Question

Question 40 Get help answering Molecular Drawing questions. Your answer is partially correct. Try again. Compound A and compound B are constitutional isomers with molecular formula C4H9Cl. Treatment of compound A with sodium methoxide gives trans-2- butene as the major product, while treatment of compound B with sodium methoxide gives a different disubstituted alkene as the major product. Draw the structure of compound A. CH3 HzĆ CI Propose two structures for compound B. H&C CH2 Edit HC H3

Answer 1

The molecular formula of compound A and B are C H,C1. Hence, they are certainly alkyl chloride with 4 carbon atoms. The possible constitutional isomers are CH3 CH3 I CH₃ H₃C . HC H2CCI H3CCI 1-chlorobutane tert-butylchloride 2-chlorobutane 1-chloro-2-methylpropane If we treat A with NaOMe, we get trans-2-butene. CH3 trans-2-butene Since the double bond formed at C2, the a must be at C2 for the dehydrohalogenation to give double bond at C2 as decribed below CH3 Na ; OMe нс trans-2-butene Note that dehydrohalogenation of other 3 possible constitutional isomer will result in double bond at Ci. Hence, the structure of Ais HC CH3 2-chlorobutane

Compound B on the other hand results in a disubstituted alkene as the major product. Lets draw the major dehydrohalogenation product of the remaining three constitutional isomers: H3C CH Na Ome (di substituted alkene) Hoc 1-chloro-2-methylpropane H:C- HECY Na OMe (monosubstituted alkene) 1-chlorobutane H₂C-gutta LCH₂ NaoMe н,с (di substituted alkene) CH2 tert-butylchloride Hence, among the remaining three possible structure of B. the first and third structure above resulted in di substituted alkenes. Hence, the possible structures of B are CH3 CH3 CH and Hąc H₃Ca