1.

The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives

Answer»

2-METHYL-1-butene
2-methyl-2-butene
2,2-dimethyl-1-butene
2-butene

Solution :`CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-CH_2-Br+underset"(ALC)""KOH"to CH_3-overset(CH_3)overset|C=CH-CH_3+KBr +H_2O`
In this reaction `1^@` carbonium ion is FORMED which rearranges to form `3^@` carbonium ion from which base obstruct proton Hence 2-methyl-2-butene is formed as a main product. `underset(1^@ "carbonium less soluble")(CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-overset+(CH_2) )overset"Methyl SHIFT"to CH_3-underset+oversetoverset(CH_3)(|)C-CH_2-CH_3 overset("Elimination of proton from" beta "carbon which is less hydrogenated")to underset"2-Methyl-2-Butene"(CH_3-oversetoverset(CH_3)(|)C=CH-CH_3)`


Discussion

No Comment Found

Related InterviewSolutions