An organic compound A (C4H9Cl) on reaction with Na/diethyl ether gives...
Organic compound A (C4H9Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative. We need to identify the compound A.
Identification of compound A:
To identify the compound A, we need to understand the reaction mechanism involved in the given reactions.
Reaction 1: Na/diethyl ether
When organic compound A (C4H9Cl) reacts with Na/diethyl ether, it undergoes a nucleophilic substitution reaction.
C4H9Cl + Na → C4H9Na + Cl
C4H9Na + C2H5OC2H5 → C4H10 + NaOC2H5
The product formed is a hydrocarbon (C4H10), which is an alkane.
Reaction 2: Monochlorination
When the hydrocarbon (C4H10) obtained from reaction 1 undergoes monochlorination, only one chloro derivative is formed. This indicates that the hydrocarbon has only one type of hydrogen atom.
C4H10 + Cl2 → C4H9Cl + HCl
Now, we need to identify the hydrocarbon (C4H10) obtained in reaction 1 to determine the compound A.
Identification of hydrocarbon (C4H10):
The molecular formula of the hydrocarbon (C4H10) obtained in reaction 1 indicates that it is an alkane with a chain of four carbon atoms.
There are two possible isomers of C4H10, which are:
1. n-Butane
2. 2-Methylpropane (Isobutane)
However, n-Butane (CH3CH2CH2CH3) has four different types of hydrogen atoms (three primary and one secondary), whereas 2-Methylpropane (CH3CH(CH3)2) has only one type of hydrogen atom (secondary).
Therefore, the hydrocarbon (C4H10) obtained in reaction 1 must be 2-Methylpropane (Isobutane).
Identification of compound A:
The organic compound A (C4H9Cl) obtained after the reaction with Na/diethyl ether must be tert-Butyl Chloride (CH3)3CCl.
Therefore, the correct answer is option A (tert-Butyl Chloride).