The maximum number of RBr producing 2-methylbutane by above sequence of reactions is __________ . (Consider the structural isomers only)

Solution

Chemistry ma’am · 3 steps

Mg then water just swaps Br for H, so count the different places Br can sit on 2-methylbutane.

1. What the sequence does. Mg in dry ether turns R-Br into the Grignard reagent R-MgBr. Water then protonates it to R-H. The carbon skeleton does not change; only Br is replaced by H.

   R-Br →(Mg, dry ether) R-MgBr →(H₂O) R-H + Mg(OH)Br

2. Count unique carbons. So every RBr that works is a monobromo-2-methylbutane. In CH₃-CH(CH₃)-CH₂-CH₃ the two methyls on C2 are equivalent, which leaves 4 different kinds of carbon carrying H.

   Br on	Bromide	Type
   C1 (either methyl on C2)	1-bromo-2-methylbutane	1°
   C2	2-bromo-2-methylbutane	3°
   C3	2-bromo-3-methylbutane	2°
   C4	1-bromo-3-methylbutane	1°

3. So. Four structurally different bromides all give 2-methylbutane, so the answer is 4, option (c).

The idea to keep. Grignard formation followed by water is a clean Br to H swap. The number of halides giving one alkane equals the number of different monohalo isomers of that alkane.

Answer from the NTA final answer key. Solution by PrepWiser.