene
4. Ethane