Reduction by LiAlH4 of hydrolyzed product of an ester gives:
1. Two alcohols
2. Two aldehyde
3. One acid and one alcohol
4. Two acids
Intermediates formed during the reaction of
with and KOH are:
1. RNHCOBr and RNCO
2. RCONHBr and RNCO
3. RNH-Br and RCONHBr
4. RCON
In the following reaction, product 'P' is:
1. RCH2OH
2. RCOOH
3. RCHO
4. RCH3
In a set of the given reactions, acetic acid yielded a product C.
\(\mathrm{CH_3COOH+PCl_5\rightarrow A\xrightarrow[Anh.AlCl_3]{C_6H_6}B\xrightarrow[ether]{C_2H_5MgBr}C}\)
Product C would be:
1. | CH3CH(OH)C2H5 |
2. | CH3COC6H5 |
3. | CH3CH(OH)C6H5 |
4. |
When m-chlorobenzaldehyde is treated with 50% KOH solution, the product(s) obtained is (are):
1. | |
2. | |
3. | |
4. |