Question

Amines are converted into alkenes by a two-step process called the Hofmann elimination. S2 reaction of the amine with an excess of CH I in the first step yields an intermediate that undergoes 2 reaction when treated with silver oxide as a base. Pentylamine, for example, yields 1-pentene. 1. excess CHI CH3CH2CH2CH2CH2NH2 240,0 HO CH3CH2CH2CH=CH2 Propose structures for the intermediate and the alkene produced in steps 1 & 2 when the following compound undergoes Hofmann elimination: CH3CH2CH(NH2)CH2CH3 - Intermediate - Alkene • You do not have to consider stereochemistry • You do not have to explicitly draw H atoms. • Include anionic counter-ions, c.8.1, in your answer, but draw them in their own sketcher, • Separate intermediates from products using the symbol from the drop-down menu. ChemDoodle Submit Answer Try Another Version 7 item attempts remaining < Previous Next

Answer 1

CH, OH - CH-CH-CH₂ 1) exun CH3IH -CH₂-CH-CH₂ CH3 HEN STS- CH-1 +re CHE Deprotenation HC-CH₂-CH-CH-CH₂ HC-CH-CH-CH-CH₂ SINH HETIS- PC NHA CH₃ St HC-cu-CH-CH₂-CH₃ + HI HC-N-CH₂ both pare same ТСН, Е St S - Product C HC- CH2 - CH - CH₂ - CH3] 2) Aggo HC-CH - CH - CH₂ - CH3 d. AC-NG CH₂ H₂O. HCENO (43 it o 19+ A - el B- Ontemediate Tehealkylammonium salt. elemination reaction V (Hoffman) H,OH + A + AOI (ppt) I CH HC-CH₂-CH=CH-CH₂ L + ADOH + HE-N-CH, chia Pent-2-ene Product Intemudiate Hc- chy-CH-CH₂ - CH₂ 46-NECH 10 HC-CH ₂ -CH=CH-CH2 Pent-2-ene Product tehaalkylammoniumlocide salt Alkines