- 1(current)
Q. No.The Williamson's synthesis reaction among the following is -
| 1. |   |
| 2. |   |
| 3. | \(C_2H_5I + C_2H_5ONa \rightarrow C_2H_5-O-C_2H_5 + NaI\)  |
| 4. |   |
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 |
| i | ii | iii | iv | |
| 1. | d | c | b | a |
| 2. | c | d | b | a |
| 3. | a | c | b | d |
| 4. | b | c | d | a |
Match the starting material in Column I with the products formed (Column II) in the reaction with HI.
| Column I |
Column II |
||
| A. |  |
1. |  |
| B. |  |
2. |  |
| C. |  |
3. |  |
| D. |  |
4. |  |
| 5. |  |
| A | B | C | D | |
| 1. | 2 | 3 | 4 | 1 |
| 2. | 3 | 1 | 5 | 2 |
| 3. | 5 | 4 | 3 | 2 |
| 4. | 4 | 5 | 2 | 1 |
Match the reactions from Column-I with their corresponding descriptions from Column-II and select the correct option.
|
Column-l (Name of Reaction) |
Column-ll (Description of Reaction) |
||
| A. |
Reimer-Tiemann reaction |
I. |
Conversion of phenol to o-hydroxysalicylic aid |
| B. |
Kolbe's reaction |
II. |
Heated copper at 573K |
| C. |
Conversion of 2alcohol to a ketone |
III. |
Reaction of an alkyl halide |
| D. |
Williamson's synthesis |
Iv. |
Conversion of phenol to salicylaldehyde |
Codes:
| A | B | C | D | |
| 1. | II | III | IV | I |
| 2. | III | IV | I | II |
| 3. | I | IV | III | II |
| 4. | IV | I | II | III |
Among the following ethers, the one that will produce methyl alcohol on treatment with hot concentrated HI is:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Preparation of ethers by 2 or 3-degree alcohols in an acidic medium is not a suitable method because:
| 1. | In case of secondary or tertiary alcohols, ketone is obtained as a product. |
| 2. | In case of secondary or tertiary alcohols, aldehyde is obtained as a product. |
| 3. | In case of secondary or tertiary alcohols, alkene is obtained as a product. |
| 4. | In case of primary alcohols, alkene is obtained as a product. |
CH3 — O — CH(CH3)2 + HI → B (Product)
The product 'B' in the above mentioned reaction is-
1. CH3OH + (CH3)2CHl
2. lCH2OCH(CH3)2
3. CH3OC(CH3)2
4. CH3l + (CH3)2CHOH
- 1(current)
Q. No.
© 2025 GoodEd Technologies Pvt. Ltd.