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Question

A Grignard reagent [A] and a haloalkane [B] react together to give [C]. Compound [C] on heating with KOH yields a mixture of two geometrical isomers, (D) and (E), of which (D) predominates isomers. (C) and (E) have the same molecular formula but (C) gives 1-bromo-3-phenylpropane on reaction with HBr in the presence of a peroxide. Give structures of (A), (B) and (C) and configuration of (D) and (E) with reasons.

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Solution


Step 1. To determine the structure of compound (C).

(i) Since (C) on treatment with HBr in presence of peroxide (anti-Markovnikov's addition) gives 1-bromo-3-phenylpropane, therefore, (C) must be 3-phenyl-1-propene.

C6H5CH2CH=CH2HBr/Peroxide−−−−−−−−−−−(Anti-Mark. Addn.)C6H5CH2CH2CH2Br
3-Phenyl -1-propene (C) 1-Bromo-3-phenylpropane

Step 2 To determine the structures of compounds (D) and (E).

Now (C), i.e., 3-phenyl-1-propene being a terminal alkene does not show geometrical isomerism. However, when (C) is heated with KOH, it gives a mixture of two geometrical isomers, (D and E). Since (C) has the same M.F. as (E) , therefore, (D) and (E) are isomers of (C) , i.e., on heating with KOH (C) undergoes isomerization to give the more stable 1-phenyl-1-propene.

Further since out of (D) and (E) , (D) predominates, therefore (D) must be the more stable trans isomer, i.e., trans-1-phenyl-1-propene and (E) must be the less stable cis-isomer, i.e., cis-1-phenyl-1-propene isomer. (Refer Image)

Step 3: To determine the structures of compounds (A) and (B).

Since (C), i.e., 3-phenyl-1-propene is obtained by the reaction between a Grignard reagent (A) and a haloalkane (B), therefore (A) must be phenyl magnesium bromide and (B) must be 3-bromo-1- propene.

C6H5MgBr + BrCH2CH=CH2ΔC6H5CH2CH=CH2+MgBr2
Phenyl mag. bromide (A) 3-Bromo-1-propene (B) 3-phenyl-1-propene (C)

Thus, (A) = phenyl magnesium bromide , (B) = 3-bromo-1-propene, (C) = 3-phenyl-1-propene, (D) = trans-1-phenyl-1-propene and (E) = cis-1-phenyl-1-propene.

1830177_1877949_ans_b7addc7b92894885a4b2e7a6aba2543b.png

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