Q.An alkyl bromide (A) reacts with sodium in ether to form 4,5-diethyloctane. The compound (A) is
Step 1. In a Wurtz reaction, 2 R-Br + 2Na --> R-R + 2NaBr: the new C-C bond in the product forms exactly at the carbon that used to carry Br in each R-Br monomer.
Step 2. Number 4,5-diethyloctane's chain C1-C8: it carries an ethyl branch at C4 and another at C5, so C4 and C5 -- the two carbons flanking the middle of the chain -- are the two halves' point of union.
Step 3. Splitting the molecule at the C4-C5 bond gives two identical R groups, each being C1-C2-C3-C4(with its ethyl branch), i.e. CH3-CH2-CH2-CH(C2H5)-, which written out fully as a monomer with Br on that carbon is CH3-CH2-CH2-CH(Br)-CH2-CH3 (renumbering the ethyl branch as part of a 6-carbon chain gives 3-bromohexane).
Step 4. This exactly matches option (d); options (a) and (b) are simple unbranched bromides that would just give n-octane or n-dodecane on Wurtz coupling, and option (c) would give a differently branched (and mismatched) product.
(d) CH3-(CH2)2-CH(Br)-CH2-CH3 (3-bromohexane)
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