Primary Alkyl halide \( C_4H_9Br \) (X) reacts with alcoholic KOH to give the compound (Y). Compound (Y) is reacted with HBr to give (Z) which is an isomer of (X). When (X) is reacted with Na/dry ether, it gives a compound (T), \( C_8H_{18} \), which is different from the compound formed when n-butyl bromide is reacted with Na/dry ether. The correct IUPAC name of (T) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2,5-Dimethylhexane
In the Wurtz reaction, an alkyl halide heated with sodium in dry ether gives an alkane with twice the carbon atoms: the two alkyl groups join through the carbons that carried the halogen. So the structure of T shows the alkyl group of X directly.
There are only two primary bromides \( \ce{C4H9Br} \): 1-bromobutane, \( \ce{CH3CH2CH2CH2Br} \) (n-butyl bromide), and 1-bromo-2-methylpropane, \( \ce{(CH3)2CHCH2Br} \). T is different from the product of n-butyl bromide (n-octane), so X is 1-bromo-2-methylpropane. The other clues agree with this.
Alcoholic KOH removes HBr (dehydrohalogenation), giving 2-methylpropene (Y):
\( \displaystyle \ce{(CH3)2CH-CH2Br + KOH} \xrightarrow{\text{alcohol}} \ce{(CH3)2C=CH2 + KBr + H2O} \)
HBr adds by Markovnikov’s rule, giving 2-bromo-2-methylpropane (Z), an isomer of X:
\( \displaystyle \ce{(CH3)2C=CH2 + HBr -> (CH3)3C-Br} \)
The Wurtz reaction of X joins two isobutyl groups:
\( \displaystyle \ce{2(CH3)2CHCH2Br + 2Na} \xrightarrow{\text{dry ether}} \ce{(CH3)2CH-CH2-CH2-CH(CH3)2 + 2NaBr} \)
The longest chain of T has six carbon atoms with methyl groups on C-2 and C-5, so T (\( \ce{C8H18} \)) is 2,5-dimethylhexane.
NCERT (2026-27 edition):
Wurtz Reaction – “Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction.”