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(B) In Hofmann elimination,the quaternary ammonium hydroxide undergoes thermal decomposition to form the least substituted alkene as the major product

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Ethene

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(B) In Hofmann elimination,the quaternary ammonium hydroxide undergoes thermal decomposition to form the least substituted alkene as the major product.Step $1$: The quaternary ammonium iodide is converted to the corresponding hydroxide using moist $Ag_2O$.$(CH_3CH_2CH_2)N^{+}(CH_2CH_3)_3 I^{-} + AgOH \rightarrow (CH_3CH_2CH_2)N^{+}(CH_2CH_3)_3 OH^{-} + AgI$Step $2$: Upon heating,the $OH^{-}$ ion abstracts a $\beta$-hydrogen from the alkyl groups attached to the nitrogen atom.The available $\beta$-hydrogens are on the propyl group (at the $\beta$-carbon) and the ethyl groups (at the $\beta$-carbon).Removal of $\beta$-hydrogen from the ethyl group yields ethene $(CH_2=CH_2)$.Removal of $\beta$-hydrogen from the propyl group yields propene $(CH_3CH=CH_2)$.According to the Hofmann rule,the less substituted alkene is the major product. Since ethene is less substituted than propene,ethene is the major product.

Identify the alkene obtained as the major product in the following Hofmann elimination reaction.
$(CH_3CH_2CH_2)N^{+}(CH_2CH_3)_3 I^{-}$ $\xrightarrow[\Delta]{\text{moist } Ag_2O} A$ $\xrightarrow[\Delta]{\text{heat}} \text{Alkene} + \text{Amine}$

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Identify the alkene obtained as the major product in the following Hofmann elimination reaction.$(CH_3CH_2CH_2)N^{+}(CH_2CH_3)_3 I^{-}$ $\xrightarrow[\Delta]{\text{moist } Ag_2O} A$ $\xrightarrow[\Delta]{\text{heat}} \text{Alkene} + \text{Amine}$