The following reaction textR - ( textCH) text2 ( textCH) text2 overset text+ textN (( text(CH) text3 text)) text3 ( textOH)- overset textΔ arrow textRCH text= ( textCH) text2 ( text+N(CH) text3 ( text)) text3 text+ ( textH) text2 textO is called

Question

The following reaction textR - ( textCH) text2 ( textCH) text2 overset text+ textN (( text(CH) text3 text)) text3 ( textOH)- overset textΔ arrow textRCH text= ( textCH) text2 ( text+N(CH) text3 ( text)) text3 text+ ( textH) text2 textO is called (A) Hoffmann-bromamide reaction (B) Cope elimination (C) Hoffmann elimination (D) Beckmann rearrangement

Tardigrade Admin · Accepted Answer

Trialkyl ammonium hydroxide undergoes Hoffmann elimination to give less substituted alkene. Elimination of more acidic

β

-H occurs from that

β

-carbon, which is joined to largest number of hydrogens. Thus, elimination of H occurs opposite to that of Saytzeff rules.
Solution

Q. The following reaction R−(CH)2​(CH)2​N+(((CH)3​))3​(OH)−→ΔRCH=(CH)2​(+N(CH)3​())3​+(H)2​O is called

Hoffmann-bromamide reaction

Cope elimination

Hoffmann elimination

Beckmann rearrangement

Trialkyl ammonium hydroxide undergoes Hoffmann elimination to give less substituted alkene. Elimination of more acidic β -H occurs from that β -carbon, which is joined to largest number of hydrogens. Thus, elimination of H occurs opposite to that of Saytzeff rules.

Q. The following reaction
$R - (CH)_{2} (CH)_{2} N + (((CH)_{3}))_{3} (OH)^{-} \to Δ RCH = (CH)_{2} (+N(CH)_{3} ())_{3} + (H)_{2} O$ is called

Trialkyl ammonium hydroxide undergoes Hoffmann elimination to give less substituted alkene. Elimination of more acidic $β$ -H occurs from that $β$ -carbon, which is joined to largest number of hydrogens. Thus, elimination of H occurs opposite to that of Saytzeff rules.